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Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 2 Kalca Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 2 Kalca

Kalca Questions and Answers

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 1 Peeli Kannukal Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 4 Chapter 1 Peeli Kannukal

Peeli Kannukal Questions and Answers

Peeli Kannukal Summary

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 3 Chapter 3 Anargha Nimisham Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 3 Chapter 3 Anargha Nimisham

Anargha Nimisham Questions and Answers

Anargha Nimisham Summary

Students can Download Chapter 2 Units and Measurement Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 2 Units and Measurement

Plus One Physics Units and Measurement One Mark Questions and Answers

Plus One Physics Chapter 2 Questions And Answers Question 1.
How many seconds are there in a light fermi?
(a) 10^-15
(b) 3.0 × 10⁸
(c) 3.33 × 10^-24
(d) 3.3 × 10^-7
Answer:
(c) 3.33 × 10^-24
One light fermi is time taken by light to travel a distance of 1 fermi ie. 10^-15m
1 light fermi = \(\frac{10^{-15}}{3 \times 10^{8}}\) = 3.33 × 10^-24s.

Units And Measurements Questions And Answers Pdf Hsslive Question 2.
Which of the following pairs have same dimensional formula for both the quantities?

1. Kinetic energy and torque
2. Resistance and Inductance
3. Young’s modulus and pressure

(a) (1)only
(b) (2) only
(c) (1) and (3) only
(d) All of three
Answer:
(c) (1) and (3) only

Plus One Physics Units And Measurements Questions And Answers Question 3.
Give four dimensionless physical quantities.
Answer:
Angle, Poisson’s ratio, strain, specific gravity.

Plus One Physics Units And Measurements Questions Question 4.
The dimensions of plank constant are the same as those of______.
Answer:
Angular momentum

Plus One Physics Chapter Wise Questions And Answers Question 5.
A physical quantity P = \(\frac{\sqrt{a b c^{2}}}{d^{3}}\) measuring a, b, c and d separately with the percentage error of 2% , 3%, 2% and 1% respectively. Minimum amount of error is contributed by the measurement of
(a) b
(b) a
(c) d
(d) c
Answer:
(b) a
P = \(\frac{\sqrt{a b c^{2}}}{d^{3}}\)

The minimum amount of error is contributed by the measurement of a.

Plus One Physics Second Chapter Questions And Answers Question 6.
The number of significant figures in 11.118 × 10^-6 is
(a) 3
(b) 6
(c) 5
(d) 4
Answer:
As per rules, number of significant figures in 11.118 × 10^-6 is 5.

Class 11 Physics Ch 2 Important Questions Question 7.
What is the number of significant figures in 0.06070?
Answer:
4.

Plus One Physics Important Questions And Answers Pdf Question 8.
If f = x², What is the relative error in f?
Answer:
\(\frac{2 \Delta x}{x}\).

Hsslive Plus One Physics Chapter Wise Questions And Answers Question 9.
Which of the following measurement is more accu¬rate?
(i) 7000m
(ii) 7 × 10²m
(iii) 7 × 10³m
Answer:
(i) 7000 m

Units And Measurements Questions And Answers Pdf Question 10.
Which of the following measurements is most, accurate?
(a) 5.0 cm
(b) 0.005 cm
(c) 5.00 cm
Answer:
(c) Is most accurate because it has three significant figures. Greater is number of significant figures, more accurate is the measurement.
(a) has 2 significant figures
(b) has 1 significant figure.

Units And Measurements Class 11 Numericals With Solutions Question 11.
Name three physical quantities having same dimension.
Answer:
Work, Energy, and Torque.

Plus One Physics Units and Measurement Tw0 Mark Questions and Answers

Units And Measurements Class 11 Questions Answers Question 1.
Using dimensional analysis derive the relation F = ma. Where the symbols have the usual meaning.
Answer:
Force on a body depends on mass(m), acceleration (a) an
F α m^aa^bt^c
M¹L¹T^-2 = M^a(LT^-2)^bT^c
M¹L¹T^-2 = M^aL^bT^-2a+c
Equating the powers, we get a = 1 ,b = 1, -2b + c = -2, c = 0
F = m¹a¹t⁰ = ma.

Plus One Physics Chapter 2 Previous Year Questions Question 2.
Use your definition to explain how simple harmonic motion can be represented by the equation y = a sin ωt
(a) Show that the above equation is dimensionally correct
Answer:
Y = a sin ωt
sin ωt has no dimensions. Hence we get L = L
Hence this equation is dimensionaly correct.

Questions On Error Analysis Class 11 Question 3.
Fill in the blanks.

1. The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_____m² (Write answer in 3 significant digits)
2. Im = ______ ly

Answer:
1. Curved area = 2πl
= 2 × 3.14 (2 × 10²) × 20 × 10²
= 2.51 × 10^-6m²

2. l ly= 9.46 × 10¹⁵ m
lm = \(\frac{l \mathrm{ly}}{9.46 \times 10^{15}}\) ≈ 10^-6ly.

Physics Chapter 2 Class 11 Numericals Question 4.

1. Give a physical quantity with a unit and no dimension.
2. Arrange the following in the descending order.
   1 light year, 1 parsec, 1 astronomical unit

Answer:

1. Angle has no dimension. But it has unit.
2. 1 parsec, 1 light year, 1 astronomical unit.

Dimensional Analysis Questions And Answers Pdf Question 5.
Magnitude of force F experienced by a certain object moving with speed V is given by F = KV². Where K is a constant. Find the dimensions of K.
Answer:
F = KV²

Class 11 Physics Chapter 2 Important Questions With Answers Question 6.
What is the maximum percentage error in the measurement of kinetic energy if percentage errors in mass and speed are 2% and 3% respectively?
Answer:
E = \(\frac{1}{2}\)v²

% error in KE = % error in mass + 2 × % error in speed
= 2% + 2 × 3% = 8%.

Units And Measurements Class 11 Important Questions Pdf Question 7.
Solve the following with regard to significant figures.

1. 5.8 + 0.125
2. 3.9 × 10⁵ – 2.5 × 10⁴

Answer:
1. 5.8 + 0.125 = 5.925
Rounding to first decimal point, we get 5.9

2. 3.9 × 10⁵ – 2.5 × 10⁴
= 3.5 × 10⁵ – 0.25 × 10⁴
= 3.65 × 10⁵
Rounding to first decimal place, we get 3.6 × 10⁵.

Question 8.
What is maximum fractional error in
i) (a + b)
ii) a – b
iii) ab
iv) \(\frac{a}{b}\)
Given ∆ a and ∆ b are absolute errors in measurements a and b.
Answer:

Question 9.

1. What is the fractional error in aⁿ? (Given absolute error in a is ∆ a)
2. What is absolute error in the measurements according to least count?
   • 3.0 kg
   • 25 s
   • 5.62 cm

Answer:
1. n\(\frac{\Delta a}{a}\)

2. The measurements according to least count:

• 0.1 kg
• 1 s
• 0.01 cm

Plus One Physics Units and Measurement Three Mark Questions and Answers

Question 1.
A stone is thrown upwards from the ground with a velocity ‘u’.

1. What is the maximum height attained by the stone?
2. Check the correctness of the equation obtained in (a) using the method of dimensional analysis.

Answer:
1. H = \(\frac{u^{2}}{2 g}\) ___(1)
u = u, v = o, a = -g, h = ?
We can find maximum height using the equation
u² = u² + 2as
0 = u² + 2 × -g × H
2gh = u²
H = \(\frac{u^{2}}{2 g}\)

2. Dimension of H = L
Dimension of u = (LT^-1)
Dimensions of time (t) = T
Dimension of g = (LT^-2)
substituting these values in eq(1) we get
L = \(\frac{\left(L T^{-1}\right)^{2}}{\left(L T^{-2}\right)}\)
L = L.

Question 2.
Derive an empirical relationship for the force experienced on the car in terms of mass of the car m, velocity v, and radius of the track r using dimensional analysis.
Answer:
Centripetal force may depends on mass (m),radius(r) and velocity(v)
F α m^ar^bv^c
M¹L¹T^-2 = M^aL^b(LT^-1)^c
M¹L¹T^-2 = M^aL^bL^cT^-c
M¹L¹T^-2 = M^aL^b+cT^-c
Equating we get a = 1, b + c = 1, c = 2, b = -1
Substituting these values in eq(1),we get
F = \(\frac{M V^{2}}{r}\).

Question 3.
Dimensional formula of a physical quantity indicate how many times fundamental quantity is involved in the measurement of the quantity.

1. What is the dimensional formula of coefficient of viscosity?
2. Write any two drawbacks of dimensional analysis.

Answer:
1. F = ηA\(\frac{d V}{d x}\)

2. The method of dimensional analysis has the following drawbacks:

• It gives no information about the dimensionless constant involved in the equation.
• The method is not applicable to equations involv¬ing trigonometric and exponential functions.
• This method cannot be employed to derive the • exact form of the relationship if it contains sum
  of two, or more terms.
• If the given physical quantity depends on more than three unknown quantities, the method fails.

Question 4.
Principle of homogeneity is based on the fact that two quantities of same nature can be added.

1. What do you mean by principle of homogeneity?
2. Velocity V depends on the time t as V = at² + bt + c. Find dimension of constants a, b, and c.

Answer:
1. For the correctness of an equation, the dimensions on either side must be the same. This is known as the principle of homogeneity of dimensions.

2. V = at² + bt + c
M⁰L¹T^-1 = aT² + bT + c
According to principle of homogenity, we get
aT² = M⁰L¹T^-1
a = \(\frac{\mathrm{M}^{0} \mathrm{L}^{1} \mathrm{T}^{-1}}{\mathrm{T}^{2}}\)
= M⁰L¹T^-3.

Question 5.
If x = a + bt + ct² where x is in meter and t in second.

1. Find the dimensional formula of ‘b’.
2. If error in the measurement of time is 2%. What will be the error in x?

Answer:
1. According to principle of homogeneity, the dimensions of both sides must be same.
ie. L = a + bT + cT²
ie : L = bT, b = L/T

2.

% error in x = 3 × %. error in ‘t’ = 3 × 2% = 6%.

Question 6.
A physical quantity P is related to four observables a, b, c as P = \(\frac{a^{3} b^{2}}{\sqrt{c d}}\). The % error in the measurement of a, b, c, and d are 1%, 3%, 4%, 2% are respectively.

1. What do you mean by error in a measurement?
2. What is the % error in the measurement of P?

Answer:
1. The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called error.

2.

% error in
P = 3 × 1 + 2 × 3 + 1/2 × 4 + 1/2 × 2
= 3 + 6 + 2+ 1
P = 12%

Question 7.
Rahul measured the height of Ramesh in different trials as 1.67m, 1.65m 1.64m, and 1.63m.

1. Find the mean absolute error?
2. Find the percentage error?

Answer:
1. Arithametic mean,

a_mean = 1.645m = 1.65
absolute error,
∆a₁ = a_mean – a₁
∆a₁ = 1.65 – 1.67 = -0.02
∆a₂ = 1.65 – 1.65 = 0
∆a₃ = 1.65 – 1.64 = 0.01
∆a₄ = 1.65 – 1.63 = 0.02
Mean absolute error

= 0.012

2. percentage error = \(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}\) × 100
= \(\frac{0.012}{1.65}\) × 100
= 0.75%.

Question 8.
In a particular experiment Ramu used the relation F = AB + (P + Q) Y to calculate force.

1. Which principle is used to check the correctness of the equation (1)
2. If the dimensional formula of Y is M⁰L¹T^-1, then find the dimensional formula of P

Answer:
1. Principle of homogenity

2. F = AB + (P+Q)Y
F = AB + PY + QY
MLT^-2 = AB + PY+ QY
According to principle of homogeneity
MLT^-2 = PY
M¹L¹T^-2 = P M⁰L¹T^-1
ie. P = \(\frac{M^{\prime} L^{1} T^{-2}}{M^{0} L^{1} T^{-1}}\) = M¹T^-3

Question 9.

1. Which of the following is precise
   • A vernier calliperse with 40 divisions on sliding scale
   • An optical instrument that can measure length of the order of wavelength of light.
2. Is it possible to increase the accuracy of screw gauge by increasing the number of divisions on the head scale?

Answer:
1. (i) L.C of vernier caliperse = \(\frac{1}{40}\) = 0.025mm
= 0.025 × 10^-3m
= 2.5 × 10^-5m.

(ii) L.C of optical instrument = 6000A°
= 6000 × 10^-10m
(Taking λ of visible light = 6000°A)= 6 × 10^-7m

2. Yes. Because L.C proportional to number of division on the headscale. So with the increase in number of divisions, the least count will increase. This leads to increase the accuracy of above screw guage.

Plus One Physics Units and Measurement Four Mark Questions and Answers

Question 1.
In an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate

1. The mean value of the body
2. Mean absolute error
3. Percentage error

Answer:
1. Mean value, M_mean
= \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\)
= 2.5g

2. Absolute error,
Absolute error ∆m₁ = |2.52 – 2.52| = 0
∆m₂ = |2.52 – 2.53| = 0.01
∆m₃ = |2.52 – 2.51| = 0.01
∆m₄ = |2.52 – 2.49| = 0.03
∆m₅ = |2.52 – 2.54| = 0.02
∴ Mean absolute error
\(\frac{0+0.01+0.01+0.03+0.02}{5}\)
∆m_mean = 0.014g

3. Percentage error = \(\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}}\) × 100
= \(\frac{0.014}{2.52}\) × 100 = 0.556.

Question 2.
While discussing the period of a pendulam, one of the student argued that period depends on the mass of the bob.

1. What is your opinion?
2. How will you prove your argument dimensionally?

Answer:

1. Period is independent of mass of the bob
2. The principle of homogeneity of dimensions also helps to derive a relationship between the different physical quantities involved; This method is known as dimensional analysis.

The period of the simple pendulum may possibly depend upon:

• The mass of the bob, m
• The length of the pendulum, I
• Acceleration due to gravity, g
• The angle of swing, q

Let us write the equation for the time period as t = k m^a l^b g^c q^d
where, k is a constant having no dimensions; a, b, care to be found out.
The dimensions of, t = T¹
Dimensions of. m = M¹
Dimensions of, l = L¹
Dimensions of, g = L¹T^-2
Angle q has no dimensions (since, q = arc/radius = L/L)
Equating the dimensions of both sides of the equation, we get,
T¹ = M^aL^b (L¹T^-2)^c
ie. T¹ = M^aL^b+c+ T^-2c.
The dimensions of the terms on both sides must be the same. Equating the powers of M, L and T.

a = 0; b + c = 0; -2c = 1
∴ c = ^–\(\frac{1}{2}\), b = ^– c = \(\frac{1}{2}\)
Hence, the equation becomes,
t = kl^1/2g^-1/2
ie, t = k\(\sqrt{1 / g}\)
Experimentally, the value of k is found to be 2p.

Plus One Physics Units and Measurement Five Mark Questions and Answers

Question 1.
In an experiment with a common balance the mass of a ring found to be 2.52g, 2.5g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate

1. The mean value of the mass of the ring
2. The absolute error in each measurement
3. Mean absolute error
4. Relative error
5. Percentage error

Answer:
1. The mean value of the mass of the ring.
M_mean = \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\) = 2.52g.

2. The absolute error in each measurement.
∆m₁ = M_mean – m₁ = 2.52 – 2.52 = 0.00
∆m₂ = M_mean – m₂ = 2.52 – 2.53 = -0.01
∆m₅ = M_mean – m₅ = 2.52 – 2.54 = -0.02

3. mean absolute error = |∆m₁| + |∆m₂|………..+|∆m₅|
= 0.014

4. Relative error = δm = \(\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}}=\frac{.014}{2.52}\) = 0.00555

5. Percentage error δm × 100 = 0.555%.

Plus One Physics Units and Measurement NCERT Questions and Answers

Question 1.
Fill in the blanks:

1. The volume of a cube of side 1 cm is equal to______m³.
2. The surface area of a solid cylinder of radius 2.0 cm and height 10.0cm is equal to____(mm)².
3. A vehicle moving with a speed of 18km h^-1 covers_____m in 1s.
4. The relative density of lead is 11.3.Its density is_____g cm^-3or_____kgm^-3.

Answer:
1. V = (1 cm)³
= (10^-2m)³
= 10^-6m³
So, answer is 10^-6.

2. Surface area = 2 πrh + 2 × πr²
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 2 × 10(10 × 10 + 2 × 10)mm²
= 1 .5 × 10⁴mm²
So, answer is 1.5 × 10⁴

3. 18kmh^-1 = \(\frac{18 \times 1000}{3600}\)ms^-1
= 5ms^-1
So, answer is 5.

4. 11.3, 11.3 × 10³ or 1.13 × 10⁴.

Question 2.
Fill in the blanks by suitable conversion of units:

1. 1 kgm²s^-2 = _____g cm²s^-2
2. 1 m =_____1 y
3. 3.0 ms² =______kmh^-2
4. G = 6.67 × 10^-11 Nm² (kg)^-2 =_____(cm)³s^-2g^-1

Answer:

1. 10⁷
2. 10^-16
3. 3.888 × 10⁴
4. 6.67 × 10^-8

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J, where 1 J = 1 kgm²S^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β metere, the unit of time is second, show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in terms of the new units.
Answer:
1 cal = 4.2kg m²s^-2

Dimensional formula of energy is [ML²T^-2]. Comparing with [M^aL^bT^c], we find that
a = 1, b = 2, c = -2

Question 4.
Which of the following is the most precise device for measuring length?

1. A vernier callipers with 20 divisions on the sliding scale.
2. A screw guage of pitch 1 mm and 100 divisions on the circular scale
3. An optical instrument that can measure length to within a wavelength of light?

Answer:
The most precise device is one whose least count is the least.
1. Least count = 1SD – 1 VD = 1 SD – \(\frac{19}{20}\) SD

2. Least count

3. Wavelength = 10^-5 cm = 0.00001 cm
Clearly, the optical instrument is the most precise.

Question 5.
State the number of significant figures in the following:

1. 0.007m²
2. 2.64 × 10²⁴kg
3. 0.2370gcm³
4. 6.320 J
5. 6.032 Nm^-2
6. 0.0006032 m²

Answer:

1. 1
2. 3
3. 4
4. 4
5. 4
6. 4

Question 6.
The length, breadth, and thickness of a rectangular sheet of metal are 4.234m, 1.005m and 2.01cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Area of the sheet = 2(l × b + b × t + t × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)m²
= 2 (4.255 + 0.0202 + 0.0851)m²
= 2 × 4.3603m²
= 8.7206m²
= 8.72m²
Volume = lbt
4.234 × 1.005 × 0.0201m³
= 0.0855m³

Question 7.
A Physical qunatity P is related to four observables a, b, c and d as follows:
P = \(\frac{\mathrm{a}^{3} \mathrm{b}^{2}}{\sqrt{\mathrm{c}} \mathrm{d}}\). The percentage errors of measurement in a, b,c, and d are 1 %, 3%, 4% and 2% respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Answer:
P = \(\frac{\mathrm{a}^{3} \mathrm{b}^{2}}{\sqrt{\mathrm{c}} \mathrm{d}}\)

% error in P = 3% + 6% + 2%+2% = 13%
3.763 should be rounded off to 3.8.

Question 8.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of special relativity due to Albert Einstein). Boy recalls the relation almost correctly but forgets where to put the constant c. He writes:
\(\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}\)
Guess where to put the missing e.
Answer:
From the given equation, \(\frac{m_{0}}{m}=\sqrt{1-v^{2}}\)
Since left hand side is dimensionless therefore right hand side should be also dimensionless.

The correct formula is m = m₀ \((\sqrt{1-\frac{v^{2}}{c^{2}}})^{-1 / 2}\).

Students can Download Chapter 2 Units and Measurement Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 2 Units and Measurement

Plus One Physics Notes Chapter 2 Summary
Introduction

a. Fundamental or base quantities:
Physics is based on measurement of physical quantities. Certain physical quantities are chosen as fundamental or base quantities. Length, mass, time, electric current thermodynamic temperature, amount of substance and luminous intensity are such base quantities.

b. Units: Fundamental Units and Derived Units Unit:
Measurement of any physical quantity is made by comparing it with a standard. Such standard of measurement are known as unit. If length of rod is 5 m, it means that the length of rod is 5 times the standard unit ‘metre’.

Fundamental Unit:
The unit of fundamental or base quantities are called fundamental or base units. The base units are listed in table.

Derived Unit
The units of other physical quantities can be expressed as combination of base units. Such units are called derived units.
Example: Unit of force is kgms^-2 (or Newton). Unit of velocity is ms^-1.

The International System Of UnitsDerived Unit
System of Units: A complete set of fundamental and derived units is called a system of unit.

a. Different system of units:
The different systems of units are CGS system FPS (or British) system, MKS system and SI system. A comparison of these systems of unit is given in the table below, (for length, mass and time)

Note: The first three systems of units were used in earlier time. Presently we use SI system.

b. International System Of Unit (Si Unit):
The internationally accepted system of unit for measurement is system international d’ unites (French for International System of Units). It is abbreviated as SI.

The SI system is based on seven fundamental units and these units have well defined and internationally accepted symbols, (given in table – 2.1)

c. Solid Angle and Plane Angle:
Other than the seven base units, two more units are defined.
1. Plane angle (dq): It is defined as ratio of length of arc (ds) to the radius, r.


The unit of plane angle is radian. Its symbol is rad.

2. Solid Angle (dW): It is defined as the ratio of the intercepted area (dA) of spherical surface, to square of its radius.


The unit of solid angle is steradian. The symbol is Sr.

Unit And Measurement Class 11 Notes Pdf Chapter 2 Measurement Of Length
Two methods are used to measure length

• direct method
• indirect method.

The metre scale, Vernier caliper, screwgauge, spherometer are used in direct method for measurement of length. The indirect method is used if range of length is beyond the above ranges.

1. Measurement Of Large Distances:
Parallax Method:
Parallax method is used to find distance of planet or star from earth. The distance between two points of observation (observatories) is called base. The angle between two directions of observation at the two points is called parallax angle or parallactic angle (q).

Plus One Physics Chapter 2 Notes Pdf Parallax Method
The planet ‘s’ is at a distance ‘D’ from the surface of earth. To measure D, the planet is observed from two observatories A and B (on earth). The distance between A and B is b and q be the parallax angle between direction of observation from A and B.

AB can be considered as an arch A h B of length ‘b’ of a circle of radius D with its center at S. (Because q is very small, \(\frac{b}{D}\)<<1], Thus from arch-radius relation.

Thus by measuring b and q distance to planet can be determined. The size of planet or angular diameter of planet can be measured using the value of D. If the angle a (angle between two directions of observation of two diametrically opposite points on planet) is measured using a


Where d is diameter of planet.

2. Estimation Of Very Small Distances:
Size Of Molecule
Electron microscope can measure distance of the order of 0.6A0 (wavelength of electron).

3. Range Of Lengths:
The size of the objects in the universe vary over a very wide range. The table (given below) gives the range and order of lengths and sizes of some objects in the universe.

Units for short and large lengths
1 fermi = 1f = 10^-15m
1 Angstrom = 1A° = 10^-10m
1 astronomical unit = 1AU = 1.496 × 10¹¹m
1 light year = 1/y = 9.46 × 10¹⁵m
(Distance that light travels with velocity of 3 × 10⁸ m/s in 1 year)
1 par sec = 3.08 × 10¹⁶m = 3.3 light year
(par sec is the distance at which average radius of earth’s orbit subtends an angle of 1 arc second).

Units And Measurements Questions And Answers Pdf Hsslive Measurement Of Mass
Mass is basic property of matter. The S.l. unit of mass is kg. While dealing with atoms and molecules, the kilogram •is an inconvenient unit. In this case there is an important standard unit called the unified atomic mass unit( u).
1 unified atomic mass unit = lu
= (1/12)^th of the mass of carbon-12

1. Range Of Masses:
The masses of the objects in the universe vary over a very wide range which is given in the table.

Units And Measurements Class 11 Chapter 2 Measurement Of Time
To measure any time interval we need a clock. We now use an atomic standard of time, which is based on the periodic vibrations produced in a cesium atom. This is the basis of the cesium clock sometimes called atomic clock.

Plus One Physics Note Chapter 2 Definition of second:
One second was defined as the duration of 9, 192, 631, 770 internal oscillations between two hyperfine levels of Cesium-133 atom in the ground state.
Range and Order of time intervals

Class 11 Physics Chapter 2 Notes Accuracy, Precision Of Instruments And Errors In Measurement
Error:
The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called error.

Systematic errors:
The systematic errors are those errors that tend to be in one direction, either positive or negative.

Sources of systematic errors

1. Instrumental errors
2. Imperfection in experimental technique or procedure
3. personal errors

1. Instrumental errors:
Instrumental error arise from the errors due to imperfect design or calibration of the measuring instrument.
eg: In Vernier Callipers, the zero mark of vernier scale may not coincide with the zero mark of the main scale.

2. Imperfection in experimental technique or procedure:
To determine the temperature of a human body, a thermometer placed under the armpit will always give a temperature lower than the actual value of the body temperature. Other external conditions (such as changes in temperature, humidity, velocity……..etc) during the experiment may affect the measurement.

3. Personal Errors:
Personal error arise due to an individual’s bias, lack of proper setting of the apparatus or individual carelessness etc.

Random errors
The random errors are those errors, which occur irregularly and hence are random with respect to sign and size. These can arise due to random and unpredictable fluctuations in experimental conditions (eg. unpredictable fluctuations in temperature, voltage supply, etc.)

Unit And Measurement Notes Pdf Chapter 2 Least Count Error
The smallest value that can be measured by the measuring instrument is called its least count. The least count error is the error associated with the resolution of the instrument. By using instruments of higher precision, improving experimental technique etc, we can reduce least count error.

1. Absolute Error, Relative Error And Percentage Error:
The magnitude of the difference between the true value of the quantity and the measured value is called absolute error in the measurement. Since the true value of the quantity is not known, the arithmetic mean of the measured values may be taken as the true value.

Physics Class 11 Chapter 2 Notes Explanation:
Suppose the values obtained in several measurements are a₁, a₂, a₃,………,a_n. Then arithmetic mean can be written as

The absolute error,
∆a₁ = a_mean – a₁
∆a₂ = a_mean – a₂
∆a_n = a_mean – a_n

a. Mean absolute error:
The arithmetic mean of all the absolute errors is known as mean absolute error. The mean absolute error in the above case,

b. Relative error:
The relative error is the ratio of the mean absolute error (Da_mean) to the mean value (a_mean).

c. Percentage error:
The relative error expressed in percent is called the percentage error (da).

Example:
Units And Measurements Class 11 Notes Pdf Download Question 1.
When the diameter of a wire is measured using a screw gauge, the successive readings are found to be 1.11 mm, 1.14mm, 1.09mm, 1.15mm and 1.16mm. Calculate the absolute error and relative error in the measurement.
Answer:
The arithmetic mean value of the measurement is

The absolute errors in the measurements are
1.13 – 1.14 = 0.02mm
1.13 – 1.14 = -0.01mm
1.13 – 1.09 = 0.04mm
1.13 – 1.15 =-0.02 mm
1.13 – 1.16 = 0.03mm
The arithmetic mean of the absolute errors

Percentage of relative error

2. Combination Of Errors:
When a quantity is determined by combining several measurements, the errors in the different measurements will combine in some way or other.

a. Error of a sum or a difference:
Rule: when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.
Explanation:
Let two quantities A and B have measured values A ± DA and B ± DB respectively. DA and DB are the absolute errors in their measurements. To find the error Dz that may occur in the sum z = A + B,
Consider
z + ∆z = (A ± ∆A) + B ± ∆B = (A + B) ± ∆A ± ∆B
The maximum possible error in the value of z is given by,

Similarly, it can be shown that, the maximum error in the difference.
Z = A – B is also given by

b. Error of product ora quotient:
Rule: When two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.
Explanation:
Suppose Z=AB and the measured values of A and B are A + DA and B + DB. They
Z + DZ = (A + DA) (B + DB)
= AB ± BDA ± ADB ± DADB
Dividing LHS by Z and RHS by AB, we get

c. Errors in case of a measured quantity raised to a power:
Suppose Z = A²

Hence, the relative error in A² is two time the error in A.
In general, if \(Z=\frac{A^{P} B^{q}}{C^{T}}\)
Then

Hence the rule: The relative error in a physical quantity raised to the power K is the K times the relative error in the individual quantity.

Chapter 2 Physics Class 11 Notes Significant Figures
Every measurement involves errors. Hence the result of measurement should be reported in a way that indicates the precision of measurement.

Normally, the reported result of measurement is a number that includes all digits in the number that are known reliable plus the first digit that is uncertain. The reliable digits plus the first uncertain digit are known as significant digits or significant figures.
Example:

• The length of a rod measured is 3.52cm. Here there are 3 significant figures. The digits 3 and 5 are reliable and the last digit 2 is uncertain.
• The mass of a body measured as 3.407g. Here there are four significant figures. The figure 7 is uncertain.

When the measurement becomes more accurate, the number of significant figure is increased.
Rules to find significant figures:
1. All the non zero digits are significant.
Example:
Physics Notes For Class 11 Kerala Syllabus Chapter 2 Question 1.
Find significant figure of

• 2500
• 263.25

Answer:

• In this case, there are two nonzero numbers. Hence significant figure is 2.
• In this, there are 5 nonzero numbers. Hence significant figure is 5.

2. All the zeros between two nonzero digits are significant, no matter where the decimal point is,
Example:
Units And Measurements Class 11 Notes Chapter 2 Question 2.
Find the significant figure

• 2.05
• 302.005
• 2000145

Answer:

• Significant figure is 3
• Significant figure is 6
• Significant figure is 7

3. If the number is less than 1, the zeros on the right of decimal point but to the left to the first nonzero digits are not significant.
Example:
Class 11 Physics Notes Units And Measurements Chapter 2 Question 1.
Find the significant figure of

• 0.002308
• 0.000135

Answer:

• 4 significant figures
• 3 significant figures

4. The terminal zeros in a number without a deci¬mal point are not significant.
Example:
Physics Class 11 Chapter 2 Question 1.
Find the significant figure of

• 12300
• 60700

Answer:

• 3
• 3

Note: But if the number obtained is on the basis of actual measurement, all zeros to the right of last non zero digit are significant.
Example: If distance is measured by a scale as 2010m. This contain 4 significant figures.

5. The terminal zeros in a number with a decimal point are significant.
Example:
Gvhss Payyoli Physics Notes Chapter 2 Question 1.
Find the significant figure of

• 3.500
• 0.06900
• 4.7000

Answer:

• 4
• 4
• 5

Method to find significant figures through scientific notation:
In this notation, every number is expressed as a × 10^b, where a is a number between 1 and 10 and b is any positive or negative power. In this method, we write the decimal after the first digit.
Example:
4700m =4.700 × 10³m
The power of 10 is irrelevant to the determination of significant figures. But all zeros appearing in the base number in the scientific notation are significant. Hence each number in this case has 4 significant figures.
Significant figures in numbers:-

a. Rules for Arithmetic operations with significant figures:
1. Rules for multiplication or division:
In multiplication or division, the computed result should not contain greater number of significant digits than in the observation which has the fewest significant digits.
Examples:
(i) 53 × 2.021 =107.113
The answer is 1.1 × 10² since the number 53 has only 2 significant digits.

(ii) 3700 10.5 = 352.38
The answer is 3.5 × 10² since the minimum number of significant figure is 2 (in the number 3700)

2. Rules for Addition and Subtraction:
In addition or substraction of given numbers, the same number of decimal places is retained in the result as are present in the number with minimum number of decimal places.
Examples:
(i) 76.436 +
12.5
88.936
The answer is 88.9, since only one decimal place is found in the number 12.5.

(ii) 43.6495 +
4.31
47.9595
The answer is 47.96 since only two decimal places are to be retained.

(iii) 8.624 –
3.1726
5.4514
The answer is 5.451

(iv) 6.5 × 10^-5 – 2.3 × 10^-6 = 6.5 × 10^-5 – 0.23 × 10^-5
= 6.27 × 10^-5
The answer is = 6.3 × 10^-5

Units And Measurements Class 11 Notes Pdf Chapter 2 Dimensions And Dimensional Analysis
All physical quantities can be expressed in terms of seven fundamental quantities. (Mass, length, time, temperature, electric current, luminous intensity and amount of substance). These seven quantities are called the seven dimensions of the physical world.

The dimensions of the three mechanical quantities mass, length and time are denoted by M, L and T. Other dimensions are denoted by K (for temperature), I (for electric current), cd (for luminous intensity) and mol (for the amount of substance).

The letters [L], [M], [T] etc. specify only the nature of the unit and not its magnitude. Since area may be regarded as the product of two lengths, the dimensions of area are represented as [L] × [L] = [L]².

Similarly, volume being the product of three lengths, its dimensions are represented by [L]³. Density being mass per unit volume, its dimensions are M/L³ or M¹L³.

Thus, the dimensions of a physical quantity are the powers to which the fundamental units of length, mass, time must be raised to represent it.
Note: The dimensions of a physical quantity and the dimensions of its unit are the same.

Class 11 Physics Notes Chapter 2 Dimensional Formula And Dimensional Equations
An equation obtained by equating a quantity with its dimensional formula is called dimensional equations of the physical quantities.
Examples:
Consider for example, the dimensions of the following physical quantities.
1. Velocity: Velocity = distance/ time = L/T = L¹T^-1 \The dimension of velocity are, zero in mass, 1 in length and-1 in time.

2. Acceleration:
Acceleration = \(\frac{\text { Change in velocity }}{\text { time }}=\frac{L^{1} T^{-1}}{T}=L^{1} T^{-2}\)

3. Force: Force = mass × acceleration
Dimensions of force = M × L¹T^-2 = M¹L¹T ^-2
That is, the dimensions of force are 1 in mass, 1 in length and -2 in time.

4. Momentum: Momentum = mass × velocity
Dimensions of momentum = M × L¹T^-1 = M¹L¹T ^-1

5. Moment of a force: Moment = force × distance
Dimensions of moment = M¹L¹T^-2 × L = M¹L²T ^-2

6. Impulse: Impulse = force × time
Dimensions of impulse = M¹L¹T^-2 × T = M¹L¹T ^-1

7. Work: Work = force × distance
Dimensions of work = M¹L¹T^-2 × L = M¹L²T ^-2

8. Energy: Energy = Work done
Dimensions of energy = dimensions of work = M¹L²T^-2.

9. Power: Power = work/time
Dimensions of power \(=\frac{M^{2} L^{2} T^{-2}}{T}p\) = M¹L²T^-3

Explain Parallax Method Class 11 Chapter 2 Dimensional Analysis And Its Applications
The important uses of dimensional equations are:

1. To check the correctness of an equation.
2. To derive a correct relationship between different physical quantities.
3. To convert one system of units into another.

1. Checking the correctness of an equation:
For the correctness of an equation, the dimensions on either side must be the same. This ‘ is known as the principle of homogeneity of dimensions.

If an equation contains more than two terms, the dimensions of each term must be the same. Thus, if x = y + z, Dimensions of x = dimensions of y = dimensions of z
Example :
Units And Dimensions Class 11 Chapter 2 Question 1.
Check the correctness of the equation s = ut + 1/2at² by the method of dimensions.
Dimensions of, s = L¹
Dimensions of, u = L¹T^-1
Dimensions of, ut = L¹T^-1 × T¹ = L¹
Dimensions of, a = L¹T^-2
Dimensions of, at² = L¹T^-2 × T² = L¹
The constant 1/2 has no dimensions. Each term has dimension L¹.
Therefore, dimensions of, ut + 1/2 at² = ¹
Thus, either side of the equation has the same dimen¬sion L1 and hence the equation is dimensionally correct.
Note: Even though the equation is dimensionally correct, it does not mean that the equation is necessarily correct. For instance the equation s = ut + at² is also dimensionally correct, though the correct equation, s = ut + 1/2 at².

2. Deriving the correct relationship between different physical quantities:
The principle of homogeneity of dimensions also helps to derive a relationship between the different physical quantities involved. This method is known as dimensional analysis.
Example :
Question 1.
Deduce an expression for the period of oscillation of a simple pendulum.
The period of the simple pendulum may possibly depend upon

• The mass of the bob, m
• The length of the pendulum, I
• Acceleration due to gravity, g
• The angle of swing, q

Let us write the equation for the time period as t = km^a l^b g^c θ^d
where, k is a constant having no dimensions; a, b, c are to be found out. ’
The dimensions of, t = T¹
Dimensions of m = M¹
Dimensions of, l = L¹
Dimensions of, g = L¹T^-2
Angle q has no dimensions (since, q = arc/radius = L/L) Equating the dimensions of both sides of the equation, we get,
T¹ = M^aL^b (L¹T^-2)^c
ie. T¹ = M^aL^b+cT^-2c
The dimensions of the terms on both sides must be the same. Equating the powers of M, L and T.
a = 0; b + c = 0; -2c = 1
∴ c = \(\frac{-1}{2}\), b = ^–c = \(\frac{1}{2}{/latex]
Hence, the equation becomes,
t = kl^1/2, 2g^-1/2
ie, t = k[latex]\sqrt{l/g}\)
Experimentally, the value of k is found to be 2p.
Limitations of Dimensional Analysis:
The method of dimensional analysis has the following limitations:

• It gives no information about the dimensionless constant involved in the equation.
• The method is not applicable to equations involving trigonometric and exponential functions.
• This method cannot be employed to derive the exact form of the relationship, if it contains sum
  of two, or more terms.
• If the given physical quantity depends on more than three unknown quantities, the method fails.

3. Conversion of one system of units to another:
Suppose we have a physical quantity of dimensions a, b and c in mass, length and time. The dimensional formula for the quantity is therefore, M^aL^bT^c. Let its numerical value be n, in one system in which the fundamental units of mass, length and time are M₁, L₁ and T₁ respectively. Then, the magnitude of the physical quantity
= n₁ M₁^aL₁^bT₁^c
Also, let the numerical value of the same quantity be n₂ in another system where the fundamental units of mass, length and time are M₂, L₂ and T₂respectively. Then the magnitude of the quantity
= n₂ M₂^aL₂^bT₂^c
Equating, n₂ M₂^aL₂^bT₂^c =
n₁ M₁^aL₁^bT₁^c

Example :
Question 1.
Find the number of dynes in one newton.
Answer:
Dyne is the unit of force in the C.G.S. system and newton is the S.I.unit. The dimensional formula for force is M¹L¹T^-2. In eqn. (1) let the suffix 1 refer to quantities in S.I and 2 those in the C.G.S. system.
Here, a = 1, b = 1 and c = 2

and n₁ = 1 (ie. one Newton)
By eqn. (1),
n₂ = 1 (1000)¹ (100)¹ (1)^-2 = 10⁵
ie. 1 newton = 10⁵ dynes.