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Students can Download Chapter 9 Coordination Compounds Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 9 Coordination Compounds

Plus Two Chemistry Coordination Compounds One Mark Questions and Answers

Question 1.
The ions or molecules bound to the central atom/ion in the coordination entity are called ___________.
Answer:
Ligands

Question 2.
In [Co(C₂O₄)₃]^3-, the coordination number of cobalt is _________.
Answer:
six

Question 3.
Which complex has a square planar structure?
(a) [Ni(CO)₄]
(b) [NiCI₄]^2-
(c) [Ni(H₂O)₆]²⁺
(d) [CU(NH₃)₄]²⁺
Answer:
(d) [CU(NH₃)₄]²⁺

Question 4.
Say TRUE or FALSE.
[Co(NO₃)(NH₃)₅]SO₄ and [Co(NO₃)(NH₃)₄(SO₄)](NH₃) are ionisation isomers.
Answer:
FALSE

Question 5.
The formula of Wilkinson’s catalyst is _________.
Answer:
[RhCl(PPh₃)₃]

Question 6.
The charge of Ni in [Ni(CO)₄] is
(a) +1
(b) +2
(c) 0
(d) +4
Answer:
(c) 0

Question 7.
The central metal ion present in chlorophyll?
(a) Fe²⁺
(b) Cu²⁺
(c) Mg²⁺
(d) CO²⁺
Answer:
(c) Mg²⁺

Question 8.
EDTA is a dentate ligand
(a) uni dentate
(b) bidentate
(c) Tridentate
(d) hexadentate
Answer:
(d) hexadentate

Question 9.
Which is an example for homoleptic complexes
(a) [Co(NH₃Cl₂]⁺
(b) [CoNH₃)₆]³⁺
(c) [Cr(NH₃)(H₂O)₃]Cl₃
(d) [CoCl₂(en)₂]
Answer:
(b) [CoNH₃)₆]³⁺

Question 10.
Ammonia will not form complex with
(a) Ag²⁺
(b) Pb²⁺
(c) Cu²⁺
(d) Cd²⁺
(e) Fe²⁺
Answer:
(b) Pb²⁺

Plus Two Chemistry Coordination Compounds Two Mark Questions and Answers

Question 1.
In a seminar, Jishnu argued that the “hexaflourocobaltate(III) ion is highly paramagnetic than hexacyanoferrate(III) ion.

1. Do you agree with this words?
2. Explain it.
3. Write the formulae of the given coordination compounds.

Answer:

1. Yes
2. CN^– is a strong field ligand so paring occurs. Number of unpaired electron decreases, paramagnetism decreases. F^– is a weak field ligand so no paring occurs, paramagnetism increases.
3. [CoF₆]^3-and [Co(CN)₆]^3-

Question 2.
Raju: Coordination compounds are coloured.
Ramu: No, co-ordination compounds are colourless.

1. Whose statement is correct?
2. Explain the reason for your answer.

Answer:

1. Raju’s statement is correct. Coordination compounds are usually coloured.
2. The colour of coordination compounds is due to d-d transition.

Question 3.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.

Question 4.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO₄ does not form any complex with (NH₄)₂SO₄. Instead, it forms a double salt FeSO₄. (NH₄)₂SO₄.6H₂O which dissociates completely into ions. CuSO₄ when mixed with NH₃ forms a complex [CU(NH₃)₄]SO₄ in which the complex ion [CU(NH₃)₄]²⁺ does not dissociate to give Cu²⁺ ion.

Question 5.
Write the geometrical isomers of [Pt(NH₃)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Answer:
The complex [Pt(NH3)(Br)(Cl)(py)] will form three geometrical isomers:

Square planar complexes of this type will not show geometrical isomerism.

Question 6.
Aqueous copper sulphate solution (blue in colour) gives:

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer:
Aqueous copper sulphate contains a coordination entity, [Cu(H₂O)₄]²⁺ which is blue in colour. Water molecule is a weaker ligand than Cl^– and F^–.
1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.

2. On addition of aqueous solution of KCl, an another bright green complex entity is formed which is soluble in water.

Question 7.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S(g) is passed through this solution?
Answer:
Aqueous solution of copper sulphate contains Cu²⁺ ions in the form of complex entity, [Cu(H₂O)₄]²⁺ and H₂O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)₄]^2- is formed due to the following reaction:

Cyanide ligand, CN^– is a strong field lignad and stability constant of [Cu(CN)₄]^2- is quite large and thus practically no Cu²⁺ ions are left in solution. On passing H₂S gas, no CuS is formed due to non- availability of Cu²⁺ ions in solution.

Question 8.
Optical isomerism is usually exhibited by complexes containing polydentate ligand. What do you mean by ligand?
Answer:
Ligand is a neutral molecule or charged ion which can donate a lone pair of electron to the metal.

Question 9.
Coordination complexes are of different types. Name the compounds.

1. [Cr(H₂O)₅Cl₂]
2. K₃[Cr(C₂O₄)₃]

Answer:

1. Pentaaquadichloridochromium(II)
2. Potassiumtrioxalatochromate(III)

Question 10.
Write the IUPAC names of the following compounds.

1. K_3M[Fe(CN)₆]
2. [C0(NH₃)₅(CO₃)]Cl

Answer:

1. Potassiumhexacyanoferrate(III)
2. Pentaamminecarbanatocobalt(III) chloride

Question 11.
[Fe(CN)₆]^3- is paramagnetic, while [Fe(CN)₆]^4- is diamagnetic. Explain with the help of VB theory.
Answer:
In [Fe(CN)₆]^3- iron is in +3 state and in [Fe(CN)₆]⁴, iron is in +2 state. [Fe(CN)₆]^3- contains five electrons in d-level (3d⁵). In this complex iron undergoes d²sp³ hybridisation.

Due to the presence of one unpaired electron, [Fe(CN)₆]^3-, is paramagnetic. In [Fe(CN)₆]^4- iron contains six electrons in d-level (3d⁶). It undergoes d²sp³ hybridisation and has no unpaired electrons. Hence, [Fe(CN)₆]^4- is diamagnetic.

Plus Two Chemistry Coordination Compounds Three Mark Questions and Answers

Question 1.
Look at the following two diagrams.

1. Is the diagrams I and II correct? Justify. If the figure is not correct, redraw it.
2. Which theory is related to this?
3. Explain briefly, how this theory is applicable to octahedral complexes.

Answer:
1. No, Figure (II) is wrong.

2. Crystal field theory.

3. In the case of octahedral complexes, the ligands are approaching the ‘d’ orbitals through the axis. As a result of this the energy of d_x²-y² and d_z² orbitals increases and the energy of the remaining three orbitals decreases. The orbitals which possess high energy are represented as ‘e_g’ levels and the orbitals which possess less energy are represented as “t_2g” levels.

Question 2.
A list of coordination compounds are given below:
[Cr(H₂O)₆] Cl₃, [Co(NH₃)₅Br] SO₄, [Co(NH₃)₅NO₂]²⁺ and [Pt(NH₃)₂Cl₂]. Which type of isomerism do these compounds exhibit?
Answer:
Hydrate Isomerism, Ionisation Isomerism, Linkage Isomerism, Geometrical Isomerism.

Question 3.
The following are examples of coordination compounds. Identify the type of isomerism exhibited by each of them and write their possible isomers,

1. [Cr(NH₃)₅Br]SO₄
2. CrCl₃.6H₂O
3. [PtCl₂(NH₃)₂]

Answer:

1. Cr(NH₃)₅Br]SO₄ – Ionisation isomerism – [Cr(NH₃)₅SO₄]Br
2. CrCl₃.6H₂O – Hydrate isomerism – [Cr(H₂O)₅Cl]Cl₂.H₂₎
3. PtCl₂(NH₃)₂] – Geometrical isomerism

Question 4.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
The order of the ligand in the spectrochemical series is H₂O < NH₃ < NO₂^–. Hence the wavelength of the light observed will be in the order:
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-

Thus, the wavelength of light absorbed (E = \(\frac{\mathrm{hc}}{\lambda}\)) will be in the opposite order:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Question 5.
In a classroom discussion, Sajan argued that CN^–, OH^–, Cl^– etc. are examples for neutral ligands.

1. Do you agree with his argument?
2. If not, give a reason with the help of examples.
3. What do you mean by chelating ligand and chelation?

Answer:

1. No.
2. They are charged ligands.
3. If a polydentate ligand is coordinated to the metals, a ring structure is obtained. It is called chelate and the phenomenon is called chelation.

Question 6.
Ligands can be arranged according to the magnitude of Δ₀ and the arrangement is given below:
l- < Br^– < Cl^– < F^– < OH^– < H₂O < NH₃ < (en)

1. What is this series known as?
2. Is the sequence incorrect order?
3. Identify the weak field and strong field ligands.

Answer:

1. Spectrochemical series
2. Yes
3. The ligands above water are strong ligand and the ligands below water are weak ligands.

Weak filed ligands – l^–, Br^–, Cl^–, F^–, OH^–, H₂0 Strong filed ligands – NH₃, en

Question 7.
Consider the following coordination compounds.

• [Co(NH₃)₅ Cl] SO₄
• [Co(NH₃)₅ SO₄]Cl

1. Write down the IUPAC name of these compounds.
2. Name the isomerism exhibited by these compounds.

Answer:
1. The IUPAC name of compounds

• Pentaamminechloridocobalt(III) sulphate
• Pentaamminesulphatocobalt(III) chloride

2. Ionisation isomerism

Question 8.
Consider the following compounds:
[Co (NH₃)₅ NO₂] Cl₂ and [Co (NH₃)₅ ONO] Cl₂

1. Identify the isomerism exhibited by these compounds.
2. Explain ionisation isomerism with example.

Answer:

1. Linkage isomerism
2. This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
   e.g. [Co(NH₃)₅Cl] SO₄ & [Co(NH₃)₅SO₄] Cl

These complexes ionises as:

• [Co(NH₃)₅Cl] SO₄ → [Co(NH₃)₅Cl]²⁺ + SO₄^2-
• [Co(NH₃)₅SO₄] Cl → [Co(NH₃)₅SO₄]+ + Cl^–

Question 9.
Consider the statement: Crystal Field Theory (CFT) is applicable to octahedral and tetrahedral complexes.

1. Is this statement true?
2. Explain the crystal field splitting in octahedral complexes with the help of a neat diagram.

Answer:

1. Yes
2. In an octahedral crystal field the ligands are approaching the metal along the axes. Hence, the energy of d_x²-y² and d_z² orbitals (e_g set) increases by 3/5Δ₀and that of d_xy, d_yz and d_xz orbitals (t_2g set) decreases by 2/5 Δ₀

Question 10.
When a ligand approaches to an octahedral complex, the degenerate ‘d’ orbitals undergoes splitting.

1. What will be the observation?
2. What are the factors influencing crystal field splitting energy?

Answer:
1. The energy of d_x²-y² and d_z² orbitals (e_g set) increases by 3/5Δ₀ and that of d_xy, d_yz and d_xz orbitals (t_2g set) decreases by 2/5 Δ₀

2. The factors influencing crystal field splitting energy.

• Nature of the ligand
• Geometry of the complex
• Valency of the metal

Question 11.
What will happen when a ligand approaches to a tetrahedral complex?
Answer:
the energy of d_xy, d_yz and d_xz orbitals (t_2g set) increases by 2/5Δ_t and that of d_x²-y² and d_z² orbitals (e_g set) decreases by 3/5 Δ_t.

Question 12.
Consider the complex ion [Ti (H₂O)₆]³⁺ In the case of an octahedral complex, what is the condition for the pairing of forth electron in the d- level?
Answer:
If the crystal field splitting energy is greater than the pairing energy, the fourth electron will pair at the t_2g level and if the pairing energy is greater than the crystal field splitting energy the electron will go to the on e_g level.

Question 13.
Is bidentate ligands same as the amidentate ligands? Justify.
Answer:
A bidentate ligand like (en), can form two coordinate bonds with the metal at the same time.
An amidentate ligand like -NO₂ can form only one coordinate bond with the metal at a time. But it can ligate through two different atoms.

Plus Two Chemistry Coordination Compounds Four Mark Questions and Answers

Question 1.
Match the following table:

Answer:

Question 2.
1. Write the IUPAC names of the following coordination compounds.

• [Pt(NH₃)Cl(NO₂]₂)
• K₃[Cr(C₂O₄)₃]

2. Identify the type of isomerism exhibited by the following complexes and distinguish them. [Co(NH₃)₅SO₄] Br and [Co(NH₃)₅Br] SO₄
Answer:
1. The IUPAC names of the coordination compounds:

• Amminechloridonitrito-N platinum(II)
• Potassiumtrisoxalatochromate(III)

2. Ionisation isomerism.
The first compound gives pale yellow ppt. with AgNO₃ solution whereas the second compound gives white ppt. with BaCl₂ solution.

Question 3.
1. Write the IUPAC name of the following compounds:

• [Pt(NH₃)₂Cl₂]
• K₄[Fe(CN)₆]

2. A list of coordination compounds are given below:

• [Cr(H₂O)₆]Cl₃,
• [CO(NH₃)₅ Br]SO₄,
• [CO(NH₃)₅ NO₂]²⁺,
• [CO(NH₃)₆] [Cr(CN)₆]

Which type of isomerism do these compounds exhibit?
Answer:
1. The IUPAC name of the coordination compounds:

• Diamminedichloridoplatinum(II)
• Potassium hexacyanoferrate(II)

2. Type of isomerism

• [Cr(H₂O)₆]Cl₃ – Hydrate isomerism
• [CO(NH₃)₅ Br]SO₄ – Ionisation isomerism
• [CO(NH₃)₅ NO₂]²⁺ – Linkage isomerism
• [CO(NH₃)₆] [Cr(CN)₆] – Coordination isomerism

Question 4.
1.. Write down the IUPAC name of

• K₄[Fe(CN)₆]
• [Pt(NH₃)₂Cl₂]

2. On the basis of VBT, explain why [Fe(H₂O)₆]²⁺ is strongly paramagnetic while [Fe(CN)₆]^3- is weakly paramagnetic.
Answer:
1. The IUPAC names are:

• Potassium hexacyanoferrate(II)
• Diamminedichloridoplatinum(II)

2.

In [Fe(H₂0)₆]^2-, iron undergoes sp³d² hybridisation (inner orbital complex). It has four unpaired electrons and hence it is highly paramagnetic.

In [Fe(CN)₆]³⁺. iron undergoes d²sp³ hybridisation (outer orbital complex). It has only one unpaired electron. Hence, it is less paramagnetic.

Question 5.
The names of some co-ordination compounds are given below:

• EDTA
• Haemoglobin
• cis-platin
• Vitamin B¹²
• D-penicillamine
• Chlorophyll
• Ni(CO)₄

a. Classify the above compounds on the basis of application of coordination compounds?
b. There are given some of the coordination compounds Name them.

1. K₃[Fe(C₂O₄)₃]
2. [Cr(CN)₃]³⁺
3. [CoSO₄(NH₃)₄]NO₃
4. [CO(NO₂)₃(NH₃)₃]

Answer:
a. On the basis of application of coordination compounds:

• In biological system – Haemoglobin, Vitamin B₁₂, Chlorophyll.
• Estimation of hardness of water – EDTA.
• Extraction of metals – Ni(CO)₄
• In medicine – D-penicillamine, cis-platin

b. The coordination compounds are:

1. K₃[Fe(C₂O₄)₃] – Potassiumtrioxalatoferrate(III)
2. [Cr(CN)₃]³⁺ – Trisethylenediaminechromium(III) ion
3. [CoSO₄(NH₃)₄]NO₃ – Tetraamminesulphato- cobalt(III) nitrate
4. [CO(NO₂)₃(NH₃)₃] – Triamminetrinitrito-N-cobalt(III)

Question 6.
1. Name the following compounds.

• [Pt(NH₃)₄ ] [CuCl₄]
• [PtCl₂ (NH₃)₄] Br₂

2. What type of isomerism is shown by the following coordination compounds?

• [Pt(NH₃)₄ ] [CuCl₄]
• [Cr(en)₃]³⁺

Answer:
1.

• Tetraammineplatinum(II) tetrachlorocuprate(II)
• Tetraamminedichloridoplatinum(IV) bromide

2. Type of isomerism:

• Coordination isomerism
• Optical isomerism

Question 7.

1. Write the d-orbital configuration of [Ti(H₂O₆]²⁺
2. Ti⁴⁺ is colourless. Why?
3. Write the possible isomers of [Co(NH₃)₅ Br] SO₄ and name them.

Answer:
1.

2. In Ti⁴⁺ there are no electrons in 3d orbitals. Hence ‘d-d’ transition cannot take place. Hence it is colourless.

3. Possible isomers of [Co(NH₃)₅ Br] SO₄ and names

• [CO(NH₃)₅ Br] SO₄ – Pentamminebromido- cobalt(III) sulphate
• [Co(NH3)5SO4]Br-Pentamminesulphatecobalt(III) bromide

Question 8.
‘A’ and ‘B’ are isomers. They have the same composition. But ‘B’ cannot give the test for sulphate.

1. Write two suitable coordination compounds which give the test for sulphate.
2. What are the two major classes of isomerism exhibited by coordination compounds?
3. Draw the structure of an octahedral complex that show optical isomerism.

Answer:
1. Two suitable coordination compounds which give the test for sulphate

• [CO(NH₃)₅Cl]SO₄
• [Co(NH₃)₅Br] SO₄

2. Structural isomerism, Stereoisomerism
3. [PtCl₂(en)₂]²⁺

Question 9.

1. Write the formula of the complex ion Chloridonitro tetramminecobalt(III)?
2. Identify the ligands, coordination number and coordination sphere.
3. Explain the structure of Tetracarbonylnickel(O) with the help of Valence Bond Theory.

Answer:

1. [CO(NH₃)₄Cl(NO₂^–)
2. Ligands-NH₃, Cl^– , NO₂^– Coordination number -4
3. Tetracarbonylnickel(0) – [Ni(CO)]₄] – Structure

Ni(CO₄) is diamagnetic as it does not contain any unpaired electron.

Question 10.
Consider the complex ion [Ti (H₂O)₆]³⁺

1. What is its outer electronic configuration and its shape?
2. What do you mean by crystal field splitting theory?

Answer:

1. Ti³⁺ – 3d¹ 4s⁰ Octahedral
2. In the case of an isolated gasesous metal atom/ ion all the five d-orbitals have the same energy (degenerate). Due to the presence of ligands are splitted the degeneracy of the d-orbitals is lifted.

Question 11.
Coordination compounds are those compounds which retain their identity even in solution and it is essential for all living matter.

1. Name a coordination compound containing Magnesium, which is essential for plants.
2. When we coordinate EDTAwith any metal, we get a ring structure. What is this process called?
3. Explain.

Answer:

1. Chlorophyll.
2. Chelation.
3. When a poly dentate ligand is coordinated to the metal, a ring structure is obtained, called chelate complex. Such complexes are more stable than similar complexes containing unidentate ligands.

Question 12.
Some ligands are given below. Arrange them in suitable headings.
[H₂O, NH₃, CN^–, CO, Cl^–, OH^– (en)]
1. What do you mean by the term ligand?
2. Write down the nomenclature of the coordination compounds given below.

• K₄[Fe(CN)₆]
• [Ag (NH₃)₂]Cl

Answer:
Neutral ligands – H₂O, NH₃, CO, en
Charged ligands – CN^–, OH^–, Cl^–
1. Ligand is a neutral molecule or charged ion which can donate atleast one lone pair of electron to the metal.
2. nomenclature of the coordination compounds

• Potassium hexacyanoferrate(II)
• Diamminesilver(I) chloride

Question 13.

1. What do you mean by optically active compounds? Give two examples.
2. Draw the ‘d’ and T forms of [Co(en)₃]³⁺.

Answer:
1. Optically active compounds are formed by chiral moneluces i.e., molecules which do not have plane of symmetry. These isomers are non- superimposable mirror images of each other. They are optically active and rotate the plane of polarised light equally but in opposite directions.

The isomer which rotates the plane of polarised light towards left is called leavorotatory (-) while that which rotate plane towards right is called dextrorotatory (+).
e.g. [Co(en)₃]³⁺, [PtCl₂(en)₂]²⁺
Dextro and laevo forms of these compounds are possible.

2.

Question 14.
What is the importance of the following coordination compounds is different fields?

1. EDTA
2. Gold cyanide [AU(CN)₂]
3. cis-platin
4. [Ag(S₂O₃)]^3-

Answer:

1. EDTA → Estimation of hardness of water
2. AU(CN)₂ → Metallurgy
3. cis-platin → Cancer therapy
4. [Ag(S₂O₃)]^3- → Photography

Question 15.
The d-block elemetns forms coordination compounds.

1. Name the coordination compound K₃ [CoF₆].
2. Write the electronic configuration of the central metal atom of the above complex by using CFT
3. Draw the figure to show the splitting of degenerate, ‘d’ orbitals in an octahedral field.

Answer:
1. K₃[CoF₆] – Potassiumhexafluridocobaltate(III)
2. The electronic configuration of Co (Z = 27) is [Ar]3d⁷4s² In K₃ [CoF₆], Co is in +3 state. The configuration of Co³⁺ is 3d⁶ 4s°.
3.

Question 16.

1. Name the compound K₃[Cr(C₂O₄)₃]
2. Explain on the basis of VB theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and [Ni(Cl)₄]^2- ion with tetrahedral structure is paramagnetic.

Answer:

1. K₃[Cr(C₂O₄)₃] – Potassiumtrioxalatochromate(III)
2. Ni = 1s²2s²2p63s²3p64s²3d²

It is diamagnetic due to absence of unpaired electrons.

∴ Due to presence of unpaired electrons it is paramagnetic.

Plus Two Chemistry Coordination Compounds NCERT Questions and Answers

Question 1.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO₄ does not form any complex with (NH₄)₂SO₄. Instead, it forms a double salt FeSO₄.(NH₄)₂SO₄.6H₂O which dissociates completely into ions. CuSO₄ when mixed with NH₃ forms a complex [CU(NH₃)₄]SO₄ in which the complex ion [CU(NH₃)₄]²⁺ does not dissociate to give Cu²⁺ ion.

Question 2.
Write the geometrical isomers of [Pt(NH₃)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Answer:
The complex [Pt(NH₃)(Br)(Cl)(py)] will form three geometrical isomers:

Square planar complexes of this type will not show geometrical isomerism.

Question 3.
Aqueous copper sulphate solution (blue in colour) gives:

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer:
Aqueous copper sulphate contains coordination entity, [Cu(H₂O)₄]²⁺ which is blue in colour. Water molecule is a weaker ligand than Cl^– and F^–.
1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.

2. On addition of aqueous solution of KCl, another bright green complex entity is formed which is soluble in water.

Question 4.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S(g) is passed through this solution?
Answer:
Aqueous solution of copper sulphate contains Cu²⁺ ions in the form of complex entity, [Cu(H₂O)₄]²⁺ and H₂O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)₄]^2- is formed due to the following reaction:

Cyanide ligand, CN^– is a strong field lignad and stability constant of [Cu(CN)₄]^2- is quite large and thus practically no Cu²⁺ ions are left in solution. On passing H₂S gas, no CuS is formed due to non-availabiliy of Cu²⁺ ions in solution.

Question 5.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
The order of the ligand in the spectrochemical series is H₂O < NH₃ < NO₂^–. Hence the wavelength of the light observed will be in the order:
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-

Thus, the wavelength of light absorbed (E = \(\frac{\mathrm{hc}}{\lambda}\)) will be in the opposite order:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Students can Download Chapter 9 Ray Optics and Optical Instruments Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 9 Ray Optics and Optical Instruments

Plus Two Physics Ray Optics and Optical Instruments NCERT Text Book Questions and Answers

Question 1.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is

1. a convex lens of focal length 20 cm, and
2. a concave lens of focal length 16 cm?

Answer:
Here the object is virtual and the image is real.
u = 12cm object on right and virtual.
1. f = +20 cm

i.e., v = 7.5 cm. (image on right and real). It is located 7.5 cm from the lens.

2. f = -16 cm

i.e., v = 48 cm. (image on right and real). Image will be located 48 cm from the lens.

Question 2.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length of the lens is to be 20 cm?
Answer:
\(\frac{\mu_{2}}{\mu_{1}}\) = µ = 1.55
R₁ = R₂ = R
f = 20 cm

R = 0.55 × 2 × 20 = 22 cm.

Question 3.
A small telescope has an objective lens of focal length 144 cm and eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer:
1. For normal adjustment.
M.P. of telescope = \(\frac{f_{0}}{f_{e}}=\frac{144}{6}\) = 24

2. The length of the telescope in normal adjustment
L = f_o + f_e
= 144 + 6 = 150 cm.

Plus Two Physics Ray Optics and Optical Instruments One Mark Questions and Answers

Question 1.
Fora total internal reflection, which of the following is correct?
(a) Light travel from rarer to denser medium.
(b) Light travel from denser to rarer medium.
(c) Light travels in air only.
(d) Light travels in water only.
Answer:
(b) Light travel from denser to rarer medium.
Explanation: In total internal reflection, light travel from denser to rarer medium.

Question 2.
Focal length of a convex lens of refraction index 1.5 is 2 cm. The focal length of lens, when immersed in a liquid of refractive index of 1.25, will be.
(a) 10 cm
(b) 2.5 cm
(c) 5 c
(d) 7.5 cm
Answer:
(c) 5 c

Question 3.
If the refractive index of a material of equilateral prism is \(\sqrt{3}\), then angle of minimum deviation of the prism is
(a) 60°
(b) 45°
(c) 30°
(d) 75°
Answer:
(a) 60°
Explanation: A = 60°, n = \(\sqrt{3}\), D = ?

D = 60°.

Question 4.
Which of the following is correct for the beam which enters the medium?
(a) Travel as a cylindrical beam
(b) Diverge
(c) Converge
(d) Diverge near the axis and converge near the periphery
Answer:
(c) Converge.
Explanation: Since the refractive index is less at the beam boundary, the ray at the edges of the beam move faster compared to the axis of beam. Hence, the beam converges.

Question 5.
A beam of monochromatic light is refracted from vacuum into a medium of refraction index 1.5. the wavelength of refracted light will be.
(a) Depend on intensity of refracted light
(b) Same
(c) smaller
(d) larger
Answer:
(c) smaller
Explanation: velocity of light decreases in a medium. Hence λ decrease in a medium (v ∝ λ).

Question 6.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. What is the power in diopters of the combination is
Answer:
Focal length of convex lens f₁ = 25 cm
Focal length of concave lens f₂ = -25 cm
Power of combination in diopters,

Question 7.
Fill in the blanks

Answer:
(i) n = \(\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}\) = 1.5

(ii) Optical fibre

Plus Two Physics Ray Optics and Optical Instruments Two Mark Questions and Answers

Question 1.
Match the following.

Answer:

Plus Two Physics Ray Optics and Optical Instruments Three Mark Questions and Answers

Question 1.
A hemispherical transparent paperweight of radius 5m and refractive index 1.5 is placed on a table. A beam of lazar, at a distance of 2m from the centre is directed as shown in the figure.
1. Name the law which is related to refraction.
2. Locate the position of the image by completing the ray diagram.

3. With the source of laser at the centre of the hemisphere, redraw the ray diagram.
Answer:
1. Snell’s law
2.

3. The refracted ray is undeviated.

Question 2.
Figure (a) below snows the image observed at the near point of eye by a boy through a simple microscope. Eye focused on near point.

1. Draw ray diagram which shows the image formation at infinity, so that the boy can observe it with a relaxed eye.
2. Distinguish between linear magnification and angular magnification.

Answer:
1.

2. Linear magnification is ratio of image height to object height. Angular magnifications is the ratio of angle subtended by the image and the object on the eye when both are at the least distance of distinct vision.

Question 3.
Figure shows the path of the light rays through a glass slab.

1. Name the phenomena involved here.
2. Relate the values of n_1, n₂, i and r on the basis of one figure.
3. Copy the figure of glass and draw the path of ray when n₂ < n₁.

Answer:
1. Refraction

2. \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

3.

Question 4.
A light ray travelling from one medium to another medium is given in the figure.

1. Write a mathematical relation for this refraction.

• n₂ < n₁
• n₂ > n₁
• n₂ = n₁

2. What is a relation between angle of incidence, angle of refraction and refractive index of medium.

3. A flint glass rod when immersed in carbon disulfide is nearly invisible why?

Answer:
1. n₂ < n₁

2. \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

3. Refractive index of flint glass rod and carbon disulfide are nearly equal. Hence no refraction (or reflection) take place.

Question 5.
A convex lens and concave lens are placed as shown in figure. For convex lens f = 10cm for concave it is 5 cm

1. Is it converging or diverging why?
2. If f₁ = 5cm and f₂ =10cm What change will occur in the optical nature of system?

Answer:
1.

f = -10 cm
Effective focal length is negative. Hence this lens is diverging.

2. Effective focal length becomes positive- Hence the lens will act as converging.

Plus Two Physics Ray Optics and Optical Instruments Four Mark Questions and Answers

Question 1.
The maximum possible magnification for a simple microscope is 10

1. How do you increase the magnification further(1)
2. Draw the ray diagram for compound microscope and find an expression for magnification (3)
3. What is the advantage of forming image at infinity? (1)

Answer:
1. Use two convex lens instead of single lens.

2.

The magnification produced by the compound microscope

Multiplying and dividing by I₁M₁ we get,

Where m₀ & m_e are the magnifying power of objective lens and eyepiece lens.
∴ m = m_e × m₀ ______(1)
Eyepiece acts as a simple microscope.
Therefore \(\mathrm{m}_{\mathrm{e}}=1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\) ______(2)
We know magnification of objective lens
m₀ = \(\frac{V_{0}}{u_{0}}\) ______(3)
Where v₀ and u₀ are the distance of the image and object from the objective lens.
Substituting (2) and (3) in (1), we get

for compound microscope, u₀ » f₀ (because the object of is placed very close to the principal focus of the objective) and v₀ ≈ L, length of microscope (because the first image is formed very close to the eye piece).

where L is the length of microscope, f₀ is the focal length of objective lens.

3. Strain for eye, will be minimum when image is at infinity.

Question 2.
The refraction of light travelling from glass to water is shown in the figure.
1. The snells law in the above case can be written as………..

2. Show that c = \(\sin ^{-1}\left(_{g} n_{w}\right)\). Where C is the critical angle of glass water interface. (2)

3. Three light rays, (Red, blue and yellow) incident at one side and its refractions are shown in the figure. Copy the figure and mark Red, blue and yellow in the figure. (1)

Answer:
1.

2. In this i = c using snell law, we can write

3.

Question 3.
When a point object is placed in front of a spherical refracting surface an image is formed in the refracting medium.
1.

Complete the ray diagram to locate the position of the image.

2. Obtain the expression \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\) for the position of image inside refracting medium.

3. If the refracting surface is concave in nature, with the same set up, locate the position of the image by drawing a ray diagram.
Answer:
1.

2. Refraction at a spherical surface:

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

n₁i = n₂r ……….(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i-e., i = α + θ ………(2)
Similarly, from Δ IAC,
a = r + β
r = α – β ……..(3)
Substituting the values of eq(2) and eq(3)in eqn.(1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁β = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………….(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, α = \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP=-u, PI = +v and PC = R
Substituting these values, we get


Case -1:
If the first medium is air, n₁ = 1, and n₂ = n,

3.

Question 4.
A compound microscope consists of an objective lens of focal length 2cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at

1. Least distance of distinct vision.
2. infinity

Answer:
1. v_e = -25cm

∴ u_o = 5cm
Length of the tube, L= |v_o| + |u_e|
∴ v_o = 15-5 = 10

u_e = -2.5cm

2. ∴ v_o = 15 – f_e = 15 – 6.25 = 8.75

u_o = -2.59cm

Question 5.
You may be observed that, the fish inside the aquarium appears to be raised.
1. What is the reason for this phenomenon?
2. Obtain an expression for apparent shift offish.
3. What happens to the height of the object, (That vertically stands in the aquarium) when it is observed by the fish.

• becomes taller
• becomes smaller
• Does not change the height. Justify your answer.

Answer:

1. Refraction
2. Expression for apparent shift is not included in the syllabus
3. Becomes taller. When light enters from rare to denser medium, it deviates towards the normal.

Question 6.
High precision optical instruments uses prisms instead of mirror to reflect light.

1. Name the phenomena used for reflecting light using prism.
2. What is the advantage of using prism instead of mirror for reflecting light?
3. The critical angle of water is 52°. Calculate the refractive index of water.

Answer:
1. Total internal reflection

2. Prism can be used for total internal reflection. Mirrors can’t be used for total internal reflection.

3.

n = 1.26.

Question 7.
Two lenses of focal lengths f₁ and f₂ are placed in contact

1. If the object is at principal axis, draw ray diagram of the image formation by this lens.
2. Obtain a general expression for effective focal length in terms of f₁ and f₂.
3. How will you combine a convex lens of focal length f₁ and concave lens f₂ such that combination acts as

• Converging
• diverging
• plane glass plate

Answer:
1.

2. Obtain an expression for the effective focal length of the combination of two thin convex lenses in contact.

3.

• Keep in a medium of refractive index lower than that of lens.
• Keep in a medium of refractive index higher than that of lens.
• Keep in a medium of refractive index equal to refractive index of lens.

Question 8.
The light rays travelling from rarer to denser medium is given in the figure
1. Redraw the diagram and correct it

2. State the law relating i and r for retracted ray.
3. Velocity of light in water is 2.25 × 10⁸ m/s, If angle of incidence is 30° calculate angle of refraction.
Answer:
1.

2. Snells law:
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and for the given colour of light used. This constant is known as the refractive index of second medium w.r. t. the first medium.

Explanation: If ‘i’ is the angle of incidence in the first medium and ‘r’ is the angle of refraction in the second medium, then by Snell’s law,

Where ₁n₂ is the refractive index of the second medium with respect to the first medium. If the first medium is air, then sin i/sin r is known as absolute refractive index of the second medium.

where ‘n’ is the refractive index of the second medium.

3.

Question 9.

1. An air bubble inside an ice block shine brilliant by……… (Refraction, Reflection, total internal reflection)
2. Explain the above phenomenon.
3. The light ray incident at one face of the prism is shown in figure. Copy this figure complete the path of the ray. (Take critical angle of prism C = 42°)

Answer:
1. Total internal reflection.

2. Whenarayoflightpassesfromadenserto rarer medium, after refraction the ray bends away from the normal. If we increases the angle of incidence beyond the critical angle, the ray is totally reflected back to the denser medium itself. This phenomenon is called total internal reflection.

3.

Question 10.
A convex lens produces an inverted image of size 1.4cm The size of object is 0.7cm

1. What is magnification in the case
2. What is the nature of image
3. If the object is at distance 30 cm from the lens calculate focal length of the lens

Answer:
1.

2. Real, inverted, magnified

3.

Question 11.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure.

1. Identify the type of the mirror used in the diagram. (1)
2. If focal length of the mirror is 10cm, what is the distance CF in the figure? (1)
3. Complete the ray diagram and mark the angle of incidence and angle of reflection. (2)
4. If the mirror is immersed in water its focal length will be

• less than 10cm
• 10cm
• greater than 10cm
• 20cm

Answer:
1. Concave mirror.

2. 10 cm (CF = PC – PF = R = f – 2f – f = f = 10cm).

3.

4. 10cm. Focal length of mirror is independent of medium.

Question 12.
The given figure shows a compound microscope with two lenses PQ and RS.

1. Identify Objective and eyepiece in the microscope.
2. A compound microscope has a magnification of 30. The focal length of its eyepiece is 5cm. Assuming the final image to be formed at the least distance of distinctive vision, calculate the magnification produced by the objective.
3. What is the length of a compound microscope in normal adjustment?

Answer:
1. Objective – PQ, eyepiece – RS.
2. Magnification, M = m₀ × m_e

3. The length of a compound microscope in normal adjustment is f₀ + f_e.

Question 13.
A compound microscope consists of an objective lens of focal length 2cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at

1. Least distance of distinct vision.
2. infinity

Answer:
1. v_e = -25cm

∴ u₀ = 5cm
Length of the tube, L= |v₀| + |u_e|
∴ v₀ = 15 – 5 = 10

u_e = -2.5 cm.

2. ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75

u₀ = -2.59cm

Plus Two Physics Ray Optics and Optical Instruments Five Mark Questions and Answers

Question 1.
A point object at a distance of 36 cm from the convex lens of focal length 10cm, is moved by 10cm in 2 sec along principle axis towards the lens. Then image will also change its position.

1. Write the law which relates object and image distance from the lens.
2. Find the initial and final position of the image and calculate average speed of image.
3. A man argues that the image will move uniformly at the same speed as that of object. What is your opinion? Justify.

Answer:
1. The lens equation \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

2. u = -36, f = 10
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

If object is moved 10cm towards the lens we can find position
u = -26, f=10
v = 7.2 cm
Speed = \(\frac{7.8-7.2}{2}\) = 3cm/sec.

3. Comparing speed of object and image we can arrive at conclusion that the argument of man is false, speed of image is different from speed of object.

Question 2.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure.

1. Identify the type of the mirror used in the diagram.
2. If focal length of the mirror is 10cm, what is the distance CF in the figure?
3. Complete the ray diagram and mark the angle of incidence and angle of reflection.
4. If the mirror is immersed in water its focal length will be

• less than 10cm
• 10cm
• greater than 10cm
• 20cm

Answer:
1. Concave mirror.

2. 10 cm (CF = PC – PF = R = f – 2f – f = f = 10cm).

3.

4. 10cm. Focal length of mirror is independent of medium.

Question 3.
The image formed by a thin lens is shown in the figure.

1. What is the nature of the image
2. Find out the power of the image
3. Draw the ray diagram showing above lens forming a magnified erect, virtual image
4. If a convex lens of focal length 20cm is kept in contact with above lens. What is the focal length and power of the combination

Answer:
1. Inverted.

2. P= \(\frac{1}{1}\) =ID.

3.

4. P = P₁ + P₂
in this case f₁ = 1 m, f₂ = 0.2m

Question 4.
A beam of light passing from one transparent medium to another obliquely, undergoes an abrupt change in direction. This phenomenon is known as refraction of light.

1. Name the law which satisfies during this refraction.
2. Draw a figure, which shows refraction through a parallel sided glass slab (Ray passing from air)
3. Using the figure obtained in (b), show that the incident ray and the emergent ray are parallel to each other. Redraw the same figure, if the light is entering from a medium denser than glass. Justify your answer.

Answer:
1. Snell’s law.

2.

3. This derivation is out of syllabus.

4. The light bends away from the normal if light enter from glass to water.

Question 5.
A group of students are given a project for constructing a telescope and they were supplied with two biconvex lens of power 1 diopter and 0.1 dioptre.

1. Of the two lenses, which can be used as objective?
2. Draw the ray diagram for the formation of the image by a telescope.
3. Arrive at an expression for magnification of a telescope.
4. Prepare a notice/ label about the precaution to be taken while using the telescope and limitations of the telescope constructed.

Answer:
1. Biconvex lens of power 0.1 dioptre.

2.

3. Magnification:
The magnifying power of a telescope is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.
\(\mathrm{m}=\frac{\text { angle subtented by the image at eye (eye piece) }}{\text { angle subtended by the object at the objective }}\)
i.e. m = \(\frac{β}{α}\) …….(1) [from figure]
But from ∆CIM, tanα = \(\frac{IM}{IC}\), α = \(\frac{IM}{IC}\)
(For small values tan α ≈ α)
from ∆C¹IM, tanβ = \(\frac{IM }{\mathrm{IC}^{1}}\) , β = \(\frac{IM }{\mathrm{IC}^{1}}\)
substituting α and β in eq (1) we get

But IC = f_o (the focal length objective lens) and IC^l = f_e (the focal length eyepiece lens.)

In this case the length of the telescope tube is (f₀ + f_e).
Case 1 :
When the image formed by the objective is within the focal length of the eyepiece, Then the final image is formed at the least distant of distinct vision. In this case, magnifying power.

4.

• As magnifying power is negative, the final image in an astronomical telescope is inverted.
• To have large magnifying power, f_o must be as large as possible and f_e must be as small as possible.
• As intermediate image is between the two lens, cross wire (ora measuring device) can be used.
• In normal setting of telescope, the final image is at infinity.

Question 6.
The following graph represent id curve of a optical instrument placed in air.

1. Name the device which give the above i-d curve.
2. Obtain an expression for deviation produced by such a device.
3. What is the relevance of the value ‘D’? Arrive at an expression for refractive index in terms of this value from (b).
4. How is the deviation affected if the above arrangement is immersed in a liquid of refractive index less than that of the above device.

Answer:
1. Prism.

2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.

The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠A + ∠R = 180°
[since N₁M are normal]
ie, ∠A + ∠M = 180°____(1)
In the ∆ QMR,
∴ r₁ + r₂ + ∠M = 180° ____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ____(3)
From the ∆ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d

3. At minimum deviation D = 2i – A, r₁ = r₂ = r

4. Deviation decreases.

Question 7.
\(\frac{1}{f}=\left(\frac{n_{2}}{n_{i}}-t\right)\left(\frac{1}{R_{i}}-\frac{1}{R_{2}}\right)\) is lens maker’s formula.
1. Write down lens maker’s formula for a convex lens.
2. “If a convex lens is immersed in water its converging power decrease. Do you agree with it? Justify your answer.
3. A convex lens of refractive index n₂ is placed in different media. Explain optic behavior in each. If n₁ is refractive index of surrounding media.

• in medium with n₂>n₁
• in a medium with n₂ < n₁
• in a medium n₁ = n₂

Answer:
1. For convex lens R₁ = +ve and R₂ = -ve

2. Yes.

From the above equation it is clear that, \(\mathrm{P} \alpha \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\)

In water, \(\frac{n_{2}}{n_{1}}\) is less. Hence power decreases.

3. P = +ve, converging
P = -ve, diverging
P = 0, Plane glass

Question 8.
Two convex lense are given in the figure A and figure B

1. Which has more curvature
2. Which has more power
3. Which lens produce more magnification
4. which lens has less focal length
5. Can these lenses act as diverging lenses in any condition?

Answer:

1. A
2. A
3. A
4. A
5. Yes, If we place this lens in a medium of higher refractive index than lens.

Question 9.
In the figure given below, PQ represents an incident ray falling in the side AB of a prism, when monochromatic light is used

1. Draw the refracted ray, emergant ray and mark the angle of deviation
2. Derive an equation for refractive index of the material of the prism in terms of angle of minimum deviation
3. Draw the incident ray and refracted ray, at the angle of minimum deviation

Answer:
1.

2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the faceAB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.

The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠A + ∠R = 180°
[since N₁M are normal]
ie, ∠A + ∠M = 180°____(1)
In the ∆ QMR,
∴ r₁ + r₂ + ∠M = 180° ____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ____(3)
From the ∆ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d
(i₁ + i₂) = d + A ____(4)
It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’.
At the minimum deviation position,
i₁ = i₂ =i, r₁ = r₂ = r and d = D
Hence eq (3) can be written as,
r + r = A
or r = \(\frac{A}{2}\) ____(5)
Similarly eq (4) can be written as,
i + i = A + D
n = \(\frac{A + D}{2}\) ____(6)
Let n be the refractive index of the prism, then we can write,
n = \(\frac{sin i}{sin r}\) ____(7)
Substituting eq (5) and eq (6) in eq (7),
\(n=\frac{\sin \frac{A+D}{2}}{\sin \frac{A}{2}}\)
i – d curve:

It is found that when the angle of incidence increases deviation (d) decreases and reaches a minimum value and then increases. This minimum value of the angle of deviation is called the angle of minimum deviation.

3. Refracted ray is parallel to base

Question 10.
Refraction of a ray of light at a spherical surface separating two media having refractive indices n₁ and n₂ is shown in the figure.

1. Which of the two media is more denser?
2. In the figure, show that\(\frac{\mathrm{n}_{1}}{\mathrm{OA}}+\frac{\mathrm{n}_{2}}{\mathrm{AI}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{AC}}\).
3. Using the above relation arrive at the thin lens formula.
4. An object is placed on the principal axis of a convex lens at a distance 8 cm from it. Find the magnification of the image if the focal length of the lens is 4 cm.

Answer:
1. Refractive index of medium 2 is greater than medium 1.

2. Refraction at a spherical surface

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

n₁i = n₂r ……….(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i-e., i = α + θ ………(2)
Similarly, from Δ IAC,
α = r + β
r = α – β……..(3)
Substituting the values of eq(2) and eq(3)in eqn. (1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁β = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………….(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, a= \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β, and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP=-u, PI = +v and PC = R
Substituting these values, we get


Case -1:
If the first medium is air, n₁ = 1, and n₂ = n,

3. Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁ Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₁.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁ Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,

This image I₁ will act as the virtual object for the surface ADC and forms the image at v.
Then we can write,

Dividing throughout by n₁, we get

if the lens is kept in air, \(\frac{n_{2}}{n_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when
U = ∞, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula

For convex lens,
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula From eq(4),

4. u = -8cm, f = +4cm
m = \(\frac{f}{f+u}=\frac{4}{4+-8}\)
m = -1.

Question 11.
Two lenses L₁ and L₂ are placed in contact as shown in figures. The focal length of each lens is 10cm

1. What is power of L₁
2. What is power of L₂
3. What is effective focal length of combination
4. “The power of convex is greater than that of concave and combination can act as a diverging lens”. Is this statement true in any situation? Explain?

Answer:
1.

2.

3.

The combination will act as plane glass.

4. This is true statement. If we place the above combination in a medium of refractive index greater than this condition.

Question 12.
A lens of particular focal length is made from a given glass by adjusting radius of curvature. The formula applied in this case is lens maker’s formula
1. Write down lens maker’s formula
2. Derive lens maker’s formula considering refraction at a spherical surface
3. Explain the following facts based on lens maker’s formula

• power of sun glasses is zero even though they are curved
• if a lens is immersed in water focal length increases

Answer:
1.

2. Refraction by a lens:
Lens Maker’s Formula (for a thin lens): Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁. Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R,) forms the image at l₁. Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,

Adding eq (1) and eq (2) we get

Dividing throughout by n₁, we get

if the lens is kept in air, \(\frac{n_{2}}{n_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when u = ∞, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ____(5)
For convex lens.
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula
From eq(4),

From eq(5)

From these two equations, we get

Linear magnification :
If h_o is the height of the object and h_i is the height of the image, then linear magnification.

3.
a. R₁ = R, R₂ = +R

power of lens, P = 0

b. We know

The above equation shows when n₁ increases f decreases the refractive index of water is greater than air. Hence when we place a lens in water, focul length decreases.

Students can Download Chapter 5 Market Equilibrium Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium

Plus Two Economics Market Equilibrium One Mark Questions and Answers

Question 1.
Complete the statement given below. Free entry and of firms imply that the market price will always be equal to ……………
Answer:
Minimum average cost (P = Min. AC)

Question 2.
Choose the correct answer. The imposition of price ceiling below the equilibrium price leads to ……….
Answer:
Excess demand

Question 3.
Market equilibrium of a commodity shows,
(a) excess demand
(b) quantity demanded greater than quantity supplied
(c) quantity demanded equals quantity supplied
(d) excess supply
Answer:
(c) quantity demanded equals quantity supplied

Question 4.
When there is increase in demand, the demand curve.
(a) shifts reight ward
(b) shifts leftward
(c) shifts downward
(d) remains constant
Answer:
(a) shifts reight ward

Question 5.
The government imposing upper limit on the price of a good or service is called:
(a) price floor
(b) price ceiling
(c) equilibrium price
(d) fair price
Answer:
(b) price ceiling

Plus Two Economics Market Equilibrium Two Mark Questions and Answers

Question 1.
At what price – higher or lower than the equilibrium price, there will be excess demand?
Answer:
When the market price is lower than the equilibrium price, there will be excess demand.

Question 2.
Make pairs.
Price floor, below equilibrium price, above equilibrium price, price ceiling
Answer:

• Price floor – above equilibrium price.
• Price ceiling – below equilibrium price.

Question 3.
Point out the consequences of price ceiling.
Answer:

1. Black marketing
2. Malpractices by fair price shops
3. Sale of inferior quality goods.

Question 4.
What do you mean by control price?
Answer:
Fixation of price of a commodity at a lower level than equilibrium price is called control price. Control price is determined to protect the interest of the consumers.

Question 5.
Market equilibrium of apple is given below in the diagram below.

1. Define market equilibrium
2. Find out the market price and equilibrium quantity

Answer:

1. market equilibrium is a situation where quantity demanded is exactly equal to the quantity supplied.
2. The price of apple is ₹40 and the equilibrium quantity is 30kg.

Plus Two Economics Market Equilibrium Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
The demand function and supply function of a product are given as q^D = 60 – P for 0 = P = 60 q^S = 30 + P for P > 10 Calculate equilibrium price.
Answer:
The demand and supply functions are given as
q^D= 60 – P for 0 = P = 60
q^S = 30 + P for P > 10
Equilibrium price is considered as the price at which quantity demanded is exactly equal to quantity supplied. Therefore we get.
60 – P = 30 + P
60-30 = 2 P
30 = 2P
P = 30/2 = 15
Therefore equilibrium price is ₹15.

Question 3.
Complete the following statements.

1. Long-run price under perfect competition will be equal to ……….
2. Minimum price fixed by government for a product is known as ……….
3. Maximum price fixed by government for a product is known as ………

Answer:

1. average cost
2. floor price
3. price ceiling

Question 4.
Demand curve for labour is downward sloping. Explain
Answer:
Demand curve for labour is downward sloping indicating that more and more labour is demanded at lower wages. This is due to the operation of declining marginal productivity of labour. Marginal productivity of labodr declines due to the operation of diminishing returns. That is why the demand curve for labour slopes downward.

Question 5.
The market demand function and market supply functions are given as, Find the equilibrium price and equilibrium quantity.
q^D = 200 – P for 0 = P = 200
q^S = 120 + P for P > 10
Answer:
We find equilibrium price by equating market demand function and market supply functions as shown below.
q^D = q^S
200-P = 120 + P
2P = 80
P = 80/2 = 40
Therefore equilibrium price is ₹40. Equilibrium quantity is obtained by substituting the equilibrium price into either the demand or supply function equations. Applying the value of price ₹40 in demand equation we have.
q^D= 200 – P
q^D =200 – 40 = 160
Therefore equilibrium price is ₹40 and equilibrium quantity is 160.

Question 6.
Prepare a note on market equilibrium.
Answer:
Equilibrium is defined as a situation where the plans of all consumers and firms in the market match and the market dears. In equilibrium, the aggregate quantity that all firms wish to sell equals the quantity that all the consumers in the market wish to buy; in other words, market supply equals market demand. The price at which equilibrium is reached is called equilibrium price and the quantity bought and sold at this price is called equilibrium quantity.

Therefore, q^D (P*) = q^S (P*) where P* denotes the equilibrium price and q^D (P*) and q^S (P*) denote the market demand and market supply of the commodity respectively at price P*

Question 7.
Suppose the demand and supply functions of commodity X are given by, Qd = 500 + 3P and Qs = 700 – P Qd = 500 + 3P Qs = 700 – P Find out the equilibrium price and quantity demanded and supplied.
Answer:
Equilibrium price and quantity can be determined by equating the demand and supply functions
Qd = Qs
500 + 3P = 700 – P
4P = 200
\(P=\frac{200}{4}=50\)
Equilibrium price is ₹50. Applying the price in the demand function, we get
500 + 3 × 50
500 + 150 =650
Therefore, equilibrium price is ₹50 and quantity is 650.
Qd = Qs
500 + 3P = 700 – P
4P = 200
\(P=\frac{200}{4}=50\)
Qd = 500 + 3 × 50
= 500 + 150 = 650

Question 8.
The diagram below illustrates the supply and demand for television sets. The original demand curve is D₂

Using the diagram, state the new demand curve (D₁ or D₃) which will apply after each of the following changes taken place. (The same answer may be used more than once)

1. A successful advertising campaign for television sets occurs
2. Income decreases
3. An increase in the population

Answer:

1. Demand increases (D₂ curve shifts right to D₃)
2. Demand decreases (D₂ curve shifts left to D₁)
3. Demand increases (D₁ curve shifts right to D₃)

Question 9.
Calculate equilibrium price and quantity based on the following information.
q^d= 400 – P (1)
q^s= 240 + 3 (p – 4) (2)
Answer:
At equilibrium,
q^d = q^s
Putting the values,
400 – p = 240 + 3(p – 4)
400 – p = 240 + 3p – 12
400-240 + 12 = 3p + p
172 = 4p
\(p=\frac{172}{4}\)
p =43
Putting p = 43 in the first equation, we get,
q^d =400 – 43 = 357 Therefore, equilibrium price = 43 and
equilibrium quantity is = 357 units

Question 10.
The diagram shows relationship between two commodities A and B.

1. Identify the commodities A and B
2. Explain what happens to the price and quantity demanded of A when the price of A falls.

Answer:

1. A and B are substitutes.
2. When the price of A falls people will demand more A. So the demand for B will fall. That will result in a decrease in the price of B.

Question 11.
The diagram below shows one of the government intervention programmes in the market.

1. Identify the programme and calculate the excess supply.
2. Explain how the government is monitoring the higher price fixed.

Answer:
1. Minimum price/floor price, 40 unit excess supply.

2. The government announces the minimum price above the market price. As a result of this intervention there occurs excess supply in the market. The government has to remove the excess from the market to maintain the price. So the government store the excess supply in the warehouses and redistribute it at the time of shortage.

Question 12.
Under perfect competition, a market for a good is in equilibrium. There is simultaneous “decrease” both in demand and supply, but there is no change in market price. Explain with the help of a diagram how it is possible.
Answer:
The decrease in demand and supply is the same, and hence the price remain the same. Shows this by drawing appropriate diagram

Question 13.
The diagrams below indicate four possible shifts in demand or in supply that could happen in particular markets. Relate each of the events described below to one of them. Also, give reason for the shift.

1. How does the lorry strike in Karnataka and Tamil Nadu affect the market for vegetables in Kerala?
2. People become aware of the fact that Birds Eye Chilly is very much helpful to prevent Cholesterol. What happens to the market for Birds Eye Chilly?
3. How do you think the rising income affect the market for fish?
4. A new technique is discovered for manufacturing computer that greatly lowers their production cost. What happens to the market for computers?

Answer:

1. figure C, supply falls and price rises.
2. figure A, demand increases and price rises.
3. figure B, demand increases and price rises.
4. figure D, supply increases and price falls.

Plus Two Economics Market Equilibrium Five Mark Questions and Answers

Question 1.
Mention the impact of the following.

1. Imposition of price ceiling below equilibrium price
2. Imposition of price floor above the equilibrium price

Answer:

1. Imposition of price ceiling below equilibrium price leads to excess demand.
2. Imposition of price floor above the equilibrium price leads to an excess supply

Question 2.
Complete the following table to show the impact of simultaneous shifts of demand and supply on equilibrium price and quantity.

Answer:

Question 3.
What will happen if the price prevailing in the market is?

1. above the equilibrium price
2. below the equilibrium price.

Answer:
1. If the price prevailing in the market is above equilibrium price, supply will exceed demand. Under such a situation some firms will not be able to sell their desired quantity; so they will lower their price. All other things remaining constant as price falls quantity demanded rises quantity supplied falls, and finally equilibrium price P* will be restored. At P* quantity demanded will be equal to quantity supplied.

2. If the price prevailing in the market is above equilibrium price, demand will exceed supply. Under such a situation some consumers will be ready to pay more prices to get the commodity. This will tend to increase the price. All other things remaining constant as price rises quantity demanded falls, quantity supplied rises, and finally, equilibrium price P* will be restored. At P* quantity demanded will be equal to quantity supplied.

Question 4.
Draw distinction between floor pricing and price ceiling.
Answer:
Floor price means minimum price. Floor price is fixed to protect producers like farmers from price crashes. It ensures a remunerative price to producers. In India, floor prices are fixed for a variety of agricultural commodities like paddy, wheat, coconut, rubber etc.

On the other hand, price ceiling mean maximum price. It is the maximum price fixed by the government. The aim of price ceiling is to protect consumers. Government fixes price ceiling for essential products and medicines to protect the interests of the consumers.

Question 5.

1. Identify the situations depicted in the following figures in panel A and B.
2. Why do such policies are followed and explain the impact of such policies?

Answer:
1. PANEL A – Price ceiling PANEL B –Price floor.

2. Price ceiling is fixed below equilibrium price. Imposition of price ceiling at ‘Pg’ gives rise to excess demand in the market price floor is fixed above equilibrium price. Imposition of floor price at ‘pg’ gives rise to excess supply.

Question 6.
Mention the factors that cause shift in the supply curve.
Answer:
The factors that cause shift in the supply curves are:

1. The change in the number of firms
2. The change in the price of factor inputs
3. Change in production technology
4. Change in the prices of related goods
5. Change in production tax.

Question 7.
How will a change in price of coffee affect the equilibrium price of tea? Explain the effect on equilibrium quantity through a diagram.
Answer:
Coffee and tea are substitutes. If prices of coffee are increased then its demand will decrease and demand tea would increase.lt will shift the demand curve of tea upwards. The equilibrium price and quantity will increase. This is shown in the following diagram.

In the diagram, when the price of coffee increased then the demand for tea increases. This has resulted in equilibrium price and quantity of tea.

Question 8.
How is price determined in labour market?
Answer:
The price of labour is determined by the forces of demand and supply of labour. The households are the suppliers of labour and demand for labour comes from firms. Labour means the hours of work provided by labourers. The wage rate is determined at the intersection of the demand and supply curve of labour.

The firm being a profit maximiser will always employ labour up to the point where the extra cost it incurs for the last labour is equal to the additional benefit he earns from employment that labour. The extra cost of hiring one more labour is the wage rate. For each extra unit of labour, he gets a benefit equal to marginal revenue product of labour.

Thus firm employs labour up to a point where: W = MRPL Where MRPL = MR x MPL As long as MRPL is greater than the wage rate the firm will earn more profit by hiring one more labour and if at any level of labour employment MRPL is less than the wage rate the firm can increase here profit by reducing labour employed.

Question 9.
Prepare a table showing differences between price ceiling and price floor.
Answer:

Question 10.

1. With the help of a diagram show how the wage rate is determind in a free market.
2. Analyse the impact of an increased entrance of foreign migrant labourers into the labour market.

Answer:
1. The diagram below shows how the wage rate of labour in a free market is determined. DL is the demand for labour and SL is the supply of labour, ‘e’ is the point of equilibrium, ‘ow’ is the wage rate and ‘oq’ is the quantity of labours.

2. When the foreign migrant labour enters into the labour market, the supply of labour will shift rightward and the wage rate will come down as shown in the figure. ‘ow’ is the original wage rate and ‘OQ’ is the original quantity of labour. ow1 is the new wage rate and OQ, is the new quantity of labourers.

Question 11.
Suppose we have two equations, one for demand and other for supply.
Demand equation: Qx^d = 100 – 10P_x
Supply equation : Qx^s = 60 + 10P_x

1. Calculate equilibrium price and quantity using the equations.
2. Construct demand and supply schedules by assigning various prices. Obtain equilibrium price and quantity graphically.

Answer:
1. Equilibrium price =2, equilibrium quantity=80
2. Demand & supply schedules

Question 12.
Let us take market of commodity ‘X’, which is in equilibrium. Suppose demand for the commodity increases. Explain the chain of effects of this change till the market again reaches equilibrium. Use diagram.
Answer:
Increase in demand leads to disequilibrium-price in-creases – super profit – new firms enter the industry – or existing firms expand production – increase in output – supply increases – supply curve shifts – the process continue until price returns the to the equilibrium level. Draws the diagram, and explains the process.

Plus Two Economics Market Equilibrium Eight Mark Questions and Answers

Question 1.
Observe the following table.

1. Find equilibrium price.
2. Fill the fourth column
3. Why ₹35 and ₹40 are not equilibrium prices?
4. Product surplus drives prices up and shortage drives them down. Do you agree?
5. Draw a diagram of the above table showing the equilibrium price determination

Answer:
1. The equilibrium price is ₹37. At this price both demand and supply are equal.
2.

3. At ₹35, demand exceeds the supply causing a shortage in the market. At ₹40, supply exceeds demand causing surplus. Therefore, these prices are equilibrium prices.
4. No. I do not agree with this argument.
5.

Question 2.
Discuss the impact of the factors mentioned below on equilibrium price and quantity.

1. shift in demand to right
2. shift in demand to left
3. shift in supply to right
4. shift in supply to left.

Answer:
1. When demand curve shifts to right (increase in demand), there will be increase in equilibrium price and increase in equilibrium quantity. This change is shown in the diagram.

2. When demand curve shift to left (decrease in demand), both equilibrium quantity and equilibrium price falls. This is shown in the diagram.

3. When supply curve shift to right (increase in supply), the equilibrium price deceases and the equilibrium quantity increases. This is given in the following diagram.

4. When supply curve shifts to left (decrease in supply), the equilibrium price increases and the equilibrium quantity decreases. This is given in the following diagram.

Question 3.
Suppose the demand and supply curves of salt are given by. q^D = 1000 – P q^S = 700 + 2P

1. Find the equilibrium price and quantity
2. Suppose that the price of input used to produce salt has increased so that the supply curve is q^S = 400 + 2P How does the equilibrium price and quantity change? Does the change conform to your expectation?
3. Suppose the government has imposed a tax of 3 per unit of salt. How does it affect the equilibrium price and quantity?

Answer:
1. equilibrium price and quantity
q^D = 1000 – P
q^S = 700 + 2P
For equilibrium
q^D = q^S
1000 – P = 700 + 2P
1000 – 700 = 3 P
3P = 300
P = 300/3 = 100
Put the value of P in supply equation
q^S = 700 + 2P
q^S = 700 + 2×100
q^S =700 + 200 = 900
Therefore the equilibrium price = ? 100 and the equilibrium quantity is = 900 units

2. For equilibrium
q^D = q^S
1000 – P = 400 + 2P
1000 – 400 = 3P
600 = 3 P
P = 600 / 3 = 200
Put the value of P in demand equation
Q^D= 1000 – P
Q^D = 1000  – 200 = 800
Therefore the equilibrium price = ₹200 and the equilibrium quantity is = 800 units This change confirms to our expectations, i.e., rise in input prices raises prices and lowers supply.

3. q^D= 1000 – P
q^S= 700 + 2P
When ₹3 as tax is imposed on sale of salt the new demand and supply function will change
q^D= 1000 – (P + 3)
q^S= 700 + (2P +3)
In part A equilibrium price was ₹100 which goes up to ₹103 with imposition of tax
q^D = 1000 – (100 + 3) = 1000 – 103 =897
q^S = 700 + (2P + 3)
= 700 + 2(100 + 3)
= 700 + 2×103
= 700 + 206 = 906
q^D < q^S
Therefore, new price and quantity has to be adjusted.

Question 4.
The diagram below shows how the price of wheat is determined in a free market.
a. Show in a seperate diagram the changes on price and quantity demanded of wheat due to the following factors

1. The price of fertilizers increases.
2. The price of rice a substitute of wheat increases.

b. Assess the impact of an increased demand for wheat and an increase in its production.

Answer:
a. price and quantity demanded of wheat.
1. When the price of fertilizers increases the supply of wheat decreases. Its price increases and the quantity falls.

2. When the price of rice, a substitute of wheat increases people may switch to consume wheat, this will increase the demand for wheat. Its price will increase and quantity also will increase.

b. When the demand for heat increases its demand curve will shift rightward. When production increases its supply curve will shift rightward as shown in the diagram below.

Due to these shifts, the quantity will increase anyhow. But the effect on the price will be in different forms. The price may fall, will be constant or even may increase. Whether the price will increase, decrease or remain constant is determined by the respective shifts in demand and supply.

If both demand and supply shift in the same magnitude the price will be same. If the shift in the demand is more than the supply the price will increase. And if the shift in the supply is more than the shift in the demand the price will fall.

Students can Download Chapter 10 Haloalkanes and Haloarenes Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 10 Haloalkanes and Haloarenes

Plus Two Chemistry Haloalkanes and Haloarenes One Mark Questions and Answers

Question 1.
Alkolic KOH is a specific reagent for
(a) dehydration
(b) dehalogenation
(c) dehydrohalogenation
(d) dehydrogenation
Answer:
(c) dehydrohalogenation

Question 2.
Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas called ………………
Answer:
carbonyl chloride/phosgene

Question 3.
Which of the following represents a gem di halide?
(a) ethylene dichloride
(b) 2,2-dichloropropane
(c) 1,3-dichloropropane
(d) 1,2-dichloropropane
Answer:
(b) 2,2-dichloropropane

Quesiton 4.
Here ‘A’ is
(a) phenol
(b) sodium phenoxide
(c) benzene
(d) cyclohexyl chloride
Answer:
(b) sodium phenoxide

Question 5.
There are _________ structural isomers of C₄H₉Br.
Answer:
Four

Question 6.
Name the insecticide prepared from chloral and chloro benzene.
Answer:
DDT

Question 7.
The reaction between arylhalide and alkylhalide in the presence of sodium and dry ether.
Answer:
Wurtz-Fittig reaction

Question 8.
Name the substance which is used as anaesthetic.
Answer:
Chloroform

Question 9.
Name an alkyl magnesium halide.
Answer:
Methyl magnesium chloride (CH₃MgCI) – Grignard reagent

Question 10.
Compounds in which the halogen atom is directly attached to an aromatic ring carbon.
Answer:
Aryl halides

Question 11.
Even though alkylhalides are polar in nature, they are insoluble in water. Comment on it.
Answer:
This is because alkylhalides can neither make or break hydrogen bonds with water molecules.

Question 12.
From the following select those compounds which are used for the preparation of alkyl halide?

Answer:
SOCI₂, HCI, and ZnCI₂

Question 13.
When (-) 2methyl butan-1-ol is heated with con. HCI +1-chloro-2 methyl butane is obtained. This reaction is an example of
(a) retension
(b) invission
(c) racemisation
(d) resolution
Answer:
(a) retension

Question 14.
If alkaline hydrolysis of a tertiary alkyl halide by aqeous alkali, if concentration of alkali is doubled then the reaction rate at constant temperature will be ___________
Answer:
will be tripled.

Question 15.
The organic compound used as feedstock in the synthesise of chlorofluorocarbon is ___________
Answer:
CCI₄

Question 16.
DDT is prepared from _____________
Answer:
Chlorobenzene and B.H.C

Question 17.
Chlorination of benzene in the presence of halogen is an example of ____________
Answer:
aromatic electrophilic substitution

Plus Two Chemistry Haloalkanes and Haloarenes Two Mark Questions and Answers

Question 1.
Write the preparation of extra pure alkyl halide from ethyl alcohol.
Answer:
When ethyl alcohol is heated with thionyl chloride, chloroethane is obtained. Here the byproducts are in gaseous state and hence this method is used for the preparation of extra pure alkyl halide.
CH₃-CH₂-OH + SOCI₂ → CH₃CH₂CI + HCI + SO₂

Question 2.
A student was treating iodoform with silver nitrate and he got yellow precipitation. Then he used chloroform instead of Iodoform.

1. Will he get the earlier result?
2. Why?

Answer:

1. No
2. C-I bond in iodoform is weaker than C-CI bond of chloroform. So C-I bond of iodoform is easily broken to form yellow precipitate of AgI when heated with AgNO₃ solution.

Question 3.

From the box, write the raw materials of chloroform and Iodoform.
Answer:
Chloroform – ethyl alcohol and bleaching powder Iodoform – Sodium carbonate, ethyl alcohol and iodine

Question 4.
During a class room discussion a student 1 argued: “When chlorobenzene is allowed to react with metallic sodium in the presence of dry ether medium, diphenyl is obtained”. Then student 2 countered: “This reaction will take place in the presence of ale. KOH”.

1. Whom you will support?
2. Name the reaction? Explain it.

Answer:

1. Student 1
2. Fittig reaction

When aryl halide is allowed to heat with sodium in dry ether medium, diaryl is obtained, or

Question 5.
Analyse the following statements:
Statement 1: Alkyl halides are polar compounds.
Statement 2: Alkyl halides are insoluble in water because alkyl halides are non-polar compounds. What is your opinion? Explain it.
Answer:
Even though alkylhalides are polar compounds, they are insoluble in water. Because they can neither form hydrogen bonds with water nor break the hydrogen bonds existing between water molecules.

Question 6.
Which of the following has the highest dipole moment?

1. CH₂CI₂
2. CHCI₃
3. CCI₄

Answer:
Dipole moment of CH₂CI₂ (1.60 D) is the highest. The dipole moment of CCI₄ is zero while that of CHCI₃ is 1.03 D. The dipolement of CHCI₃ is less than that of CH₂CI₂ because the bond dipole of third C-CI bond opposes the resultant of bond dipoles of the other two C-CI bonds.

Question 7.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉CI in bright sunlight. Identify the hydrocarbon.
Answer:
Since the hydrocarbon gives only one monochloro compound, it indicates that all the hydrogen atoms in the hydrocarbon are equivalent. Thus, the compound is cyclopentane.

Question 8.
Which of the following has the highest dipole moment? CH₂CI₂, CHCI₃, CCI₄. Justify.
Answer:
The dipole moment of CH₂CI₂ (1.6 D) is the highest. The dipole moment of CCI₄ is zero while that of CHCI₃ is 1.03 D. The dipole moment of CHCI3 is less than that of CH₂CI₂ because the bond dipole of third C-CI bond opposes the resultant of bond dipoles of the other two C-CI bonds.

Question 9.
Which alkyl halide from the following pair would you expect to react more rapidly by SN² reaction mechanism? Justify your answer. CH₃-CH₂-CH₂-CH₂-Br or CH₃-CH₂-CH(Br)-CH₃
Answer:

Presence of bulky groups around carbon atoms induce an inhibitory effect. Bulky group hinders the approaching nucleophiles in SN² mechanisam.

Question 10.
Explain the stereochemical aspects of SN¹ and SN² reactions selecting suitable example.
Answer:
S_N¹ reaction:
In the case of optically active alkyl halides, S_N¹ reactions are accompanied by racemisation. This is because the attack of the nucleophile may be accomplished from either side resulting in a mixture of products, one having the same configuration and the other having opposite configuration.

S_N² reaction:
In the case of optically active alkyl halides, there is inversion of configuration. This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present

Question 11.
Arrange the following compounds in the order of reactivity towards SN² displacement. 2-Bromo-2-methyl butane, 1-Bromopentane 2-Bromopentane
Answer:
1-Bromopentane > 2-Bromo-pentane > 2-Bromo-2- methyl butane

Question 12.
What happens when methyl bromide is treated with sodium in presence of dry ether? Write the chemical equation and name the reaction.
Answer:
Ethane is formed.

called Wurtz reaction.

Question 13.
Chloroform is stored in black coloured bottles. Why?
Answer:
In presence of sunlight chloroform undergoes oxidation to form carbonyl chloride (phosgene).

Plus Two Chemistry Haloalkanes and Haloarenes Three Mark Questions and Answers

Question 1.
Raju heated the test tube containing ale. KOH and primary amine with one compound. A foul smell is obtained.

1. What is the compound?
2. Name the foul smelling product obtained.
3. Name the reaction.

Answer:

1. Chloroform
2. Isocyanide/Carbyl amine
3. Carbyl amine reaction/lsocyanide test

Question 2.
Nitration is one of the ring substitution reactions.

1. How it will be carried out?
2. What are the products obtained when chloro-benzene is heated with the nitrating mixture?
3. What is the difference if the same compound is heated with fuming H₂SO₄?

Answer:
1. By heating with nitrating mixture (mixture of conc.HN0₃ and conc.H₂O₄)

2. A mixture of 1-Chloro-2-nitrobenzene (minor product) and 1-Chloro-4-nitrobenzene (major product) is obtained.

3. If chloro benzene is heated with fuming H₂So₄ sulphonation will take place resulting in the formation of a mixture of 2-Chlorobenzene sulphonic acid (minor product) and 4- Chlorobenzene sulphonic acid (major product).

Question 3.
Consider the reaction: RX + Mg → RMgX

1. Identify the compound ‘RMgX’.
2. Explain the reaction.

Answer:

1. Grignard Reagent
2. When alkyl halides are treated with magnesium in the presence of dry ether, alkyl magnesium halide is obtained.

Question 4.
Chloroform kept in brown coloured bottles filled up to the neck.

1. What is the reason for this?
2. Few drops of 1% ethyl alcohol is added to chloroform to be kept for long. Give reason.

Answer:
1. In the presence of sunlight chloroform undergoes oxidation to form carbony chloride or phosgene, a highly poisonous gas.

This reaction can be avoided by storing it in dark bottles, completely filled upto brim. The use of brown bottles cuts off active light radiations and filling upto brim keeps out air. So chloroform is kept in brown bottles.

2. Addition of a little ethanol fixes the toxic COCI₂ as non-poisonous diethyl carbonate.

Question 5.
Fill in the blanks:

Answer:

1. CH₃CH₂I
2. Ethyl Iodide
3. CH₃-CHBr-CH₃
4. 2-Bromopropane
5. Iso-butyl chloride
6. 1-Chloro-2-methylpropane

Question 6.
In a Chemistry class, teacher explained that when benzene diazonium chloride is allowed to react with cuprous chloride and HCI, Chloro benzene is obtained. Then teacher asked

1. If we use copper powder instead of cuprous chloride. What will be the product?
2. Write the name of the reaction.
3. Write the equation which shows the reaction between benzene diazonium chloride with copper powder and HCI.

Answer:
1. Chlorobenzene is obtained.

2. Gattermann reaction.

3.

Question 7.
In a Lab, one student took a compound in test tube and he added iodine and alkali. He notices a yellow precipitate.

1. Write the name of the test.
2. Which type of compounds give this test?
3. According to the above answer, ethyl alcohol or methyl alcohol, which one gives this test?

Answer:

1. Iodoform test
2. Iodoform test is given by those compounds which are having CH₃CO – group or CH₃CHOH group.
3. Ethyl alcohol gives Iodoform test becuase it has CH₃CHOH group.

Question 8.

For the preparation of chlorobenzene, Nikhil wrote the equation ‘A’ and Nishanth wrote the equation ‘B’.

1. Which of the above equation is correct? Why?
2. Write the name of the reactions ‘A’ and ‘B’
3. Explain any one reaction.

Answer:

1. 1. Both the reactions (A) and (B) are correct.
   2. (A) -Sandmeyer reaction (B) – Gattermann reaction
   3. Sandmeyer reaction – When benzene diazonium chloride is allowed to react with cuprous chloride and HCI, chlorobenzene is obtained, benzene diazonium chloride + cuprous chloride
      

Question 9.
a. In the following pairs of compounds which would undergo SN₂ reaction faster?
(CH₃)₃C-Brand CH₃-CH₂-Br
b. i) CH₃ – CH₂ – CH₂ – OH + SOCI₂ → ……..

Answer:
a. CH₃CH₂Br. SN₂ reaction is faster in the case of primary alkyl halides since the transition state is more stable.
b. i) CH₃CH₂CH₂CI
ii) CH₃-CH₂ -CH₂ – CH₂Br (Anti-Markownikoff’s addition)

Question 10.

1. Nucleophilic substitution of haloalkane takes place through two different mechanisms, SN¹ and SN². Why do inversion of configuration take place in SN²?
2. What is racemic mixture?
3. Comment on the optical activity of recemic mixture.

Answer:

1. is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present.
2. Equimolar mixture of ‘d’ and T forms of an optically active compound is called racemic mixture.
3. Racemic mixture is optically inactive due to external compensation. A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer.

Question 11.
Write equations for the preparation of 1-iodobutane from
i) 1 – butanol
ii) 1 – chlorobutane
iii) but -1- ene
Answer:
i) CH₃CH₂CH₂CH₂OH+ HI → CH₃CH₂CH₂CH₂I + H₂O

Question 12.
Which compound in each of the following pairs will react faster in S_N2 reaction with – OH?

1. CH₃Br or CH₃I
2. (CH₃)₃CCI or CH₃CI

Answer:

1. CH₃I will react faster because C-I bond undergoes cleavage more easily as compared to C-Br bond.
2. CH₃CI will react faster because in it the carbon carrying halogen is sterically less hindered.

Question 13.
Identify the product X, Y, and Z in the following reaction.

Answer:

• X = CH₃CH₂CI
• Y = CH₃CH₂CN
• Z = CH₃CH₂CONH₂

Question 14.
Identify the compounds X, Y, and Z in the following.

Answer:

• X = CH₃ – CH₂Br
• Y = CH₃CH₂OH
• Z = CHI₃

Plus Two Chemistry Haloalkanes and Haloarenes Four Mark Questions and Answers

Question 1.
CH₃-CH = CH₂ + HCI

1. What are the possible products?
2. Of these which one is the major product?
3. Name the rule which helps you to answer the above question.
4. Explain the rule.

Answer:
1.

2. 2-Chloropropane is major.
3. Markownikoffs rule.
4. When a hydrogen halide is added to an unsymmetrical alkenethe halogen atom of alkyl halide will go to double-bonded carbon atom containing lesser number of hydrogen atom.

Question 2.
Two compounds are given

1. What is the difference between these two compounds?
2. Write the name of reaction they undergo and explain it.

Answer:

1. 1^st compound is cyanide. 2^nd compound is isocyanide
2. Nucleophilic substitution reactions.

The reactions in which a stronger nucleophile substitutes or displaces a weaker nucleophile are called nucleophilic substitution reactions.

Question 3.

1. Name ‘A’.
2. What is the role of ‘A’?
3. Name the product obtained.
4. What is this reaction called?

Answer:

1. Dry ether
2. Dry ether is used to prevent explosion
3. Diphenyl
4. Fittig reaction

Question 4.


Answer:

Question 5.
a) Haloalkanes give β-elimination.
i) Prepare CH₂ = CH₂ from CH₃ – CH₂X and alchoholic KOH.
ii) Give the major product of the β-elimination of

Name the related rule.

b) In the following pairs of halogen compounds which would undergo SN² reaction faster? Justify.

Answer:


It is primary halide and therefore undergoes S_N² reaction faster.

An iodide is a better leaving group because of its large size. It will be released at a faster rate in the presence of attacking nucleophile.

Question 6.

1. Which among the following compounds undergo SN¹ substitution easily – 3° or 1° alkyl halide? Give reason. Justify.
2. Grignard reagents should be prepared under anhydrous conditions. Give reason.

Answer:

1. 3°- alkyl halide. Because the 3° carbocation is more stable than a 1° carbocation.
2. Grignard reagents react with water and get decomposed (hydrolysed). Hence they should be prepared under anhydrous conditions.

Question 7.
a. In the following pair of halogen compounds which undergo SN² reactions faster.

1. C₆H₂ – CH₂ – CI and C₆ H₃ -CI
2. CH₃ – CH₂ – CH₂ – CI and CH₃ – CH₂ – CH₂ – I

b.

Answer:
a. Undergo SN² reactions faster

1. C₆H₅CH₂CI
2. CH₃ -CH₂ -CH₂ -I (R-I bond is weaker than R-CI bond)

b.

i) CH₃ -CH₂ -CH₂ -CH₂ Br
ii) CH₃-CH₂-CN

Question 8.

Complete the reaction.
b) R- CH₂– CI, R₂CHCI, R₃CCI. Arrange these alkylhalide in the order of reactivity towards SN¹ and SN² mechanism.

Answer:

Question 9.
Write the isomers of compound having molecular formula C₄H₉Br.
Answer:
For the molecular formula C₄H₉Br four isomers are possible.
(i) CH₃-CH₂ – CH₂ – CH₃ (1-Bromobutane)

Plus Two Chemistry Haloalkanes and Haloarenes NCERT Questions and Answers

Question 1.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₁₀CI in bright sunlight. Identify the hydrocarbon.
Answer:
Since the hydrocarbon gives only one monochloro compound, it indicates that all the hydrogen atoms in the hydrocarbon are equivalent. Thus, the compound is cyclopentane.

Question 2.
Write the isomers of compound having molecular formula C₄H₉Br.
Answer:
For the molecular formula C₄H₉Br four isomers are possible.
(i) CH₃-CH₂ – CH₂ – CH₃ (1-Bromobutane)

Question 3.
Write equations for the preparation of 1-iodobutane from
i) 1 – butanol
ii) 1 – chlorobutane
iii) but – 1 – ene
Answer:
i) CH₃CH₂CH₂CH₂OH+ HI → CH₃CH₂CH₂CH₂I + H₂O

Question 4.
Which compound in each of the following pairs will react faster in S_N2 reaction with OH^–?
Answer:

• CH₃I will react faster because C-I bond undergoes cleavage more easily as compared to C-Br bond.
• CH₃CI will react faster because in it the carbon carrying halogen is sterically less hindered.

Question 5.
Out of C₆H₅CH₂CI and C₆H₅CHCIC₆H₅which is more easily hydrolysed by aqueous KOH?
Answer:
C₆H₅CH₂CIC₆H₅ is more easily hdrolysed because in this case the reaction proceeds through more stable intermediate carbocation.

The intermediate carbocation in this case is stabilised by resonance effect of two phenyl groups whereas in the other case it is stabilised by resonance effect of only one phenyl group.

Students can Download Chapter 8 Electro Magnetic Waves Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 8 Electro Magnetic Waves

Plus Two Physics Electro Magnetic Waves NCERT Text Book Questions and Answers

Question 1.
What physical quantity is the same for X-rays of wavelength 10^-10, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Answer:
The speed in vacuum is the same for all.
c = 3 × 10⁸ms^-1

Question 2.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
\(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) are in x-y plane and are mutually perpendicular. Given v = 30 MHz = 30 × 10⁶ Hz, c = 3 × 10⁸ms^-1.
∴ \(\frac{c}{v}=\frac{3 \times 10^{8}}{30 \times 10^{6}}\) = 10m.

Question 3.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
For v₁ = 7.5 MHz = 7.5 × 10⁶ Hz

For v₂ = 12 MHz = 12 × 10⁶ Hz

So the wavelength band: 40m – 25m.

Question 4.
Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad s^-1t] \(\hat{i}\)

1. What is the direction of propagation?
2. What is the wavelength X?
3. What is the frequency v?
4. What is the amplitude of the magnetic field part of the wave?
5. Write an expression for the magnetic field part of the wave.

Answer:

1. \(-\hat{j}\)
2. λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{1.8}\) = 3.48 = 3.5m
3. v = C/λ = 3 × 10⁸/ 3.5 = 85.7 × 10⁶ = 86 MHz.
4. B₀ = \(\frac{E_{0}}{c}=\frac{31}{3 \times 10^{8}}\) = 10 × 10^-8T = 100nT.
5. 100nT) cos (1.8 rad/m) y + (5.4 × 10⁶ rad/s)t \(\hat{k}\)

Plus Two Physics Electro Magnetic Waves One Mark Questions and Answers

Question 1.
The structure of solids is investigated by using
(a) cosmic rays
(b) X-rays
(c) y-rays
(d) infra-red radiations
Answer:
(b) X-rays

Question 2.
Wavelength of light of frequency 100Hz.
(a) 4 × 10⁶m
(b) 3 × 10⁶m
(c) 2 × 1o⁶m
(d) 5 × 10^-5m
Answer:
(b) 3 × 10⁶m
λ = \(\frac{3 \times 10^{8}}{100}\) = 3 × 10⁶m.

Question 3.
What is the biological importance of ozone layer
(a) It stops ultraviolet rays
(b) ozone layer reduces green house effect
(c) ozone layer reflects radio waves
(d) ozone layer controls O₂/H₂ ratio in atmosphere
Answer:
(a) It stops ultraviolet rays
The ozone layer absorbs the harmful ultraviolet rays coming from sun.

Question 4.
Pick the odd one out from the following x-rays, visible light, matter waves, radio waves.
Answer:
Matter waves, (matter-wave is not a em wave).

Question 5.
What physical quantity is the same for X-rays of wavelength 10^-10, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Answer:
The speed in vacuum is the same for all.
c = 3 × 10⁸ms^-1.

Plus Two Physics Electro Magnetic Waves Two Mark Questions and Answers

Question 1.
State whether True or False.

1. Electromagnetic waves propagate in the direction of electric field.
2. For an electromagnetic wave the ratio of E to B is equal to speed of light.
3. In an electromagnetic wave, electric field leads by π/2.
4. Electromagnetic waves can be produced by accelerating electric charge.

Answer:

1. False
2. True
3. False
4. True

Plus Two Physics Electro Magnetic Waves Three Mark Questions and Answers

Question 1.
Match the following:

Answer:

• Radiowave – accelerated motion of charges in wires – cellular phones.
• Infrared waves – Hot bodies and molecule – Green house effect.
• X-rays – High energy electrons – diagnostic purpose
• Gama-rays – Radio active nuclei – destroy cancer cells.

Plus Two Physics Electro Magnetic Waves Four Mark Questions and Answers

Question 1.
In an electro magnetic wave electric and magnetic field vectors are given by E = 120 sin (ωt + kz) i, B = 40 × 10^-8sin (ωt +kz)j

1. What is the direction of propagation of electro magnetic wave?
2. Determine the ratio of amplitude of electric field to magnetic field in the case of the above em wave?
3. How can you relate the above ratio with m₀ and e₀.

Answer:

1. A direction or perpendicular to the direction of variation of electric field and magnetic field, ie.z direction.
2. \(\frac{E_{0}}{B_{0}}=\frac{120}{40 \times 10^{-8}}\) = 3 × 10⁸m/s.
3. \(\frac{E_{0}}{B_{0}}=\sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}\) = C.

Question 2.
Match the following.

Answer:

• γ-ray – Radioactivity – Nucleus
• X-ray – Diagnosis- Photon emission by fast moving e’s
• uv-ray- Sunburn – Electronicde excitation
• Microwave – Remote sensing – Oscillating current.

Question 3.
The magnetic field of a plane electromagnetic wave is given by B = 2 × 10^-7 sin (0.5 × 10³x + 1.5 × 10¹¹t)

1. What is the maximum amplitude of this magnetic field?
2. What is wave length and frequency?
3. If this electro magnetic move in z-direction. Write the equation of it’s electric field.

Answer:
1. 2 × 10^-7.

2. Comparing with standard equation
B = B_x Sin (kx + ωt)
we get, kx = 0.5 × 10³x.

m = 12.56 × 10^-3m
ω = 1.5 × 10¹¹, 2πf = 1.5 × 10¹¹
f = 2.3 × 10¹¹hz

3. Amplitude E₀ = B₀ C
= 2 × 10^-7 × 3 × 10⁸ = 60 N/C
If magnetic field vibrate in x direction,
then E_y = 60 sin (0.5 × 10³z + 1.5 × 10¹¹t).

Question 4.
The diagram shows the main regions of the electro magnetic spectrum.

1. Write the names of the blank regions.
2. Which radiation shown in the table has the shortest wavelength?
3. Which ray among these are responsible for green house effect? Explain green house effect.

Answer:
1. Ultra violet, Infrared.

2. Gamma-ray

3. Infrared
Earth is heated by infrared radiations from the sun. At the same time earth radiates longer wavelength to the space. This radiation (outgoing) of longer wavelength is reflected back by CO₂ and other gases.

Hence radiation (from earth) can’t escape from earth’s surface. Thus the temperature at earth’s surface increases. This phenomenon is called Green house effect.

Question 5.

1. Sun is a very powerful source of ultraviolet rays. But we do not receive much ultraviolet rays on the surface of the earth. Why?
2. The electric field of an ultraviolet ray is given by E = 60 sin (0.5 × 10³ – 1.5 × 10¹¹t) What is the amplitude of the magnetic field vector in the ultraviolet ray?

Answer:
1. Ozone prevents the UV ray.

2. \(B=\frac{E}{C}=\frac{60}{3 \times 10^{8}}\)
= 2 × 10^-7wb/m².

Students can Download Chapter 4 The Theory of The Firm Under Perfect Competition Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 4 The Theory of The Firm Under Perfect Competition

Plus Two Economics The Theory of The Firm Under Perfect Competition One Mark Questions and Answers

Question 1.
TR curve under perfect competition passes through the origin. Do you agree?
Answer:
Yes, I agree.

Question 2.
Pick the odd one out.

1. Homogeneous products, free entry, and exit, price maker.
2. Homogeneous product, freedom of entry and exit, large number of buyers and sellers, product differentiation.
3. MC = MR, MR = AFC, MC cuts MR from below, P = AC.

Answer:

1. Price maker. Others are features of perfect competition.
2. Product differentiation. Others are features of perfect competition.
3. MR = AFC. Others are conditions for firm’s equilibrium under perfect competition.

Question 3.
Pick out the odd one and justify it.
Free entry, profit maximization, perfect knowledge, price discrimination.
Answer:
Price discrimination. Others are features of perfect competition.

Question 4.
Which of the following represents normal profit?
(a) MR = MC
(b) AR = AC
(c) TR > TC
(d) AR < AC
Answer:
(b) AR = AC

Question 5.
Firm is price taker under:
(a) perfect competition
(b) monopoly
(c) monopolistic competition
(d) duopoly
Answer:
(a) perfect competition Question

6. Shut down point occurs at:
(a) rising part of
(b) following part of AVC
(c) minimum point AVC
(d) none of these
Answer:
(c) minimum point AVC

Question 7.
A firm is able to sell any quantity of the good at a given price. The firm’s marginal revenue will be
(a) Greater than AR
(b) Less than AR
(c) Equal to AR
(d) Zero
Answer:
(c) Equal to AR

Question 8.
The short run shut-down point of a firm in a perfectly competitive firm is.
(a) P = AVC
(b) P = AC
(c) P > AVC
(d) P < AVC
Answer:
(a) P = AVC

Question 9.
If P exceeds AVC but is smaller than AC at the best level of output, the firm in perfect competition is
(a) Making profit
(b) Minimizing losses in the short run
(c) Incurring a loss and should stop producing
(d) Breaking even
Answer:
(b) Minimizing losses in the short run

Plus Two Economics The Theory of The Firm Under Perfect Competition Two Mark Questions and Answers

Question 1.
A firm cannot make supernormal profit in the long run under perfect competition. Do you agree? Substantiate your answer.
Answer:
Yes, I do agree to the statement that a firm cannot make supernormal profit in the long run under perfect competition. This is because freedom of entry will prevent supernormal profit in the long run.

Question 2.
Define ‘Break even point’.
Answer:
The point on the supply curve at which a firm earns normal profit is called the break even point. The point of minimum average cost at which the supply curve cuts the LRAC curve is, therefore, the break-even point of a firm.

Question 3.
Examine the difference between the short run price and the long run price of a firm under perfect competition.
Answer:
There is a major difference between the short run and long run price under perfect competition. In the short run, price should be equal to or greater than the minimum AVC. If the price falls below this level, the firm will shut down production. On the other hand, the long run price should be equal to or greater than the minimum AC. Below this level, the firm will shut down production.

Question 4.
What is the supply curve the firm in the long run?
Answer:
In the long run, a firm will produce output only when its price is at least equal to the average cost of production. Therefore, average cost curve represents the supply curve of the firm.

Question 5.
How does technological progress affect the supply curve of the firm?
Answer:
The supply curve of the firm slopes upward from left to right indicating a direct relationship between price and quantity supplied. When there is technological progress, it will affect the supply. Now the firm will be in a position to produce and supply more output at the same price. Therefore the supply curve will shift towards the right side.

Question 6.
How does an increase in the number of firms in a market affect the market supply curve?
Answer:
As the number of firms changes the market supply curve shifts. When the number of firms increases the market supply curve shifts to the right. On the other hand, if the number of firms decreases, the market supply curve shifts to the left.

Question 7.
Define ‘shut down point’
Answer:
Shut down point refers to a situation where average revenue is equal to average variable cost. If the price fails to cover even average variable cost, firm will stop its production.

Question 8.
Make pairs:
Perfect competition, price marker, oligopoly, Cournot model, price taker, monopoly.
Answer:

• Perfect competition – price taker
• Monopoly – price maker
• Oligopoly – Cournot model

Question 9.
The diagram below shows two curves faced by a firm under perfect competition. Name them.

Answer:

1. Total revenue curve
2. Average/Marginal revenue curve

Question 10.
Choose the correct answer from the given multiple choices.
a. Identify the equilibrium condition of a firm under perfect competition.

1. AC=MR, & AC cuts MR from above.
2. MC=MR, & AC cuts MR from below.
3. AC=MR, & MC cuts MR from below.
4. MC=MR, &MC cuts MR from above.
5. MC=MR, & MC cuts MR from below.

b. The demand for the product of a firm is perfectly elastic in one of the following markets.
Identify the market.

1. monopoly
2. monopolistic competition
3. perfect competition
4. monopsony
5. oligopoly

Answer:
a. 5
b. 3

Plus Two Economics The Theory of The Firm Under Perfect Competition Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
A firm maximizes profit when the difference between total revenue and total cost is the maximum. Profit is maximized when certain conditions are satisfied. Do you agree?
Answer:
Yes.
A firm maximizes profit when the following three conditions are satisfied.

1. The market price, p, is equal to the marginal cost.
2. The marginal cost is nondecreasing.
3. In the short run, the market price must be greater than or equal to the average variable cost. In the long run, the market price must be grater than or equal to the average cost.

Question 3.
Write the economic terms

1. Cost vary with output
2. Total Cost divided by quantity
3. TR_n – TR_n-1
4. TR = TC point
5. A case where an increase in all the inputs lead to a just proportionate increase in output.

Answer:

1. Variable cost
2. Average cost
3. Marginal revenue
4. Breakeven point
5. Constant returns

Question 4.
Perfect competition does not exist in the real world. Do you agree? Substantiate your view.
Answer:
Yes, I agree to the statement that in the real world perfect competition does not exist.

This is because, it is rare to find the features of perfect competition especially the features like perfect knowledge, perfect mobility of factors and product, etc. What we really find in the world is monopolistic competition which is a mix of perfect competition and monopoly.

Question 5.
A firm’s supply curve in the short run is the rising part of the SMC curve. Why?
Answer:
A firm under perfect competition in the short run will start supplying only if the price is equal to or greater than the short run average variable cost. Therefore, the rising part of the short run marginal cost which begins with the minimum SAVC is the supply curve of the firm in the short run.

Question 6.
Imagine that S° is the original supply curve of the firm. If a unit tax is imposed, what happens to the supply curve? Show the change in a diagram.
Answer:
If a unit tax is imposed, firm’s long run supply curve shifts to the left. It is shown in the following diagram.

Question 7.
Fill in the blanks.

1. Long run price under perfect competition will be equal to ……………….
2. Maximum price fixed for a product by the government is called ………..
3. The point denoted by the minimum of AVC is called …………

Answer:

1. Average cost
2. Price ceiling
3. Shutdown point

Question 8.
Choose the appropriate answer.
a. A fall in supply caused by a fall in price is known as:

1. Contraction of supply
2. Expansion of supply
3. Increase in supply ‘

b. When supply curve is a vertical straight line, supply is

1. Perfectly elastic
2. Perfectly inelastic
3. Unitary elastic

c. The price under perfect competition. short run should be at least equal to

1. Short run MC
2. Short run AC
3. Short run AVC

Answer:
a. Contraction of supply
b. Perfectly inelastic
c. Short run AVC

Question 9.
List the factors affecting elasticity of supply.
Answer:
The factors affecting elasticity of supply are

1. nature of the commodity
2. cost of production
3. time period
4. technique of production

Question 10.
Observe the following figures and answer the questions.

1. Point out the elasticities on the above supply curves.
2. Which method is applied here?

Answer:
1. supply curves:

• S₁ Elastic supply
• S₂ Unitary elastic supply
• S₃ Inelastic supply

2. Geometric method is applied here.

Question 11.

1. Identify the market structure that is represented by their curve in the diagram.
2. Explain why the AR curve is horizontal

Answer:
1. Perfect competition

2. It is assumed that under perfect competition compared to the industry the share of each firm is meager. No firm can influence the market supply. So even if a firm doubles the quantity supplied the market supply will not change. The price remains the same. So theARorthe demand curve is horizontal.

Question 12.
Information about a firm is given in the following

Find out the equilibrium level of output in terms of MC & MR. Give reasons for your answer.
Answer:

Equilibrium quantity is 4. At this level of output MC = MR & MC cuts MR from below.

Question 13.
A firm can sell any quantity of the output it produces at a given price. If so, what is the behaviour of marginal revenue and average revenue? Draw the two curves in a single diagram.
Answer:
The demand curve facing the firm is perfectly elastic. At this condition AR=MR=Price

Plus Two Economics The Theory of The Firm Under Perfect Competition Five Mark Questions and Answers

Question 1.
Consider the following table

1. At which production level there will be no profit or loss to the producer?
2. Comment on the profit and loss conditions as TC 1000 and TR750.

Answer:
1. At the production level of 100 units of output, the producer incurs 1500 TC and 1500 TR. So there will be no profit or loss to the producer

2. When TC =1000
TR = 750
Loss = TC-TR
= 1000-750 = 250

Question 2.
How would each of the following affect the market supply curve for wheat?

1. A new and improved technique is discovered.
2. The price of fertilizers falls
3. The government offers new tax concessions to farmers
4. Bad weather affects the crops.

Answer:

1. A new and improved technique is discovered: increase market supply
2. The price of fertilizers falls: increase market supply
3. The government offers new tax concessions to farmers: increase market supply
4. Bad weather affects the crops: decrease market supply

Question 3.
The following table shows the total cost schedule of a competitive firm. It is given that the price of the good is ₹10. Calculate the profit at each output level. Find the profit maximizing level of output.

Answer:

Question 4.
Explain briefly break even point with the help of an example.
Answer:
XUS Break even point refers to a situation when total revenue is equal to total cost assuming a given selling price per unit of output.
TR = TC
This can be explained with the help of an example.

If the firm produces less than 200 units, itsTR< TC. So it bears loss. If the firm produces more than 200 units, its TR > TC. So it earns profit.

If the firm produces 200 units, its TR = TC. This is the breakeven point. Break even point is a normal profit point as beyond it the firm earns super normal profits and below it, the firm incur losses.

Question 5.
Choose the correct answer
a. profit of a firm is the revenue earned:

1. zero of cost
2. net of cost
3. gross of cost
4. none of these

b. TMC curve cuts LAC curve :

1. at minimum point
2. at maximum point
3. below the LAC curve
4. none of these

c. under perfect competition, firm is :

1. price taker
2. price maker
3. both 1 and 2
4. none of the above

d. MR can be negative but AR is:

1. negative
2. positive
3. either positive or negative
4. none of the above

Answer:
a. net of cost
b. at minimum point
c. price taker
d. positive

Question 6.
Observe the following diagram.

Plot the break even point and shut down point.
Answer:

• Point ‘A’ represents shut down point
• Point ‘B’ represents the break even point.

Question 7.
Complete the following tables and identify the market structure


Answer:

This market represents a perfectly competitive market because in this market, P = AR = MR.

This market represents a monopoly market. In this market, AR and MR are different and AR > MR.

Question 8.
At the market price of ₹10, a firm supplies 4 units of orange. The market price rises to ₹30. The price elasticity of the firm’s supply is 1.25. What quantity will the firm supply at the new price?
Answer:
In the given example,
P = 10 Q = 4 P1 = 30 es = 1.25
ΔP = 30-10 = 20
Applying these values in the formula, we get, \(e_{s}=\frac{\Delta Q}{\Delta P} \times \frac{P}{Q}\)

ΔQ =1.25×8=10

Question 9.
Suppose that the market demand in a perfectly competitive industry is given by, Qd = 7000 – 500 p and the market supply function is given by, Qs = 4000 +250 P. Find the market equilibrium price.
Answer:
Equilibrium is determined by the condition,
Qd = Qs.
In this example,
7000-500 p = 4000 +250 p
7000-4000 = 250 p + 500 p
3000 = 750 P
∴ \(P=\frac{3000}{750}=4\)
Therefore, the equilibrium price in the market is ₹4.
7000 – 500 p = 4000 + 250 p
7000 – 4000 = 250 p + 500 p
3000 = 750 P
∴ \(P=\frac{3000}{750}=4\)

Question 10.
Point out the features of perfect competition.
Answer:
Features of perfect competion are pointed out below.

1. Large number of buyers and sellers
2. Homogeneous product
3. Freedom of entry and exit
4. Free movement of product and factors of production.
5. Profit motive.
6. Perfect knowledge of market conditions
7. Absence of transport cost.

Question 11.
Identify from the diagram below.

1. Shutdown point and breakeven point.
2. Distinguish between these two.

Answer:

1. c is the break even point. b is the shut even point

2. Break down point shows a situation where a firm earns no profit or no loss. It is the point where AR = AC. Shut down point shows a situation where a firm is compelled to stop the production since it is not able to cover its variable cost. This is the situation where the firm’s P > AVC.

Question 12.
Identify the profit maximizing level of output from the diagram below. List out the condition for profit maximization.

Answer:
b is the profit maximizing level of output.
MR = MC
At the profit maximizing level of output MC is non decreasing, that is the slope of MC is positive.
P > AVC.

Question 13.
Match the commodities given below with the diagram. Justify your answer.

Answer:
a. Mobile Phone

b. Coconut
The supply of mobile phones are relatively price elastic. This is because the purchase of mobile phones can easily react to a change in price. It is a manufactured good. All the raw materials needed to produce a mobile phone can easily be available. The producers can pile up stock and when price increases they can increase the supply.

The supply of coconut can be price inelastic. This is because the producers cannot easily react to a change in the price of coconut. Years needed for a coconut tree to get mature and start to produce coconuts. It is an agricultural product. The producers cannot hold the stock of coconut fora longer duration.

Question 14.
The following table gives you certain information about a firm.

1. Find the price at which output is sold and identify the form market,
2. Is this firm a price-taker or price-maker? Give reasons.
3. Find the firm’s equilibrium level of output in terms of MC &MR. Give reasons.
4. Also find profit of the firm at this level of output.

Answer:

1. Price = 8
2. Price-taker, a firm has no influence on price determination under perfect competition.
3. Equilibrium quantity is 3. At this level of output MC=MR.
4. Firm is at no profit-no loss condition, i. e., he is at break even point.

Plus Two Economics The Theory of The Firm Under Perfect Competition Eight Mark Questions and Answers

Question 1.
Prepare a seminar report on ‘perfect competitive markets’
Answer:
Respected teachers and dear friends,
The topic of my seminar paper is perfect competitive markets. The term market refers to all the places in which buyers and sellers are in contact with each other for the purchase and sale of any commodity or service. There are different kinds of markets based on their characteristics – say perfect competitive markets and noncompetitive markets.

Introduction:
Perfect competition is a market situation characterized by the existence of large number of buyers and sellers, homogeneous products, free mobility of factors of production, freedom of entry and exit, perfect knowledge of market conditions and absence of transportation costs.

Contents:

1. Profit Maximisation
2. Profit Maximisation in short-run: Diagrammatic representation
3. Long run profit maximisation

1. Profit Maximisation:
The main objective of the producer is to maximize the profit levels of his firm. The output level at which the firm maximizes the profit is called the equilibrium of the firm. The profit level of the firm is the difference between Total Revenue and Total Cost. Symbolically it is represented as X = TR – TC.

The firm under perfect competition maximizes its profit under three conditions. They are:

• The MC must be equal to MR (MC = MR)
• MC must be non-decreasing. It means that MC curve should cut the MR from below.
• Third condition has two parts, one for short-term and the other for long-term.

a. In the short run, price should be more than or equal to the minimum point of Average Variable Cost (AVC). It can be denoted as P> AVC.

b. In the long run, price should be more than or equal to the minimum point of Average Cost (AC). It can be denoted as P > AC.

2. Profit Maximisation in short run: Diagrammatic representation:
The profit maximizing condition of a firm in short-run can be understood from the figure. All the three profit maximizing conditions of a firm in short run are satisfied at point of output level q0.

• The price must be equal to MC (P = MC)
• MC must be non-decreasing.
• P > AVC

All the profit maximizing conditions are satisfied in the figure given above.

1. The prices become equal to MC. At point E, the firm reaches equilibrium at the output level q (MC = MR).
2. At q MC is non-decreasing at point E.
3. At point E, prices have become more than AVC (P > AVC at point E).

Therefore, three conditions for the equilibrium level of output are depicted in a single figure.

3. Long run profit maximisation:
The profit maximization level of the firm is reached when the long run supply curve of the firm is that portion of LRMC curve which lies over and above the minimum point of LRAC curve (P > minimum of LRAC). The supply curve of a firm, in the long run, is the rising portion of minimum point of LMRC. It can be explained with the help of the following figure.

Conclusion:
Thus it can be concluded that perfect competition is a market structure characterized by complete absence rivalry among individual sellers. Sellers in the market do not hold any freedom of influencing market prices. Therefore, they are known as price takers. However, it should be admitted that it is a rare form of market.

Question 2.
The diagram shows one of the short run equilibrium situations of a firm under perfect competition.

1. List out any four features of such market condition.
2. Identify the equilibrium situation of the firm profit, normal profit, loss diagrammatically. Show a situation in which the firm makes a profit.
3. With the help of a diagram explain the long run situation of a firm under perfect competition.

Answer:
1. Market Conditions:

• Large number of sellers
• Homegeneous product
• Firm is a price taker
• Free entry and exit

2. The firm is making a normal profit.
The firm produces ‘oq’ level of output and charges a price ‘op’. The shaded area in the diagram shows profit.

3. Firm under perfeert competition in the long run makes only normal profit. This is because if firms are making profit, new entrants will be attracted into the industry.

The price falls due to increase in supply and the extra profit will be taken away. If the firms are making loss some of them will leave the industry, price rises and the loss will be turned into profit. This is shown in the diagram below.

The firm under perfect competition in the long run will produce ‘oq’ level of output and charges a price ‘op’.

Question 3.
How are price and output determined under perfect competition in the short run? Compare the profit of a firm in the short run and long run and bring out the difference. Give suitable diagram.
Answer:
Under perfect competition, market determines the price – price taker and not price maker firms produce the output that maximises its profit – a firm may get abnormal profit or normal profit or loss in the short run – only normal profit will prevail in the long run – due to free entry and exit – Draws separate diagram for short run and long run-explains price and output determination as well as profit in short run and long run.

PROFIT MAXIMIZATION IN SHORT-RUN:
DIAGRAMMATIC REPRESENTATION:
The profit maximizing condition of a firm in short-rn can be understood from the figure. All the three profit maximizing conditions of a firm in short run are satisfied at point of output level q₀.

1. The price must be equal to MC (P = MC)
2. MC must be non-decreasing.
3. P > AVC

All the profit maximizing conditions are satisfied in the figure given above.

1. The prices become equal to MC. At point E, the firm reaches equilibrium at the output level q (MC = MR). 0
2. At q MC is non-decreasing at point E.
3. At point E, prices have become more than AVC (P > AVC at point E).

Therefore, three conditions for the equilibrium level of output are depicted in a single figure.

LONG RUN PROFIT MAXIMISATION:
The profit maximization level of the firm is reached when the long run supply curve of the firm is that portion of LRMC curve which lies over and above the minimum point of LRAC curve (P>minimum of LRAC). The supply curve of a firm, in the long run, is the rising portion of minimum point of LMRC. It can be explained with the help of the following figure.

Students can Download Chapter 7 Alternating Current Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 7 Alternating Current

Plus Two Physics Alternating Current NCERT Text Book Questions and Answers

Question 1.
A 100Ω resistor is connected to a 220V, 50 Hz ac supply.

1. What is the rms value of current in the circuit?
2. What is the net power consumed over a full cycle?

Answer:
Given R = 100Ω, E_ν = 220V, ν = 50 Hz.
1. Since l_ν = \(\frac{E_{ν}}{12}\)
So l_ν = \(\frac{220}{100}\) = 2.2 A

2. P_aν = E_ν I_ν = 220 × 2.2
or P_aν = 484 W.

Question 2.

1. The peak voltage of an ac supply is 300V. What is the rms voltage?
2. The rms value of current in an ac circuit is 10A. What is the peak current?

Answer:
1. Given E₀ = 300 V, E_aν = ?
Since E_ν = \(\frac{E_{0}}{\sqrt{2}}\) = 0.707 × 300
or E_ν = 212.1V.

2. Given I_ν = 10A, I₀=?
Since l₀ = \(\sqrt{2}\) E_ν = 1.414 × 10
or I = 14.14 A.

Question 3.
A 44 mH inductor is connected to 220V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Given L = 44 mH = 44 × 10^-3H
E_ν = 220V, ν = 50Hz, I_ν = ?
Since I_ν = \(\frac{E_{ν}}{x_{L}}=\frac{220}{\omega L}\)

or Iν = 15.9A.

Question 4.
A 60µF capacitor is connected to a 11OV, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Given C = 60µF = 60 × 10^-6F, E_ν = 110V, ν = 60 Hz.
Since I_ν = \(\frac{E_{ν}}{x_{c}}\)
∴ I_ν = ωCE_ν
= 2pνCE_ν = 2 × 3.142 × 60 × 60 × 10^-6 × 110
= 2.49A or
I_ν = 2.49V.

Question 5.
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer:
In both the cases the net power consumed is zero because in both the cases.
Net power consumed P = E_ν l_νcosΦ
and Φ =90°
∴ P = 0 (in each case).

Question 6.
Obtain the resonant frequency of a series LCR circuit with L=2.0H, C=32µV and R = 10?. What is the Q-value of this circuit?
Answer:
Given L = 2.0H, C = 32µF = 32 × 10^-6F
R = 10Ω, Q = ?, ω₀ = ?
Resonant frequency

Question 7.
A charged 30µF capacitor is connected to a 27mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
Given C = 30µF=30 × 10^-6F,L = 27mH = 27 × 10^-3H
ω₀ = ?

or ω₀ = 1.1 × 10^-3S^-1.

Plus Two Physics Alternating Current One Mark Questions and Answers

Question 1.
Which type of transformer you use to operate the coffee maker at 220 V?
Answer:
Step down transformer.

Question 2.
In an A C. circuit, I_rms and I_o are related as.

Answer:
(d) I_rms = \(\frac{I_{0}}{\sqrt{2}}\)

Question 3.
A capacitor of capacitance C has reactance X. If capacitance and frequency become double then reactance will be
(a)  4X
(b) X/2
(c) X/4
(d) 2X
Answer:
(c) Explanation

Question 4.
Fill in the blanks

• Impedance: admittance
• ……………..: conductance

Answer:
Resistance

Question 5.
Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp?
Answer:
No power is developed across the inductor as heat.

Question 6.
In an a.c circuit with phase voltage V and current I, the power dissipated is
(a) V.l
(b) Depends on phase angle between V and I
(c) \(\frac{1}{2}\) × V.I
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(b) Depends on phase angle between V and I

Plus Two Physics Alternating Current Two Mark Questions and Answers

Question 1.
Fill in the blanks.

Answer:
(i) Current lags by π/2
(ii) X_c = 1/cω
(iii) R
(iv) Phase difference is zero.

Question 2.
A.C. adaptor converts household ac into low voltage dc. A stepdown transformer is a essential part of ac adapter.

1. What is the use of step down transformer?
2. What is the principle of a transformer? Explain.

Answer:

1. To decrease voltage
2. It works on the principle of mutual induction.

Plus Two Physics Alternating Current Three Mark Questions and Answers

Question 1.
Match the following

Answer:

Question 2.

The following figure is a part of a radio circuit.

1. Identify the circuit.
2. What happens to this circuit if X_L = X_C
3. lf X_C > X_L draw the phaser diagram.

Answer:
1. LCR circuit.

2. When X_L = X_C, the impedance of circuit becomes minimum and the current corresponding to that frequency is maximum.

3.

Plus Two Physics Alternating Current Four Mark Questions and Answers

Question 1.
Figure below shows a bulb connected in an electrical circuit.

1. When the key is switched ON the bulb obtains maximum glow only after a shorter interval of time which property of the solenoid is responsible for the delay?

• Self-induction
• Mutual Induction
• Inductive reactance
• None of the above

2. If the flux linked with the solenoid changes from 0 to 1 weber in 2 sec. Find the induced emf in the solenoid.

3. If the 3v battery is replaced by an AC source with the key closed, what will be observation? Justify your answer.
Answer:
1. Self-induction.

2. \(\frac{d \phi}{d t}=\frac{1}{2}\) = 0.5V.

3. When AC is connected the brightness of bulb will be decreased. This is due to the back emf in the circuit.

Question 2.
A graph connecting the voltage generated by an a.c. source and time is shown.

1. What is the maximum voltage generated by the source?
2. Write the relation connecting voltage and time
3. This a.c. source, when connected to a resistor, produces 40J of heat per second. Find the equivalent d.c. voltage which will produce the same heat in this resistor.

Answer:

1. 200v.
2. V = V₀sin ωt
3. \(V_{\max }=\frac{V_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}\) = 141.8v.

Question 3.
An inductor, capacitor, and resister are connected in series to an a.c. source V = V₀sin ωt.

1. Draw a circuit diagram of L.C.R. series circuit with applied a.c. voltage.
2. Find an expression for impedance of L.C.R. series circuit using phasor diagram.
3. What is impedance of L.C.R. series circuit at resonance?

Answer:
1.

2.

For impedance of LCR circuit.
From the right angled triangle OAE,
Final voltage, V = \(\sqrt{v_{n}^{2}+\left(v_{L}-v_{c}\right)^{2}}\)

Where Z is called impedance of LCR circuit

3. Impedance, Z=R.

Question 4.
A voltage source is connected to an electrical component X as shown in figure.

1. Identify the device X.
2. Which of the following equations can represent the current through the circuit?

• i = i_m sin(ωt + π/2)
• i = i_m sin(ωt – π/2)
• i = i_m sin ωt
• i = i_m sin(ωt + π/4)

3. Draw the phasor diagram for the circuit. (2)
Answer:
1. Resistor
2. i = i_m sin ωt
3.

Question 5.
A friend from abroad presents you a coffee maker when she visited you. Unfortunately, it was designed to operate at 110V line to obtain 960W power that it needs.

1. Which type of transformer you use to operate the coffee maker at 220V? (1)
2. Assuming the transformer you use as ideal, calculate the primary and secondary currents. (2)
3. What is the resistance of the coffee maker? (1)

Answer:
1. Step down transformer.

2. Since the transformer is ideal
V_pI_p = V_s I_s = 960W, V_p = 220v, V_s = 110v
V_pI_p =960

3.

Plus Two Physics Alternating Current Five Mark Questions and Answers

Question 1.
The voltage-current values obtained from a transformer constructed by a student is shown in the following table.

1. Identify the transformer as step up or step down.
2. How much power is wasted by the transformer?
3. What are the possible energy losses in a transformer?
4. If the input voltage is 48v and input current is 1 A, is it possible to light 240v, 100w bulb using the above transformer. Justify.

Answer:
1. Step down transformer.

2. Power loss = 200w -10w = 190w

3. The possible energy losses in a transformer:

• Eddy current loss
• Copper loss
• Hysteries loss
• Flux leakage

4. In this case input power = VI = 48 × 1 = 48W.
If transformer does not waste energy, input power =out put power.
Hence maximum output power 48W. But bulb requires 100w. Hence the bulb does not glow with this low input power.

Question 2.
The current through fluorescent lights are usually limited using an inductor.

1. Obtain the relation i = i_m sin(ωt – π/2)for an inductor across which an alternating emf v = v_m sin ωt is applied. (2)
2. Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp? (1)
3. When 100 V DC source is connected across a coil a current of 1 A flows through it. When 100V, 50 Hz AC source is applied across the same coil only 0.5 A flows. Calculate the resistance and inductance of the coil. (2)

Answer:
1. AC voltage applied to an Inductor

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage. Let the applied voltage be
V = V₀ sinωt…………(1)
Due to the flow of alternating current through coil, an emf, \(\mathrm{L} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\) is produced in the coil. This induced emf is equal and opposite to the applied emf (in the case of ideal inductor).

2. No power is developed across the inductor as heat.

3. Resistance of the coil R = \(\frac{v}{I}=\frac{100}{1}\) = 100Ω. Current through the coil when ac source is applied.

R² + X²_L = 200²
X²_L = 200² – 100²
X_L = 173.2Ω
L_ω = 173.2

Question 3.
An alternating voltage is connected to a box with some unknown circuital arrangement. The applied voltage and current through the circuit are measured as v = 80 sinωt volt and i = 1.6 sin(ωt + 45°) ampere.

1. Does the current lead or lag the voltage?
2. Is the circuit in the box largely capacitive or inductive?
3. Is the circuit in the box at resonance?
4. What is the average power delivered by the box?

Answer:
1. Leads.

2. Capacitive

3. No Hint: Current and voltage are not in the same phase.

4. P = V_rmsl_rmsCosΦ, V_m = 80v, i_m = 1.6A

p = 45.25W

Students can Download Chapter 11 Alcohols, Phenols and Ethers Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 11 Alcohols, Phenols and Ethers

Plus Two Chemistry Alcohols, Phenols and Ethers One Mark Questions and Answers

Question 1.
Which of the following is most acidic?
(a) H₂0
(b) CH₃OH
(c) C₂H₅OH
(d) CH₃CH₂CH₂OH
Answer:
(a) H₂O

Question 2.
Propan-1 -ol on reaction with conc.H₂SO₄ at 413 K gives _____________
Answer:
1 -Propoxy propane

Question 3.
Which of the following alcohols can be obtained from HCHO?
(a) CH₃OH
(b) C₂H₅OH
(c) CH₃CH₂CH₂CH₂OH
(d) All of these
Answer:
(d) All of these

Question 4.
Phenol can be distinguished from ethyl alcohol by all reagents except
(a) NaOH
(b) Na
(c) Br₂/H₂O
(d) FeCl₃
Answer:
(b) Na

Question 5.
Anisole on reaction with HI forms
(a) phenol and methyl iodide
(b) iodobenzene and methanol
(c) benzene and methyl iodide
(d) phenol and methanal
Answer:
(a) phenol and methyl iodide

Question 6.
Arrange the following compounds in the increasing order of acidity: Phenol, Alcohol, and Water.
Answer:
Alcohol < Water < Phenol

Question 7.
Phenol is distinguished from ethanol by the following reagents except.
(a) Iron
(b) Sodium
(c) Bromin
(d) NaOH
Answer:
(b) Sodium

Question 8.
Phenol can be converted to o-hydroxy benzaldehyde by
Answer:
Reimer-Tiemann reaction.

Question 9.
4-methoxy acetophenone can be prepared from anisol by ______________
Answer:
Friedel crafts reaction.

Question 10.
Which of the following does not answer idoform test?
(a) 1-propanol
(b) Ethanol
(c) 2 propanol
(d) Ethanal
Answer:
(a) 1- propanol

Question 11.
Chlorination of toluene in presence of light and heat followed by treatment with aqeous KOH gives ________
Answer:
Benzyl alcohol.

Plus Two Chemistry Alcohols, Phenols and Ethers Two Mark Questions and Answers

Question 1.
Reaction of alcohol with metallic sodium is used as a test for alcohol. Substantiate this statement with the help of an equation.
Answer:
Alcohol reacts with metallic Na to form sodium alkoxide with the liberation of H₂.
ROH + Na → RONa + 1/2 H₂

Question 2.
Ethyl alcohol gives iodoform test. Methyl alcohol does not give iodoform test.

1. Do you agree with this?
2. If yes or no, substantiate your view.
3. How can you distinguish between 1 -butanol and 2-butanol?

Answer:
1. Yes.

2. Compounds containing CH₃CO- group and CH₃CH(OH)- group on reaction with iodine and alkali give yellow colour of iodoform. Ethanol contains CH₃CH (OH) – group.

3. 2-Butanol gives iodoform test as it contains CH₃CH(OH)-group, whereas 1-Butanol does not answer iodoform test.

Question 3.
When an organic compound is treated with neutral ferric chloride a violet colour is obtained. What will be the compound? Explain.
Answer:
Phenol. Phenol forms a violet-coloured water soluble complex with ferric chloride.

Question 4.
– O – is the functional group of ether. Classify the following into two groups with appropriate heading.
CH₃-O-CH₃, CH₃-O-C₂H₅, C₂H₅-O-C₂H₅, C₆H₅– O – CH₃.
Answer:

Question 5.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution reaction?
Answer:
In phenol, the -OH group is directly attached to a carbon of benzene ring. The lone pair of oxygen participates into resonance with the benzene ring. As a result, electron density on benzene ring increases making it more easy to attack by an electrophile.

Question 6.
Write down the equations for the following conversions using Grignard reagent?
Methanal → Ethanol
Ethanal → 2-Propanol
Answer:

Question 7.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol has higher boiling point than butane because it has stronger interparticle forces. In propanol intermolecular hydrogen bonding is present whereas in butane intermolecular forces are weak van der Waals’ forces. A lot of heat is required to break intermolecular hydrogen bonding among propanol molecules.

Question 8.
While separating a mixture of ortho and para nitrophenols by stream distillation, name the isomer which will be steam volatile. Give reason.
Answer:
Ortho-Nitrophenol is steam volatile because in it there is intramolecular hydrogen bonding.

Due to intramolecular hydrogen bonding, the intermolecular forces in ortho-nitrophenol are weaker than that in para-nitrophenol (which has intermolecular hydrogen bonding) and hence it undergoes less association.

Question 9.
Give reason for higher boiling point of ethanol in comparison to methoxy methane.
Answer:
In ethanol the intermolecular forces are hydrogen bonds whereas in methoxymethane the intermolecular forces are dipole-diple forces. Since the intermolecular forces in ethanol are stronger than those in methoxymethane it has higher boiling point than methoxymethane.

Question 10.
While separating a mixture of ortho and para nitro phenols by steam distillation, name the isomer, which will be steam volatile. Give reason.
Answer:
Ortho-Nitro phenol exhibit intramolecular hydrogen bonding and para-Nitro phenol exhibit intermolecular hydrogen bonding. The ortho isomer is steam volatile because there is not intermolecular association.

Plus Two Chemistry Alcohols, Phenols and Ethers Three Mark Questions and Answers

Question 1.
Write a notes on:

1. Rectified Spirit
2. Power Alcohol
3. Denatured Spirit

Answer:
1. 95.6% ethyl alcohol is known as a rectified spirit.

2. A mixture of ethyl alcohol and gasoline can be used as a fuel in the internal combustion engine. It is known as power alcohol.

3. Commercial alcohol is made unfit for drinking by adding certain substances like pyridine, methanol etc. Spirit thus obtained is called denatured spirit.

Question 2.
Fill in the blanks:

Answer:

Question 3.
Ethyl alcohol can be prepared by fermentation of molasses.

1. What do you mean by fermentation?
2. Explain the process.
3. What do you mean by ‘wash’?

Answer:
1. Fermentation is the process of breaking up of large molecules into small molecule in the presence of biological catalyst called enzymes.

2. Molasses is the mother liquor left behind after the crystallisation of cane sugar from sugar cane juice. It is 40% sucrose solution. First it is diluted to 10% solution. Then yeast is added. Temperature of the system is kept at 305 K. The following reactions will take place.

3. 8-10% ethyl alcohol is known as ‘wash’.

Question 4.
Raju prepared soap in the chemistry lab. A liquid remained after the preparation. He argued that it was useless.

1. Do you agree with him? Why?
2. How can you prepare glycerol from spentlye?

Answer:

1. No. It can be used to prepare glycerol.
2. Spentlye contains unreacted alkali, glycerol, water, NaCI, and soluble soap. It is first treated with acid to remove alkali, then with aluminium sulphate to remove soluble soap. It is then evaporated under reduced pressure to remove NaCI.

The resulting solution is then decolorised using animal charcoal, Then the solution is distilled under reduced pressure to remove water. In this way we get glycerol.

Question 5.
When an old sample of ether was heated, it exploded.

1. What is the reason for this phenomenon?
2. How can you detect peroxide content in ether?
3. How can we remove peroxide from old sample of ether?

Answer:

1. Due to the formation of peroxide.
2. Presence of peroxide can be tested by adding ferrous salt solution followed by addition of KCNS solution. Formation of blood red colour indicates the presence of peroxide.
3. The peroxide can be removed by washing with ferrous salt solution.

Question 6.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Because of the presence of O-H group in them, alcohols are capable of forming hydrogen bonds with water molecules whereas hydrocarbons cannot form hydrogen bonds with water. As a result, alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 8.
Explain why ortho-nitrophenol is more acidic than ortho-methoxyphenol?
Answer:
ortho-Nitrophenol is more acidic than ortho- methoxyphenol because nitro group by its electron withdrawing resonance effect stabilises the phenoxide ion whereas methoxy group by its electron releasing effect destabilises the phenoxide ion. Greater the stability of the phenoxide ion, greater is the dissociation of phenol and greater is its acid strength.

Question 9.
With the help of a mixture of con. HCI and ZnCI2 how can you distinguish between 1°, 2°, 3° alcohols.
Answer:

• 1° alcohol + Lucas reagent → no reaction
• 2° alcohol + Lucas reagent → turbidity within five minutes
• 3° alcohol + Lucas reagent → turbidity occurs immediately

Lucas reagent anhydrous ZnCI₂/HCI

Plus Two Chemistry Alcohols, Phenols and Ethers Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.

a) Alcohols are having high boiling points than corresponding alkyl halides and ethers. Why?
b) Phenol is more acidic than ethanol. Give the reason.
c) Predict the products :

Answer:
a) Due to the presence of polar hydroxyl group alcohols can associate through intermolecular hydrogen bonding.
b) Phenoxide ion is stabilized by resonance while alkoxide ion has no resonance stabilisation.
c)

Question 3

1. Phenol is acidic even though it has no carboxylic group, why?
2. Convert phenol to salicylic acid and name the reaction.

Answer:
1. By the removal of a proton from phenol a phenoxide ion is obtained. It is stabilised by resonance. Hence phenol acts as an acid.
2.

.

Question 4.
1. Compare the solubility of diethyl ether and n- butane in water.
2. Give the product(s) from the reaction of one mole of diethyl ether with

• one mole of conc. HI and
• excess of HI.

Answer:
1. Diethyl ether being weakly polar is capable of forming intermolecular hydrogen bonding with water. Hence, diethyl ether is soluble in water while n-butane is not.

2. The product(s) from the reaction of one mole of diethyl ether.

• C₂H₅OH and C₂H₅I
• 2C₂H₅I

Question 5.
Identify X and Y.

Answer:

Question 6.
1. Boiling point depends on the inter molecular hydrogen bonding.

• Ethanol and propane have comparable molecular masses but their boiling points differ widely. Why?

2. Williamson synthesis is an important method of ether synthesis. Which of the following reactions is better for ether synthesis? Justify.

Answer:
1. Ethanol molecules are associated by intermolecular hydrogen bonding.

• Ethanol molecules are associated by intermolecular hydrogen bonding. This is absent in propane. So Ethanol has higher boiling point.

2. (CH₃)₃C-ONa + CH₃-Br is better. The tertiary alkyl halide undergoes elimination in presence of base to form alkene.

Question 7.
Chloro methane reacts with aqueous sodium hydroxide to form methanol.

1. What happens when chlorobenzene reacts with aqueous sodium hydroxide? Justify.
2. Write the reaction by which chlorobenzene can be converted to phenol.

Answer:

1. No reaction. This is due to sp² state of carbon to which CI is attached, less polarity of C-X bond and resonance stabilisation.
2. Dow’s process.
   When chlorobenzene is heated with sodium hydroxide solution at 623 K under a pressure of 200 atm in the presence of copper catalyst, sodium phenoxide is obtained. This on hydrolysis gives phenol.

Question 8.
Phenol exhibit acidic character.

1. Why phenol exhibit acidic character?
2. Explain it with the help of resonating structure of phenoxide ion.

Answer:

1. Phenol can donate proton and phenoxide ion thus formed is stabilized by resonance.
2. The resonating structure of phenoxide ion is given below.

Question 9.
What is the action of phenol with

1. Aqueous Br₂?
2. Br₂ in CS₂?
3. Nitrating mixture?
4. dil. HNO₃?

Answer:
1. Phenol on action with aqueous Br₂ gives 2, 4, 6,- tribromophenol.

2. Phenol on action with Br₂ in CS₂ gives o- bromophenol and p-bromophenol.

3.

4.

Question 10.
Write notes on
a) Kolbe’s reaction
b) Reimer-Tiemann reaction
Answer:
a) Kolbe’s reaction: When Sodium phenoxide is heated with CO₂ at 400 K and under a pressure 6-7 atm, sodium salicylate is obtained. This on hydrolysis gives ortho-hydroxy benzoic acid or salicylic acid.

Reimer – Tiemman reaction: When phenol is heated with CHCI₃ at 340 K, o-hydroxy benzaldehyde or salicylaldehyde is obtained.

Question 11.

1. Fill in the blanks:
   CH₃CH₂OH + SOCI₂ → CH₃CH₂CI+ …?… +….?….
2. This method is used to prepare extra pure alkyl halide. Do you agree? Why?

Answer:

1. CH₃CH₂OH + SOCI₂ -> CH₃CH₂CI + SO₂ + HCI
2. Yes. The by products are escapable gases. Hence, the reaction gives pure alkyl halides.

Question 12.
1. Arrange the compounds in the increasing order of their strength.

• 4-Nitro phenol
• Phenol
• Propan-1-ol
• 4-Methyl phenol.

2. You are given benzene, conc.H₂SO_4, and NaOH. Prepare phenol using these compounds.
Answer:
1. 4-Nitrophenol > Phenol > 4-Methylphenol > 1- Propanol. Presence of electron with darwing groups at ortho and para positions increases the acidic strength of substituted phenols.

2. By the action of benzene with conc.H₂SO₄ & NaOH, sodium benzene sulphonate is formed. This on acidification gives phenol.

Plus Two Chemistry Alcohols, Phenols and Ethers NCERT Questions and Answers

Question 1.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol has higher boiling point than butane because it has stronger interparticle forces. In propanol intermolecular hydrogen bonding is present whereas in butane intermolecular forces are weak van der Waals’ forces. A lot of heat is required to break intermolecular hydrogen bonding among propanol molecules.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Because of the presence of O-H group in them, alcohols are capable of forming hydrogen bonds with water molecules whereas hydrocarbons cannot form hydrogen bonds with water. As a result, alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 4.
Explain why ortho-nitrophenol is more acidic than ortho-methoxy phenol?
Answer:
ortho-Nitrophenol is more acidic than ortho methoxy phenol because nitro group by its electron with drawing resonance effect stabilises the phenoxide ion whereas methoxy group by its electron releasing effect destabilises the phenoxide ion. Greater the stability of the phenoxide ion, greater is the dissociation of phenol and greater is its acid strength.

Question 5.
Give IUPAC names of the following ethers:
i) C₂H₅OCH₂-CH(CH₃)CH₃
ii) CH₃OCH₂CH₂CI
iii) O₂N-C₆H₄-OCH₃(p)
iv) CH₃CH₂CH₂OCH₃

Answer:
i) 1-Ethoxy-2-methylpropane
ii) 2-Chloro-1-methoxyethane
iii) 4-Nitroanisole
iv) 1-Methoxypropane
v) 1-Ethoxy-4, 4-dimethyl cyclohexane
vi) Ethoxybenzene

Students can Download Chapter 6 Electro Magnetic Induction Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 6 Electro Magnetic Induction

Plus Two Physics Electro Magnetic Induction NCERT Text Book Questions and Answers

Question 1.
A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0ms^-1 at right angles to the horizontal component of the earth’s magnetic field 0.3 × 10^-4Wbm².

1. What is the instantaneous value of the e.m.f? induced in the wire?
2. What is the direction of the e.m.f.?
3. Which end of the wire is at the higher electrical potential?

Answer:
I – 10m, v = 5.0ms^-1, B = 0.30 × 10^-4Wbm²
1. Instantaneous value of e.m.f. induced
e = B lv
= 0.30 × 10^-4 × 10 × 5.0
or e = 1.5 × 10^-3V.

2. The direction of e.m.f. is from eastto west.

3. Since the current is flowing from east to west so the eastern end is at higher potential.

Question 2.
Current in a circuit falls from 5.0Ato 0.0A in 0.1s. If an average e.m.f. of 200V induced, give an estimate of the self-inductance of the circuit.
Answer:
dl = l₂ – l₁ = 0.0 – 5.0 = -5.0A,
dt = 0.1s, e = 200V, L=?
Since

or L = 4H

Question 3.
A pair of adjacent coils has a mutual inductance of 1 5H. If the current in one coil changes from 0 to 20A in 0.5s. What is the charge of flux linkage with the other coil?
Answer:
M= 1.5H, dl = l₂ – l₁ = 20 – 0 = 20A, dt = 0.5s, dΦ = ?
Since

And also

or dΦ = Mdl
or dΦ = 1.5 × 20 = 30 Wb.

Question 4.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 25m long if the earth’s magnetic field at the location has a magnitude of 5.0 × 10^-4 and the dip angle is 30?
Answer:
v = 1800 km/h = 500ms^-1 I = 25m
Vertical component of B along horizontal
B = -Bsinθ = 5.0 × 10^-4 sin30° = 2.5 × 10^-4T
e = B/v sinθ
= 500 × 2.5 × 10^-4 × 25
= 31 V
The direction ofthe wing is immaterial.

Question 5.
Suppose the loop in Q.6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3T at the rate of 0.02TS^-1. If the cut is joined and the loop has a resistance of 1.6Ω, how much power is dissipated by the loops as heat? What is the source of this power?
Answer:
Induced e.m.f.

= 8 × 2 × 10^-4 × 0.02 = 3.2 × 10^-5V
Induced current,

= 2 × 10^-5A
Power Loss = I²r = (2 × 10^-5)² × 1.6 = 6.4 × 10^-10 W. Source of this power is the external agent responsible for changing the magnetic field with time.

Plus Two Physics Electro Magnetic Induction One Mark Questions and Answers

Question 1.
Eddy current are produced when.
(a) a metal is kept in varying magnetic field
(b) a metal is kept in steady magnetic field
(c) a circular coil is placed in a magnetic field
(d) through a circular coil, current is passed
Answer:
(a) a metal is kept in varying magnetic field

Question 2.
A100 milli henry coil carries a current of 1 A. energy stored in its magnetic field is
(a) 0.5 J
(b) 1 A
(c) 0.05 J
(d) 0.1 J
Answer:
(c) 0.05 J
Explanation : E = \(\frac{1}{2}\) LI² = \(\frac{1}{2}\) × (100 × 10^-3) × 1²
= 0.05 J.

Question 3.
A straight line conductor of length 0.4 m is moved with a speed of 7m/s perpendicular to a magnetic field of intensity 0.9 Wb/m². The induced e. m. f. across the conductor is.
(a) 5.04 V
(b) 25.2 V
(c) 1.26 V
(d) 2.52
Answer:
(d) 2.52
Induced e.m.f (s) = B/V
= 0.9 × 0.4 × 7 = 2.52V.

Question 4.
The core of a transformer is laminated because
(a) ratio of voltage in primary and secondary may be increased
(b) energy losses due to eddy currents may be minimized
(c) the weight of the transformer may be reduced
(d) rusting of the core may be prevented
Answer:
(b)

Plus Two Physics Electro Magnetic Induction Four Mark Questions and Answers

Question 1.

A conductor XY is moving through a uniform magnetic field of intensity B as shown in figure.

1. Name the emf.
2. The motion of XY towards right side causes an anticlockwise induced current. What will be the direction of magnetic induction in the region A?
3. The length of the conductor XY is 20cm. It is moving with a velocity 50m/s perpendicular to the magnetic fie Id. If the induced emf in the conductor is 2V find the magnitudes of magnetic field?

Answer:
1. Motional emf/induced emf.

2. Applying right hand grip rule at A direction of magnetic field is away from the paper.

3. ε = Blvsinθ, I = 20cm = 20 × 10^-2m, v
= 50m/s, ε = 2v

= 0.2 T.

Question 2.
1. When S₁ close bulb glows instantaneously when S₂ closes there is a delay in glowing the bulb.(1) This is due to………

• resistance of the coil
• back emf in the coil
• mutual induction
• none

2. Explain the phenomenon self-induction regarding above experiment (2)
3. Draw the graph with energy and current for a inductor. (1)
Answer:
1. Back emf in the coil.

2. When a.c. current is passed through the coil, a change in flux is produced. This change in flux produces a back e.m.f. in the coil. The phenomena of production of back emf is called self induction.

3. The graph with energy and current for a inductor:

Plus Two Physics Electro Magnetic Induction Five Mark Questions and Answers

Question 1.
When AC is switched on the thin metallic disc is found to thrown up in air.

1. Which makes the disc to thrown?
2. How will you explain the mechanism behind the movement of disc
3. Write the working principle of induction heater.

Answer:

1. Eddy current produced in the coil thrown disc into air.
2. Whenever the magnetic flux linked with a metal block changes, induced currents are produced due to this current, disc becomes a magnet. Hence disc thrown up in to air.
3. The change in flux produces eddy current in a metal. The heat produced by eddy current is used for cooking in induction heater.

Question 2.
AC generator is a device based on electromagnetic induction used to convert mechanical energy into electrical energy.

1. Draw a graph showing variation of induced emf over a cycle. Also indicate peak value of emf. (2)
2. How peak value of emf is related to its rms value? (1)
3. A rectangular wire loop of side 4cm × 6cm has 50 turns uniformly from 0.1 Tesla to 0.3 Tesla in 6 × 10^-2 second. Calculate induced emf in the coil. (2)

Answer:
1.

2. Erms = \(\frac{E_{0}}{\sqrt{2}}\)

3.

and Φ = BAN
Φ₁ = 0.1 × (24 × 10^-4) × 50
= 120 × 10^-4wb
Φ₂ = 0.3 × (24 × 10^-4) × 50
= 360 × 10^-4wb

= 40 × 10^-2V.

Students can Download Chapter 3 Production and Costs Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 3 Production and Costs

Plus Two Economics Production and Costs One Mark Questions and Answers

Question 1.
Which among the following cost curves is not ‘U’ shaped?
Answer:
AFC curve is rectangular hyperbola.

Question 2.
Identify the shape of the following AFC curve?
Answer:

The shape of AFC curve is rectangular hyperbola.

Question 3.
When a firm increased the number of labour from 10 to 11, keeping the capital fixed the total product increased from 120 to 130. Which of the following statement is correct in this regard?
(a) The total product fell
(b) This is a long run production
(c) The average product is rising
(d) 10 is the marginal product of an increased unit of labour.
Answer:
(d) 10 is the marginal product of an increased unit of labour.

Question 4.
Slope of an isoquant is:
(a) marginal cost
(b) DMRS
(c) DMRTS
(d) None of the above
Answer:
(c) DMRTS

Question 5.
When MP becomes zero, TP:
(a) increases
(b) decreases
(c) becomes maximum
(d) becomes negative
Answer:
(c) becomes maximum

Question 6.
Which among the following is not ‘u’ shape
(a) MC
(b) AC
(c) AVC
(d) AFC
Answer:
(d) AFC

Question 7.
When AC is minimum
(a) MC > AC
(b) MC = AC
(c) MC < AC
(d) None of these
Answer:
(b) MC = AC

Question 8.
All the following curves are U shaped except
(a) the AVC curve
(b) the AFC curve
(c) the AC curve
(d) the MC curve
Answer:
(b) the AFC curve

Plus Two Economics Production and Costs Two Mark Questions and Answers

Question 1.
Let the production function of a firm be Q = 5 L^1/2 K^1/2
Find the maximum output that the firm can produce with 100 units of L and 100 units of K.
Answer:
The production function is
Q = 5 L^1/2 K^1/2
Since the firm uses 100 units of L and 100 units of
K, we get the production function as
Q = 5 × 100^1/2 100^1/2
Q = 5 × 10x 10
Q = 5 × 100 = 500

Question 2.
Derive marginal product and average product from the total product schedule.

Answer:

Question 3.
Find the maximum possible output for a firm with zero units of labour and 10 units of capital when its production function is
Q = 5L + 2K
Answer:
The production function is Giving the values of L and K, we get
Q = 5 × 0 + 2 × 10
Q = 0 + 20 = 20

Question 4.
“In the long run, all costs are variable”. Do you agree? Justify your answer.
Answer:
Yes. In the long run, all costs are variable. This is because there is sufficient time available in the long run for any factor of production to get increased. As there are only variable factors, in the long run, all costs are variable.

Question 5.
“The AVC is U shaped” explain the reasons for the ‘U’ shape of AVC curve. Also, draw a diagram to clarify your points.
Answer:
The SMC and AVC curves start rising when production starts. As output increases, SMC falls. AVC being the average of marginal costs also falls but falls less than SMC. Then after a point, SMC start rising. AVC, however, continues to fall as long as value of SMC remains less than the prevailing value of AVC.

Once the SMC has risen sufficiently its value becomes greater than the value of AVC. The AVC starts rising. Therefore, the AVC curve is “U” shaped. It is given in the following diagram.

Question 6.
TFC of a firm is ₹2,000 and TVC is ₹3,000. It produces 20 units. Calculate the AVC and AC of the firm.
Answer:
TC = TFC + TVC
AC = 5000/20 = 250
AVC = 3000/20 = 150

Question 7.
Can there be some fixed cost in the long run?
Answer:
No, there cannot be some fixed cost in the long run. This is because, in the long run all factors of production can be adjusted and variable.

Question 8.
Let the production function of a firm be Q = 10 L^1/2 K^1/2 Find out the maximum possible output that the firm can produce with 100 units of L and 100 units of K.
Answer:
The production function is,
Q = 10 L^1/2 K^1/2
L =100 units
K =100 units
∴ Q = 10 × 1001/2 × 1001/2
= 10 × (10²)^1/2 × (10²)^1/2
= 10 × 10 × 10
= 1,000 units

Question 9.
The table given below shows the total product schedule of labour. Determine the AP and MP of labour?

Answer:

Question 10.
Prove that AC is the sum of AFC and AVC.
Answer:

Question 11.
A bus company started its operation with two buses, four labourers. It has paid a one-time road tax. As the passengers increased the company arranged more tripes, with the help of additional labour and for using more fuels. Classify the costs incurred by the bus company into fixed and variable.
Answer:

Question 12.
Both short run and long run average and marginal cost curves are U-shaped. But reasons for the U-shape are not the same. Bring out the differences.
Answer:
Behaviour of output in the short run as explained by the law of variable proportions Behaviour of output in the long run as explained by the laws of returns to scale.

Plus Two Economics Production and Costs Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
Categorize the following into fixed cost
(a) wages of temporary workers
(b) cost of raw materials
(c) salary of permanent staff
(d) cost of transportation
(e) cost of plant
Answer:

Question 3.
Identify the following curves.

Answer:

1. TC
2. TVC
3. TFC

Question 4.
Identify the shapes of the following cost curves.

1. AFC
2. AC
3. TFC
4. TVC
5. TC

Answer:

1. AFC – Rectangular hyperbola
2. AC – ‘U’shape
3. TFC – Straight line parallel to X axis
4. TVC – Inverse‘S’ shape
5. TC – Inverse‘S’ shape

Question 5.
There exists vertical distance between 1. TVC and TC and (b) TC and TFC. What does this distance indicate?

1. TVC and TC
2. TC and TFC

Answer:

1. The vertical distance between TVC and TC represents the Total Fixed Costs (TFC).
2. The vertical distance between TC and TFC represents the Total Variable Costs (TVC)

Question 6.
Complete the following table

Answer:

Question 7.
Name the following curves. ,

Answer:

Question 8.
What do you mean by an ‘isoquant’?
Answer:
An isoquant is the set of all possible combinations of the two inputs that yield the same maximum possible level of output. Each isoquant represents a particular level of output and is labelled with that amount of output. The shape of isoquant is drawn below.

Question 9.
With the help of a diagram show the relation between average exists and average variable cost.

Answer:
Both AC and AVC are U shaped. As the output increases the gap between AC and AVC narrows.

Question 10.
In the short run, Total Variable Cost is zero when output is zero. When output rises, Total Cost also rises. Draw a suitable diagram and explain the relationship between Total Fixed Cost, Total Variable Cost and Total Cost.
Answer:

Question 11.
“Average Fixed Cost (AFC) curve is a continuously falling curve.”

1. Substantiate this statement by giving the reasons.
2. Graphically represent the AFC curve

Answer:
1. TFC is constant & hence AFC falls
2.

Plus Two Economics Production and Costs Five Mark Questions and Answers

Question 1.
The marginal product and total product of an input are related’. Prove this statement.’
Answer:
The marginal product (MP) and total product (TP) of an input are related. The points of relationship are given below.

1. When MP increases, TP also increases
2. When MP is zero, TP becomes maximum
3. When MP becomes negative, TP turns negative

Question 2.
Differentiate between returns to a factor and returns to scale.
Answer:
Returns to a factor refers to the effects on output of changes in one input with all other inputs are hold constant. On the other hand, returns to scale refers to the effect on total output of charges in some constant rate in all the inputs simultaneously.

Question 3.
The marginal cost and average cost are related to each other. Prove this using diagram and a numerical example?
Answer:

Marginal cost are addition made to the total costs by the production of an additional unit of the commodity.
MC = TC_n – TC_n-1

Average lost is per unit cost of production which is obtained by dividing the total cost by number of units of output produced.
\(A C=\frac{T C}{Q}\)

Since AC is obtained by dividing TC by all units, MC is the addition to TC by producing an additional unit so MC brings a change in AC.

From above:

1. When MC is less than AC, then MC will fall
2. When MC is equal to AC, AC remains costant.
3. When MC is more than AC, AC rises

Question 4.
Which of the following represent the long run production function?
1. MC is less than AC, then MC will fall

• Law of variable proportion
• Returns to scale

2.  Explain what may happen when all inputs vary simultaneously?
Answer:
Returns to scale.
When all inputs are varied simultaneously the output may change in three ways.
1. Constant returns to scale:
It is when a proportional increase in inputs result in an increase in output by the same proportion.

2. Increasing returns to scale:
It is when a proportional increase in all inputs result in an increase in output by more than proportion.

3. Decreasing returns to scale:
It is when a proportional increase in all inputs result in an increase in output by less than proportion

Question 5.
State whether the statement are true or false.

1. TC never becomes zero.
2. AC is sum total of AFC and AVC.
3. When AC rises, AC and MC are equal
4. Real cost is the cost in money terms
5. TFC curve is U shaped

Answer:

1. True
2. True
3. False
4. False
5. False

Question 6.
The following table shows the total cost schedule of a firm. What is the total fixed cost schedule of the firm? Calculate the JVC, AFC, AVC, SAC and SMC of the firm

Answer:

Question 7.
The diagram shows the total product curve of a factor in the law of variable proportion.

1. Explain the law
2. With the help of a diagram illustrate the average and marginal product curve and their relation.

Answer:
1. Law of variable proportion says that if one variable input is added with other fixed inputs the marginal product of a factor input initially rises but after reaching a certain level of output it starts falling.

2. The relation between the average and marginal product is given below. BothAPand MPare ‘n ’shaped. MPwill always pass through the maximum of AP.

Question 8.
Theory of production deals with producer’s behaviour, and according to this theory output produced by a firm passes through three stages in the short run.

1. Which are the three stages of production?
2. Analyse and bring out the salient features of each stage.
3. Which stage of production is very important as far as a firm is concerned? Why?
4. Give suitable diagram by drawing the Total Prod-uct, Average Product & Marginal Product curves.

Answer:

1. Increasing, diminishing and negative
2. First output increases at increasing rate, then at diminishing rate & finally starts falling.
3. Second stage, profit maximisation occurs in this Stage.
4.

Question 9.
If other things remaining same, graphically explain what happens to the supply curve for readymade shirts if there is

1. An increase in the wages paid to the tailors.
2. An increase in the price of ready-made shirts

Answer:

Plus Two Economics Production and Costs Eight Mark Questions and Answers

Question 1.
Prepare a seminar report on the topic “Production Function.”
Answer:
Respected teachers and dear friends:
The topic of my seminar paper is “Production Function”. The production function of a firm shows relationship between inputs and output. In this seminar paper, I would like to present the different production functions such as short run production function and the long run production function.

Introduction:
As we know the production function shows the transformation of inputs into output. The production function can be of two types – short run production function known as Law of Variable proportion and the long run production function known as returns to scale. Q =F(F₁ F₂… F_n)

Contents:
A. Law of Variable proportion

1. Increasing returns
2. Diminishing returns
3. Negative returns

A. The Law of Variable Proportions
When more and more units of a variable input are added with the fixed input, the marginal product would increase only upto a certain point. Therefore, the marginal product declines. This phenomenon is known as the Law of Variable Proportions. It is also known as returns to a factor.

The shape of TP, AP and MP suggests that they are specifically passing through three phases.
They are:
1. First phase:
In the first stage, both AP and MP increase. As a result TP also increases at an increasing rate. This stage is known as the stage of increasing return to a factor. AP reaches the maximum level in this stage.

2. Second phase:
Both AP and MP decrease at this stage. The TP increases at a decreasing rate. More importantly, TP reaches maximum and MP touches zero. This stage is also known as the stage of diminishing returns to a factor.

3. Third phase:
At this stage, the MP becomes negative. As a result, TP also starts declining. The decline of AP is continuous. In the graph, when TP reaches maximum and MP touches zero. When MP becomes negative, TP starts declining. This stage is known as the stage of negative returns to a factor.

B. Returns to scale

1. Increasing returns
2. Constant returns
3. Decreasing returns
   (for details, refer summary part)

B. Returns to scale
As stated earlier, all the factor inputs become variable in the long run and thereby no distinction can be made between fixed inputs and variable inputs. Return to scale is associated with long run production function.

Returns to scale refers to the change in output when all inputs are variable and the proportion between inputs remains constant. When all the inputs vary in the same proportion, the output (TP) behaves in different manner, which can be clubbed into three categories.

1. Increasing Returns to Scale (IRS):
When a proportionate change in all the inputs leads to more than proportionate change in output, it is known as the stage of increasing returns to scale. For instance, a 10% change in inputs results in more than 10% change in output.

2. Constant Returns to Scale (CRS):
When a proportionate change in all the inputs leads to change in output in the same proportion is known as the stage of constant returns to scale. It indicates that a 10% change in inputs leads to exactly 10% change in output.

3. Decreasing Returns to Scale (DRS):
When a proportionate change in all the inputs leads to less than proportionate change in output is known as the stage of decreasing returns to scale. It indicates that a 10% change in inputs leads to less than 10% change in output.

Conclusion:
Thus it can be concluded that there are two types of production function depending upon the use of inputs and the time period.

Question 2.
Prepare a seminar paper on Law of Returns to Scale.
Answer:
Returns to scale:
As stated earlier, all the factor inputs become variable in the long run and thereby no distinction can be made between fixed inputs and variable inputs. Return to scale is associated with long run production function.

Returns to scale refers to the change in output when all inputs are variable and the proportion between inputs remains constant. When all the inputs vary in the same proportion, the output (TP) behaves in different manner, which can be clubbed into three categories.

1. Increasing Returns to Scale (IRS):
When a proportionate change in all the inputs leads to more than proportionate change in output, it is known as the stage of increasing returns to scale. For instance, a 10 percentage change in inputs results in more than 10 percentage change in output.

2. Constant Returns to Scale (CRS):
When a proportionate change in all the inputs leads to change in output in the same proportion is known as the stage of constant returns to scale. It indicates that a 10 percentage change in inputs leads to exactly 10 percentage change in output.

3. Decreasing Returns to Scale (DRS):
When a proportionate change in all the inputs leads to less than proportionate change in output is known as the stage of decreasing returns to scale. It indicates that a 10 percentage change in inputs leads to less than 10 percentage change in output.

Question 3.
From the table given below

1. Fill the blank columns
2. Using the information in the table draw TC, TVC, TFC in one and AC, AVC, AFC, MC in other and prepare a short note on them.

Answer:
1.

2. Use the data to draw the diagram.
TC, TVC and TFC represent total costs. TC = TFC + TVC. TFC is the cost incurred on fixed factors. TVC is the cost incurred on variable costs. Even though the level of output is zero there will be some costs. This is known as fixed cost. When the level of output is zero total variable costs also will be zero.

It will increase as the level of output increases. AFC is the cost that always falls. It is a rectangular hyperbola. AVC, AC, MC are ‘U’ shaped. The MC will always pass through the minimum oftheAVCand AC.

Students can Download Chapter 12 Aldehydes, Ketones and Carboxylic Acids Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids One Mark Questions and Answers

Question 1.
The reaction is called
(a) Sandmeyer’s reaction
(b) Rosenmund’s reduction
(c) HVZ reaction
(d) Cannizaro’s reaction
Answer:
(b) Rosenmund’s reduction

Question 2.
Say TRUE or FALSE:
Aldol condensation is given by all aldehydes and ketones.
Answer:
False

Question 3.
A 40% solution of _________ is called formation
Answer:
formaldehyde (HCHO)

Question 4.
Benzamide on heating with bromine and caustic alkali gives
(a) benzene
(b) methylamine and benzene
(c) aniline
(d) m-Bromobenzaldehyde
Answer:
(c) aniline

Question 5.
In the reaction the product ‘B’ is
(a) acetanilide
(b) glycine
(c) ammonium acetate
(d) methane
Answer:
(b) glycine

Question 6.
Arrange the following in the decreasing order of acidity.
ClCH₂COOH, Cl₃CCOOH, CH₃COOH, Cl₂HCOOH
Answer:
Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH

Question 7.
Reaction of butanone with methyl magnesium bromiode followed by hydrolysis gives_________
Answer:
2 methyl -2- butanol

Question 8.
The major product of the addition of water molecule to propyne in the presence of mercuric sulphate and dil sulphuric acid is ________
Answer:
Propanone

Question 9.
One mole of propanone and one mole of formalde¬hyde are the products of ozonolysis of one mole of an alkene. The alkene is ________
Answer:
2 methyl propene

Question 10.
Which of the following is a better reducing agent for the following reduction.
RCOOH → RCH₂OH
(a) SnCI₂/HCI
(b) NaBH₄/ether
(c) H₂/pd
(d) N₂H₄/C₂H₅ONa
(e) B₂H₆/H₃O⁺
Answer:
(e) B₂H₆/H₃O⁺

Question 11.
The total number of acyclic structural isomers possible for compound with molecular formula C₄H₁₀O is ________
Answer:
7

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Two Mark Questions and Answers

Question 1.
Reactivity of ketone is less than that of aldehyde. Why?
Answer:
Due to steric hindrance and inductive effect of alkyl group.

Question 2.
Carboxylic acid decompose into carboxylate ion and H⁺ ion.

1. Explain this on the basis of resonating structure of carboxylic acid.
2. Arrange the following in the increasing order of acidity. HCOOH, CH₃COOH
3. Substantiate.

Answer:

1. Carboxylic acid decomposes to give proton and carboxylate ion and is stabilized by resonance. This explains the acidic character of carboxylic acid.
2. CH₃COOH < HCOOH
3. In acetic acid the electron donating effect (+l – effect) of -CH₃ group destabilises the carboxylate anion and decreases the acid strength. Whereas in formic acid the H atom has not electron withdrawing or electron donating effect.

Question 3.
What is Etard’s reaction?
Answer:
Etard’s reaction:
When toluene is oxidized using chromyl chloride, benzaldehyde is obtained.

Question 4.
What is HVZ reaction? Explain.
Answer:
HVZ reaction – When a carboxylic acid is treated with red P and halogen, the α-H atoms are successively replaced by halogen.

This reaction has great synthetic importance as the halogen atom can be replaced by a number of other groups giving useful products.

Question 5.
Predict the product and name the reaction:

1. HCHO + NaOH → B + C
2. CH₃COOH + CH₃OH → E

Answer:

1. CH₃OH + HCOONa – Cannizzaro reaction
2. CH₃COOCH₃ – Esterification

Question 6.
Write the name of any two tests to distinguish between acetaldehyde and acetone.
Answer:
Benedict’s test, Fehling’s test – Both tests are answered by acetaldehyde and not by aceotne.

Question 7.
An organic compound with molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
Answer:
From the given data it is clear that as the compound forms2, 4-DNP derivative it has /CO group. Since it reduces Tollens’ reagent therefore -CHO group is present. As it can also undergo Cannizzaro reaction therefore α -hydrogen is absent

The oxidation product suggests that the compound has a benzene ring. One of the – COOH groups have been obtained by the oxidation of – CHO group and the other from alkyl group. Hence on these basis, the structure of C₉H₁₀O is

Question 8.
Write a notes on

1. Reimer – Tiemann reaction
2. Rosenmund Reduction

Answer:
1. Reimer – Tiemman reaction: When phenol is heated with CHCI₃ at 340 K, o-hydroxy benzaldehyde or salicylaldehyde is obtained.

2. Rosenmund reduction: When an acid chloride is reduced by using hydrogen gas in presence of Pd and BaSO₄, an aldehyde is obtained.
OR

Question 9.
Distinguish the following compounds using any one test.
H₃C – CO – CH₃ and CH₃CH₂CHO
Answer:
CH₃COCH₃ give Iodoform test which CH₃CHO does not answer this test.

Question 10.
Aspirin is commonly used in medicine. How it is prepared? Give the equation.
Answer:
Aspirin is acetyl salicyclic acid. When salicyclic acid is treated with acetyl chloride, aspirin is obtained.

Salicylic acid + acetyl chloride → Aspirin

Question 11.
How will you prepare CH₃-CO-NH₂ and CH₃COOCH₃ from CH3COOH?
Answer:

Question 12.
Give the IUPAC name of the following compounds.
i) C₆H₅CH = CHCHO
ii) (CH₃)₃CCH₂COOH
Answer:

Question 13.
Give a test to distinguish between acetaldehyde and acetone.
Answer:
CH₃ – CO – CH₃ contains CH₃CO – group and hence it gives iodoform test.
CH₃ – CH₂ – CHO does not give iodoform test.

Question 14.
Predict the major product in the following reactions:

Answer:

Question 15.
Distinguish between formaldehyde & acetaldehyde.
Answer:

Question 16.
Which is more acidic, 2-chloropropanoic acid or 3- chloropropanoicacid? Why?
Answer:
2-chloropropanoic acid. Becasue the electron with drawing -Cl group is more closer to -COOH group in this compound.

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Three Mark Questions and Answers

Question 1.
Fill in the blanks:

Answer:

Question 2.
A student is given Tollens’ reagent for oxidation of aldehyde.

1. What is Tollens’ reagent?
2. Can you help him to do the experiment?
3. What is the result of the experiment?

Answer:

1. Tollens’ reagent is ammoniacal silver nitrate solution
2. Yes. To a little of the solution add Tollens’ reagent
3. A black precipitate of silver or silver mirror is obtained

Question 3.

1. What is the role of LiAIH₄?
2. Give one example of an oxidising agent?
3. What is the action of a carboxylic acid with alcohol?

Answer:

1. LiAIH₄ is a strong reducing agent. It reduces RCOOH to 1° alcohol (RCH₂OH)
2. Acidified KMnO₄.
3. When carboxylic acid is treated with an alcohol in the presence of dehydrating agent like conc.H₂SO₄, an ester is formed. This is called esterification reaction.

Question 4.
What happens when primary, secondary and tertiary alcohols are passed over red hot copper? Give equations.
Answer:
1° alcohol hot copper Aldehyde
2° alcohol over hot copper – Ketone
3° alcohol over hot copper- alkene

Question 5
Fill in the blanks.

Answer:

1. HVZ reaction
2. CH₃CH₃
3. Hoffmann bromamide degradation reaction
4. HCOONa + CH₃OH
5. Reimer-Tiemann reaction
6. C₆H₅Cl

Question 6.
Draw the structure of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Answer:

Question 7.

i) 2HCHO + NaOH → CH₃OH + HCOONa
ii)

(a) Identify Cannizzaro and Aldol condensation.
(b) What is the difference between the above two reactions?

Answer:
(a) Cannizzaro reaction:
2HCHO + NaOH CH₃OH + HCOONa
Aldol condensation:

(b) Cannizzaro reaction is given by aldehydes having no α-H atom whereas aldol condensation is given by aldehydes containing α-H atom.

Question 8.
a) In a practical class a group of students heated ethanal with NaOH. Another group heated methanal with conc.NaOH.
i) Identify the products in each reaction with equation.
ii) Name the reactions.
b) Aldehydes are more reactive than ketones. Comment on the statement.
Answer:
a)

ii) The first reaction is Cannizarro reaction and the second reaction is aldol condensation.

b) Due to the ‘+l’ effect and steric hindrance of surrounding alkyl group ketones are less reactive than aldehydes.

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Four Mark Questions and Answers

Question 1.
i) Arrange the following in the increasing order of acidic strength and justify your answer.
CH₃COOH, CHCI₂COOH, CH₂CICOOH, CCI₃COOH

ii) Suggest a method to convert acetic acid to chloroacetic acid. Name the reaction and write the chemical equation.
Answer:
i) CH₃COOH < CH₂CI-COOH < CHCI₂COOH < CCI₃COOH This is due to the electron withdrawing character of chlorine.

ii) HVZ reaction – When a carboxylic acid is treated with red P and halogen, the a-H atoms are successively replaced by halogen.

This reaction has great synthetic importance as the halogen atom can be replaced by a number of other groups giving useful products.

Question 2.
a) Which of the following carbonyl compounds answer aldol condensation reaction and give equation.
HCHO, CH₃CHO, CCI₃CHO, C₆H₆-CHO

b) Arrange the following compounds in the increasing order of acidity:
CH₃COOH, CH₂CICOOH, CH₃-CH₂-COOH, C₆H₅-COOH
Answer:
a) Among these compounds only CH₃CHO answer aldol condensation reaction. Others will not answer this reaction because they have no α – hydrogen atom.

Question 3.

1. Aldehydes and ketones are carbonyl compounds Give a test for identification of aldehydes.
2. Acidic strength is related to the stability of carboxylate anion. Which acid of each pair is stronger?

Answer:
1. Benedict test. Benedict reagent is a mixture of sodium carbonate and sodium citrate. This on reaction with aldehyde gives red precipitate of Cu₂S.

2. Acidic strength is related to the stability of carboxylate anion. Acid of each pair is stronger:
i) CH₂FCOOH. This is due to the high electron with drawing effect of F.

This is due to the high electron with drawing effect of the -CF₃ group.

Question 4.
Substituents on carboxylic acids have much effect on their acidity. Substantiate the statement with the following examples.
a) HCOOH, CH₃COOH, CH₂CICOOH

Answer:
CH₂CICOOH > HCOOH > CH₃COOH
The methyl group will intensify the negative charge on the carboxylate ion and destabilise it as compared to formate ion. So HCOOH is stronger than CH₃COOH. The electron withdrawing effect of a Cl makes chloroacetic acid stronger than HCOOH and CH₃COOH.

In the case of aromatic carboxylic acids, presence of electron withdrawing groups at ortho and para position increases their acidity while presence of electron donating groups decreases their acidity.

In 4-nitro benzonic acid acid strength is greater than that of benzoic acid due to the electron withdrawing nature of -NO₂ group while in 4-methoxy benzoic acid acid strength decreases due to the electron donating nature of the methoxy group.

Question 5.
Give a chemical test to distinguish between each of
the following pair of organic compounds.

1. Propanal and Propanol
2. Propanone and Propanal

Answer:

1. Propanal is an aldehyde and it gives a silver mirror with Tollens’ reagent while propanol is an alcohol and will not answer Tollens’test.

2. Propanone gives yellow precipitate of iodoform on reaction with I₂ and NaOH while propanal does not give iodoform test. OR Propanal gives silver mirror with Tollens’ reagent while propanone does not give silver mirror test.

Question 6.
What is meant by the following terms? Give an example of the reaction in each case.

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP-derivative
10. Schiff’s base

Answer:
1. Cyanohydrin – It is a compound which contain both OH and CN groups. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

2. Acetal – compounds formed by the reaction of aldehydes with monohydric alcohols in presence of dry HCI gas.

3. Semicarbazone – the product of carbonyl compounds with semicarbazide is known as semicarbazone.

4. Aldol – It is a condensation product of aldehydes or ketones having atleast one α – hydrogen atom with dilute alkali as catalyst.

5. Hemiacetal – It is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

6. Oxime – Addition compound formed by the reaction of aldehyde or ketone with hydroxylamine.

7. Ketal – It is a cyclic compound obtained by reaction of aceone with ethylene glycol.

8. Imine – Addition compound formed by the reaction of aldehyde or ketone with ammonia.

9. 2, 4-DNP derivative – 2, 4-phenylhydrazone (DNP) is the addition compound formed by the reaction of aldehydes and ketones with 2, 4-dinitrophenylhydrazine.

10. Schiff’s base – Addition compound formed by the reaction of aldehyde or ketone with amine.

Question 7.
Name the following compounds according to IUPAC system of nomenclature.

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂CI
3. CH₃CH=CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₂CCH₂COOH
7. OHCC₆H₅CHO-p

Answer:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3, 3, 5-Trimethylhexane-3-one
6. 3, 3-Dimethylbutanoicacid
7. Benzene 1, 4-dicarbaldehyde

Question 8.

1. What is the relation between an electron donating group and acidic character?
2. How carboxylic acids maintain their acid character?

Answer:

1. Electron donating group decreases the acid character.
2. Carboxylic acid decomposes to give proton and carboxylate ion and is stabilized by resonance. This explains the acidic character of carboxylic acid.

Question 9.
Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:
i) PhMgBr and then H₃O⁺
ii) Tollens’reagent
iii) Semicarbazide and weak acid
iv) Excess ethanol and acid
v) Zinc amalgam and dilute hydrochloric acid
Answer:

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids NCERT Questions and Answers

Question 1.
What is meant by the following terms? Give an example of the reaction in each case

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP-derivative
10. Schiff’s base

Answer:
1. Cyanohydrin – It is a compound which contain both OH and CN groups. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

2. Acetal – compounds formed by the reaction of aldehydes with monohydric alcohols in presence of dry HCI gas.

3. Semicarbazone – the product of carbonyl compounds with semicarbazide is known as semicarbazone.

4. Aldol – It is a condensation product of aldehydes or ketones having atleast one α – hydrogen atom with dilute alkali as catalyst.

5. Hemiacetal – It is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

6. Oxime – Addition compound formed by the reaction of aldehyde or ketone with hydroxylamine.

7. Ketal – It is a cyclic compound obtained by reaction of aceone with ethylene glycol.

8. Imine – Addition compound formed by the reaction of aldehyde or ketone with ammonia.

9. 2, 4-DNP derivative – 2, 4-phenylhydrazone (DNP) is the addition compound formed by the reaction of aldehydes and ketones with 2, 4-dinitrophenylhydrazine.

10. Schiff’s base – Addition compound formed by the reaction of aldehyde or ketone with amine.

Question 2.
Name the following compounds according to IUPAC system of nomenclature.

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂CI
3. CH₃CH = CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₃CCH₂COOH
7. OHCC₆H₅CHO-p

Answer:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3, 3, 5-Trimethylhexane-3-one
6. 3, 3-Dimethylbutanoicacid
7. Benzene 1, 4-dicarbaldehyde

Question 3.
Draw the structure of the following compounds
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-methylbenzaldehyde
(iv) 4-methylpent-3-en-2-one
(v) 3-bromo-4-phenylpentanoicacid
(vi) 4-Chloropentan-2-one
(vii) p, p-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoicacid
Answer:

Question 4.
An organic compound with molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
Answer:
From the given data it is clear that as the compound forms 2, 4-DNP derivative it has >CO group. Since it reduces Tollens’ reagent -CHO group is present. As it can also undergo Cannizzaro reaction α- hydrogen is absent.

The oxidation product suggests that the compound has a benzene ring. One of the – COOH groups have been obtained by the oxidation of – CHO group and the other from alkyl group. Hence on these basis, the structure of C₉H₁₀O is

Question 5.
Write structural formulas and names of the four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde served as nucleophile and which as electrophile.
Answer:
i) Propanal as electrophile as well as nucleophile

ii) Propanal as electrophile and butanal as nucleophile

iii) Butanal as electrophile and propanal as nucleophile

iv) Butanal as electrophile as well as nucleophile

Question 6.
Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:
i) PhMgBr and then H₃O⁺
ii) Tollens’reagent
iii) Semicarbazide and weak acid
iv) Excess ethanol and acid
v) Zinc amalgam and dilute hydrochloric acid
Answer:

Question 7.
Give simple chemical tests to distinguish between

1. Propanal and propanone
2. Acetophenone and Benzophenone
3. Phenol and Benzoic acid
4. Benzaldehyde and acetophenone
5. Ethanal and propanal

Answer:
1. Propanal and propanone:
Propanal and propanone can be distinguished by iodoform test as it is given by propanone and not by propanal
Propanone reacts with hot NaOH/I₂ to give yellow precipitate of iodoform.

2. Acetophenone and benzophenone:
Acetophenone gives iodoform test but benzophenone does not respond.

3. Phenol and benzoic acid:
This can be distinguished by treating FeCI₃ solution. Phenol gives violet colour with FeCI₃ solution while benzoic acid gives buff colured precipitate.

4. Benzaldehyde and acetophenone:
Acetophenone responds to iodoform test while benzaldehyde does not.

5. Ethanal and propanal:
Ethanal gives yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test)

Propanal does not give yellow ppt.

Students can Download Chapter 5 Magnetism and Matter Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 5 Magnetism and Matter

Plus Two Physics Magnetism and Matter NCERT Text Book Questions and Answers

Question 1.
Answer the following questions regarding earth’s magnetism:

1. A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
2. The angle of dip at a location in southern India is about 18°. Would you expect a greater or lesser dip angle in Britain?
3. If you make a map of magnetic field lines at Melbourne in Australia, would the lines seen to go into the ground or come out of the ground?
4. In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
5. The earth’s field is claimed roughly approximately, the field due to a dipole magnetic moment 8 × 10²² JT^-1 located at the centre. Check your or¬der of magnitude of this number same way.
6. Geologists claim that besides the main magnetic NS poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

Answer:
1. Magnetic declination, angle of dip, horizontal component of earth’s magnetic field.

2. Greater angle of dip in Britain, (it is about 70°), because Britain is closer to the magnetic north pole.

3. Field lines of B due to the earth’s magnetism would seem to come out of the ground.

4. A compass is free to move in a horizontal plane, while the earth’s field is exactly vertical at the magnetic poles. So the compass can point in any direction there.

5. Use the formula for field B on the normal bisector of a dipole of magnetic moment m.
B = \(\frac{\mu_{0}}{4 \pi} \frac{M}{r^{3}}\)
Take M = 8 × 10²² JT^-1
r = 6.4 × 10⁶ m;
one gets B = 0.3G
which checks with the orde of magnitude of the observed field on the earth.

6. The earth’s field is only approximately a dipole field. Local N-S poles may arise due to, for instance, magnetized minerals deposits.

Question 2.
A short bar magnet of magnetic moment m = 0.32JT^-1 is placed in a uniform external magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientations would correspond to its

• stable and
• unstable equilibrium? What is the potential energy of the magnet in each case?

Answer:
Given M=0.32JT^-1 B = 0.15T, U = ?

• If \(\overrightarrow{\mathrm{M}} \| \overrightarrow{\mathrm{B}}\), then we have stable equilibrium and U = -MB = -0.32 × 0.15T = -4.8 × 10^-2 J
• If \(\overrightarrow{\mathrm{M}}\) is anti-parallel to \(\overrightarrow{\mathrm{B}}\) , we have unstable equilibrium and
  U = MB = 0.32 × 0.15J = 4.8 × 10^-2J

Question 3.
A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10^-4 m² carries a current of 3.0A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Given N = 800, A = 2.5 × 10^-4 m², I = 3.0A, M = ?
Since M = NIA
∴ M = 800 × 3 × 2.5 × 10^-4
or M = 0.60JT^-1 along the axis of the solenoid.

Question 4.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0A, is suspended through its centre allowing it to turn in a horizontal plane.

1. What is the magnetic moment associated with the solenoid?
2. What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 is set up at an angle of 30° with the axis of the sole¬noid?

Answer:
1. Magnetic dipole moment,
M = nIA
= 2000 × 4.0 × 1.6 × 10^-4 = 1.28JT^-1

2. Net force = 0
Torque, τ = MB sinθ
= 1.28 × 7.5 × 10^-2 × sin 30°
= 1.28 × 7.5 × 10^-3 × 1/2 = 4.8 × 10^-2 Nm.

Question 5.
Answer the following questions:

1. Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?
2. Why is diamagnetism, in contrast, almost independent of temperature?
3. If a toroid used bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
4.  If the permeability of a ferromagnetic material independent of the magnetic field? If not is it more for lower or higher fields?
5. Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point.
6. Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Answer:
1. The tendency to disrupt the alignment of dipoles (with the magnetising field) arising from random thermal motion is reduced at lower temperatures.

2. The induced dipole moment in a diamagnetic sample is always opposite to the magnetising field, no matter what the internal motion of the atom is.

3. Slightly less, since bismuth is diamagnetic.

4. No, as is evident from the magnetisation curve. From the slope of the magnetisation curve, it is clear that μ is greater for lower fields.

5. Proof of the important fact (of much practical use) is based on boundary conditions of magnetic fields (B & H) at the interface of two media. (When one of the media has μ>>1, the field lines meet this medium nearly normally).

6. Yes. Apart from minor differences in the strength of the individual atomic dipoles of two different materials, a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. But of course, saturation requires impractically high magnetising fields.

Plus Two Physics Magnetism and Matter One Mark Questions and Answers

Question 1.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show
(a) anti ferromagnetism
(b) no magnetic property
(c) diamagnetism
(d) paramagnetism
Answer:
(d) paramagnetism

Question 2.
According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature T is proportional to
(a) 1/T
(b) T
(c) 1/T²
(d) T²
Answer:
(a) 1/T
Explanation : According to Curie’s law X ∝ \(\frac{1}{T}\)

Question 3.
Above Curie temperature
(a) a paramagnetic substance becomes ferromagnetic substance
(b) a ferromagnetic substance becomes paramagnetic
(c) a paramagnetic substance becomes diamagnetic
(d) a diamagnetic substance becomes paramagnetic
Answer:
(b) a ferromagnetic substance becomes paramagnetic

Plus Two Physics Magnetism and Matter Two Mark Questions and Answers

Question 1.
Classify the following properties into diamagnetic, paramagnetic and ferromagnetic.

1. Susceptibility -1≤x<0
2. In uniform magnetic field it acquires a large magnetisation in the direction of the field.
3. When it is suspended in a magnetic field, it will come to rest perpendiculartothe magnetic field.
4. Susceptibility of the substance varies inversely as temperature of the substance upto curie temperature ie. x_m∝ \(\frac{1}{T}\)

Answer:

• Dia – a,c
• Para – d
• Ferro – b

Question 2.
Classify the following properties into diamagnetism and ferro magnetism.

1. In non uniform magnetic field, it more from high to law field.
2. Magnetic field lines are repelled from this material, If we place in a external magnetic field.
3. susceptibility greater than one, <0, +ve
4. Iron, Nickel, Cobalt are the examples

Answer:

1. Diamagnetism
2. Diamagnetism
3. Ferro magnetism
4. Ferro magnetism

Question 3.
Fill in the blanks.

Answer:

1. negative
2. Weak magnetic field to strong magnetic field.
3. Individual atoms have tiny magnetic moments
4. µ_r >1

Plus Two Physics Magnetism and Matter Three Mark Questions and Answers

Question 1.
The figure shows hysteresis curves for soft iron and stell.

1. Which among the two is the hysteresis loop of soft iron?
2. Which one among the two materials is preferred for use in transformers and galvanometers?
3. When steel is once magnetized, the magnetiza-tion is not easily destroyed even if it is exposed to strong reverse fields. Give reason.

Answer:

1. Fig. b
2. Soft iron
3. When steel is magnetised the magnetic domains in the material is permanently set and magnetised permanently in the direction of the applied magnetic field.

Question 2.
The figure shows a liquid placed on the pole pieces of two magnets.

1. Which magnetic behaviour is exhibited by the liquid? (1)
2. Write any two characteristics of this magnetic behaviour? (1)
3. Does this behaviour transform with temperature. Why? (1)

Answer:
1. Diamagnetism
2. characteristics of this magnetic behaviour:

• Permeability of a diamagnetic material is less than one.
• Susceptibility is small and negative.

3. No. The magnetic dipole moment induced in the diamagnetic material is opposite to the magnetising field and hence does not affect the thermal motion of atoms. Hence change in temperature has no effect on diamagnetism.

Question 3.
A magnetic material contained in a curved glass plate, when placed in a nonuniform magnetic field, exhibits a property as shown in figure.

1. Which type of magnetic material is this? Explain the property
2. Write two examples for such a magnetic material. Explain how the property relates with temperature?
3. “The susceptibility of a material is small” what do you mean by this statement.

Answer:

1. Diamagnetism: Diamagnetic materials are repelled from external magnetic field.
2. Bismuth, Copper, Lead, Nitrogen, Water. Diamagnetism is independent of temperature.
3. Diamagnetism developed by external magnetic field is very small.

Question 4.
A place where dip = 90°, B_H = 0

1. This place is at…………
2. What is the value of B_u at this place?
3. Can a magnetic needle align in the N-S direction at this place?

Answer:
1. Magnetic pole.

2. B = \(\sqrt{B_{v}^{2}+B_{h}^{2}}\)
But B_h = 0
∴ B = B_v

3. No. Magnetic needle align vertically

Question 5.
Classify into diamagnet, Paramagnet and Fero magnet.

1. Feebly magnetized in the same direction
2. m_r slightly more than none
3. does not allow lines of force
4. temperature independent
5. exhibit hysteresis
6. strongly attracted by a bar magnet

Answer:

1. Paramagnet
2. Para magnet
3. Dia magnet
4. Diamagnet
5. Ferromagnet
6. Ferromagnet

Plus Two Physics Magnetism and Matter Four Mark Questions and Answers

Question 1.
Magnetic field lines are the visualization of magnetic field.

1. Write any three properties of magnetic field lines.
2. The arrangement shows two bar magnets placed near each other. Draw the magnetic field lines of the system.

Answer:
1. properties of magnetic field lines:

• The magnetic field lines of a magnet form continuous closed loops.
• The tangent to the field line at a given point represents the direction of magnetic field at that point.
• Flux density of magnetic field represents the strength of magnetic field.
• The magnetic field lines do not intersect

2.

Question 2.
The figure shows the magnetic field of earth.

1. Identify the labels A, B, C. (1)
2. The lines drawn on a map through places that have the same declination are called………(1)
3. The horizontal component of earth’s magnetic field at a place is 0.25 × 10^-4T and the resultant magnetic field is 0.5 × 10^-4 T. Find the dip and the vertical component of the earth’s magnetic field at the place. (2)

Answer:
1. the labels A, B, C:

• A – Magnetic equator
• B – Magnetic axis
• C – Declination

2. Isogonic lines

3. Horizontal component of earth’s magnetic field is
B_H = B cosδ
0.25 × 10^-4 = 0.5 × 10^-4 δ
δ = 60°
Vertical component of earth’s magnetic field is
Bv = B sin δ = 0.5 × 10^-4 sin 60 = 0.43 × 10^-4T.

Plus Two Physics Magnetism and Matter Five Mark Questions and Answers

Question 1.
When a magnetic needle is placed in a non-uniform magnetic field it experiences
1.

• a force but no torque
• a torque but no force
• force and torque
• neither a force nor a torque (1)

2. A bar magnet held perpendicular to a uniform magnetic field as in the figure. If the torque acting on it is to be reduced to 1/4^th by rotating the magnet towards the direction of the field, find the angle through which the magnet is to be rotated. (2)

3. State whether the potential energy of the magnet increases or decreases after rotation. Justify your answer. (2)
Answer:
1. force and torque

2. When the bar magnet is perpendicular to the field Torque is maximum
τ = MBsinθ = MBsin(90) = MB
When rotated through an angle θ, torque is

Angle through which the magnet is to be rotated is 90 – θ = 75.53°

3. Potential energy decreases. Potential energy is minimum when the magnet is parallel to the field. U = MBcosθ. When rotated to 0° (to make magnet parallel to the field) Potential energy decreases.

Students can Download Chapter 4 Moving Charges and Magnetism Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 4 Moving Charges and Magnetism

Plus Two Physics Moving Charges and Magnetism NCERT Text Book Questions and Answers

Question 1.
A Circular coil of wire consisting of 100 turns, each of radius 8.0cm carries a current of 0.40A. What is the magnitude of the magnetic field B at the centre of the coil?
Answer:
Given n = 100, r = 8.0cm = 8 × 10^-2
I = 0.4A, B = ?
At the centre of circular coil

= π × 10⁴T = 3.1 × 10^-4T.

Question 2.
A uniform field equal to 1.5T exists in a cylindrical region of radius 10.0cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if

1. The wire intersects the axis?
2. The wire is turned from N-S to north-east-north-west direction.
3. The wire in the N-S direction is lowered from the axis by a distance of 4.0cm?

Answer:
1. F = \(\mathrm{Bl} \ell\) = 1.5 × 7 × \(\frac{20}{100}\) or F = 2.1 N acting vertically downwards.

2. Force will again be 2.1N.

3. F = \(\frac{1.5 \times 7 \times 16}{100}\) = 1.68N.

Question 3.
Two long and parallel straight wires A and B carrying currents of 8.0A and 5.0A in the same direction are separated by a distance of 4.0cm. Estimate the force on a 10cm section of wire A.
Answer:
Given I₁ = 8.0A, l₂ = 5.0A, r = 4.0cm = 0.04m
l = 10cm = 0.10m
Since

(direction is given by Fleming left-hand rule).

Plus Two Physics Moving Charges and Magnetism One Mark Questions and Answers

Question 1.
A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. Find the force on the wire is
Answer:
F = i/B = 1.2 × 0.5 × 2 = 1.2 N

Question 2.
To convert a galvanometer into a ammeter, one needs to connect a.
(a) low resistance in parallel
(b) high resistance in parallel
(c) low resistance in series
(d) high resistance in series
Answer:
(a) low resistance in parallel

Question 3.
A coil carrying electric current is placed in uniform magnetic field.
(a) torque is formed
(b) e.m.f is induced
(c) both a and b are correct
(d) none of these
Answer:
(a) torque is formed

Question 4.
Direction of motion of unit the positive test charge gives direction electric field. Direction of motion of…….gives direction of magnetic field.
Answer:
Unit north pole.

Question 5.
A magnetic system with zero dipole moment
(a) Solenoid
(b) current carrying coil
(c) current loop
(d) toroid
Answer:
(d) Toroid.

Question 6.
Find odd one regarding polarity solenoid, torroid, current carrying loop, bar magnet.
Answer:
Torroid (In this case North and South pole are absent).

Question 7.
The nature of path when a charged particle is projected 30° to the direction of magnetic field. (Helix, cycloid, straight line, parabola).
Answer:
Helix

Question 8.
What is solenoid?
Answer:
An insulated copper wire wound in the form of cylinder is called solenoid.

Question 9.
Write mathematical expression for ampere’s theorem.
Answer:
∫B.dl = µ₀I

Plus Two Physics Moving Charges and Magnetism Two Mark Questions and Answers

Question 1.
The figure shows a long straight conductor carrying a current I. A magnetic field is produced around the conductor.

What is the magnitude of the magnetic induction at a point ‘P’ which is at a distant ‘x’ from the conductor?
What is the shape of the magnetic line of force?

Answer:
1. B = \(\frac{\mu_{0} I}{2 \pi x}\). Magnetic field is directed in to the plane of paper.

2. Circular.

Question 2.
A particle of mass 6.65 × 10^-27 kg having positive charge equal to two times of electron, moves with a speed of 6 × 10⁵ m/s in a direction perpendicular to that of a given magnetic field of flux density 0.4 weber/m². Find the acceleration of the particle.
Answer:

q = 2e, v = 6 × 10⁵ m/s
B = 0.4, m = 6.65 × 10^-27 kg
∴ acceleration,

a = 1.15 × 10¹³ m/s².

Question 3.
Classify in to true or false.

1. The magnetic field in the middle of current carrying solenoid depends up on cross sectional area.
2. The magnetic field depends up on current.
3. The magnetic field depends up on the material of the core.
4. The magnetic field depends up on total numbers of turns per unit length.

Answer:

1. False
2. True
3. True
4. True

Question 4.
A short straight conductor carries current I

1. Write the expression for magnetic field due to this conductor.
2. Represent graphically the variation of magnetic field with distance from the wire.

Answer:
1. dB = \(\frac{\mu_{0}}{4 \pi} \frac{|d| \sin \theta}{r^{2}}\)

2.

Plus Two Physics Moving Charges and Magnetism Three Mark Questions and Answers

Question 1.
The magnetic field along the axis of a circular coil is found to be B = \(\frac{\mu_{\mathrm{o}} \mathrm{Ia}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)^{3 / 2}}\)
1. What is the magnetic field along the axis if r>>a
2.

• Compare the above magnetic field with the electric field along the axis of an electric dipole
• What is the equation of magnetic dipole moment?

Answer:
1. B = \(\frac{\mu_{0} / a^{2}}{2 r^{3}}\)

2.

• Electric field due to electric dipole, E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{P}{r^{3}}\) magnetic field due to magnetic dipole B = \(\frac{\mu_{0}}{4 \pi} \frac{m}{x^{3}}\).
• Magnetic moment m = IA

Question 2.
The circuit diagram for verifying Ohm’s Law is given below, A student unknowingly connects a galvanometer in the place of the ammeter.

1. What will happen to the galvanometer?
2. What modification has to be made in the galvanometer if he still wants to use the galvanometer in place of the ammeter?
3. Assuming voltmeter to be an ideal one, what will happen if the student interchanges the position of the voltmeter and ammeter?

Answer:

1. Galvanometer will be damaged.
2. use a shunt resistance.
3. No current flows, because an ideal voltmeter has infinite resistance.

Question 3.
Consider a galvanometer with a full scale deflection of 1 m A and resistance 100Ω.

1. How is the device connected in the circuit?
2. How can it be converted to an ammeter with full scale deflection 1 ampere?

Answer:
1. Connected in series.

2.

by connecting a shunt resistance. 0.1Ω in parallel with galvanometer, we can convert galvanometer in to ammeter.

Question 4.
Two infinitely long straight parallel wires carry currents I each as shown in fig.

1. Which law helps to find direction of magnetic field around a current carrying conductor?
2. What is the magnitude, direction of the magnetic fields at A, and C?

Answer:
1. Right hand grip rule.

2. The magnitude, direction of the magnetic fields at A, and C:

• A – outward to the plane of paper
• C-Inward to the plane of paper.

Question 5.
Analyze the figure and answer the following questions.

1. What is the nature of force between these conductors is……….
2. What is the field due to I₁ at second conductor?
3. What is the force experienced per unit length of II^nd conductor?

Answer:

1. Attractive
2. B= \(\frac{\mu_{0} d_{1}}{2 \pi r}\)
3. f = \(\frac{\mu_{0} l_{1} l_{2}}{2 \pi r}\)

Question 6.
“I am very light and present in every matter. When I move along the equator from east to west, I am pushed up. When I am stationary, no force”.

1. Name the force and write its expression.
2. Who am I?
3. What would happen if moved at the poles?

Answer:

1. Lorentz force in earth’s magnetic field F =qvB
2. Electron
3. The earth’s magnetic field at poles is perpendicular to the earth surface. When an electron move, it is pushed to one side, parallel to earth’s surface.

Plus Two Physics Moving Charges and Magnetism Four Mark Questions and Answers

Question 1.
The internal connections of a moving coil galvanometer is given in the fig (i) and fig (ii)

1. What is the use of moving coil galvanometer?
2. Which figure(i) or figure(ii) is used to measure Voltage?
3. Write an expression for resistance required to convert the moving coil galvanometer in to voltmeter.
4. If a very small resistance (eg, copper wire) is used to convert moving coil galvanometer in to voltmeter, will it work properly?

Answer:

1. Moving coil galvanometer is used to detect the presence of current
2. The instrument shown in figure (2) is used to measure voltage.
3. Resistance is connected in series with galvanometer. R = \(\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}\) – G
4. No, A high resistance is required to convert galvanometer in to voltmeter.

Question 2.
When a charged particle enter normal to a uniform magnetic field, it take a circular path.

1. Name the particle accelerator using this principle.
2. Explain the working of that particle accelerator with relevant theory.
3. The neutrons can’t be accelerated using this partide accelerator. Why?

Answer:
1. Cyclotron.

2. Cyclotron:
a. Uses:
It is a device used to accelerate particles to high energy.

b. Principles:
Cyclotron is based on two facts

• An electric field can accelerate a charged particle.
• A perpendicular magnetic field gives the ion a circular path.

c. Working:

At certain instant, let D₁ be positive and D₂ be negative. Ion (+ve) will be accelerated towards D₂ and describes a semicircular path (inside it). When the particle reaches the gap, D₁ becomes negative and D₂ becomes positive.

So ion is accelerated towards D₁ and undergoes a circular motion with larger radius. This process repeats again and again. Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate.

3. Charge of neutron is zero. Only charged particle can be accelerated using particle accelerated.

Question 3.
A current flows through a circular loop of radius r is shown in the figure.

1. What is the direction of magnetic field at ‘o’? (1)

2. Derive an equation for magnetic field at ‘o’ due to the circular loop carrying current i? (2)

3. If the loop splits into two equal halves as shown in figure. (1)

What will be the magnetic field at the center ‘o’?
Answer:
1. B = \(\frac{\mu_{0} I}{2 r}\) in t0 the plane of paper

2. Magnetic field on the axis of a circular current loop:
Consider a circular loop of radius ‘a’ and carrying current T. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dl,
dB =\(\frac{\mu_{0} \mathrm{Idl} \sin 90}{4 \pi \mathrm{x}^{2}}\)

[since sin 90° -1]
The dB can be resolved into dB cosΦ (along Py) and dB sinΦ (along Px). Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinΦ (along px) and dB cosΦ (along py¹)
dB cosΦ components cancel each other, because they are in opposite direction. So only dB sinΦ components are found at P, so total filed at P is

from ∆AOP we get x = (r² + a² )^1/2

Let there be N turns in the loop then,

Point at the centre of the loop:
When the point is at the centre of the loop, (r = 0) Then,

3. Zero

Question 4.
A boy connects a galvanometer directly to a cell of emf 1.5v to measure a current through a load 1Ω.

1. Which instrument can be used to measure the current in such a circuit?
2. What changes should be made in the galvanometer to measure such a high current? Explain using a circuit diagram.
3. The boy connected the galvanometer into a high current measuring device and he connected the device parallel to the load. What will be the observation. Justify.

Answer:
1. Ammeter

2.

A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it.
Theory:
Let G be the resistance of the galvanometer, giving full deflection fora current I_g. To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In this arrangement, Ig current flows through Galvanometer and remaining (I-I_g) current flows through shunt resistance.
Since G and S are parallel
P.d Across G = p.d across S
I_g × G =(I-I_g)S
S = \(\frac{\lg \mathrm{G}}{\left(1-\mathrm{l}_{\mathrm{g}}\right)}\)
Connecting this shunt resistance across galvanometer we can convert a galvanometer into ammeter.

3. Ammeter is a low resistance device. Hence it draws high current. This high current will damage it.

Question 5.
An electric charge will experience a force in uniform electric field. Similarly a moving charge experience a magnetic force (Lorentz) in magnetic field. The SI unit of magnetic field intensity is defined in terms of Lorentz Force.

1. Write the expression for magnetic Lorentz force.
2. Mention any two difference between electric field arid magnetic field.
3. Give an account of work done by Magnetic Lorentz force on moving charge and corresponding change in K.E.

Answer:
1.

2. Electric field is due to a charge, either in motion or at rest. Magnetic field is due to the motion of charge. Direction of electric force is colinear to electric field. Direction of magnetic force is perpendicularto magnetic field.

3. If velocity (displacement) is perpendicular to Lorentz force the work done will be zero and hence there will no change in K.E.

Question 6.
“Moving coil Galvanorrieter is a device used for detecting very feeble current”.

1. What is the working principle of a moving coil galvanometer?
2. Describe the construction and working of a moving coil galvanometer.
3. When a high current is passed through a moving coil galvanometer, it will get destroyed. How?

Answer:
1. Principle:
A conductor carrying current when placed in a magnetic field experiences a force, (given by Fleming’s left hand rule), τ = NIAB.

2.

A moving coil galvanometer consists of rectangular coil of wire having area ‘A’ and number of turns ‘n’ which is wound on metallic frame and is placed between two magnets. The magnets are concave in shape, which produces radial field.
Working :
Let T be the current flowing the coil, Then the torque acting on the coil. τ = NIAB, Where A is the area of coil and B is the magnetic field.

This torque produces a rotation on coil, thus fiber is twisted and angle (Φ). Due to this twisting a restoring torque (τ = KΦ) is produced in spring. Under equilibrium, we can write
Torque on the coil = restoring torque on the spring
NIAB = KΦ
Φ \(=\left(\frac{\mathrm{BAN}}{\mathrm{K}}\right)\) I
The quantity inside the bracket is constant for a galvanometer.
Φ ∝ I
The above equation shows that the deflection depends on current passing through galvanometer.

3. High current will produce large amount of heat. This heat will destroy coils in the galvanometer.

Question 7.
When a current carrying conductor is placed in a magnetic field it experiences a force.
1. Arrive at the expression forthe force experienced by the conductor. (2)
2. A conductor carrying current I direct out of the plane of the paper is lying in the magnetic field as in Fig. Draw the direction of force experienced by the conductor. (1)

3. If the conductor is lying parallel to the field what will be the force? (2)
Answer:
1. Consider a rod of uniform cross section ‘A’ and length ‘l’ Let ‘n’ be the number of electrons per unit volume (number density). ‘v_d’ be the drift velocityof electrons for steady current ‘I’.
Total number of electrons in the entire volume of rod =nAl
Charge of total electrons = nA l .e
‘e’ is the charge of a single electron.
The Lorentz force on electrons,

2.

3. Zero

Question 8.
A long straight conductor carrying current is placed near a current carrying circular loop as in the figure.

1. If B₁ is the field of the ring and B₂ the field due to straight conductor what will be the direction of B₁ and B₂ at O. (1)

2. The current through the loop and the conductor are 2A and the conductor is at as distance 20cm from the centre of loop. What should be the diameter of the loop so that the net field at O is zero. (3)
Answer:
1. B₁ into the plane and B₂ out of the plane.

2. B₁ is the field of the ring and B₂ the field of due to straight conductor.
B = B₁ – B₂ = 0
B₁ = B₂

2a = 40π × 10^-2m = 1,256m
Diameter d = 1.256 m.

Question 9.
A charged particle is travelling in the figure.

1. Name the force experienced by the charge in the region II. (1)
2. Give the expression for the net force experienced by the charge in the region II. (1)
3. If the charge reaches the region III without any change in its initial direction of motion find the velocity of the charged particle in terms of E and B. (2)

Answer:
1. Lorentz Force

2. F = q (E + v × B)

3. Electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle. If the total force on the charge is zero and the charge will move in the fields undeflected. This happens when
qE = qvB or v = \(\frac{E}{B}\).

Question 10.
The medical diagnostic technique called magnetic resonance imaging (MRI) requires that patient lie in a strong magnetic field. It consists of two large solenoids, placed above and below.

1. What is solenoid?
2. Which law help you to find magnetic field due to solenoid? State the law.
3. Obtain an expression for magnetic field due to solenoid using the above law.
4. If the diameter of one of the MRI coil is increased without changing the current, does the magnetic field that it produce at its centre increases, decreases or stay the same? Justify.

Answer:
1. An insulated conducting wire wound in the form of cylinder is called solenoid.

2. Ampere’s circuital law:
Ampere’s circuital theorem states that the line integral of the magnetic field around any closed path in free space is equal to µ₀ times the net current passing through the surface.

3.

Consider a solenoid having radius Y. Let ‘n’ be the number of turns per unit length and I be the current flowing through it.

In order to find the magnetic field (inside the solenoid ) consider an Amperian loop PQRS. Let ‘l‘ be the length and ‘b’ the breadth

Applying Amperes law, we can write

Substituting the above values in eq (1), we get
Bl = µ₀ l_enc (2)
But l_enc = n l I
where ’nl ’ is the total number of turns that carries current I (inside the loop PQRS)
∴ eq (2) can be written as
Bl = µ₀nIl
B = µ₀ nI
If core of solenoid is filled with a medium of relative permittivity µ_r, then
B = µ₀µ_rnl

4. No change. Magnetic field is independent of radius.

Plus Two Physics Moving Charges and Magnetism Five Mark Questions and Answers

Question 1.
You are supplied with a galvanometer, resistor, and some connecting wires.

1. Using a circuit diagram, show how will you convert the given galvanometer into an ammeter.
2. Find the expression for the shunt resistance in the circuit.
3. A galvanometer is to be converted into an ammeter of range 0 – 1 A. Galvanometer has resistance 100Ω and the current for full scale deflection is 10mA. Find the length of the nichrome wire to be used as shunt.

Given, Resistivity ρ= 1.1 x 10^-6Ωm
Diameter of the wire = 1mm
Answer:
1.

2. Let I_g be the current through the galvanometer of resistance R_G and the shunt resistance be r_s.
Let I be the current to be measured by the converted ammeter.
We can write,
I_gR_G = (I – I_g)r_s
∴ r_s = \(\frac{I_{g} R_{G}}{\left(1-I_{g}\right)}\)

3. Given I = 1A
I_g = 10mA
R_G = 100Ω
Diameter = 1mm