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Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership

April 13, 2021March 16, 2020 by Prasanna

Students can Download Chapter 5 Dissolution of Partnership Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership

Plus Two Accountancy Dissolution of Partnership One Mark Questions and Answers

Question 1.
Which is not correct in the case of Dissolution of Partnership
(a) The original partnership agreement is terminated
(b) Some partners continue in the business
(c) No partner to continue in the business
(d) A new partnership comes in to existence
Answer:
(c) No partner to continue in the business

Question 2.
Dissolution of partnership does not lead to
(a) Termination of the original partnership agreement
(b) Dissolution of the existing partnership
(c) Coming in to existence of a new partnership
(d) Dissolution of the firm
Answer:
(d) Dissolution of the firm

Question 3.
Realisation Account is a
(a) Nominal Account
(b) Real Account
(c) Personal Account
(d) None of these
Answer:
(a) Nominal Account

Question 4.
The Account prepared at the time of dissolution of a partnership firm
(a) Revaluation Account
(b) P&L Adjustment A/c
(c) P&L Appropriation A/c
(d) Realisation Account
Answer:
(d) Realisation Account

Question 5.
The Realization account is closed by transferring the profit or loss to
(a) Partner’s Capital Accounts
(b) Partner’s Loan Account
(c) Bank Account
(d) Balance Sheet
Answer:
(a) Partner’s Capital Account

Question 6.
The Loan from Mrs.of a partner is credited to
(a) Her Capital Account
(b) Husband’s Capital Account
(c) Husband’s Loan Account
(d) Realisation Account
Answer:
(d) Realisation Account

Question 7.
On dissolution of partnership Firm, amount realised from unrecorded asset is credited to.
(a) Realisation A/c
(b) Re-Valuation A/c
(c) Capital A/c
(d) Goodwill A/c
Answer:
(a) Realisation A/c.

Question 8.
Entry for closing Provision for Bad debts at the time of dissolution of firm is_______.
Answer:
Provision for baddebt a/c Dr. To Realisation

Question 9.
Should you pass any entry for the payment of creditors worth Rs. 5000 on dissolution. If they accept stock of the same value? If yes, what is the journal entry?
Answer:
Creditors takes over stock of the same value. So no journal entry is need to be passed.

Question 10.
A firm is compulsorily dissolved when all partners or when all except one partner become insolvent – True or False
Answer:
True.

Question 11.
Unrecorded liabilities when paid by a partner are shown in_______.
Answer:
Debit of realisation a/c

Question 12.
On dissolution of a firm, bank overdraft is transferred.
(a) cash a/c
(b) bank a/c
(c) Realisation a/c
(d) capital a/c
Answer:
(c) Realisation a/c

Question 13.
On dissolution of the firm, partners capital accounts are closed through_______account.
Answer:
Cash/Bank Account

Plus Two Accountancy Dissolution of Partnership Two Mark Questions and Answers

Question 1.
What do you mean by Dissolution of partnership?
Answer:
Dissolution of partnership means termination of the existing partnership agreement between the partners. This may due to admission, retirement or death of a partner. In the case of dissolution of partnership, the firm continues to exist.

Question 2.
What is meant by Dissolution of firm?
Answer:
Termination of the partnership agreement between all the partners is known as dissolution of firm. In the case of dissolution of firm, the firm ceases to exist and the business of the firm is closed.

Question 3.
Why the balance at bank is not transferred to the Realisation A/c on the dissolution of a Partnership? Answer:
On the dissolution of a partnership the balance at bank is not transferred to the Realisation A/c be cause, there is no need to realise the same.

Question 4.
How will you settle firm’s debts and private debts of partner’s on the dissolution of a firm?
Answer:
Since the liability of partners is unlimited their private assets can be used to pay off the firm’s debts. But they will have the right to use their assets for paying their private debts first. They need to contribute only the remaining assets.

Question 5.
What is Realisation Account?
Answer:
Realisation Account is an account prepared at the time of dissolution of a partnership firm. It is to close the assets and liabilities and to find out the profit or loss and for the payment of liabilities.

Question 6.
What is the Accounting treatment of settlement with the creditors through transfer of an asset?
Answer:
Settlement with the creditors through transfer of assets require no entry. It is because the liability to the creditors has already been closed by transferring the same to realization account. The asset account also was closed by transferring to the same account.

Question 7.
How goodwill is treated on dissolution of the firm ?
Answer:
On dissolution of firm goodwill is treated like the other assets. It is transferred to realization account at its balance sheet amount. The amount realized for goodwill if any, is credited to realization account.

Question 8.
Complete the series

Sacrificing ratio: admission: Gainining ratio:?
Dissolution: Realisation A/c: Reconstitution: ?
Trading A/c: Profit and Loss A/c: Profit and Loss A/c:?
Balance of capital A/c: Balance sheet Balance of profit and loss appropriation a/c:?

Answer:

S.R. (Sacrificing Ratio) : Admission, G.R (Gaining Ratio) : Retirement
Dissolution : Realisation a/c Reconstitution: Revaluation a/c
Trading a/c : P&La/c, P & L a/c : P & L appropriation a/c
Balance of capital a/c : B/S, Balance of P & L appropciation a/c : Capital a/c.

Plus Two Accountancy Dissolution of Partnership Three Mark Questions and Answers

Question 1.
Which are the cases where a partnership is dissolved?
Answer:
Following are the cases in which a partnership is dissolved.

Change in the profit sharing ratio
Admission of a partner
Retirement, death of a partner
Insolvency of a partner
Completion of the venture for which it is formed
Expiry of the period

Question 2.
Toya and Soya are partners sharing profits and losses equally. They decided to dissolve the firm on 15^th March, 2005 which resulted in a loss of Rs. 30,000. The capital accounts of Toya and Soya was Rs. 20,000 and Rs. 30,000 respectively. The cash account showed a balance of Rs. 20,000. You are required to pass journal entries for

Transfer of loss to the capital accounts of partners.
Making final payments to the partners.

Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 1

Plus Two Accountancy Dissolution of Partnership Five Mark Questions and Answers

Question 1.
Distinguish between dissolution of partnership and dissolution of firm.
Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 2

Question 2.
How the accounts are settled on dissolution?
Answer:
On dissolution of a firm, the assets are realized (disposed) and the liabilities are paid off. Balance if any is shared among the partners. According to the Partnership Act, the following rules can be followed for the settlement of accounts.
1. Loss is to be paid first out of profits, next out of capital and last out of the private assets of partners in their ratio.

2. Amount realized from the assets of the firm shall be used in the following order

Paying the realisation expenses
Paying the liabilities to outsiders
Paying the loans from partners
Paying the capital of the partners
Surplus if any is to be distributed to partners

Question 3.
Explain the different modes of dissolution of a partnership firm.
Answer:
Different modes of dissolution of a partnership firm are the following
(i) Dissolution by Agreement (section 40)
A firm is formed by an agreement between the partners. So it may be dissolved by the partnership agreement or with the consent of all the partners.

(ii) Compulsory Dissolution (Section 41)
A firm is dissolved compulsorily in the following cases

When all the partners or all except one become insane or insolvent.
When the business of the firm becomes illegal.
When all the partners except one retire.
When all the partners or all except one die.
- When the period of partnership expires,
- When the venture for which the partnership was formed becomes complete.

(iii) Dissolution on the happening of contingencies(42)
A firm may be dissolved on the happening of the following contingencies or events

On the death of a partner
On the insolvency of a partner

(iv) Dissolution by Notice (Sec 43)
If the partnership is a partnership at will, it can be dissolved by a partner by giving a notice to the other partners showing his will to dissolve the firm.

(v) Dissolution by Court (Sec 44)
A Court may issue an order to a partnership firm to dissolve the same on the suit of a partner in the following circumstances

If a partner becomes insane
If a partner becomes in capable of performing his duties
If a partner is found guilty of misconduct affecting the firm
If a partner intentionally and continuously commits breach of contract
If a partner transfers his interest in the firm to an outsider.
If the business of the firm cannot be carried on except at a loss
lf the court thinks it just and equitable to dissolve the firm

Question 4.
Differentiate between Realisation Account and Revaluation Account.
Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 3

Question 5.
What entry would you pass for the following transaction on the dissolution of a firm having partners Vishal and Rakesh.

An unrecorded asset realised Rs. 6200.
Dissolution expenses amounted to Rs. 3200.
Creditors already transferred to realisation account were paid Rs. 88,000.
Stock worth Rs. 5400 already transferred to realisation account was sold for Rs. 4100.
Profit on realisation Rs. 48000 to be distributed between partners, Vishal and Rakesh?

Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 4

Question 6.
Boby, Jestin, and Sudheer are in partnership in the ratio of 3:2:3. They have decided to dissolve the firm. On the date of dissolution total creditors were Rs.16,000; Bills discounted Rs. 2,650 during the year, has become a real liability which has to be paid, through this has not been recorded anywhere in the books of accounts. Their capital account balances were Boby Rs. 12000; Jestin Rs. 10000; Sudheer Rs. 8000 respectively. Boby advanced Rs. 14000 besides his capital account.
Find out

Total Sundry Assets
Realisation Account
Capital accounts of partners

Answer:
Balance Sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 5
Realisation A/c

Capital A/cs

Plus Two Accountancy Dissolution of Partnership Eight Mark Questions and Answers

Question 1.
Appu and Chinku were partners in a firm sharing profits and losses in the ratio of 4: 3. Their Balance Sheet as on 31^st December 2005 was as follows.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 8

The firm is dissolved as on the Balance sheet date. The assets were realized as follows.

Sundry Debtors	Rs. 14,000
Stock-in-trade	Rs. 21,000
Furniture	Rs. 17,500
Machinery	Rs. 25,000

Sundry Creditors were paid at a discount of 15%. The expenses on realisation amounted to Rs. 2,500. Pass journal entries and prepare ledger accounts on dissolution of the firm.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 10

Realisation Account

Partners’ Capital Account
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 13
Bank Account

General Reserve Account

Question 2.
The following is the Balance Sheet of Felix, Edwin, and Abel sharing profits and losses in the ratio of 2: 1: 1 as on 31^th March 2005.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 16

The firm is dissolved. Sundry debtors realized Rs. 25.0 and stock Rs. 17,000. Trade mark and goodwill became valueless. Edwin agrees to discharge the bank loan. Creditors are paid Rs. 25,000 in full settlement, realisation expenses amounted to Rs. 6.0 paid by Felix. Pass journal entries and prepare ledger accounts.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 18

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 19
Realisation Account

Partner’s Capital Account

Bank Account

Notes:

If nothing is given in the question, it is assumed that the assets(real) are realized and liabilities are paid off at their book value.
Partners loan is not transferred to realisation account but paid directly.

Question 3.
Anu and Binu were partners sharing profits and losses in the ratio of 1/2 and 3/4. Their Balance Sheet as on 30^th June 2004 was as follows.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 23
The firm is dissolved. Furniture and Machinery realized 10% less than their book values. Rs.20,000 is collected from debtors. Anu took over the stock at Rs. 25,000. The firm had an unrecorded liability on outstanding expenses Rs.2,500. Realisation expenses amounted to Rs. 2,000. Record journal entries and prepare ledger accounts.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 24

Realisation Account
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 25
Partner’s Capital Account

Bank Account

Question 4.
S.Raj, Narchison, and Boby are partners sharing profits in the ratio of 2:2:1, Whose ledger accounts on 31.03.06 reveal the following balances.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 28
The firm dissolved on the above date.

During the years S.Raj withdraw from Bank Rs. 6,000 for his personal purpose which has not been brought into the records.
Rs. 10,000 was realised on account of unrecorded investments which was totally utilised for a liability on account of a claim payable to customers and the balance has been paid in cash.
Fixed Assets realised more than 10% of book value.
Sundry debtors could be collected only to the extend of 90% of the book value.

Prepare necessary accounts.
Answer:
1. Realisation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 29

Capital Account
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 30
Bank A/c

Question 5.
Mr. White and Mr. Black are partners sharing profits in the ratio of 3:2. They decided to close the firm and their Balance sheet is given below.
Balance sheet as on 31.03.2005
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 32
Assets realised as follows:
Building – 32,000, Debtors – 28,000, Furniture – 36,000 Liabilities settled as follows. Plant has been taken over by bank at Rs. 66,000 in respect of the loan granted by the bank and the rest has been paid in cash. Creditors are settled at Rs. 30,000. Realisation expenses came to Rs. 1,000 which have been met by Mr. Black. Prepare necessary accounts to dissolve the firm.
Answer:
Realisation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 33

Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 34
Bank A/c

Question 6.
Ali, Banu, and Cini were in partnership who have dissolved their firm on 31.10.2006 on which date their Balance sheet stood as follows.
Balance sheet as on 31.10.2006.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 36

Bank passbook shows its balance to be Rs. 21,300. The difference is due to realisation of a claim directly credited to Bank Account.
Bills Receivable collected Rs. 300 less.
Stock has been utilized to settle the loan with Biju.
Unrecorded electronic equipments worth Rs.5000 has been utilised for settling the liability on account of Bills payable.
Land and Buildings was realised at Rs. 2,40,000.
All other assets were realised and liabilities were settled at book value.

Prepare necessary accounts.
Answer:
Realisation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 37

Capital A/cs
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 38
Cash A/c

Question 7.
Sam, Zen, and Jhony are in partnership sharing profits and losses in the ratio 3:2:1. Their Balance sheet as on 31^st December, 2004 was as follows.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 40
The firm was dissolved on the above date with the following terms.

Building was taken over by Sam at book value and he agreed to discharge the creditors.
Accured interest was not collected, where as there was a contingent liability of Rs. 600 which was met.
Assets realised as follows: Plant – 25000, Stock – 5000, Debtors – 4600
Realisation expenses amounted to Rs. 600 You are required to prepare
- Realisation account
- Capital accounts
- Cash account

Answer:
Realisation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 41
Capital A/cs

Cash A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 43

Question 8.
Joe, Maggi, and Hassan were partners sharing profits and losses in the ratio of 1:2:2. Their Balance sheet as on 31 December 2004 was as follows.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 44
The partners agreed to dissolve the firm on the following terms.

Assets realised as follows:
Land and Building Rs. 1,20,000 Stock 40,000
Accounts receivable 15,000
Expenses on dissolution is Rs. 3000
A creditor accepts office equipments for Rs. 7000 and the remaining creditors were paid in full by cheque.
The joint life insurance policy was surrendered for Rs.9000. Prepare realisation a/c, capital accounts and bank account.

Answer:
Realisation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 45

Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership - 46
Bank A/c

Plus One Physics Notes Chapter 13 Kinetic Theory

April 13, 2021March 16, 2020 by Prasanna

Students can Download Chapter 13 Kinetic Theory 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 13 Kinetic Theory

Summary
Introduction
The kinetic theory was developed in the nineteenth century by Maxwell, Boltzmann and other. It gives a molecular interpretation of pressure and temperature of a gas. It also explains gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It help us to find molecular sizes and masses.

Molecular Nature Of Matter
The scientific atomic theory is credited to John Dalton. Atomic theory is not the end of quest, but the beginning. Atoms consist of a nucleons and electrons. The nucleous itself is made up of protons and neutrons. The protons and neutrons are again made up of quarks. Even quarks may not be the end of the story.

There may be string-like elementary entities. In this chapter we shall limit ourselves to understanding the behavior of gases.

Behavior Of Gases
Gas Laws
1. Boyles law
The law states that at a given temperature, the volume of a given mass of gas varies inversely as its pressure.
It can be written as
\(P a \frac{1}{V}\) (at constant T)
PV = constant
PV = μ RT
μ → No. of moles, R → universal gas constant, T → temperature
Boyles law is not obeyed by gasses at all temperatures and pressure. Usually, Boyles, law is obeyed by gases at high temperature and low pressure (Graph is given below).

A real gas which obey this law is called ideal or perfect gas.

Variation of ‘R’ \(\left(=\frac{\mathrm{PV}}{\mathrm{T}}\right)\) with Pressure (for different temperatures)
Plus One Physics Notes Chapter 13 Kinetic Theory 1

Variation of R \(\left(=\frac{\mathrm{PV}}{\mathrm{T}}\right)\) with Pressure for temperature T₁ > T₂ > T₃ is shown in the above graph. The above graph shows that, all real gases approach ideal gas behavior at low pressure and high temperature.

At low pressure or high temperature, the molecules are for apart and molecular interactions are negligible. Without interactions the gas behaves like an ideal gas.

Note:
R = 8.324 J mol^-1 k^-1.
Variation of V with P for different temperature:
Plus One Physics Notes Chapter 13 Kinetic Theory 2

The above graph shows experimental PV curves for steam at three temperature The dotted line are the theoretical curves. (According to Boyles law). The theoretical value and experimental value comes in agreement at high temperatures and low pressures.

2. Charles law
Charles law states that the volume of a given mass of gas is proportional to its temperature when its pressure is kept constant.
V a T (P is constant)
i.e., Plus One Physics Notes Chapter 13 Kinetic Theory 3

The graph between V and T:
Plus One Physics Notes Chapter 13 Kinetic Theory 4

The above graph shows experimental T-V curves (solid lines) for Co₂ at three pressures compared with Charles law (dotted lines). T is in unit of 300 k and V in units of 0.13 litres.

Question 1.
Why theoretical value does not agree with experimental value?
Answer:
According to Charles law the graph between T and V is straight line. It means that when temperature decreases, the volume of gas decreases and finally becomes zero.

Practically volume will not be zero. Because the molecules require some finite space to exist. This implies that we cannot reduce its temperature to zero value. A zero kelvin temperature is only an idealized concept.

Dalton’s law of Partial Pressures:
It states that, the total pressure of a mixture of ideal gases is the sum of partial pressures.

Proof:
Consider a mixture of non-interacting ideal gases. μ₁, moles of gas 1, μ₂ moles of gas 2 etc. in a vessel of volume V at temperature T and pressure P. Using Boyles law, we can write
PV = (μ₁ + μ₂ +…………..)RT
P = \(\frac{\mu_{1} \mathrm{RT}}{\mathrm{V}}+\frac{\mu_{2} R T}{V}+\ldots \ldots \ldots\)
P = P₁ + P₂ +……………………

Kinetic Theory Of An Ideal Gas
The kinetic theory of gases has been developed by Clausius, Maxwell, Boltzmann and others. The theory is based on the following postulates.

The gas is a collection of large number of molecules. The molecules are perfectly elastic hard spheres.
The size of a molecule is negligible compared with the distance between the molecules.
The molecules are always in random motion
During their motion, the molecules collide with each other and with the walls of the containing vessel.
The collisions are elastic and hence the total K.E energy and the total momentum of the colliding molecules before and after collisions are the same.
The kinetic energy of a molecule is proportional to the absolute temperature of the gas.
There is no force of attraction or repulsion between molecules.

The pressure of an ideal gas:
Plus One Physics Notes Chapter 13 Kinetic Theory 5

Consider molecules of gas in a container. The molecules are moving in random directions with velocity V. This is the velocity of a molecule in any direction. The velocity V can be resolved along x, y and z directions as V_x, V_y, and V_z respectively.

If we assume a molecule hits the area A of the container with velocity V_x and rebounds back with -V_x. (The velocities V_x and V_y do not change because this collision is perfectly an elastic one).

Therefore, the change in momentum imparted to the area A by the molecule
= mv_x^– – mV_x
= 2mV_x

To find the total number of collisions taking place in a time t, consider the motion of the molecules towards the wall. The molecules covers a distance V_xt along the x direction in a time t. All the molecules within the volume AV_xt will collide with the area in a time t.

If ‘n’ is the number of molecules per unit volume, the total number of molecules hitting the area A,
N = AV_xt n.

But on an average, only half of those molecules will be hitting the area, and the remaining molecules will be moving away from the area. Hence the momentum imported to the area in a time t
Q = 2mv_x × \(\frac{1}{2}\) AV_xtn.
= nmV_x²At
The rate of change of momentum,
\(\frac{Q}{t}\) = nmV_x²A
But rate of change of momentum is called force, ie. force F = nmV_x²A
∴ Pressure P = nmV_x² (P = \(\frac{F}{A}\))
Different molecules move with different velocities. Therefore, the average value V_x² is to be taken. If \(\overline{\mathbf{v}}_{\mathbf{x}}^{2}\) is the average value then the pressure.

\(p=n m \bar{v}_{x}^{2}\) ……………….. (1)

\(\overline{\mathbf{v}}_{\mathbf{x}}^{2}\) is known as the mean square velocity. Since the gas is isotropic (having the same properties in all directions), we can write
Plus One Physics Notes Chapter 13 Kinetic Theory 6

Kinetic Interpretation of gas laws:

Question 2.
Derive the ideal gas equation from P = \(\frac{1}{3} \mathrm{nm} \overline{\mathrm{v}}^{2}\)
Answer:
The average kinetic energy of the molecule is
KE = \(\frac{1}{2} m \bar{v}^{2}\) …………………….(3)
The eq (2) can be modified as
Plus One Physics Notes Chapter 13 Kinetic Theory 7

The average Kinetic energy of a molecule remains constant when the temperature is constant. That is when the temperature varies, \(\overline{\mathrm{KE}}\) also varies accordingly. The kinetic energy of a molecule is related to its absolute temperature by an equation
\(\overline{\mathrm{KE}}\) = \(\frac{3}{2}\)K_BT
Substitute equation (5) in equation (6); we get
Plus One Physics Notes Chapter 13 Kinetic Theory 8
N = μ N₀
P V = μ N₀K_BT
P V = μ R T…………………… (8)
(N_BK_B = R)
This is the ideal gas equation

Deduction of Boyles law:
If the temperature is constant for gas, the eq (8) can be written as
PV = Constant
This is called Boyles law

Deduction of Charles law:
If pressure of a gas is constant, the eq (8) can be written as
V a T

This is called Charles law.

Deduction of Avogadro’s Hypothesis:
If P₁, T₁, and V constant, N₀ will be constant, ie. equal Volumes of all gases, under the same conditions of pressure and temperature will contain the same number of molecules. This is known as Avogadro’s hypothesis.

Law Of Equipartion Of Energy
Degrees of freedom:
Degrees of freedom is number of independent ways by which a molecule can possess kinetic energy of translation, rotation and vibration.

Law of equipartition energy:
The total kinetic energy of a molecule is equally divided among the different degrees of freedom.

K.E. per degree of freedom:
The average energy per degree of freedom
Plus One Physics Notes Chapter 13 Kinetic Theory 10
Where K_B is Boltzman constant.

Degrees of freedom and energy of monoatomic gas:
A monoatomic atom has 3 degrees of freedom, ie; it can move in x, y and z-direction. The average energy of single monoatomic gas in the x-direction,
Plus One Physics Notes Chapter 13 Kinetic Theory 11

Note:
1. If a molecule is restricted to move in plane. It has 2 degrees of freedom.
2. If a molecule is restricted to move in a line, it has only 1 degrees of freedom.

Degrees of freedom and energy of single diatomic molecule (rigid rotator)
Plus One Physics Notes Chapter 13 Kinetic Theory 12

Consider a diatomic molecule as a rigid rotator (Rigid rotator means that the molecule does not vibrate). A rigid diatomic molecule has 3 translation degrees of freedom and 2 rotational degrees of freedom.
(Rotational degrees of freedom is shown in the above figure).
∴ Total average energy of diatomic rigid rotator,
Plus One Physics Notes Chapter 13 Kinetic Theory 13

Note:
A diatomic molecule has 3 rotational degrees of freedom. But we consider only 2 degrees of freedom. We neglect rotation along the line joining the atoms. Because it has very small moment of inertia.

Degrees of freedom and energy of single diatomic molecule of nonrigid rotator:
Molecules like ‘co’ even at moderate temperatures have a mode of vibration. The vibration energy of a diatomic molecule.
Plus One Physics Notes Chapter 13 Kinetic Theory 14
Where K is the force constant of the oscillator and y the vibrational coordinate. The vibration energy mode contain two terms (1) Kinetic energy (2) Potential energy. Hence a single mode of vibration of molecule is considered as 2 degrees of freedom (potential energy and kinetic energy)
∴ The total vibrational energy of a single-mode
= 2 × \(\frac{1}{2}\)K_BT
= K_BT
∴ The total energy of diatomic nonrigid rotator
Plus One Physics Notes Chapter 13 Kinetic Theory 15

Degrees of freedom and energy of polyatomic molecule, (non- rigid rotator):
If a polyatomic molecule has ‘f’ modes of vibration, total number of degrees freedom = 3 vibration + 3 rotator+f vibration.
∴ Total average energy of single polyatomic molecule,
Plus One Physics Notes Chapter 13 Kinetic Theory 16

Specific Heat Capacity
Monoatomic Gases (Molar specific heat capacity):
The energy of a single monoatomic gas = 3 × \(\frac{1}{2}\) K_BT
The energy of one mole monoatomic gas = 3 × \(\frac{1}{2}\) K_BT × N_A
[one-mole atom contain Avogadro number (N_A) of atoms]
Plus One Physics Notes Chapter 13 Kinetic Theory 17

A Diatomic Gas (Molar specific heat capacity):
A rigid diatomic molecule has 5 degrees of freedom : 3 translational and 2 rotational.
∴ energy of single diatomic (rigid) molecule = 5 × \(\frac{1}{2}\) K_BT
for one mole of diatomic molecule, energy U = 5 × \(\frac{1}{2}\) K_BT × N_A
Plus One Physics Notes Chapter 13 Kinetic Theory 18

Nonrigid diatomic molecule having a vibrational mode (Molar specific heat capacity):
If the diatomic molecule is not rigid but has a vibration mode. The energy of one mole,
Plus One Physics Notes Chapter 13 Kinetic Theory 20

Polyatomic Gas (Molar specific heat capacity):
In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (t) of vibrational modes.

The energy of one mole polyatomic gas,
Plus One Physics Notes Chapter 13 Kinetic Theory 21

Note: The experimental value of C_P and C_V of polyatomic gases are greater than the predicted values. The theoretical value and experimental value will be equal when we include vibrational modes of motion in the calculation.

Specific heat capacity of solids:
Consider a solid of N atoms. Each atom is vibrating about its mean position. Each vibration mode has two degrees of freedom (corresponding to potential energy and kinetic energy). Hence an oscillation in one dimension has average energy of

2 × \(\frac{1}{2}\) K_BT = K_BT
∴ Total energy in three dimension
= 3 × K_BT
= 3K_BT.

For one mole of solid, total energy,
U = 3K_BT × N_A
[N_A = Avagadro number]
= 3N_AK_BT
U = 3RT …………………. (1) [∴ R = K_BN_A].
We know specific heat capacity,
Plus One Physics Notes Chapter 13 Kinetic Theory 23

Note:
In solids, we do not consider the translational and rotational degrees of freedom. We consider only vibrational degrees of freedom.

Specific heat capacity of water:
We treat water like a solid. A water molecule has 3 atoms. Each atom in the molecule is vibrating about its mean position. A single vibration mode has 2 degrees of freedom (1) potential energy (2) kinetic energy.

ie; The energy of one atom in one dimensional vibration mode =
2 × \(\frac{1}{2}\) K_BT = K_BT

The energy of one atom having 3 dimensional vibration mode = 3 × K_BT
The energy of one H₂0 molecule having 3 dimension vibration mode
U = 3 × 3K_BT × N_A
U = 9RT [∵ R = K_B N_A]
∴ Specific heat capacity,
\(\mathrm{c}=\frac{\mathrm{dU}}{\mathrm{dT}}=\frac{\mathrm{d}}{\mathrm{dT}}(\mathrm{RT})\)
C = 9R.

Mean Free Path
Molecules in a gas have large speeds. Yet a gas leaking from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room. Why?

The molecules in a gas have a finite size. So they collide with other molecules during their motion. As a result, they cannot move straight like path. Their paths are continuously deflected.

Mean free path:
The mean free path is the average distance covered by a molecule between two successive collisions.

Expression for mean free path:
Plus One Physics Notes Chapter 13 Kinetic Theory 24

Suppose the molecules of a gas are spheres with diameterd. Let 〈v〉 be the average velocity of the molecule.
The volume covered by a molecule during its motion, in a time Δt = πd² 〈v〉 Δt.
If ‘n’ is the number of molecules per unit volume, the total number of molecules in the above volume
= πd² 〈v〉 Δt n.
The number collisions in a time Dt,
= πd² 〈v〉 Δt n.
Number of collisions in one second,
= n π d² 〈v〉
∴ The time between two successive collisions,
\(\tau=\frac{1}{n \pi d^{2}\langle v\rangle}\).
The average distance between two successive collisions,
Plus One Physics Notes Chapter 13 Kinetic Theory 25

In this derivation, we imagined the other molecules to be rest. But actually all other molecules are moving. Hence we must take relative velocity 〈v_r〉 instead of 〈v〉. A more exact treatment gives
\(\ell=\frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^{2}} \dots(1)\)
The mean free path given by the above equation depends inversely on the number density and the size of the molecule.

Plus One Zoology Notes Chapter 5 Digestion and Absorption

April 13, 2021March 16, 2020 by Prasanna

Students can Download Chapter 5 Digestion and Absorption Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 5 Digestion and Absorption

What is digestion?
This process of conversion of complex food substances to simple absorbable forms is called digestion.

DIGESTIVE SYSTEM:
It includes

Alimentary canal
Associated glands.

Alimentary canal:
Plus One Zoology Notes Chapter 5 Digestion and Absorption 1

The human digestive system
1. The alimentary canal begins with the mouth, and it leadsto the buccal cavity or oral cavity.

2. The oral cavity has a number of teeth and a musculartongue. Each tooth is embedded in a socket of jaw bone. This type of attachment is called thecodont.

3. Human being forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Dental formula of adult human
An adult human has 32 permanent teeth which are of four different types (Heterodont dentition).

incisors (I)
canine (C)
premolars (PM)
molars (M).

Plus One Zoology Notes Chapter 5 Digestion and Absorption 2
Arrangement of teeth in each half of the upper and lower jaw in the order I, C, PM, M is represented by a dental formula which in human is
\(\frac{2123}{2123}\)

The hardest part of teeth is made up of enamel, helps in the mastication oWood.
The tongue is attached to the floor of the oral cavity by the frenulum.
The upper surface of the tongue has small projections called papillae, some of which bear taste buds.
The oral cavity leads pharynx which serves as a common passage for food and air.
The oesophagus and the trachea (wind pipe) open into the pharynx.
A cartilaginous flap called epiglottis prevents the entry of food into the glottis – opening of the wind pipe – during swallowing.
The oesophagus is a thin, long tube which passing through the neck, thorax and diaphragm and leads to a ‘J’ shaped bag like structure called stomach.
A muscular sphincter (gastro-oesophageal) regulates the opening of oesophagus into the stomach.

The stomach, located in the upper left portion of The abdominal cavity, has three major parts:

A cardiac portion into which the oesophagus opens
A fundic region and
A pyloric portion which opens into the first part of small intestine

Small intestine is distinguishable into three regions:

A ‘U’ shaped duodenum
A long coiled middle portion jejunum and
A highly coiled ileum

The opening of the stomach into the duodenum is guarded by the pyloric A sphincter. Ileum opens into the large intestine.
Plus One Zoology Notes Chapter 5 Digestion and Absorption 3

Large intestine:
It consists of

caecum
colon
rectum.

Plus One Zoology Notes Chapter 5 Digestion and Absorption 4
Caecum:
It is a small blind sac consists of some symbiotic micro-organisms. A narrow finger-like tubular projection, the vermiform appendix which is a vestigial organ, arises from the caecum. The caecum opens into the colon.

colon:
It is divided into three parts-an ascending, a transverse and a descending part. The descending part opens into the Rectum.

Rectum: It opens out through the anus.
Wall layers of alimentary canal:
It consists of four layers namely

Serosa
muscularis
sub-mucosa &
mucosa.

Plus One Zoology Notes Chapter 5 Digestion and Absorption 5

Serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues.
Muscularis is formed by smooth muscles
The submucosal layer is formed of loose connective tissues containing nerves, blood and lymph vessels.
The innermost layer lining of the alimentary canal is the mucosa.
- In duodenum, glands are also present in sub-mucosa.
- Mucosa forms irregular folds (rugae) in the stomach and small finger-like foldings called villi in the small intestine
- The cells lining the villi produce numerous projections called microvilli giving a brush border appearance.
- These modifications increase the surface area.
- Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal.
- Mucosal epithelium has goblet cells which secrete mucus that help in lubrication.
- Mucosa also forms glands in the stomach (gastric glands) and crypts in between the bases of villi in the intestine (crypts of Lieberkuhn).

Digestive Glands:
It includes

salivary glands
liver
pancreas.

1. Salivary glands:
Saliva is mainly produced by three pairs of salivary glands

parotids (cheek)
sub-maxillary/sub-mandibular (lower jaw)
sublinguals (belowthe tongue).

The duct systems of liver, gall bladder and pancreas These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.

2. Liver:

It is the largest gland of the body weighing about 1.2 to 1.5 kg in an adult human.
It is situated in the abdominal cavity, just below the diaphragm and has two lobes.
The hepatic lobules are the structural andfunctional units of liver containing hepatic cells arranged in the form of cords.
Each lobule is covered by a thin connective tissue sheath called the Glisson’s capsule.
The bile secreted by the hepatic cells passes through the hepatic ducts and is stored in gall bladder.
The duct of gall bladder (cystic duct) along with the hepatic duct from the liver forms the common bile. duct
The bile duct and the pancreatic duct open together into the duodenum which is guarded by sphincter of Oddi.

Plus One Zoology Notes Chapter 5 Digestion and Absorption 6
3. Pancreas:
It is a heterocrine (both exocrine and endocrine) elongated organ situated between the limbs of the ‘U’ shaped duodenum.

Secretions of exocrine and endocrine:

The exocrine portion secretes an alkaline pancreatic juice containing enzymes
The endocrine portion secretes hormones, insulin and glucagons.

DIGESTION OF FOOD:

The process of digestion is accomplished by mechanical and chemical processes.
The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. *Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus.

Contents of saliva:
It contains electrolytes (Na⁺, K+⁺, Cl^–, HCO^–“) and enzymes, salivary amylase and lysozyme.
The chemical process of digestion is initiated in the oral cavity by the carbohydrate splitting enzyme, the salivary amylase( pH 6.8).

Digestion of starch:
About 30 percent of starch is hydrolysed here by salivary amylase (optimum pH 6.8) into a disaccharide -maltose.

Enzyme for preventing infections:
Lysozyme present in saliva acts as an antibacterial agent that prevents infections.

The bolus is then passed into the pharynx and then into the oesophagus by swallowing or deglutition.
The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis.
The gastro-oesophageal sphincter controls the passage of food into the stomach.

Mucosa of stomach and gastric glands:
Gastric glands have three major types of cells namely

(i) mucus neck cells which secrete mucus
(ii) peptic or chief cells which secrete the proenzyme pepsinogen and
(iii) parietal or oxyntic cells which secrete HCI and intrinsic factor (factor essential for absorption of vitamin B12).

Digestion in stomach:
1. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme.

2. The proenzyme pepsinogen, in the presence hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach.

3. Pepsin converts proteins into proteoses and peptones (peptides).

4. The mucus and bicarbonates play an important role in lubrication and protection of the mucosal epithelium from concentrated hydrochloric acid- pH (pH 1.8).

Special type of proteolytic enzyme in infants:
Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins. The bile, pancreatic juice and the intestinal juice are the secretions released into the, small intestine.

Pancreatic juice:
It contains inactive enzymes

1. Trypsinogen,
2. chymotrypsinogen,
3. procarboxypeptidases,
4. amylases, lipases and
5. nucleases.

Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into aqtive trypsin.

Contents of Bile and functions:
The bile released into the duodenum contains bile pigments (bilirubin and bili-verdin), bile salts, cholesterol and phospholipids but no enzymes Bile helps in emulsification of fats, i.e., breaking down of the fats into very smalfmicelles.

Bile also activates lipases. The intestinal mucosal epithelium has goblet cells which secrete mucus. The secretions of the brush border cells of the mucosa alongwith the mucus constitute the intestinal juice or succu sentericus.

Intestinal juice/succus entericus:
It contains enzymes like

disaccharidases (eg: maltase)
dipeptidases
lipases,
nucleosidases etc.

How does the intestine protect itself from digestion?
1. The mucus alongwith the bicarbonates from the pancreas protects the intestinal mucosa from acid as well as provide an alkaline medium (pH 7.8) for enzymatic activities.

2. Sub-mucosal glands (Brunner’s glands) also help in this.
Plus One Zoology Notes Chapter 5 Digestion and Absorption 8

The breakdown of biomacromolecules occurs in the duodenum region of the small intestine. The simple substances thus formed are absorbed in the jejunum and ileum regions of the small intestine. The undigested and unabsorbed substances are passed on to the large intestine.
Plus One Zoology Notes Chapter 5 Digestion and Absorption 9

Functions of large intestine:

1. Absorption of some water, minerals and certain drugs
2. Secretion of mucus which helps in adhering the waste (undigested) particles together and lubricating it for an easy passage.

The undigested, unabsorbed substances called faeces enters into the caecum of the large intestine through ileo-caecal valve, which prevents the back flow of the faecal matter. It is temporarily stored in the rectum till defaecation.

The sight, smell and the presence of food in the oral cavity can stimulate the secretion of saliva.
Gastric and intestinal secretions are also stimulated by neural signals.
The muscular activities of different parts of the alimentary canal controlled by neural mechanisms, both local and through CNS.

ABSORPTION OF DIGESTED PRODUCTS:

The end products of digestion passthrough the intestinal mucosa into the blood or lymph.
It is carried out by passive, active or facilitated transport mechanisms.
Small amounts of monosacharides like glucose, amino acids and some of electrolytes like chloride ions are generally absorbed by simple diffusion.
Fructose and some amino acids are absorbed with the help of the carrier ions like Na⁺. This mechanism is called the facilitated transport.
Transport of water depends upon the osmotic gradient.
Various nutrients like amino acids, monosacharides like glucose,electrolytes like Na⁺ are absorbed into the blood by active transport and hence requires energy.

How does fat absorption occur?
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. Then, the fat globules are coated with small protein called as chylomicrons which are transported into the lymph vessels (lacteals) in the villi.

These lymph vessels ultimately release the absorbed substances into the blood stream. The maximum absorption occurs in the small intestine. The absorbed substances finally reach the tissues which utilise them for their activities. This process is called assimilation.

The digestive wastes, solidified into faeces in the rectum initiate a neural reflex causing an urge for its removal. The egestion of faeces to the outside through the anal opening (defaecation) is a voluntary process and is carried out by a mass peristaltic movement.
Plus One Zoology Notes Chapter 5 Digestion and Absorption 10

DISORDERS OF DIGESTIVE SYSTEM:
Jaundice:
The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.

Vomiting:
It is the ejection of stomach contents through the mouth. This reflex action is controlled by the vomit centre in the medulla. „

Diarrhoea:
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of food.

Constipation:
In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly. ‘

Indigestion:
In this condition, the food is not properly digested leading to a feeling of fullness. The causes are inadequate enzyme secretion, anxiety, food poisoning, overeating, and spicy food.

NCERT SUPPLEMENTARY SYLLABUS
Calorific value of carbohydrate, protein and fat:
Carbohydrates, proteins and fats are chief sources of energy in humans. These are oxidized and released energy stored in ATP, it is used for many activities of the cell.

calorific value kcal = 4.184kJ):
It is defined as the amount of heat produced in calories (cal) or in joules (J) from complete combustion of 1 gram food in a bomb calorimeter (a closed metal chamber filled with 02). One kilocalorie is the amount of heat energy needed to raise the temperature of one kilogram of water through 100C(1.80F).

The calorific values of carbohydrates, proteins and fats are 4.1 kcal /g, 5.65 kcal /g and 9.45 kcal/g, respectively. The actual amounts of energy liberated in the body by these nutrients referred to as the physiologic value of the food, and are 4.0 kcal /g, 4.0 kcal Ig and 9.0 kcal /g respectively.

DEFICIENCY DISEASES:
The low amount of nutrients (Vitamin A, iron and iodine) in the diet cause deficiency disorders The important among them is protein energy malnutrition (PEM). lt is major health and nutritional problems in India.

Young children (0-6 years) require more protein for each kilogram of body weight than adults. So they are more vulnerable to malnutrition. Malnutrition leads to permanent impairment of physical and mental growth and childhood mortality and morbidity. The details of the disorders are given below.
Plus One Zoology Notes Chapter 5 Digestion and Absorption 11
The child suffering from PEM can recover if adequate quantities of protein and carbohydrate rich food are given.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement/Death of a Partner

April 13, 2021March 16, 2020 by Prasanna

Students can Download Chapter 4 Reconstitution of a Partnership Firm – Retirement/Death of a Partner Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement/Death of a Partner

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner One Mark Questions and Answers

Question 1.
A partner who severs his connection with his firm is known as
(a) Retiring partner
(b) Outgoing partner
(c) Incoming partner
(d) None of these
Answer:
(b) Outgoing partner

Question 2.
On retirement of a partner
(a) The partnership is dissolved
(b) The firm is dissolved
(c) The business is dissolved
(d) None of these
Answer:
(a) The partnership is dissolved

Question 3.
The retiring partner is not liable for
(a) The losses of the firm
(b) The losses of the firm till the date of retirement
(c) The losses at the time of his retirement
(d) The losses after his retirement
Answer:
(d) The losses after his retirement

Question 4.
If the firm is not in a position to pay the amount due to the retiring partner, the amount is transferred to
(a) The retiring partner’s capital account
(b) The retiring partner’s loan account
(c) All the partner’s capital account
(d) None of these
Answer:
(b) The retiring partner’s loan account

Question 5.
The amount due to the deceased partner is transferred to
(a) His capital account
(b) His loan account
(c) His executor’s capital account
(d) His executor’s loan account
Answer:
(d) His executor’s loan account

Question 6.
When premium paid on Joint Life Policy is treated as an investment not as a business expense, it is transferred to
(a) Trading account
(b) Profit and Loss Account
(c) Joint Life Policy Account
(d) Joint Life Policy account and Balance Sheet
Answer:
(d) Joint Life Policy account and Balance Sheet

Question 7.
When partner retiring from the firm, the ratio relevant is_______.
(a) Sacrificing ratio
(b) Gaining ratio
(c) New ratio Gaining ratio
Answer:
(b) Gaining ratio

Question 8.
Write the narration of the given journal entry
Continuing Partners Capital A/c Dr.
To Retiring Partners capital A/c
(______________________)
Answer:
Give share of goodwil to retiring partners.

Question 9.
P/L Suspense A/c Dr.
To Deceased partners capital A/c What is the entry stands for?
Answer:
Credit of deceased partners share of profit in the in terim period.

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Two Mark Questions and Answers

Question 1.
What do you mean by retirement of a partner?
Answer:
Withdrawal of a partner from a partnership firm either by giving a notice of retirement or with the consent of the other partners or as per the provisions of the partnership agreement is called retirement.

Question 2.
What do you mean by Gaining Ratio? How it is calculated?
Answer:
At the time of retirement of a partner, the ratio in which the continuing partners share the profit of out going partner’s profit is called gaining ratio.
Gaining ratio = New ratio – Old ratio.

Question 3.
How a retiring partner’s share of goodwill is compensated?
Answer:
A retiring partner has the right to get his share of goodwill, because the goodwill of the firm has been earned with his efforts too. So he should be compensated by the other partners in their gaining ratio.

Question 4.
What is the treatment of accumulated profits or losses on the retirement of a partner?
Answer:
The general reserve and accumulated profits or losses are transferred to all partners’ capital accounts in their profit sharing ratio. The general reserve and accumulated profits are transferred to the credit side of the account and the accumulated losses to the debit side.

Question 5.
What are the differences between retirement and death, from the accounting point of view?
Answer:

Retirement is a known thing, so usually takes place at the end of an accounting period, but death may take place at any time.
On retirement, a partner severs his connection with the firm Voluntarily. But in death, it is automatic.
On retirement, the amount due to the retiring partner is transferred to his Loan Account, while in death; the total amount due to the deceased partner is transferred to his Executor’s Loan Account.

Question 6.
X, Y, and Z were sharing profits in the ratio of 3:2:1. Z retires from the firm. X and Y decide to share future profits in the ratio of 7:5. Calculate the gaining ratio.
Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 1

Question 7.
A, B and C are partners sharing profits in the ratio of 5 : 3: 2. C retires and the goodwill is valued at Rs. 40,000. Give entries in the books of the firm regarding treatment of goodwill.
Answer:
C’s share of goodwill = 40,000 × \(\frac{2}{10}\) = Rs. 8,000
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 2

Question 8.
X, Y, Z are partners sharing profits in the ratio of 5:3:2. X retires and for this purpose goodwill is valued at Rs. 25,000. Continuing partners agree that their new profit sharing ratio shall be equal. Record necessary journal entry.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 3
NOTES:
1. Gain of partner = New share – Old share
Y = 1/2 – 3/10 = 5 – 3/10 = 2/10
Z = 1/2 – 2/10 = 5 – 2/10 = 3/10
Gaining ratio is 2 : 3

2. X’s share of goodwill = 25,000 × 5/10 = 12,500

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Three Mark Questions and Answers

Question 1.
What are the problems that arise with regard to the accounting treatment on the retirement of a partner?
Answer:

Change in the Profit sharing ratio.
Adjustment of goodwill.
Treatment of accumulated profits and losses.
Revaluation of the assets and liabilities.
Calculation of the profit and loss up to the date of retirement.
Ascertainment of the total amount due to the retiring partner.
Payment of the amount due to the retiring partner.
Adjustment of the capitals of the continuing partners.

Question 2.
A, B and C were partners sharing profits in the ratio of 3: 5: 7. C retires and his share is taken up by A and B in the ratio of 3: 2. Find out the new profit sharing and gaining ratio of A and B.
Answer:
New share = Old share + Gain
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 4
A’s gain is 3/5 of 7/15 = 21/75
B’s gain is 2/5 of 7/15 = 14/75

Gaining ratio is the proportion in which they have acquired C’s share of profit, i.e., 3 : 2. This can be checked by working out in the following way.
Gaining ratio = New share – old share
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 6

Question 3.
Observe the following journal entry which has been passed at the time of retirement of Ganga Prasad. The other partners were Sheena and Rajani.

Which account would you prepare to share the profit on revaluation?
Prepare that account and give a journal entry to share the profit?

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 33
Answer:
1. Revaluation a/c

Journal Entry

Question 4.
X, Y, and Z are partners in a firm and they close their books on December 31 every year. They are sharing profits and losses in the ration of 3: 2: 1. The partnership deed provides that if a partner retires from the firm during the course of an accounting year, his share of profit from the date of last balance sheet to the date of retirement should be calculated on the basis of the average profits of the last three completed years.
On 1^st April 2004 Y retired from the firm. The profits of the firm during the years 2001, 2002 and 2003 were Rs. 12,500, Rs. 8,500 and Rs. 6,000 respectively. Write the journal entry to record the share of profit of the retiring partner for the year 2004.
Answer:

(Being Y’s share of profit for 2004 brought into A/c)
Notes: Profits for the last 3 years
= 12,500 + 8.500 + 6,000 = Rs. 27,000
Average profit = 27,000/3 = Rs. 9,000
Profit from the date of last balance sheet to the date . of retirement.
= 9,000 × 3/12 = Rs. 2,250
Y’s share thereof = 2,250 × 2/ 6 = Rs. 750.

Question 5.
Mr. Raj died on 25.08.2006 who was an active partner in a firm. The other partners were Mr. Das and Mrs.Das. The books of accounts reveal the following:

General Reserve	Rs. 12,000
Capital-Raj	Rs. 30,000
Profit and Loss A/c (Dr)	Rs. 18,000
Drawings of Raj	Rs. 10,000
Mr.Raj’s loan to the firm	Rs. 20,000

Interest on loan payable to Raj upto the date of death Rs. 1,000
Value of goodwill estimated at Rs. 24,000
Calculate the amount due to the legal heirs of Mr.Raj.
Answer:
Raj’s Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 8
R’s loan to the firm: 20000
R’s interest on loan: 1000
Total amount due to R’s legal heirs = His capital a/c
balance and R’s loan to the firm and interest on loan = 26000 + 20000 + 1000 = 47000.

Question 6.
X, Y, and Z are partners in a firm. Y retires from the firm on 1^st January, 2002. On his date of retirement, Rs. 60,000 is due to him. X and Z promise to pay in three equal annual instalments together with interest at 12% per annum. Prepare Y’s loan account for the three years.
Answer:
Dr. Y’s Loan Account Cr.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 9
Amount of instalment each year = 60,000/3 = 20,000
Amount paid each year = Rs. 20,000 + Interest.

Question 7.
X,Y, and Z were partner sharing profit in proportion to 5:3:2. Good will does not appear in the books, but it is agreed to be worth Rs. 1,00,000. X retires from the firm and Y and Z decide to share future profits equally. You are required to make adjustment entry for good will without opening good will account at all. Show your working clearly.
Answer:
X’s share of goodwill adjusted through capital accounts in the gaining ratio.
Old ratio = 5:3:2
New ratio =1:1
Gaining ratio = New ratio – Old ratio
Gain of Y = 1/2 – 3/10 = 2/10
Gain of Z= 1/2 – 2/10 = 3/10
Gaining ratio of Y and Z = 2 : 3
Value of goodwill of the firm = 1,00,000
X’s share of goowill = 1,00,000 × 5/10 = 50,000
Journal Entry:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 10
[X’s share of goodwill adjusted through the capital accounts of remaining partners in the gaining ratio of 2:3].

Question 8.
Aby, Suby and Minu are partners sharing profits in the ratio of 5:3:2. Minu retired on 31.09.06. The capital account balance and share of reserve due to Minu together amounted to Rs. 1,80,000. But Aby and Suby agreed to pay him Rs. 2,40,000. The new profit sharing ratio of Aby and Suby have been fixed at 3:2.

Why has Minu been paid over and above the actual amount due to him?
Give a journal entry to record this through capital a/c adjustments.

Answer:
A:S:M = 5:3:2
New ratio of A & S = 3:2
∴ Gaining ratio = New ratio – Old ratio
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 11
Gaining ratio = 1:1
Amount payable to Minu = 2,40,000
Capital + Reserve of Minu = 1,80,000
∴ Share of Goodwill due to Minu
= 240000 – 180000 = 60000
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 12

Question 9.
Debee, Sedee and Nedee are in partnership, who were sharing profits in the ratio of 3:2:1. On 31.03.05, Nedee left the firm as per agreement. The following details are available.
Balance Sheet:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 13

Depreciate fixed assets @10%.
Only General Reserve is to be credited to the extent of Nedee’s share through capital adjustment of the partners.
Receivables are sold to a debt collection agency at Rs. 5,400/-

Nedee’s accounts are to be settled soon either by paying off or bringing in necessary cash as the case may be. Prepare the necessary a/c to show the amount due to Nedee.
Answer:
Capital A/cs
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 14
Amount to be brought in by Nedee is Rs. 5,800.

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Five Mark Questions and Answers

Question 1.
How will you calculate the amount payable to a retiring partner?
Answer:
If retirement takes place on the closing date of the accounting year ascertainment of profit or loss is easy. But if the retirement occurs during an accounting year, the profit or loss from the date of last Balance Sheet to the date of retirement is also to be determined. Partnership deed provides the method of calculating the profit or loss of that period. It is calculated by any of the following methods.

On the basis of the last year’s profit.
On the basis of the average profit of a certain » number of past years.
By providing interest on the capital .of the retiring partner at. a certain rate.
By finding out the correct profit till the retirement date.
On any other basis as provided in the partnership agreement.

The profit or loss calculated above is only an estimate. So the journal entry for the same is; In case of profit.
Profit and Loss Suspense A/c Dr. To Retiring Partner’s Capital A/c:
P&L suspense A/c is shown on the asset side of the Balance Sheet prepared immediately after retirement. At the end of the accounting year; it is closed by transferring to Profit and Loss Account. Reverse is done in case of loss.

For calculating, the total amount payable to the retiring partner, his capital account is prepared.
It is started with the balance in the same on the last.
Balance Sheet and credited with:

His share of goodwill.
His share of revaluation profit.
His share of accumulated profits and reserve.
His share of profits upto the retirement date since the last Balance Sheet.
Interest, salary or commission if any due to him.

The capital account is debited with:

His share of accumulated losses.
His share of revaluation loss.
His drawings if any during the period.
Interest on such drawings.
His share of loss upto the retirement date since the last Balance Sheet.

If the account shows a credit balance, it is the total amount payable to him. On the other hand, if the account has a debit balance, it represents the amount payable by him to the firm.

Question 2.
X, Y, Z are partners in a firm sharing profits and losses equally. X retired from the firm on which date the balance sheet stood as follows:
Balance Sheet as on____
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 15
On the date of retirement it was found that

Patents have on value.
Furniture is to be depreciated by 15%.
Machinery is to be brought down to Rs. 10,000.

Pass the necessary journal entries to give effect to the revaluation of assets and liabilities at the time of retirement.
Answer:
Journal:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 16

Question 3.
A, B and C were Partner’s sharing Profits and losses in the ratio of 3:2:1. Their capitals were as under as per the balance sheet as on 31-Dec-2010.
A-Rs. 30,000; B-Rs. 20,000; C-Rs. 15,000. On 31 March 2011, C died, and you are asked to prepare deceased partners Capital account after considering the following facts.

Capital carried interest at 12% p.a.
C’ drawings from 1^st Jan 2011 to the date of his death amounted to Rs. 4,500.
C’s share of Profits for the portion of current financial year for which he lived was to betaken at the sum calculated on the average Profit of the last three completed years and good will was to be raised on the basis of two years Purchase of the average Profit of those three years.

The annual Profits were Rs. 19,000, Rs. 16,000 and Rs. 19,000 respectively. Show C’s Capital account, Answer:
C’s Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 17
Working Note:
1. Share of profit to the date of death.
Average profit for past 3 years

Therefore, C’s share (1/6) for 3 months = 18.000 × 1/6 × 3/12 = 750.

2. Goodwill calculation
Average Profit = 18,000
Goodwill = Average profit × 2’years purchase = 18,000 × 2 = 36,000
C’s share of goodwill = 36,000 × 1/6 = 6,000
C’s share of goodwill adjusted through the capital account of remaining partners in the gaining ratio of 3:2.

Question 4.
X, Y, and Z were partners in a firm with capitals of Rs. 15,000, Rs. 9,500, Rs. 10,000 respectively and sharing profits in proportions of 1/2, 1/4 and 1/4. On 31st December 2005, Y retires and for the purpose of his retirement, the goodwill of the firm has been valued at Rs. 12,000. Pass the necessary entries assuming that ’Y’ has been paid and show the Capital Accounts of all partners.
Answer:
Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 19
Working Note:
Old ratio = 1/2 : 1/4 : 1/4
1 × 2/2 × 2 = 2/4

Goodwill of the form = 12,000
Y’s share of goodwill = 12,000 × 1/4 = 3,000
Gaining ratio = New ratio – Old ratio
X’s Gain = 2/3 – 2/4 = 8 – 6/12 = 2/12
Z’s Gain = 1/3 – 1/4 = 4 – 3 /12 = 1/12
Gaining ratio = 2 :1
Journal:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 21

Question 5.
X, Y, and Z carried on business in partnership, shar¬ing profits in the ratio 3:2:1. The balance sheet on 31st December, 2003 showed their capitals to be Rs. 8,400; Rs. 6,800 and Rs. 7,400.
On 31^st March, 2004 X died. Write journal entries and prepare an account for presentation to his legal representatives having regard to the following facts:

Capital earned interest at 5 percent per annum.
X’s drawings from 1st January, 2004 to the date of his death amounted to Rs. 800; interest on drawings for the period Rs. 45.
X’s share of profit for the portion of current financial year for which he lived was to be taken a sum calculated on the average of the last three completed years.
Goodwill was to be raised on the basis of two year’s purchase of average profits of those three years.

The annual profits for the three years were Rs. 4,800; Rs. 3,500 and Rs. 4,300 respectively.
Answer:
Journal:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 22

Notes
1. Share of profit to date of death:
Average profit for past 3 years
\(\frac{4,800+3,500+4,300}{3}\)
= Rs. 4,200
∴ X’s share (3/6) for 3 months 3
= 4,200 × 3/6 × \(\frac{3}{12}\) = RS. 525 12.

2. Calculation of goodwill:
Average profit for past 3 years = Rs. 4,200
2 years purchase = Rs. 4,200× 2 = Rs. 8,400
X’s share (3/6) of goodwill = Rs. 8,400 × 3/ 6 = Rs. 4,200.

Question 6.
Heisal, Roy and A Gomez are in partnership sharing profits in their capital ratio. The Balance Sheet on 15^th March, 2006 is given below.
Balance sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 25
Further information on retirement of Roy on 15-6-06.

Profit for 3 months	Rs. 9000
Drawings: Heisal	R.s 1000
Roy	Rs. 2000
A.Gomez	Rs. 3000

Interest on capital @ 5% p.a.
Salary to Roy Rs. 300 p.m.
The firm had a fixed deposit worth Rs. 3000 which has not accounted so far, has to be brought into the books. Marketable scrips were valued at Rs. 23,000. Prepare:

Profit and Loss appropriation a/c
Capital A/c’s
Balance sheet

Answer:
P/L Appropriation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 26

Revaluation A/c

Capital A/c

Balance Sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 30

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Eight Mark Questions and Answers

Question 1.
The Balance Sheet of X, Y. Z on 31st March 2003 is given below.
Balance sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 31
They were sharing profits and losses in the ratio of 2:2:1. Y decided to retire from the firm. It was agreed that:

X and Z would share the profit in the ratio of 5:3
Goodwill was valued at Rs. 1,05,000
Machinery to be taken at Rs. 75,000
Buildings should be valued at Rs. 1,50,000
The value of stock should be Rs. 30,000
An amount of Rs. 1,500 should be written off as bad debt.

Pass the necessary journal entries and prepare the balance sheet of the new firm.
Answer:
Note: Calculation of gaining ratio
X = 5/8 – 2/5 – 25 – 16/40 = 9/40
Z = 3/8 – 1/5 = 15 – 8/40 = 7/40
Therefore, gaining ratio = 9:7

Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 32

Balance Sheet of X and Z as on 1^st April 2003

Question 2.
Abey, Neha, and Anil are partners, who share profits and losses in 5:3:2 ratio. The following information is extracted from the books of accounts on 31.03.06.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 39
On the above date Anil decided to retire from the firm as agreed upon. Fixed assets to be revalued at Rs. 86,000. Average profit calculated based on the past 5 year was Rs 15,000. Ascertain the amount due to the retiring partner.
Answer:
Profit and Loss Appropriation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 40

Anil’s Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 41
Working Note
Here, Goodwill is caculated on the basis of capitalization of Average profit method.
∴ Goodwill = Total value of business – Net tangible Asset.
Total value =
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 42
= 15,000 × \(\frac{100}{10}\) = 1,50,000
Net tangible asset = Fixed Asset + Current Asset
= 86,000 + 24,000 = 1,10,000
Goodwill = 1,50,000 – 1,10,000 = 40,000
Anil’s share of goodwill = 40,000 × 2/10 = 8,000
Anil’s share of goodwill adjusted through capital accounts in the gaining ratio.
Old ratio = 5:3:2
New ratio = 5:3
Gaining ration = 5 : 3
Abey’s share = 8,000 × 5/8 = 5,000
Neha’s share = 8,000 × 3/8 = 3,000
Journal entry:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 43

Question 3.
P, Q and R partiners sharing profits and losses in the ratio 3:2:1. The Balance sheet as on 31^st December 2003 is given below:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 44
On 31^st March 2004, Q decided to retire from the business due to ill-health subject to the following conditions.

That the goodwill should be valued at two year’s. purchase of the average profits of the preceding three years. The profits for the three preceding years were, 2001 – Rs. 9,000, 2000 – Rs. 15,000 and 2003-Rs. 12,000.
The profits for the three months ending 31 st March, 2004 be estimated on the basis of the profits for the year 2003.
That the motor car is to be given to Q, at a value of Rs. 16,000 and the balance due to him is to be paid immediately in cash by bringing the required amount by P and R in their profit sharing ratio which is 3: 1.

Answer:
Revaluation a/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 45

Calculation of Goodwill
Average profit of 3 years
\(=\frac{9,000+15,000+12,000}{3}\) = 12,000
Total goodwill = 12,000 × 2 = Rs. 24,000
Q’s share = 24,000 × 2/6 = Rs. 8,000
Q’s share of profit = 12,000 × 3/12 × 1/3 = 1,000
Balance Sheet as on 31^st March 2004
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 50

Question 4.
The balance sheet of X, Y, and Z on 31^st December, 2003, the date of X’s retirement was as follows:
Balance Sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 51
The following terms have been agreed upon:

Goodwill was valued at Rs. 18,000.
The value of land and buildings should be appreciated by Rs. 10,000
Plant and Machinery should be reduced to Rs. 23,000.
Create provision @ 5% on debtors for bad and doubtful debts and Rs. 700 on creditors.
The entire sum payable to X is to be brought by Y and Z in such a manner that their capital accounts are in proportion to their profit sharing ratio which is to be equal.

Prepare:

Revaluation account.
Partner’s capital accounts
Bank account, and
Balance sheet after X’s retirement.

Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 52

Balance Sheet of Y and Z as on 1^st January 2004

Notes:
Calculation of goodwill
Total goodwill ‘ = 18,000
X’s share 1/3 = 18,000 × 1/3 = Rs. 6,000
Working Notes:
(i) Gaining Ratio:
X: 3/4 – 3/6 = 9 – 6/12 = 3/12
Z: 1/4 – 1/6 = 3 – 2/12 = 1/12
Y’s share of goodwill of Rs. 12,000 (Rs. 36,000 × 2/6) will be contributed by X Rs. 9,000 and Z Rs. 3,000

(ii) Since the new profit sharing ratio between X and Z being 3:1, they will have to maintain their capitals at Rs. 90,000 and Rs. 30,000 respectively.

Question 5.
Anil, Bhanu, and Chandu were partners in a firm sharing profits in the ratio of 5:3:2. On March 31, 2007, their Balance Sheet was as under:
Books of Anil, Bhanu, and Chandu Balance Sheet as on March 31, 2007
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 56
Anil died on October 1, 2007. It was agreed between his executors and the remaining partners that:

Goodwill to be valued at 21/2 year’s purchase of the average profits of the previous four years which were:
Year2003-04-Rs. 13,000, Year 2004-05-Rs. 12,000
Year 2005-06-RS.20,000, Year 2006-07 – Rs. 15,000
Patents be valued at Rs. 8,000. Machinery at Rs. 28,000 and Building at Rs. 25,000
Profit for the year 2007-08 be taken as having accrued at the same rate as that of the previous year.
Interest on capital be provided at 10% p.a.
Half of the amount due to Anil be paid immediately.

Prepare Revaluation Account, Anil’s Capital Account and Anil’s Executor’s Account as on October 1, 2007.
Answer:
Revaluation Account
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement Death of a Partner - 57
Anil’s Capital Account

Anil’s Executor’s Account

Working Note:
1. Goodwill = 21/2 years purchase × Average Profit
Average Profit

Anil’s share of Goodwill = \(\frac{5}{10}\) × Rs.37,500 = 18750

2. Profit from the date of last balance sheet to date of death (April 1,2007 to October 1,2007) = 6 months
Profit for 6 months = Rs.15,000 × \(\frac{6}{12}\) = Rs.7,500
Anil’s share of profit = Rs.7,500 × \(\frac{5}{10}\) = Rs.3,750.

3. Interest on Capital
(April 1, 2007 to October 1, 2007)
= Rs. 30000 × \(\frac{10}{100}\) × \(\frac{6}{12}\) = Rs. 1,500.

Kerala SSLC Physics Model Question Paper 5 English Medium

April 13, 2021March 16, 2020 by Prasanna

Students can Download Kerala SSLC Physics Model Question Paper 5 English Medium Pdf, Kerala SSLC Physics Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Physics Model Question Paper 5 English Medium

General Instructions:

The first 15 minutes is the cool off time. You may – use the time to read and plan your answers.
Answer the questions only after reading the instructions and questions thoroughly.
Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
Five questions are given in each section. Answer any four from each section.
Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

Answer any four questions. Each question carries 1 score. [4 × 1 = 4]

Question 1.
Fill suitably
Generator → Armature → Induced e.m.f
Microphone → _____ → Induced e.m.f
Answer:
Voice coil

Question 2.
Name two solar devices.
Answer:
Solar panel, Solar cooker.

Question 3.
Give any one factor that influences the direction of motion of a current carrying conductor in a magnetic field.
Answer:
Direction of electric current / Direction of magnetic field.

Question 4.
Which effect of electric current is made used in safety fuse?
Answer:
Heating effect.

Question 5.
What is the critical angle of glass?
answer:
42°

Section – B

Answer any 4 questions. Each question carries 2 score. [4 × 2 = 8]

Question 6.
Give two practical applications of total internal reflection.
Answer:

In medical field – Endoscope
In field of Tele communication – Optical fibre cables

Question 7.
a) What is refraction?
b) What is the reason for refraction?
Answer:
a) When a ray of light entering obliquely from one medium to another, its path undergoes a deviation at the surface of separation,
b) Refraction is due to the difference in the optical densities of different media.

Question 8.
The statements given below are related to a step down transformer. Tabulate them as those related to primary and secondary.
i) Winding with thick wire.
ii) Current flowing at higher voltage
iii) Winding with thin wire
iv) current flowing at a low voltage.
Answer:
Primary:

Winding with thin wire
Current flowing at higher voltage

Secondary:

Winding with thick wire
Current flowing at low voltage.

Question 9.
A child sitting at the back bench of a classroom is unable to see the letters on the board clearly. What is the defect of the eye of the child? How can it be remedied? Draw its ray diagram.
Answer:
Nearsightedness.
This can be overcome by using concave lens of suitable focal length.

Question 10.
Why is hydrogen not used as a domestic fuel?
Answer:
Hydrogen is highly inflammable and explosive. It is difficult to be stored and transported.

Section – C

Answer any 4 questions. Each question carries 3 score. [4 × 3 = 12]

Question 11.
Classify the energy from the following sources as green energy and brown energy. Solar cells, atomic reactor, tidal energy, hydroelectric power, diesel energies, windmills, thermal power.
Answer:

Green energy	Brown energy
Solar cells	Atomic reactors
Tidal energy	Diesel engines
Hydro electric power	Thermal power station
Windmills

Question 12.
Observe the circuit diagram given below and answer the following questions.
Kerala SSLC Physics Model Question Paper 5 English Medium 1
a) Which are the instruments labelled as P and Q in the diagram?
b) If you replace the copper wire AB with a Nichrome wire of same length and area of cross section.
i) What change would you notice in the reading on the device Q? Why?
ii) What will happen to the heat produced in the conductor? Explain with references to Joule’s Law.
Answer:
a) P – Rheostat
Q – Ammeter

b) i) Reading will decrease. Due to high resistance current decreases
ii) Because of Nichrome’s high resistance the current in the circuit will decrease.
According to Joules law H = I²Rt, decrease in the amount of current will reduce the amount of heat.

Question 13.
Roshan observed a beautiful rainbow in the western Sky from his school ground.
a) When did Roshan observe the rainbow? [Morning, Noon, Evening, Prediction of time is impossible]
b) Draw the diagram of dispersion taking place in a water droplet during the formation of a rainbow.
Answer:
a) Morning.
b)
Kerala SSLC Physics Model Question Paper 5 English Medium 2

Question 14.
Observe the picture and answer the questions.
Kerala SSLC Physics Model Question Paper 5 English Medium 3
a) What happens to the air column around the paper core when the voice coil vibrates?
b) Is the voice coil vibrates if d.c. reaches the voice coil?
c) What is the working principle?
d) Mention the energy change taking place.
Answer:
a) Air column also vibrates. As a result sound is produced.
b) No. Because there is no flux change occurs.
c) Motor principle
d) Electrical energy sound energy

Question 15.
Wrap a rubber ball of diameter 12 cm completely with an aluminium foil and make the surfaces smooth. Where will be the image of an object kept 12 cm away from the centre of the ball? Is the image real or virtual?
Answer:
Diameter = 12 cm. R = 6 cm
Kerala SSLC Physics Model Question Paper 5 English Medium 4

Section – D

Answer any 4 questions. Each question carries 4 score. [4 × 4 = 16]

Question 16.
A motor cyclist observes a car coming from behind with a magnification 1/6. If the actual distance between the car and the bike is 30 m calculate the radius of curvature of the mirror.
Answer:
Kerala SSLC Physics Model Question Paper 5 English Medium 5

Question 17.
Observe the circuit A and B shown below:
Kerala SSLC Physics Model Question Paper 5 English Medium 6
a) If current through 40 W bulb in the circuit A is 0.6 A What is the current through 100 W bulb in the same circuit.
b) Which among the above two circuit is suitable for house hold connection?
c) What are the advantages of using above circuit for house hold electric connection?
Answer:
a) 0.6 A
b) Parallel connection.
c)

Bulbs can be controlled individually, using switches.
All the appliances works as per the power marked on them.
The same voltage is available for all the bulbs.
Total resistance decreases.

Question 18.
a) A bulb of power 40 W is designed to operate at 240 V. Calculate resistance of the filament in the bulb.
b) What are the characteristics required for the material chosen for making filament of incandescent lamp?
Answer:
a) V = 240 V
P = 40 W
R = ?
Kerala SSLC Physics Model Question Paper 5 English Medium 7

High resistance
High melting point
High ductility
Ability to emit white light in the white hot condition.

Question 19.
Blue colour of sky is due to the phenomenon of scattering
a) What is meant by scattering
b) How is Tyndal effect related to above phenomenon?
c) What are the advantage of using infrared photography?
d) In which colour does sky appear when viewed from moon?
Answer:
a) Irregular and partial reflection of light is scattering.
b) When rays of light pass through a colloidal fluid or suspension, the tiny particles get illuminated due to scattering. Because of this, the path of light is made visible. This phenomenon is Tyndat effect.
c) To take clear photographs of distant objects.
d) The sky is seen dark.

Question 20.
Primary of a transformer has 20,000 turns and the secondary has 30,000 turns. 160 V AC is applied at the primary of the transformer?
a) What is the voltage available at secondary of the above transformer?
b) If the number of turns in the secondary transformer is greater than that in the primary then more voltage is induced in the secondary. Why?
c) How much power must be supplied to the primary of the transformer so that 500 W power is obtained from secondary?
Answer:
a) N_p = 20000
N_s = 30000
V_p = 160 V
V_s = ?
Kerala SSLC Physics Model Question Paper 5 English Medium 8
b) Induced emf is directly proportional to number of turns.
c) 500 W

Plus One Zoology Notes Chapter 3 Structural Organisation in Animals

April 13, 2021March 16, 2020 by Prasanna

Students can Download Chapter 3 Structural Organisation in Animals Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 3 Structural Organisation in Animals

What is a tissue?
In multicellular animals, a group of similar cells along with intercellular substances perform a specific function. Such an organization is called tissue.

ANIMAL TISSUES:
The tissues are different and are broadly classified into four types:

Epithelial
Connective
Muscularand
Neural.

Epithelial Tissue:
This tissues are found in the covering ora lining for some part of the body. The cells are compactly packed with little intercellular matrix.
There are two types of epithelial tissues namely:

SIMPLE EPITHELIUM
COMPOUND EPITHELIUM

1. Simple epithelium:

It is composed of a single layer of cells and functions as a lining for body cavities, ducts, and tubes.
The compound epithelium consists of two or more cell layers and has protective function as it does in our skin
On the basis of structural modification of the cells, simple epithelium is further divided into three types. These are

Squamous
Cuboidai
Columnar

Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 1
1. Squamous epithelium:
It forms single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in a functions like forming a diffusion boundary.

2. Cuboidai epithelium:
It is composed of a single layer of cube-like cells. This is commonly found in ducts of glands and tubular parts of nephrons in kidneys. Its main functions are secretion and absorption.

3. Columnar epithelium:
It is composed of a single layer of tall and slende cells. Their nuclei are located at the base. Free surface may have microvilli. They are found in the lining of stomach and intestine and help in secretion and absorption.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 2

Ciliated epithelium:
If the columnar or cuboidai cells bear cilia on their free surface they are called ciliated epithelium Their function is to move particles or mucus in a specific direction overthe epithelium. They are present in bronchioles and fallopian tubes.

Glandular epithelium:
The modified columnar or cuboidal cells perform secretion and are called glandular epithelium.
They are mainly of two types:

Unicellular (goblet cells of the alimentary canal)
Multicellular(salivary gland).

On the basis of the mode of pouring of their secretions, glands are divided into two categories namely exocrine and endocrine glands.
Exocrine glands:
They secrete mucus, saliva, earwax, oil, milk, digestive enzymes and other cell products. These products are released through ducts or tubes.

Endocrine glands:
They do not have ducts. Their products called hormones are secreted directly into the blood.

2. Compound epithelium:
It is multi-layered of cells.
a. Their main function is to provide protection against chemical and mechanical stresses.

b. They cover the dry surface of the skin, the moist surface of buccal cavity, pharynx, inner lining of ducts of salivary glands and of pancreatic ducts.

Three types of cell junctions are found in the epithelium and other tissues. These are called as tight, adhering and gap junctions.

Tight junctions help to stop substances from leaking across a tissue.
Adhering junctions perform cementing to keep neighbouring cells together.
Gap junctions facilitate the cells to communicate with each other by connecting the cytoplasm of adjoining cells for rapid transfer of ions and molecules.

Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 3

Connective Tissue:
Connective tissues helps to linking and supporting othertissues/organs of the body. They include

cartilage
bone
adipose
blood.

Collagen or elastin:
In all connective tissues except blood, the cells secrete fibres of structural proteins called collagen or elastin.
The fibres provide strength, elasticity and flexibility to the tissue.
These cells also secrete modified polysaccharides, which accumulate between cells and fibres and act as matrix (ground substance).
Connective tissues are classified into three types:

Loose connective tissue
Dense connective tissue and
Specialised connective tissue

Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 4
Loose connective tissue:
They are loosely arranged in a semi-fluid ground substance. For example,
Areolar tissue:
It is present beneath the skin. It contains fibroblasts (cells that produce and secrete fibres),macrophages and mast cells.

Adipose tissue:
It is the loose connective tissue located mainly beneath the skin. The cells of this tissue are specialised to store fats.

Dense connective tissues.
In this Fibres and fibroblasts are compactly packed and are called dense regular and dense irregular tissues. In the dense regular connective tissues, the collagen fibres are present injows between bundles of fibres.
eg:

Tendons, which attach skeletal muscles to bones
ligaments which attach one bone to another.

Dense irregular connective tissue has fibroblasts and many fibres (mostly collagen) that are present in the skin.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 5

Specialized connective tissues: They are Cartilage, bones and blood.

Cartilage:
Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix. Most of the cartilages in vertebrate embryos are replaced by bones in adults. Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column, limbs and hands in adults.

Bones:
They have a hard ground substance rich in calcium salts and collagen fibres which give bone its strength. Bones support and protect softer tissues and organs. The bone cells (osteocytes) are present in the spaces called lacunae.

Limb bones, such as the long bones of the legs, serve weight-bearing functions. They also interact with skeletal muscles attached to them to bring about movements. The bone marrow in some bones is the site of production of blood cells.

Blood:
It is a fluid connective tissue containing plasma, red blood cells (RBC), white blood cells (WBC) and platelets. It also helps in the transport of various substances.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 6

Muscle Tissue:
It consists of long, cylindrical fibres arranged in parallel arrays. These fibres are composed of numerous fine fibrils, called myofibrils. Muscle fibres contract (shorten) in response to stimulation, then relax (lengthen) and in their uncontracted state. Muscles play an active role in all the movements of the body Muscles are of three types, skeletal, smooth, and cardiac.
1. Skeletal muscle:
It is the tissue is closely attached to skeletal bones. In a typical muscle such as the biceps, striated (striped) skeletal muscle fibres are bundled together in a parallel fashion.

2. Smooth muscle:
These fibres taper at both ends (fusiform) and do not show striations. The wall of internal organs such as the blood vessels, stomach and intestine contains this type of muscle tissue. Smooth muscles are ‘involuntary’ as their functioning cannot be directly controlled.

3. Cardiac muscle tissue:
It is a contractile tissue present only in the heart. Cell junctions fuse the plasma membranes of cardiac muscle cells and make them stick together In Communication when one cell receives a signal to contract, its neighbours are also stimulated to contract.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 7

Neural Tissue:
Neurons, the unit of neural system are excitable cells The neuroglial cell which constitute the rest of the neural system protect and support neurons. Neuroglia make up more than one half the volume of neural tissue in our body.

Nerve impulse transmisssion:
When a neuron is suitably stimulated, an electrical disturbance is generated which swiftly travels along its plasma membrane and reaches at the neuron’s endings, or output zone, triggers events that may cause stimulation or inhibition of adjacent neurons and other cells.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 8

ORGAN AND ORGAN SYSTEM:
Each organ in our body is made of one or more type of tissues. For example, our heart consists of all the four types of tissues, i.e., epithelial, connective, muscular and neural. In animals morphology refers to the external appearance of the organs or parts of the body.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 9

EARTHWORM:
Earthworm is a reddish brown terrestrial invertebrate The common Indian earthworms are Pheretima and Lumbricus.

Morphology:
They have long cylindrical body.lt is divided into more than hundred short segments which are similar. The dorsal surface of the body is marked by a dark median mid dorsal line (dorsal blood vessel). The ventral surface is distinguished by the presence of genital openings (pores).

Anterior end consists of the mouth and the prostomium, a lobe which serves as a covering for the mouth . The prostomium is sensory in function. The first body segment is called the peristomium (buccal segment) which conjoins the mouth.

In a mature worm, segments 14-16 are covered by a prominent dark band of glandular tissue called clitellum. Thus the body is divisible into three prominent regions.

preclitellar
clitellar &
postclitellar segments

Four pairs of spermathecal apertures are situated 5^th – 9^th segments. A single female genital pore is present in the mid-ventral line of 14^th segment. A pair of male genital pores are present on 18th segment.

Numerous minute pores called nephridiopores open on the surface of the body. In each body segment, except the first, last and clitellum, there are rows of S-shaped setae, Setae plays an important role is in locomotion.

Anatomy:
The body wall of the earthworm is covered externally by a thin non-cellular cuticle below which is the epidermis, two muscle layers and an innermost coelomic epithelium. The epidermis is made up of a single layer of columnar epithelial cells which contain secretory gland cells. A terminal mouth opens into the

buccal cavity (1 – 3 segments) which leads into muscular pharynx.
Oesophagus (5 – 7 segments), continues into a muscular gizzard (8 – 9 segments).

Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 10

It helps in grinding the soil particles and decaying leaves etc. The stomach extends from 9 – 14 segments. The food of the earthworm is decaying leaves and organic matter mixed with soil. Calciferous glands, present in the stomach, neutralise the humic acid present in humus.

Intestine starts from the 15^th segment onwards and continues till the last segment. A pair of short and conical intestinal caecae project from the intestine on the 26^th segment. The characteristic feature of the intestine between 26 – 35 segments is the presence of internal median fold of dorsal wall called typhlosole.

This increases theeffective area of absorption in the intestine. The alimentary canal opens to the exterior by a small rounded aperture called anus. These simpler molecules are absorbed through intestinal membranes and are utilised. Pheretima shows closed type of blood vascular system, consisting of blood vessels, capillaries and heart.

Blood glands are present on the 4^th, 5^th and 6^th segments. They produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature.

In Earthworms respiratory exchange occurs through moist body surface into their blood stream. The excretory organs occur as segmentally arranged coiled tubules called nephridia They are of three types:

septal nephridia, present on both the sides of intersegmental septa of segment 15 to the last that open into intestine,
integumentary nephridia, attached to lining of the body wall of segment 3 to the last that open on the body surface and
pharyngeal nephridia, present as three paired tufts in the 4^th, 5^th and 6^th segments

Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 11

Nephridia:
It regulate the volume and composition of the body fluids. A nephridium starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular part of the nephridium which delivers the wastes.

Nervous system:
It is represented by ganglia arranged segment wise on the ventral paired nerve cord (3^rd and 4^th segments). The cerebral ganglia along with other nerves in the ring integrate sensory input as well as command muscular responses of the body.

Sensory system does not have eyes but does possess light and touch sensitive organs to distinguish the light intensities and to fee the vibrations in the ground. Worms have specialised chemoreceptors (taste receptors) which react to chemical stimuli. Earthworm is hermaphrodite (bisexual), i.e., testes and ovaries are present in the same individual.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 12
There are two pairs of testes present in the 10^th and 11^th segments. Their vasa deferentia run up to the 18^th segment where they join the prostatic duct. Two pairs of accessory glands are present one pair each in the 17^th and 19^th segments. The common prostrate and spermatic duct opens to the exterior by a pair of male genital pores on the ventro-lateral side of the 18^th segment.

Four pairs of spermathecae are located in 6^th – 9^th segments (one pair in each segment).They receive and store spermatozoa during copulation. One pair of ovaries is attached at the 12^th and 13^th segments. Ovarian funnels are present beneath the ovaries which continue into oviduct, join together and open on the ventral side as a single median female genital pore on the 14^th segment.

A mutual exchange of sperm occurs between two worms during mating. Mature sperm and egg cells and nutritive fluid are deposited in cocoons produced by the gland cells of clitellum.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 13

Fertilisation:
Fertilisation and development occur within the cocoons which are deposited in soil. The ova (eggs) are fertilised by the sperm cells within the cocoon. The cocoon holds the worm embryos. After about 3 weeks, each cocoon produces two to twenty baby worms. Earthworms development is direct, i.e., there is no larva formed.

Earthworms are known as ‘friends of farmers’ because they make burrows in the soil and make it porous which helps in respiration and penetration of the developing plant roots. The process of increasing fertility of soil by the earthworms is called vermicomposting. They are also used as bait in game fishing

COCKROACH:
Cockroaches are are included in class Insecta of Phylum Arthropoda. They are nocturnal omnivores that live in damp places throughout the world. They are found in human homes and thus are serious pests and vectors of several diseases.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 14

Morphology:
The adults of the common species of cockroach, Periplaneta americana are about 34 – 53 mm long with wings that extend beyond the tip of the abdomen in males. The body of the cockroach is segmented and divisible into three distinct regions – head, thorax and abdomen. The body is covered by a hard chitinous exoskeleton.

Exoskeleton has hardened plates called sclerites that are joined to each other by a thin and flexible articular membrane (arthrodial membrane). Head is formed by the fusion of six segments and shows great mobility in all directions due to flexible neck It has compound eyes. A pair of antennae arise from sockets lying in front of eyes. They help in monitoring the environment.

Anterior end of the head bears appendages forming biting and chewing type of mouth parts. It consists of a labrum (upper lip), pair of mandibles, a pair of maxillae and a labium (lower lip). A median flexible lobe, acting as tongue (hypopharynx), lies within the cavity enclosed by the outhparts.
Thorax consists of three parts:

prothorax
mesothorax
metathorax.

Each thoracic segment bears a pair of walking legs. The first pair of wings arises from mesothorax and the second pair from metathorax. Forewings (mesothoracic) called tegmina are opaque dark and leathery and cover the hind wings when at rest. The hind wings are transparent, membranous and are used in flight.

The abdomen in both males and females consists of 10 segments. In females, the 7^th sternum is boat shaped and together with the 8^th and 9^th sterna forms a genital pouch whose anterior part contains female gonopore, spermathecal pores and collateral glands.

In males, genital pouch lies at the hind end of abdomen bounded dorsally by 9^th and 10^th terga and ventrally by the 9^th sternum. It contains dorsal anus, ventral male genital pore and gonapophysis. Males bear a pair of short, threadlike anal styles which are absent in females. In both sexes, the 10^th segment bears a pair of jointed filamentous structures called anal cerci.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 15

Anatomy:
The alimentary canal is divided into three regions: foregut, midgut and hindgut The mouth opens into a short tubular pharynx, leading to a narrow tubular passage called oesophagus.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 16
Oesophagus opens into a sac like structure called crop used for storing of food. The crop is followed by gizzard or proventriculus. Gizzard helps in grinding the food particles. A ring of 6 – 8 blind tubules called hepatic or gastric caecae is present at the junction of foregut and midgut, which secrete digestive juice.

At the junction of midgut and hindgut is present another ring of 100 – 150 yellow coloured thin filamentous Malphigian tubules. They help in removal of excretory products from haemolymph. The hindgut is differentiated into ileum, colon and rectum. The rectum opens out through anus.

Blood vascular system of cockroach is an open type Blood vessels are open into space (haemocoel). Visceral organs located in the haemocoel are bathed in blood (haemolymph).The haemolymph is composed of colourless plasma and haemocytes.

Heart of cockroach consists of elongated muscular tube lying along mid dorsal line of thorax and abdomen. Blood from sinuses enter heart through ostia and is pumped anteriorly to sinuses again. The respiratory system consists of a network of trachea, that open through 10 pairs of small holes called spiracles present on the lateral side of the body.

Thin branching tubes carry oxygen from the airto all the parts. Exchange of gases take place at the tracheoles by diffusion. During excretion Malpighian tubules absorb nitrogenous waste products and convert them into uric acid which is excreted out through the hindgut. Therefore, this insect is called uricotelic.

In addition, the fat body, nephrocytes and urecose glands also help in excretion. The nervous system of cockroach consists of segmentally arranged ganglia joined by paired longitudinal connectives. Three ganglia lie in the thorax, and six in the abdomen.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 17

The nervous system of cockroach is spread throughout the body. If the head of a cockroach is cut off, it will still live for as long as one week. In the head region, the brain is represented by supra-oesophageal ganglion which supplies nerves to antennae and compound eyes. In cockroach, the sense organs are antennae, eyes, maxillary palps, larfial palps, anal cerci, etc.

Compound eves of cockroach:
The compound eyes are situated at the dorsal surface of the head. Each eye consists of about 2000 hexagonal ommatidia. With the help of several ommatidia, a cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity but less resolution, being common during night (hence called nocturnal vision).

Cockroaches are dioecious Male reproductive system consists of a pair of testes lying one on each lateral side in the 4^th – 6^th abdominal segments. From each testis arises a thin vas deferens, which opens into ejaculatory duct through seminal vesicle. The ejaculatory duct opens into male gonopore situated ventral to anus.

A characteristic mushroom shaped gland is present in the 6^th – 7^th abdominal segments which functions as an accessory reproductive gland. The external genitalia are represented by male gonapophysis or phallomere The sperms are stored in the seminal vesicles and are glued together in the form of bundles called spermatophores which are discharged during copulation.

The female reproductive sysytem consists of two large ovaries, lying laterally in the 2^nd – 6^th abdominal segments. Each ovary is formed of a group of eight ovarian tubules or ovarioles, containing a chain of developing ova. Oviducts of each ovary unite into a single median oviduct (also called vagina) which opens into the genital chamber.

Sperms are transferred through spermatophores. Their fertilised eggs are stored in capsules called oothecae. On an average, females produce 9 – 10 oothecae, each containing 14 – 16 eggs. The development of P. Americana is paurometabolous, meaning there is development through nymphal stage.

The nymph grows by moulting about 13 times to reach the adult form. The next to last nymphal stage haswing pads but only adult cockroaches have wings.

FROGS:
Frogs are belong to class Amphibia of phylum Chordata. The most common species of frog found in India is Rana tigrina. Their body temperature varies with the temperature of the environment. Such animals are called cold blooded or poikilotherms They have the ability to change the colour to hide them from their enemies (camouflage).

This protective coloration is called mimicry. During peak summer and winterthey take shelter in deep burrows to protect them from extreme heat and cold. This is called as summer sleep (aestivation) and wintersleep (hibernation).
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 19

Morphology:
The skin is maintained in a moist condition. The frog never drinks water but absorb it through the skin. Body of a frog is divisible into head and trunk. A neck and tail are absent. Above the mouth, a pair of nostrils is present. Eyes are bulged and covered by a nictitating membrane that protects them while in water.

On either side of eyes a membranous tympanum (ear) receives sound signals. The forelimbs and hind limbs help in swimming, walking, leaping and burrowing. The hind limbs end in five digits and they are larger and muscular than fore limbs that end in four digits.

Feet have webbed digits that help in swimming. Frogs exhibit sexual dimorphism. Male frogs can be distinguished by the presence of sound producing vocal sacs and also a copulatory pad on the first digit of the fore limbs which are absent in female frogs.

Anatomy:
The body cavity consists of organ systems such as digestive, circulatory, respiratory, nervous, excretory and reproductive systems. The digestive system consists of alimentary canal and digestive glands. The alimentary canal is short because frogs are carnivores and hence the length of intestine is reduced.

The mouth opens into the buccal cavity that leads to the oesophagus through pharynx. Oesophagus is a short tube that opens into the stomach which in turn continues as the intestine,rectum and finally opens outside by the cloaca. Liver secretes bile that is stored in the gall bladder. Pancreas, a digestive gland produces pancreatic juice containing digestive enzymes.

Digestion of food takes place by the action of HCI and gastric juices secreted from the walls of the stomach. Partiailv digested food called chyme is passed from stomach to the first part of the intestine, the duodenum. The duodenum receives bile from gall bladder and pancreaticjuicesfrom the pancreas through a common bile duct.

Bile emulsifies fat and pancreatic juices digest carbohydrates and proteins. Final digestion takes place in the intestine. Digested food is absorbed by the numerous finger-like folds in the inner wall of intestine called villi and microvilli. The undigested solid waste moves into the rectum and passes out through cloaca.

In water, skin acts as aquatic respiratory organ (cutaneous respiration). Dissolved oxygen in the water is exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs. The respiration by lungs is called pulmonary respiration.

The lungs are a pair of elongated, pink coloured sac- like structures present in the upper part of the trunk region (thorax). Air enters through the nostrils into the buccal cavity and then to lungs. During aestivation and hibernation gaseous exchange takes place through skin. The blood vascular system involves heart, blood vessels and blood.

The lymphatic system consists of lymph, lymph channels and lymph nodes. Heart is a muscular structure situated in the upper part of the body cavity. It has three chambers, two atria and one ventricle and is covered by a membrane called pericardium. Atriangularstructure called sinus venosus joins the right atrium.

It receives blood through the major veins called vena cava. The ventricle opens into a saclike conus arteriosus on the ventral side of the heart. The blood from the heart is carried to all parts of the body by the arteries (arterial system).The veins collect blood from different parts of body to the heart and form the venous system.

Special venous connection between liver and intestine as well as the kidney and lower parts of the body are present in frogs. The former is called hepatic portal system and the latter is called renal portal system.

The blood is composed of plasma and cells. The blood cells are RBC (red blood cells) or erythrocytes, WBC (white blood cells) or leucocytes and platelets. RBC’s are nucleated and contain red coloured pigment namely haemoglobin. The lymph is different from blood. It lacks few proteins and RBCs.

The excretory system consists of a pair of kidneys, ureters, cloaca and urinary bladder. Each kidney is composed of several structural and functional units called uriniferous tubules or nephrons. Two ureters emerge from the kidneys in the male frogs. The ureters act as urinogenital duct which opens into the cloaca.

In females the ureters and oviduct open seperately in the cloaca. The thin-walled urinary bladder is present ventral to the rectum which also opens in the cloaca. The frog excretes urea and thus is a ureotelic animal.

The chemical coordination of various organs of the body is achieved by hormones which are secreted by the endocrine glands. The endocrine glands found in frog are pituitary, thyroid, parathyroid, thymus, pineal body, pancreatic islets, adrenals and gonads.

The nervous system is organized into a central nervous system (brain and spinal cord), a peripheral nervous system (cranial and spinal nerves) and an autonomic nervous system (sympathetic and parasympathetic). There are ten pairs of cranial nerves arising from the brain. Brain is enclosed in a bony structure called brain box(cranium).

The brain is divided into fore-brain, mid-brain and hind-brain. Forebrain includes olfactory lobes, paired cerebral hemispheres and unpaired diencephalon. The midbrain is characterised by a pair of optic lobes. Hind-brain consists of cerebellum and medulla oblongata.

The medulla oblongata passes out through the foramen magnum and continues into spinal cord, which is enclosed in the vertebral column. Frog has different types of sense organs, namely organs of touch (sensory papillae), taste (taste buds), smell (nasal epithelium), vision (eyes) and hearing (tympanum with internal ears).

Eyes in a frog are a pair of spherical structures situated in the orbit in skull. External ear is absent in frogs and only tympanum can be seen externally. The ear is an organ of hearing as well as balancing (equilibrium).

Male reproductive organs consist of a pair of yellowish ovoid testes which are found adhered to the upper part of kidneys by a double fold of peritoneum called mesorchium. Vasa efferentia are 10 – 12 in number that arise from testes. They enter the kidneys on their side and open into Bidder’s canal.

Finally it communicates with the urinogenital duct that comes out of the kidneys and opens into the cloaca. The cloaca is a small, median chamber that is used to pass faecal matter, urine and sperms to the exterior.

The female reproductive organs include a pair of ovaries. They are situated near kidneys and there is no functional connection with kidneys. A pair of oviduct arising from the ovaries opens into the cloaca separately. A mature female can lay 2500 to 3000 ova at a time.
Plus One Zoology Notes Chapter 3 Structural Organisation in Animals 20
Fertilisation is external and takes place in Water. Development involves a larval stage called tadpole. Tadpole undergoes metamorphosis to form the adult. Frogs are beneficial for mankind because they eat insects and protect the crop.

Frogs maintain ecological balance because these serve as an important link of food chain and foodweb in the ecosystem. In some countries the muscular legs of frog are used as food by man.

Kerala SSLC Physics Model Question Paper 4 English Medium

April 13, 2021March 16, 2020 by Prasanna

Students can Download Kerala SSLC Physics Model Question Paper 4 English Medium Pdf, Kerala SSLC Physics Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Physics Model Question Paper 4 English Medium

General Instructions:

The first 15 minutes is the cool off time. You may – use the time to read and plan your answers.
Answer the questions only after reading the instructions and questions thoroughly.
Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
Five questions are given in each section. Answer any four from each section.
Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

Answer any four questions. Each question carries 1 score. [4 × 1 = 4]

Question 1.
Find the relation between the first pair and complete the second pair.
Electric oven : Heating effect
Mixie : _______.
Answer:
Mechanical effect.

Question 2.
What is. the voltage of electricity supplied to the distribution transformer?
Answer:
11 KV.

Question 3.
Select the odd one
Solar cells, tidal energy, atomic reactor, Hydroelectric power.
Answer:
Atomic reactor

Question 4.
Arrange the given media in the increasing order of their optical densities
Water, Diamond, Glass, Air
Answer:
Air < Water < glass< diamond.

Question 5.
Which mirror always forms diminished and erect image?
Answer:
Convex mirror.

Section – B

Answer any 4 questions. Each question carries 2 score. [4 × 2 = 8]

Question 6.
a) What is the relation between the deviation of component colours of white light through a prism and their wavelength?
b) Classify the colours green, red, indigo and yellow in the descending order of their wavelengths.
Answer:
a) When wavelength increases deviation decreases. When wavelength decreases deviation increases.
b) Red, yellow, green, Indigo.

Question 7.
Write down 2 limitations of wind mills?
answer:

This can be established only at those places where wind is available for most time of the year.
We may require storage systems to use electricity when there is no wind.

Question 8.
The telescope called ‘Chandra X – ray Observatory’ is placed in the outer space. What is the advantage of placing it there? Explain with, reference to the scattering of light in the atmosphere.
Answer:
In the outer space, there is no atmosphere and scattering of light does not take place. So clear images can be captured.

Question 9.
Give reasons for the following.
a) Red light is used as signal lamps.
b) Sky in the moon appears dark.
Answer:
a) Wave length of red is greater. So rate of scattering is less.
b) Due to the absence of atmosphere, light does not undergo scattering.

Question 10.
In an AC generator which part is kept stationary, Why?
Answer:
Armature is kept stationary and field magnet is allowed to rotate. The armature is too heavy to rotate. Also this helps to eliminate the graphite brushes and thereby avoid sparks.

Section – C

Answer any 4 questions. Each question carries 3 score. [4 × 3 = 12]

Question 11.
In a house 5 lamps of 60 w used in 3 hours and 6 lamps of 40 w are used in 5 hours daily.
a) Which is the device used to measure the used electric current?
b) Find the amount of current used for 30 days?
Answer:
a) Kwh meter
b) The amount of current used 60 w of 5 lamps for 3 hours = \(\frac{60 \times 5 \times 3}{1000}\) = 0.9 unit
The amount of current used 40 w of 6 lamps for 5 hour’s = \(\frac{40 \times 6 \times 5}{1000}\) = 1.2 unit
Total current used in one day = 0.9 + 1.2 = 2.1 unit
Amount of current used in one month = 2.1 × 30 = 63 Unit

Question 12.
The following statements in the boxes are related to working of a loudspeaker. Arrange them in the correct order.
Kerala SSLC Physics Model Question Paper 4 English Medium 1
Answer:

Question 13.
Write down the advantages and limitations of solar cooker.
Answer:
Advantages of solar cooker:

Renewable source of energy is used in it
It does not cause environmental pollution
Low expense

Limitations:

Not practical in rainy seasons and during night.
Fried items can not be prepared
Long time is needed.

Question 14.
A motor cyclist observes a car coming from behind with a magnification 1/6. If the actual distance between the car and the bike is 30 m calculate the radius of curvature of the mirror.
Answer:
Kerala SSLC Physics Model Question Paper 4 English Medium 3

Question 15.
When Newton’s colour disc rotates fast it appears white.
a) Mention the phenomenon related to this?
b) Define this phenomenon.
c) Write another example related to this?
Answer:
a) Persistence of vision.
b) When a person sees an object, its image remains in the retina for a time interval of a 1/16 second. This phenomenon is called persistence of vision.
c) At the time of rain fall drops, we feel that they are glass rods.

Section – D

Answer any 4 questions. Each question carries 4 score. [4 × 4 = 16]

Question 16.
All the constituent colours of sunlight do not have same rate of scattering.
a) Write the reason for this? (1)
b) Describe an experiment to demonstrate that scattering of all colours are not equal. (2)
c) Under what condition all the colours are scattered equally. (1)
Answer:
a) Rate of scattering is directly proportional to the wavelength of the wave and the size of the particles.

b) Allow light from a torch to fall on the water form one side of the beaker. The light emerging form the beaker is focussed on a white screen. Sodium thiosulphate is dissolved in water. Add, hydochloric acid to the water. Blue colour spreads at the beginning. Then the colours emerging from the solution in the order of VIBGYOR.

c) If the size of the particles is greater than the wave¬length of light, then the scattering is same for all colours.

Question 17.
a) What is the principle of a transformer?
b) When 240 V was applied to the primary of a transformer, the voltage in the secondary was 12 V. If the number of turns in the primary is 4,800 what will be the number of turns in the secondary?
c) To which coil of this transformer thick wire is to be used?
Answer:
a) Mutual induction
b) V_p = 240 v
V_s = 12 v
N_p = 4800
N_s = ?
Kerala SSLC Physics Model Question Paper 4 English Medium 4
c) Secondary

Question 18.
Two electric irons working on 230 V AC. The resistance passed by one is 800 W and by the second is 1200 W. If so
a) Which electric iron posses more intensity of current?
b) Which one posses more power consumption?
c) Calculate the amount of consumed energy while the electric-iron of resistance 800 Ω is to be worked for 2 hours?
Answer:
a) Electric iron having resistance 800 W
b) Electric iron having less resistance
c) H =I²RT
V = 230 V, t = 2 × 60 × 60s, R = 800 Ω
Kerala SSLC Physics Model Question Paper 4 English Medium 5

Question 19.
The calorific value of hydrogen is 1,50,000 KJ/Kg and that of LPG is 55,000 KJ/Kg.
a) What do you mean by calorific value?
b) What is the unit of calorific values?
c) Which among the above is a good fuel?
d) Which one is selected as a fuel in home? Illustrate your answer.
Answer:
a) The amount of heat liberated by the complete combustion of 1 kg of fuel is.its calorific value.
b) Its unit is kilojoule/kilogram.
c) Hydrogen, because its calorific value is greater than that of LPG.
d) LPG. There is a chance for explosion when hydrogen is burnt. Moreover, it is difficult to store hydrogen safely. This is why we can’t select hydrogen as a fuel in homes.

Question 20.
Observe the figures and answer the following questions:
Kerala SSLC Physics Model Question Paper 4 English Medium 6
a) Of the circuit A and B which is the one used for household electrical circuit?
b) Write three advantages of making circuits in this manner.
Answer:
a)

b)

We can control electric appliances with separate switches.
Every instrument in the circuit gets maximum and equal voltage. All instruments can separate switches, low effective resistance.

Kerala SSLC IT Theory Model Question Paper 2 English Medium

April 13, 2021March 16, 2020 by Prasanna

Students can Download Kerala SSLC IT Theory Model Question Paper 2 English Medium Pdf, Kerala SSLC IT Theory Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC IT Theory Model Question Paper 2 English Medium

Time: 1½ Hours
Total Score: 40 Marks

Section – I

Choose one correct answer

Question 1.
To change the animation speed or time using the tool
a) Toggle night N
b) Adjust progress value G
c) Time backward B
d) Time forward A
Answer:
b) Adjust progress value G

Question 2.
According to you, what is the biggest advantage of data base management system?
a) We can made one report
b) can’t make a report
c) We can make different types of reports
d) We can’t save the reports
Answer:
c) We can make different types of reports

Question 3.
Name the extension of project file in synfig studio?
a) svg
b) pdf
c) sif z
d) ods
Answer:
c) sif z

Question 4.
Which is the first working computer?
a) Desktop
b) Analytical Engine
c) ENIAC
d) Laptop
Answer:
c) ENIAC

Question 5.
The tool used to display the map of the world showing time zones
a) Toggle Tropic tool
b) Clock and map tool
c) Toggle meridian
d) Sun/moon toggle menu
Answer:
b) Clock arid map tool

Question 6.
More than two pictures are placed on over the other, the option to raise lower to top by
a) Object → Lower to bottom
b) Object → Raise to top
c) Insert → Lower to bottom
d) Object → group
Answer:
b) Object → Raise to top

Question 7.
A correct statement related to clone formatting is
a) Use clone formatting for a large report
b) Formattings of headings can be changed by changing the formatting of each heading separately
c) Formattings of headings can be changed by changing the formatting of one heading
d) Formattings of heading cannot be copped to another
Answer:
b) Formattings of headings can be changed by changing the formatting of each heading separately

Question 8.
What is the full form of CSS?
a) Computer System Style
b) Cascading Style Sheet
c) Cascading Style System
Answer:
b) Cascading Style Sheet

Question 9.
What is the use of the command fd()
a) Move the Turtle to the right ,
b) Move the Turtle in a circle
c) Mark a point on the graphic screen
d) Specify the colour of the line to be drawn on the screen
Answer:
a) Move the Turtle to the right

Question 10.
In computer, data is processed and stored in the form of signal.
a) Digital
b) Analog
c) Hybrid
d) None of these
Answer:
a) Digital

Section – II

Choose two correct answers.

Question 11.
From the following name the people who considered to be started grid computing
a) Ian foster
b) Carl Kesselman
c) Steve Tuecke
d) Bert bos
Answer:
a) I an foster, b) Carl Kesselman & c) Steve Tuecke

Question 12.
What are the phenomena to explain through sun clock software
a) Concept of time zones of earth
b) India
c) The features of the earth surface
d) World map
Answer:
a & c

Question 13.
Bell Laboratories build Unix operating system. These versions used later, with a lot of modifications and addition in other operating systems made by.
a) GNU/Linux
b) Berkeley
c) Microsoft
d) Apple Corporation
Answer:
b & d

Question 14.
Give the two activities done before starting animation from first key frame
a) Current time is 60 f
b) Current time is-0 f
c) Before editing the motion, animate the edit mode button is active
d) press the play button to see the animation
Answer:
b & c

Question 15.
What are the differences of a server computer compared to a normal computer
a) can give one IP address
b) can give morethan one IP address
c) can host a website
d) can host different websites to different IP addresses
Answer:
b & d

Kerala SSLC IT Theory Model Question Paper 1 English Medium

April 13, 2021March 16, 2020 by Prasanna

Students can Download Kerala SSLC IT Theory Model Question Paper 1 English Medium Pdf, Kerala SSLC IT Theory Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC IT Theory Model Question Paper 1 English Medium

Time: 1½ Hours
Total Score: 40 Marks

Section- I

Choose one correct answer

Question 1.
Name the utility used in inkscape to give a 3D effect to 2D images?
a) Radial Gradient
b) Blur
c) opacity
d) Difference
Answer:
a) Radial Gradient

Question 2.
Anu was prepared a style named main head. Then he select the all headings in a report and click on the style named main head. What happens for the headings?
a) All headings are changed to same type
b) No change to the headings
c) Headings are changed to different sizes
d) Headings are seen in different colours
Answer:
a) All headings are changed to same type

Question 3.
A loop statement contains another loop statement then it is called
a) Nested loop
b) Direct loop
c) Indirect loop
d) None of these
Answer:
a) Nested loop

Question 4.
The idea of was suggested by Hakon Wium Lie and Bert Bos.
a) CSS
b) HTML
c) Python
Answer:
a) CSS

Question 5.
What is the network related icon displayed on the screen of the computer connected to a Network.
a) Network Manager Applet
b) Network Neighbourhood
c) Network system
d) None of these
Answer:
a) Network Manager Applet

Question 6.
The land that is being used for widening the road. The buildings that are get affected by this. How much can be saved by changing the width of the road? Above things are easily determind by a technology available in QGIS. Name that technology available in QGIS.
a) vector
b) Layer
c) Attributes
d) Suffer(s)
Answer:
d) Buffer (s)

Question 7.
are web sites to publish our works like stories, poems and articles.
a) blogs
b) www
c) Internet
d) HTTP
Answer:
a.

Question 8.
What are the activities can conduct through queries
a) To add records
b) To create tables
c) To change the records
d) To create a database file
Answer:
a & c

Question 9.
Give the order of activity to create a new user
a) Application → graphics → administration → user and group
b) Application → Office → User and groups → User accounts
c) Application → System Tools → User and groups user accounts
d) Application → system tools → User and groups
Answer:
c) Application → System Tools → User and groups user accounts

Question 10.
Which tool is used to fill colours to the objects in synfig studio?
Kerala SSLC IT Theory Model Question Paper 1 English Medium 1
Answer:

Section – II

Choose two correct answers

Question 11.
Correct statements for proprietary softwares
a) Computer programms are made free to copy and share
b) Able to examine the programms for study purposes
c) No freedom to copy and share
d) No freedom to examine them for study purposes
Answer:
d) No freedom to examine them for study purposes

Question 12.
What are the uses of multi-coloured buttons on the handle of a image?
a) To switch on the Animate editing mode
b) To adjust size
c) To take copy
d) To rotate if needed
Answer:
b & d

Question 13.
Who are the email service providers
a) www.gmail.com
b) IETF
c) www.yahoo.com
d) ICANN
Answer:
a & c

Question 14.
Select the order activity to get open attribute table in QGIS.
a) Select a layer from the layer palette
b) Right click on the layer
c) Right click on the layer and select open attribute table
d) Right click on the layer and select delete
Answer:
a & c

Question 15.
The different ways to select all created images in a canvas
a) Click and drag the mouse in such a way that it covers all images
b) While pressing on the shift key and click on all images separately
c) Select using mouse
d) Selectthrough the menu
Answer:
a & b

Kerala SSLC Physics Model Question Paper 3 English Medium

April 13, 2021March 16, 2020 by Prasanna

Students can Download Kerala SSLC Physics Model Question Paper 3 English Medium Pdf, Kerala SSLC Physics Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Physics Model Question Paper 3 English Medium

General Instructions:

The first 15 minutes is the cool off time. You may – use the time to read and plan your answers.
Answer the questions only after reading the instructions and questions thoroughly.
Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
Five questions are given in each section. Answer any four from each section.
Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

Answer any four questions. Each question carries 1 score. [4 × 1 = 4]

Question 1.
Find the relation and complete the word pair
Transformer: Mutual induction
Inductor : __________.
Answer:
Self induction

Question 2.
Select the odd one
Cock, ammonia, coal tar, Naphtha
Answer:
Naphtha

Question 3.
Write down 2 properties of fuse wire.
Answer:
Low melting point, high resistance

Question 4.
Which of the given does not indicate power
I²R, 1/I, IR², V²/R
Answer:
IR²

Question 5.
What is the relation between anlage of incidence and angle of reflection?
Answer:
Equal.

Section – B

Answer any 4 questions. Each question carries 2 score. [4 × 2 = 8]

Question 6.
Write down two advantages and two limitations of Hydrogen as a fuel
Answer:
Advantages:

Greater calorific value
Easily available

Limitations:

Highly inflammable and explosive
Difficult to store and transport.

Question 7.
The electricity generated at the power stations is transmitted to distant places in high voltage.
a) What is the voltage at which electricity is generated in the power station?
b) What are the problems facing in long distant power transmission?
Answer:
a) 11 KV
b) Power loss, Voltage drop

Question 8.
Match the following

A	B
AC Generator	Self Induction
Transformer	Slip Ring
Bio mass	Mutual Induction
Inductor	Green Energy
	Methane

Answer:

A	B
AC Generator	Slip Ring
Transformer	Mutual Induction
Bio mass	Green Energy
Inductor	Self Induction

Question 9.
Kerala SSLC Physics Model Question Paper 3 English Medium 1
a) What is the intensity of current through the circuit?
b) Calculate the heat developed in the circuit for 20 minutes?
Answer:
a) I = \(\frac{V}{R}=\frac{10}{30}\)
= 0.3 A

b) H = I²Rt
= 0.3² × 30 × 20 × 60
= 3240 J

Question 10.
Mention the phenomena responsible for the following
a) Formation of rainbow
b) Tyndal effect
Answer:
a) Dispersion
b) Scattering

Section – C

Answer any 4 questions. Each question carries 3 score. [4 × 3 = 12]

Question 11.
Observe carefully the following diagram which show the splitting up of a composite light into its constituent coloures.
Kerala SSLC Physics Model Question Paper 3 English Medium 2
a) Which among the above figure is correct ? Give reason. (2)
b) Write the name of the phenomenon. (1)
Answer:
a)

Light undergoes refraction and the extend of deviation depends on the wavelength. Violet having least wavelength undergoes maximum deviation and red rays having highest wavelength undergoes minimum deviation.
b) Dispersion.

Question 12.
Thickness of insulated copper wire used in the primary turns and secondary turns of a transformer are not equal.
a) In a step up transformer which set of turns (primary/secondary) is made of thicker copper wire? Justify your answer. (2)
b) Write down the working principle of transformer. (1)
Answer:
a) Primary
Reason: In a transformer if there is no power loss, power in the primary coil is equal to the power in the secondary P = Iv, in a step up transformer, in the primary coil, voltage will be low and .current will be high. To carry high current, thick wires are needed.
b) Mutual induction.

Question 13.
Wind mill, Hydro electric power, Nuclear reactor and Solar cell are four energy sources.
a) Classify the above four energy sources as green energy and brown energy. (2)
b) What is meant by green energy? (1)
Answer:
a)

Green energy	Brown energy
Wind mill	Nuclear reactor
Hydro electric power
Solar cell

b) Green energy is the energy produced from natural sources which does not cause environmental pollution.

Question 14.
An electric bulb is labelled 40W, 200V.
a) Calculate the current flowing through it?
b) What is the resistance of the filament?
Answer:
a)
Kerala SSLC Physics Model Question Paper 3 English Medium 4

Question 15.
Kerala SSLC Physics Model Question Paper 3 English Medium 5
a) Identify the figure in which bulbs are being connected in series?
b) Select the figure with which the bulbs connected in parallel?
c) Which circuit is suitable for household circuit? Justify your answer.
Answer:
a)

b)

c) Parallel connection.
Each devices can be controlled by separate switches. Voltage remains same. Effective voltage will be very less.

Section – D

Answer any 4 questions. Each question carries 4 score. [4 × 4 = 16]

Question 16.
White light splits into its constituent colours when passed through a glass prism.
a) Name the process of splitting
b) Give the reasons for this type of splitting
c) Draw the diagram for recombining these colours produced by the prism.
Answer:
a) Dispersion
b) Each colour differs in their wavelength. Red having greater wavelength deviates least. Violet having less wavelength deviates most. So colour splitted up.
c)
Kerala SSLC Physics Model Question Paper 3 English Medium 8

Question 17.
Analyse the circuit and answer the following questions.
Kerala SSLC Physics Model Question Paper 3 English Medium 9
1) What is the current through A?
2) What is the current through B?
3) What will be the ammetdr reading?
4) To decrease the q reading in the ammeter, how should the resistors be connected?
Answer:
1. Current through A = I₁ = \(\frac{v}{R}=\frac{12}{3}\) = 4A
2. Current through B = I₂ = \(\frac{v}{R}=\frac{12}{2}\) = 2A
3. Ammeter reading = I₁ + I₂ = 4 + 2 = 6A
4. Resistors should be connected in parallel. Then effective resistance = 3 + 6 = 9W
I = \(\frac{12}{9}\) = 1.33 A

Question 18.
There are 10000 turns in a transformer. Voltage in the primary is 240 V and current is 0.2 A. The transformer is wound so as to get a current 0.4 A in the secondary.
a) What type of transformer is this?
b) What is the secondary voltage. Also find the number of turns in the secondary.
c) What is the maximum output power that received from this transformer?
Answer:
a) Step up transformer
b)
Kerala SSLC Physics Model Question Paper 3 English Medium 10
c) P = VI = 240 × 0.2
= 48 w

Question 19.
When an object is placed infront of a spherical mirror at a distance 30 cm, the magnification is -1.
a) Write down the properties of the image?
b) What kind of mirror is this?
c) If the object is placed at 10 cm. What changes occur to the properties of the image?
d) Illustrate the conclusions.
Answer:
a) Image will be real, inverted, same size of the object.
b) Concave mirror
c) Image will be erect, virtual and enlarged.
d) 10 cm means object is in between F and P, then the image is erect, enlarged, and virtual behind the mirror.

Question 20.
State the Motor Rule. If the directions of current in the conductor and the magnetic field are the same, in which way will the conductor move?
Answer:
A conductor, which can move freely and which is kept in a magnetic field experiences a force when current passes through it and it moves.
If the direction of current in the conductor and the magnetic field are the same, the conductor will not move.

Kerala SSLC IT Theory Model Question Paper 2 Malayalam Medium

April 13, 2021March 13, 2020 by Prasanna

Students can Download Kerala SSLC IT Theory Model Question Paper 2 Malayalam Medium Pdf, Kerala SSLC IT Theory Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC IT Theory Model Question Paper 2 Malayalam Medium

Time: 1½ Hours
Total Score: 40 Marks

Kerala SSLC IT Theory Model Question Paper 2 Malayalam Medium 1

Kerala SSLC IT Theory Model Question Paper 2 Malayalam Medium 3

Kerala SSLC IT Theory Model Question Paper 2 Malayalam Medium 7

Kerala SSLC Chemistry Model Question Paper 4 English Medium

April 13, 2021March 13, 2020 by Prasanna

Students can Download Kerala SSLC Chemistry Model Question Paper 4 English Medium Pdf, Kerala SSLC Chemistry Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Chemistry Model Question Paper 4 English Medium

General Instructions:

The first 15 minutes is the cool off time. You may use the time to read and plan your answers.
Answer the questions only after reading the instructions and questions thoroughly.
Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
Five questions are given in each section. Answer any four from each section.
Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

(Answer any 4 questions from 1 to 5. Each question carries 1 score) (4 × 1 = 4)

Question 1.
The outermost subshell of inert gases except helium
contains electrons. (1)
(2,6,10,14)
Answer:
6

Question 2.
How much is the volume of 32 g of O₂ at STP? (1) (Atomic mass O = 16)
Answer:
22.4 L

Question 3.
The functional group present in organic acids is (1)
Answer:
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 4

Question 4.
To which category of medicines does paracetamol belong? (1)
Answer:
Antipyretic

Question 5.
What is the major constituent of LPG? (1)
Answer:
Butane

Section – B

(Answer any 4 questions from 6 to 10. Each question carries 2 scores)(4 × 2 = 8)

Question 6.
The order of reactivity of certain metals are as follows:
Mg > Zn > Fe> Cu
a) Which among these does notreactwith dil.HCl?(1)
b) A piece of Mg ribbon is dipped in ZnSO₄ solution.
Write down the equation showing the redox reaction taking place. (1)
Answer:
a) Cu
b) Mg + ZnSO₄ → MgSO₄ + Zn / Mg + Zn⁺ → Mg²⁺ + Zn

Question 7.
Magnetite (Fe₃O₄), haematite (Fe₂O₃) and copper pyrites are some ores.
a) Which of the ores is concentrated by froth floatation? (1)
b) Which one is concentrated by magnetic separation? (1)
Answer:
a) Copper pyrites
b) Magnetite (Fe₃O₄)

Question 8.
Some samples of certain elements are given.
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 1
a) Which among the above has the highest number of molecules? (1)
b) Which samples occupies a volume of 22.4 L at STP? (1)
[Atomic mass: H=1, He = 4, N = 14 and O = 16]
Answer:
a) 8g H₂
b) 28 g N₂

Question 9.
An organic compound has 3 carbon atoms and an OH functional group on the second carbon atom.
a) Give its IUPAC name. (1)
b) Write down the structure of a position isomer of this compound. (1)
Answer:
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 5

Question 10.
The third shell of an atomX has 6 electrons.
a) Write the subshell electronic configuration of X. (1)
b) What will be the normal valency of X? (1)
Answer:
a) 1s²2s²2p⁶ 3s² 3p⁴
b) -2

Section – C

(Answer any 4 questions from 11 to 15. Each question carries 3 scores) (4 × 3 = 12)

Question 11.
Cement is an important building material.
a) What are the main raw materials of cement production? (1)
b) What is the role of gypsum in cement manufacute? (1)
c) Cement is not kept in moist places. Why? (1)
Answer:
a) Limestone, Clay
b) To regulate the setting time of cement
c) Cement absorbs moisture and hardens permanently,

Question 12.
Atomic number of iron is 26.
a) To which group of the periodic table does iron belong? (1)
b) What is the oxidation state of iron in Fe₂O₃? (1)
c) Write down the subshell electronic configuration of the ion of iron with this oxidation state. (1)
Answer:
a) 26Fe- 1s²2s²2p⁶3s²3p⁶3d⁶4s²(group-8)
b) +3
c) Fe³⁺ -1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Question 13.
Consider the equilibrium given
N₂O₄ + Heat ⇌ 2NO₂
a) When does a reversible reaction attain equilibrium? (1)
b) What will happen to the rate of forward reaction if
a high pressure is applied on the system? (1)
(c) At low temperature N₂O₄ decomposes only slowly. Why? (1)
Answer:
a) When the speed of forward reaction and backward reaction becomes equal.
b) Rate of foward reaction decreases because due to forward reaction volume (no. of- mols) . increases.
c) Forward reaction is endothermic.,

Question 14.
Complete the following equations:
a) C₂H₆+ Cl₂ → ………+ HCI (1)
b) C₃H₆ + Cl₂ → …………(1)
c) nCH₂ = CH₂ → …………(1)
Answer:
a) C₂H₅Cl
b) C₃H₆Cl₂
c) [CH₂ – CH₂]_n

Question 15.
The equation showing the reaction of hydrogen with oxygen to form water is given below:
2H₂ + O₂ → 2H₂O
a) What is the ratio of reactant molecules hydrogen and oxygen of the reaction? (1)
b) Suppose 32 g oxygen and 10 g of’hydrogen are allowed to react together.
i) How many molecules of water will be for¬med? (1)
ii) Which reactant will be left behind after the reaction? Hoyv’many moles? (1)
Answer:
a) 2 : 1
b) i) 2 molecules.
ii) 6 g of H₂ will be left behind after the reaction.

Section – D

(Answer any 4 questions from 16 to 20. Each question carries 4 scores)(4 × 4 = 16)

Question 16.
Aluminium is the most abundant metal on the earth’s crust.
a) Name the ore of aluminium. (1)
b) This ore is concentrated by Leaching. What is meant by Leaching? (1)
c) What is the role of cryolite in a aluminium production? (1)
d) Give the name and any one use of an alloy of aluminium (1)
Answer:
a) Bauxite (Al₂O₃.2H₂O)
b) The ore is dissolved in a suitable solvent and the impurities are filtered out.
c) During the electrolytic refining of aluminium, alumina is dissolved in molten cryolite. Cryolite lowers the melting point of alumina and increase electrical conductivity.
d) Alnico. Used to produce strong magnets.

Question 17.
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 2
Answer the following questions based on the above . structure.
a) Correctly number the chain. CO
b) Name the side chains. (1)
c) Write the IUPAC name of the compound. (1)
d) Give the structure of the straight chain isomer of this hydrocarbon. (1)
Answer:
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 6
b) Methyl
c) 2, 4- dimethyl hexane
d) CH₃-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₃ (Hectane)

Question 18.
a) How is ethanol manufactured? (2)
b) Complete the equation (1)
CH₃ – CH₂ – OH + CH₃COOH →………….. + H₂O
c) To which category of organic compounds does the product of the above reaction belong? (1)
Answer:
a) Ethanol is manufactured by the fermentation of molasses (sugar solution)
b) CH₃ – CH₂ – OH + CH₃-COOH → CH₃ – CO – CH₂– CH₃ + H₂O
c) Esters

Question 19.
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 3
A Galvanic cell is represented above .
a) At which metal electrode does oxidation take place? (1)
b) From which metal to which metal do the electrons flow? (1)
c) Which metal acts as the cathode. (1)
d) Write down the equation showing the redox reaction taking place in the.cell. (1)
(Hint: Reactivity Mg>Fe)
Answer:
a) Magnessium (Mg)
b) Mg to Fe
c) Iron(Fe)
d)
Kerala SSLC Chemistry Model Question Paper 4 English Medium - 7

Question 20
Sodium thiosulphate and hydrochloric acid are given.
a) Write an experiment to demonstrate the effect of temperature on reaction rate. (2)
b) What happens to the rate of reaction on increasing the temperature? (1)
c) Write another method to increase the rate of the reaction. (1)
Answer:
a) Take equal volume of dilute sodium thiosulphate solution in two boiling tubes. Heat one of them. Add equal volume of dil.HCl in to both the boiling tubes. Reaction takes place faster on heated boiling tube.
b) When temperature increases rate of reaction also increases.
c) Use concentrated solutions of sodium thiosulphate and HCl.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner

April 13, 2021March 13, 2020 by Prasanna

Students can Download Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner One Mark Questions and Answers

Question 1.
Total value of business-Net tangible assets’ is the value of goodwill under.
(a) Superprofit method
(b) Present value of super profit method
(c) Capitalization of average profit method
(d) Weighted average profit method
Answer:
(c) Capitalization of average profit method.

Question 2.
Which of the following does not lead to reconstitution of a partnership firm?
(a) Admission of a new partner
(b) Retirement of a partner
(c) Death of a partner
(d) Dissolution of a partnership
Answer:
(d) Dissolution of a partnership

Question 3.
Change in agreement (relationship among the partners) lead to
(a) Reconstitution
(b) Dissolution
(c) Reconstruction
(d) Amalgamation
Answer:
(d) Reconstitution

Question 4.
Change in profit sharing ratio of the existing partners result in
(a) Gain to all partners
(b) Sacrifice to ail partners
(c) Gain to some partners and sacrifice to others
(d) None of these
Answer:
(c) Gain to some partners and sacrifice to others.

Question 5.
Unless otherwise mentioned, sacrificing ratio will be
(a) Equal ratio
(b) New ratio
(c) Old ratio
(d) None of these
Answer:
(c) Old ratio

Question 6.
The profit or loss arising from revaluation of assets and liabilities is transferred to
(a) Old partners’ capital account
(b) All partners’ capital account
(c) New partners’ capital account
(d) Profit and Loss Appropriation account
Answer:
(a) Old partners’ capital account

Question 7.
The profit or loss on revaluation is transferred to the old partners capital a/c in
(a) Old ratio
(b) Sacrificing ratio
(c) New ratio
(d) In the ratio of capital
Answer:
(a) Old ratio

Question 8.
At the time of admission of a new partner the re-serves and accumulated profits in P & L account is transferred to
(a) Profit and Loss appropriation account
(b) Profit and Loss Adjustment Account
(c) Old Partner’s capital account
(d) Revaluation account
Answer:
(c) Old partner’s capital account

Question 9.
The amount of goodwill brought in by the new partner is shared among the old partners in
(a) Old ratio
(b) Sacrificing ratio
(c) New ratio
(d) None of these
Answer:
(b) Sacrificing ratio

Question 10.
The goodwill brought in kind (assets) by the new partner is transferred to
(a) Revaluation account
(b) Profit and Loss Account
(c) Sacrificing partners’ capital account
(d) All partners’ capital account
Answer:
(c) Sacrificing partners’ capital account

Question 11.
When the new partner is not able to bring his share of goodwill, his account will be
(a) Debited
(b) Credited
(c) Omitted
(d) Closed
Answer:
(a) Debited

Question 12.
In partnership, a minor
(a) Cannot be a partner
(b) Can be a partner
(c) Can be admitted only to the benefits of a partnership.
(d) Can be a partner and share profit & losses along with other partners.
Answer:
(c) Can be admitted only to the benefits of a partnership.

Question 13.
Complete the following on the basis of the hint given

Premium – Sacrificing ratio.
Revaluation profit – _______.

Answer:
Old profit sharing.ratio

Question 14.
Joy’s capital A/c
Dr. Saju’s capital A/c Dr
To Profit and Loss A/c.
What is the entry stands for?
Answer:
Accumulated losses transferred to old partners capital a/c.

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Two Mark Questions and Answers

Question 1.
‘Goodwill is an asset, but is not visible’. Describe.
Answer:
Goodwill is the value of the reputation of a firm. As such it is an asset to the firm. But it is an intangible asset and is not visible.

Question 2.
A firm has an average profit of Rs. 50,000 during the last certain years. The normal rate of return is 10%. The firm has net tangible assets of Rs. 3,00,000. Calculate the value of goodwill using capitalization method.
Answer:
Average profit = Rs. 50,000
Normal rate of return = 10%
Capitalised value of
average profit = \(\frac{50,000 \times 100}{10}\) = Rs. 5,00,000
Goodwill = Capitalised value of average profit – Total of net tangible assets.
= Rs. 5,00,000 – 3,00,000 = Rs. 2,00,000.

Question 3.
When a new partner is admitted into a firm?
Answer:
Inclusion of a new partner into an existing firm is called admission of a partner. A new partner is admitted, when a firm needs more capital, managerial skill, etc.

Question 4.
What are the rights acquired by a new partner?
Answer:

Right to share the assets of the firm – For this the new partner has to bring a certain amount of capital.
Right to share the profits of the firm – For this he has to bring his share of goodwill.

Question 5.
What do you mean by sacrificing ratio?
Answer:
At the time of admission of a new partner, the old partners have to sacrifice a certain portion of their profits for the incoming partner. The ratio in which they give up or sacrifice their profit is called sacrificing ratio.

Question 6.
What is a revaluation account?
Answer:
Revaluation account is a nominal account prepared at the time of admission of a new partner. This is prepared to find out the profit or loss on revaluing the assets, if they are overstated or understated.

Question 7.
What is meant by premium or goodwill?
Answer:
At the time of admission, the new partner has to bring in a certain amount for getting a share in future profit. This amount is called premium or goodwill.

Question 8.
What treatment is made of accumulated profits and losses on the admission of a new partner?
Answer:
Accumulated profits and losses are distributed amongst the old partner’s in their old profit sharing ratio. The new partner should not share such profits or losses because these arose before his admission.

Question 9.
What is Memorandum Revaluation Account?
Answer:
A memorandum revaluation account is prepared when the partners decide to record the effect of revaluation of assets and liabilities without affecting the old figures of assets and liabilities in the balance sheet.

Question 10.
A new partner is admitted into a firm; but he is not in a position to bring his share of goodwill. What the firm will do?
Answer:
When the new partner is not able to bring his share of goodwill, his account is debited and the sacrificing partners capital account is credited. The following is the journal entry.
New Partners’ Capital A/c Dr.
To Sacrificing Partners’Capital A/c.

Question 11.
An equipment having a book value of Rs.2,600 was sold at Rs. 3,000 on the date of admission of a partner.

How much amount will be credited to Revaluation A/c?
Give journal entry for the above.

Answer:

Cash a/c Dr.	3000
To Equipment	2600
To Revaluation (Being equipment sold)	400

Question 12.
In connection with the admission of Mr.Santhosh Kumar as equal partner, one of the existing partners
of the firm Mrs. Sreema has taken over the plant and equipments worth Rs. 15000 at Rs. 18000 on the date of admission. Give a journal entry to this effect.
Answer:

Sreema’s capital Dr.	18000
To Plant & Equipment	15000
To Revaluation (Being P & E taken over by Sreema)	3000

Question 13.
X and Y are partners sharing profits and loses in the ratio of 2:1. They admit Z into the firm for a fourth, share. Calculate new ratio and sacrificing ratio.
Answer:
Old ratio = 2:1
New ratio =2:1:1
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 1
Here old ratio and sacrificing ratio are the same.

Question 14.
P and Q are partners in a firm sharing profits in the ratio of 5 : 3. They admit R for 1/6 share. The total goodwill of the firm is Rs. 50,000. Goodwill existing in the books is Rs. 25,000. Pass the journal entry for the share of goodwill to be brought in by R.
Answer:
Amount of goodwill to be brought in by R
= 1/6 of (50,000 – 25,000)
= 1/6 of 25,000
= 1/6 × 25,000 = Rs. 4,167
The Journal entry is

Cash A/c Dr.	4,167
To Goodwill (Premium) (Share of Good will brought in by R)	4,167

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Three Mark Questions and Answers

Question 1.
Calculate the value of goodwill at 2 years, purchase from the following 3 years average profits.

1995	Rs. 27,000
1996	Rs. 28,000
1997	Rs. 29,000

Answer:
Average profit = \(\frac{27,000+28,000+29,000}{3}\)
= 28,000
Goodwill = 2 yeas purchase of the average profit = 2 × 28,000 = 56,000.

Question 2.
A business has earned average profits of Rs. 1,00,000 during the last few years and the normal rate of return in a similar business is 10%. Ascertain the value of goodwill by capitalisation of superprofits method, given that the value of net assets of the business is Rs.8,20,000.
Answer:
Goodwill = superprofit × 100/ Normal rate of return Super profit = Actual/Average profit – Normal profit Normal profit = Capital employed × Normal rate of return = 820000 × 10/100 = 82000
Super profit = 100000 – 82000 = 18000
Goodwill = 18000 × 100/10 = Rs. 180000.

Question 3.
Ram and Rahim are partners in a firm sharing profits in the ratio of 3:2. Their capitals were Rs.80000 and Rs.50000 respectively. They admitted Syam on January 1^st 2014 as a new partner for 1/5 share in the future profits. Syam bought Rs.60,000 as his capital. Calculate the value of goodwill of the firm.
Answer:
Syam’s capital = 60000
Syam’s share of capital = 1/5
Total capital of new firm = 60000 × 5/1 = 300000
Total capital of Ram, Rahim & Syam
= 80000 + 50000 + 60000 = 190000
Goodwill of the firm = 300000 – 190000 = 110000
Syam’s share of goodwill = 110000 × 1/5 = Rs.22,000.

Question 4.
L and M are partners sharing profits in the ratio of 5: 4. On 1^st July 2005 they admit N into the firm for 1/10 share in future profits. N contributed the following assets for his capital and share of goodwill. Stock-in-trade Rs. 50,000, Furniture Rs. 25,000 and Land and Buildings Rs. 75,000 and machinery Rs. 50,000. Goodwill of the firm was valued at Rs. 45,000. Give the journal entries.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 2

Question 5.
P and Q are partners sharing profits and losses in the ratio of 3:2. They admit R into the firm with 2/5 share which he gets equally from P & Q. Calculate the new ratio and sacrificing ratio.
Answer:
Old ratio = 3 : 2 = 3/5 : 2/5
R’s share = 2/5 ie. (1/5 from P 1/5 from Q)
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 4
Here the sacrificing ratio is equal (1: 1) as R gets equally from P&Q.

Question 6.
Which are the matters on which accounting adjustments are required at the time of the admission of a new partner?
Answer:
At the time of the admission of a new partner, accounting adjustments are required on the following

Capital of the new partner
Ascertainment of profit sharing ratios – new and sacrificing
Revaluation of assets and liabilities
Adjustment of accumulated profits (including reserves) or losses.
Calculation of goodwill
Adjustment of capital accounts of partners.

Question 7.
Explain the premium method of treatment of goodwill.
Answer:
Under premium method, the new partner brings his share of goodwill in cash. The amount so brought in by him is shared among the old partners in the sacrificing ratio.
The journal entries here are:

Cash a/c Dr.
To premium for goodwill a/c (cash brought in by the new partner for goodwill)
Premium for good will a/c
To old partners capital a/c (Goodwill shared among the old partners)

“If the amount of premium is paid privately to the old partners, no need of entering the same in the books.”

Question 8.
A new partner instead of bringing his share of good¬will in cash brought the same as assets. How will you treat it?
Answer:
When an incoming partner brings his share of goodwill in kind (as assets), the assets account will be debited. Credit is given to premium for good will account with the share of goodwill and new partner’s capital account with the share of capital.
The journal entries here are:

Assets a/c Dr. To New partner’s Capital A/c
To Premium (goodwill) A/c (Assets brought in by the new Partner)
New partners’ capital a/c Dr. To Sacrificing partners Capital A/c
(Share of goodwill brought in by the new partner transferred to old partners capital).

Question 9.
A and B are partners sharing profits and losses equally (1:1). They admit C for 1/6 share in future profits. Calculate the new ratio and sacrificing ratio.
Answer:
Old Ratio = 1:1 C’s
Share = 1/6
Remaining portion = 1 – 1/6 = 5/6
This 5/6 is to share among A & B in their old ratio.
So their new shares will be
A’s share 1/2 of 5/6 = 1/2 × 5/6 = 5/12
B’s share 1/2 of 5/6 = 1/2 × 5/6 = 5/12
The new ratio between
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 5
Old ratio and sacrificing ratio are the same here.

Question 10.
Roshi and Riya are partners sharing profits and losses in the ratio of 5 : 3. Maria is admitted into the firm. Roshi sacrifices 1 /5 of her share and Riya sacrifices 1/6 in favour of Maria. Calculate the new ratio.
Answer:
Old ratio = 5:3
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 6

Question 11.
Ansa and Valsa are partners in a firm sharing profits and losses in the ratio of 3:2. They admit Sona into the partnership fora sixth share for which she brings in Rs. 35000 as capital and Rs. 20,000 for good will. Pass journal entries in the books of the firm.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 8
Note: Old ratio itself is the sacrificing ratio here as the ratio between old partners is not changed.

Question 12.
X and Y are partners sharing profits and losses equally. They admit Z for a third share for which he brings Rs. 15,000 for good will. X and Y withdrew the full amount of goodwill immediately. Pass journal entries in the books of the firm.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 9

Question 13.
P and Q are partners sharing profits and losses in the ratio of 2:1. They admit R for a third share. He brings in Rs. 30,000 for goodwill, half of which is withdrawn by the old partners. Pass journal entries.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 10

Question 14.
X and Y are partners sharing profits in the ratio of 4:3. Z is admitted for 1/6 share in profits. Their capitals were Rs. 50,000 and 40,000 respectively. It is also agreed that Z’s capital should be proportionate to her profit sharing ratio. Find out the amount to be brought in by Z as capital.
Answer:
Share of Profit of Z = \(\frac{1}{6}\)
Share of profit of X and Y = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Total capital of X & Y= 50,000+ 40,000 = 90,000
Capital of X and Y for 5/6 share = 90,000
∴ Total capital of X,
Y and Z = 90,000×6/5 = 1,08,000
∴ Capital to be brought in by Z = 1,08,000 – 90,000
or 1,08,000 × 1/6 = 18,000.

Question 15.
A trading firm has in its ledger book, an accumulated profit balance of Rs. 30,000 in general reserve. The partners Smitha, Neha, and Anila who share profits in the ratio of 3:2:1. They have decided to become equal partners. Show journal entry to adjust the existing general reserve through capital accounts.
Answer:
Old ratio = 3:2:1 = 3/6 : 2/6: 1/6
New ratio = 1:1:1 = 1/3:1/3:1/3

Share of general reserve = 30,000 × \(\frac{1}{6}\) = 5000
Anila’s capital A/c Dr. 5000
To Smitha’s capital 5000
(Being goodwill adjusted between capital a/cs of Smitha and Anila, Neha’s profit sharing ratio remains
unchanged (Neha’s old ratio = \(\frac{2}{6}\) – \(\frac{1}{3}\), New ratio = \(\frac{1}{3}\)).

Question 16.
The Profits of firm for the last five years were as follows

Year	Profits
2002-03	20,000
2003-04	24,000
2004-05	30,000
2005-06	25,000
2006-07	18,000

Calculate the value of goodwill on the basis of 3 year’s Purchase of weighted average profit based on weights 1, 2, 3, 4, & 5 respectively to the profits for 2002, 2003, 2004, 2005 and 2006.
Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 13
Weighted Average Profit = \(\frac{3,48,000}{15}\) = 23,200
Goodwill = 23,200 × 3 = 69,600

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Five Mark Questions and Answers

Question 1.
List the factors affecting goodwill.
Answer:
Following are the important factors affecting the goodwill of a firm

Nature of business -A firm producing goods having constant demand will have more goodwill.
Suitable location – A firm which is situated in a favourable locality will have more goodwill.
Efficiency of management – if the management of a firm is efficient, it wil have high goodwill.
Running period – a firm which is running for a long period of time, will have more goodwill.
Requirement of capital – if a firm requires a lesser amount of capital, it will have high goodwill.
Market situation – if competition in the market is limited, it helps a firm to have more goodwill.

Question 2.
Describe the methods of valuing goodwill.
Answer:
The following are the common methods used for valuing good will

Average profit method/simple average profit method.
Super profit method
capitalisation method.

1. Average Profit method:
Under this method, the goodwill is valued at agreed number of years purchase of the average profits of the past few years.
Goodwill = Average profits × No. of years purchased.
Average profit = \(\frac{\text { Total profits }}{\text { No. of years }}\)
Weighted Average Profit method:
Goodwill = weighted Average Profit × No. of years purchase. Weighted average is based on specified weights like 1, 2, 3, 4 for respective year’s profit.

2. Superprofit Method:
Under this method, goodwill is calculated by multiplying the super profit with the agreed num-ber of years.
Goodwill = Super Profit × No.of years purchase
Super Profit = Actual or Average Profit – Normal Profit
Normal profit = Capital employed × Normal Rate of Return
Capital employed = Total Assets – Total Liabilities or outside liabilities
Average profit = \(\frac{\text { Total profits }}{\text { No. of years }}\).

3. Capitalisation Method:
Under this method, the goodwill can be calcu-lated in two ways

by capitalising the average profits.
by capitalising the super profits.

(i) Capitalisation of average profits
Under this method, the value of goodwill is calculated by deducting the capital employed (net assets) in the business from the capitalized value of average profits on the basis of normal rate of return.
Good will = capitalised value – capital employed (net asset)
Capitalised value of average profit
Average Profit × \(\frac{100}{\text { Normal Rate of Return }}\)

(ii) Capitalisation of Super Profits
Under this method the goodwill can be ascer tained by capitalising the super profit directly.
Goodwill = Super Profits × \(\frac{100}{\text { Normal Rate of Return }}\).

Question 3.
The following are the particulars in respect of two partnership firms.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 14
Manu wishes to join in any one of the above firm which can make better profit. He seeks your advice as to which firm is more worth while and reputed.
Answer:
Capital Exployed = Assets – Liabilities = (10,000 + 15,000 + 20,000 + 20,000) – 5,000 = 60,000
Normal profit = Capital employed × Normal rate of return
60,000 × 10/100 = 6,000
Actual profit = 5,500
Super profit = Actual profit – Normal Profit
= 5,500 – 6,000 = -500
Firm Y
Capital Employed= (2,000 + 8,000 + 10,000 + 20,000) – 5,000 = 35,000
Normal profit = 35.000 × 10/100 = 3,500
Actual profit = 4,000
Superprofit = 4,000 – 3,500 = 500
Conclusion: Firm ‘Y’ earns Rs. 4,000 which is above normal profit.
Firm Y’s performance is better. So select Firm Y’.

Question 4.
How the capital accounts of the partners are adjusted at the time of admission of a new partner?
Answer:
At the time of admission of a new partner the capital accounts of the partners may be adjusted in the following ways.
1. Asking the new partner to bring in the capital on the basis of the existing partner’s capitals. Here new partners’ capital is calculated as follows.

Totalling the capitals of the existing partners left after making all adjustments.
Totalling the new profit sharing rights of the old partners.
Total capital as per (a) is treated as the capital for the total rights as per (b)
On the basis of the above, calculating the amount of capital to be brought in by the new partner.

2. Adjusting the capital of the old partners on the basis of the capital brought in by the new partner. This is done as follows.

Comparing the capital of the incoming partner with the capitals of old partners.
Asking the partner to bring in the required amount whose capital is less.
Allowing the partner to withdraw the surplus amount whose capital is more.

Question 5.
A and B are partners sharing profits in the ratio of 3:2. On 1^st April 2005 they admit C into the firm. C brought in Rs. 1,00,000 for his capital but he was not in a position to bring his share of goodwill. The goodwill of the firm was valued at Rs. 1,50,000. Goodwill existing in the books of the firm is Rs. 2,75,000. The new profit sharing ratio is 2: 1: 1. Pass the journal entries.
Answer:
Working Note:
Sacrificing Ratio = Old Ratio – New ratio
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 15
C’s share of goodwill = 1,50,000 × 1/4 = 37,500 This Rs. 37,500 is to be debited to the new partner’s capital account and credited to old partners’ capital as C (new partner) cannot bring in the same.
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 16

Question 6.
R & S are partners sharing profits and losses in the ratio of 3:2. They admitted T into the firm. They have agreed to share the future profits – equally. T brought in Rs. 45,000 as his capital and Rs. 40,000 for his share of goodwill. The goodwill of the firm as in the books is Rs. 12,500. Write the journal entries.
Answer:
Notes
Sacrificing Ratio = Old ratio – New ratio
Old ratio = 3:2
New ratio = 1:1:1
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 18
Journal

Question 7.
Terry and C.L. Stephen are in partnership engaged in software development on accounting packages to various companies.
Ledger balances as shown by the books of accounts are:

Capital: Terry	3,00,000
C.L. Stephen	2,50,000
Plant and Machinery	2,00,000
Office Fixtures	1,00,000
Current Assets	1,50,000
General Reserve	80,000
Bank loan	1,60,000
Stock in trade	40,00,000
Land and Building	3,00,000

They have decided to admit Francis who is the son of Mr. Terry on the following terms.
Fixed Assets valued 10% more than the book value.
Interest payable on Bank loan Rs 16,000
Office Fixtures was taken over by C.L. Stephen.

The new partner’s capital A/c is to be credited with half of Mr. Terry’s capital A/c before making any adjustments. Prepare capital accounts of the partners.
Answer:
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 20
Capital Accounts

Note: The new partners capital A/c is to be credited with half of Mr. Terry’s capital a/c before making Adjustment.
Journal entry is:

Question 8.
A and B are partners sharing profit & losses in the ratio of 3:2. They admit C as a partner who is unable to bring goodwill in cash, but pays Rs. 16,000 as his capital. A Goodwill Account is raised in the books of the firm. Goodwill of the firm is valued at two year’s purchase of average three year’s profits. The profits for the three years were Rs. 10,000, Rs. 8,000 and Rs. 9,000. The net profit sharing ratio will be 5:2:2. The partners decided to write off goodwill after C’s admission. Make the Journal Entries, write up the Capital A/c. of partners & Goodwill calculation.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 23
Capital A/c

Calculation of Goodwill
Average profit = \(\frac{10,000+8,000+9,000}{3}\) = 9,ooo
Goodwill = Average profit × No. of years purchase
= 9,000 × 2 = 18,000
C’s share of Goodwill = 18,000 × 2/9 = 4,000
Sacrificing ratio = Old ratio – New ratio
Old ratio = 3:2
New ratio = 5:2:2
Sacrifice of A = 3/5 – 5/9 = 27 – 25 / 45 = 2/45
Sacrifice of B = 2/5 – 2/9 = 18 – 10 / 45 = 8/45
Sacrificing ratio of A and B = 2 : 8 = 1 : 4

Question 9.
You are given the following information on a reconstitution of a firm.
Partners capital

Ammu – Rs. 20,000
Beena – Rs. 30,000
Ceema – Rs. 20,000
Old profit sharing ratio – 2:3:1
New Ratio – 1:2:3
Revaluation profit – 22,500

State the reason for reconstitution.
Give a journal entry to adjust the revaluation profit through capital accounts of partners.

Answer:
Reconstitution refers to a change in the nature of relationship among partners due to

Change in profit sharing ratio
Admission
Retirement
Death or Amalgamation of two partnership firms.

In the firm of Ammu, Beena and Ceema reconstitution of firm takes place because they decided to change their profit sharing ratio:
Old ratio = 2:3:1 = 2/6 : 3/6: 1/6
New ratio = 1:2:3 = 1/6 : 2/6: 3/6
Ammu’s sacrifice = old ratio – new ratio
= 2/6 – 1/6 = 1/6
Beena’s sacrifice = old ratio-new ratio
= 3/6 – 2/6 = 1/6
Ceema’s gain = new ratio-old ratio = 3/6 – 1/6 = 2/6
Ammu’s revaluation profit = 22,500 × \(\frac{1}{6}\) = 3750
Beena’s revaluation profit = 22,500 × \(\frac{1}{6}\) = 3750
Ceema’s revaluation profit = 22,500 × \(\frac{1}{6}\) = 7500.

Question 10.
Ann and Gopu were doing sole proprietorship business of same nature. Both are close friends. On 1^st April 2005, they have decided to start a partnership business and have brought their existing assets into the new firm. They share profits in the ratio of 3:2. Details of existing assets and liabilities.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 25
On 31^st December, 2006, they have changed their profit sharing ratio and become equal partners. The assets were then revalued as follows:

Building is up by 10%
Plant is down by 10%
Furniture is up by 10%
Stock is valued at 1,50,000
Goodwill valued at Rs. 10,000

Give journal entries at to bring capital into the records on 01.04.2005.
Prepare :
- Revaluation A/c
- Capital A/c
New Balance Sheet

Answer:
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 26
Capital A/c

New B/S

Question 11.
Haridas and Sudheer Raj are partners in a firm sharing profits in the ratio 3:2. On 1.04.2004, they admit Ramdas into the firm for a 5^th share in profits. Ramdas contributed the following in respect of his capital and goodwill.

Stock	Rs. 10,000
Furniture	Rs. 20,000
Plant	Rs. 30,000
Building	Rs. 40,000

Goodwill has been valued at 2 years purchase of super profit of past 3 years.

1.4.2002	profit Rs. 18,000
1.4.2003	profit Rs. 25,000
1.4.2004	profit Rs. 32,000

Capital employed is Rs. 2,00,000 and normal rate of return is 10%.
Give journal entries in respect of:

Capital contributed by Ramdas.
Goodwill brought in by him.

Answer:
Super profit = (Actual profit) Average profit – Normal profit
Average profit =
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 29
Normal Profit = Capital employed × Normal rate of
return =200000 × \(\frac{10}{100}\) = 20000
Super Profit = 25000 – 20000 = 5000
Value of Goodwill = Super profit × No. of years purchase = 5000 × 2 = 10000
Ramdas’s (New Partner) Share of goodwill = Total goodwill of firm × Ramdas’ share
= 10000 × \(\frac{1}{5}\) = 2000 5
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 30

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Eight Mark Questions and Answers

Question 1.
Jo and Sony are partners sharing profits and losses in the raito of 2: 1. Their Balance Sheet as on 31^st December 2004 was as follows.
Balance Sheet As on 31^st December 2004
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 31
Ebo is admitted into the partnership on the Balance Sheet date on the basis of the following.

Ebo will bring Rs. 50,000 as his capital.
Stock in trade should be decreased by Rs. 5,000
Plant and Machinery should be increased to Rs. 35,000 and Land and Building should be appreciated by 10%.
Bills payable and creditors be decreased by 5% and 10% respectively.

Record necessary journal entries, prepare revaluation account and partner’s capital account and also prepare the Balance sheet after Ebo’s admission.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 32

Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 33
Partner’s Capital A/c

Balance Sheet as on 1^st January 2005

Question 2.
Sunu and Jinu are partners in a firm sharing profits and losses equally. The following is their Balance Sheet as on 31.12.2005.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 36
On the balance sheet date Tinu is admitted into the partnership on the following terms.

Tinu should bring in Rs. 60,000 as his capital
Furniture should be revalued at Rs. 50,000 and machinery at 25% less.
Bank overdraft should be decreased to Rs. 75,000
A provision of 10% is to be made for bad debts.
An unrecorded liability of Rs. 5,000 on rent is to be recorded.

Give journal entries, prepare revaluation account, capital accounts of partners and the Balance Sheet after Tinu’s admission.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 37

Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 39
Partner’s capital A/c

Balance Sheet as on 1^st January 2006

Question 3.
The following is the Balance sheet of L & M as on 30^th June 2005.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 43
L & M were sharing profits and losses in the ratio of 2:1. N is admitted into the firm for a fourth share. The following are the conditions agreed upon.

Provision for bad and doubtful debts be increased to Rs. 2,500
Land and Buildings was to be depreciated by Rs. 10,000
The firm had an unrecorded machinery of Rs. 10,000 which is to be recorded.
N is asked to bring Rs. 50,000 as his capital and Rs. 15,000 for good will.
L & M had to withdraw half of the goodwill brought in by N.

Record journal entries, prepare Profit Loss adjustment account, capital accounts, and the New Balance Sheet
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 44

Profit & Loss Adjustment (Revaluation) a/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 46
Partner’s Capital A/c

Balance Sheet as on 1^st July 2005

Question 4.
Edwin and Abel are partners sharing profits and losses in the ratio of 4:3. Their Balance sheet as on 30^th June 2005 is as follows.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 49
Jerin is admitted into the firm with 2/7^th share. The following are the terms and conditions.

Jerin should bring in Rs. 40,000 as his capital and share of goodwill. The value of the goodwill of the firm is fixed at Rs. 35,000.
The amount of furniture and fittings should be written down by Rs. 5,000.
The full amount of goodwill should be withdrawn by old partners.
Creditors should be reduced by Rs. 2,000.
The new profit sharing ratio should be 3: 2:2.

Prepare necessary accounts and the Balance Sheet after the admission of Jerin.
Answer:
Working notes :
1. Jerin’sshare of goodwill = Goodwill of the firm × 2/7
= 35,000 × 2/7 = 10,000

2. Sacrificing ratio = Old share – New share
Edwin’s sacrifice = 4/7 – 3/7 = 1 / 7
Abel’s sacrifice = 3/7 – 2/7 = 1 / 7
Ratio = 1/7 : 1/7 ie. = 1 :1
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 50

Partner’s capital a/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 51
Balance Sheet as on 1^s July 2005

Question 5.
Ram and Gopal are partners in a firm sharing profit and loss in the ratio of 3:1 respectively. The following is their Balance Sheet as on 31 /12/2008.
Balance Sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 53
They admit Menon into partnership on 1-1-2009 on the following terms.

Menon should bring Rs. 10,000 as his capital for 1/5^th share and 18,000 as her share of good will.
Liability for workmen compensation estimated at Rs. 1500.
Value of land and building be appreciated by Rs. 5000.
Stock reduced by 5%
A provision of 5% should be made for doubtful debts in debtors.

Prepare revaluation a/c, capital account of partners and the Balance sheet of new firm.
Answer:
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 54

Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 56
Balance Sheet

Working Note:
Menon’s share of goodwill = 18,000 Sacrificing ratio = 3:1
Goodwill credited in Ram = 18,000 × 3/4 = 13,500 Gopal = 18,000 × 1/4 = 4,500.

Question 6.
The following is the Balance Sheet of A and B as on 31^st Dec. 2004.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 58
A & B share profits and losses in the ratio of 5 : 3. They admit C into the partnership for equal share. C brings in Rs. 50,000 as capital and Rs. 5,000 for his share of goodwill. Goodwill of the firm is valued at Rs. 30,000. The following conditions were agreed upon.

Land and buildings are appreciated by 20%.
Stock is decreased by Rs. 5,000/-
Creditors include Rs. 2,500/- not become payable.
Unexpired insurance or insurance paid in advance Rs. 2,500 is to be recorded.

Record journal entries, prepare ledger accounts and the new Balance Sheet.
Answer:
Notes:
Calculation of sacrificing ratio
Sacrificing ratio = old ratio – new ratio
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 59
Journal

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 61
Revaluation A/c

Partner’s Capital A/c

Balance sheet as on 1^st January 2005

Question 7.
P and Q are partners sharing profits and losses in the ratio of 4:3. Their balance sheet as on 30^th June 2004 is as follows.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 65
R is admitted into the firm on the basis of the follow-ing conditions.

Sundry debtors should be revalued at Rs. 1,00,000.
R should bring in Rs. 15000 as capital and Rs. 10000 as his share of goodwill. He will get 1/6 share in future profits.
The capital accounts of all partners should be adjusted on the basis of their profit sharing ratio by bringing in or paying off the cash as the case may be.

Prepare necessary ledger accounts and the new Balance sheet of the firm.
Answer:
Working notes:
Calculation of new ratio:
Old Ratio between P & Q = 4 : 3
R’s share = 1/6
Remaining portion = 1 – 1/6 = 5/6
P’s new share = 4/7 of 5/6 = 4/7 × 5/6 = 20/42
Q’s new share = 3/7 of 5/6 = 3/7 × 5/6 = 15/42
R’s share = 1/6 = 7/42
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 66

Calculation of capital required:
R’s capital for
1/6 share in profits = 15,000
Total capital of the firm= 15,000 × 6/1 = 90,000
P’s capital = 90,000 × 20/42 = 42,857
Q’s capital = 90,000 × 15/42 = 32,143
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 67
Partners capital a/c

Cash A/c

Balance sheet as on 1^st July 2004

Question 8.
A and B are partners in a firm sharing profits and losses as 3/5 and 2/5.C, comes in for 1/5^Th share of profit. He pays Rs. 8,000 as goodwill premium and 50% of the adjusted capitals of A and B. Balance Sheet of A and B on the date of Cs’ admission stand as follows:
Balance Sheet
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 71
Land and Buildings are to be valued at Rs. 40,000. Plant is to be depreciated by 10% and stock by Rs. 500.Sundry Debtors is worth Rs. 31,750. A liability of Rs. 1,750 for outstanding expenses has been omitted to be recorded in the books. A and B have a joint life policy of Rs. 15,000 not shown in the books, the premium for which has been charged to Profit & Loss Account. The surrender value of the policy on the date of admissions is Rs. 2,000, and is agreed to raise a life policy account in the books at this value. Give Journal Entries.
Answer:
Journal
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 72

Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 73
Dr. Partner’s Capital A/c Cr.

Capital to be brought in by C = 50% of the adjusted capital of A and B.
i.e., = 50% of 57,500+ 40,000 = 50% of 97,500 = 48,750.

Question 9.
Below given the details related to a firm.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 76

Can you analyse the adjustment on admission of a new partner and show the balance sheet after admission.
Answer:
Revaluation A/c

Capital Accounts
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 79
Balance Sheet

Working Note
Goodwill Calculation: Here, Goodwill is calculated on the basis of super profit method.
Goodwill = Superprofit × No. years purchase
Superprofit = Actual or Average profit – Normal profit
Normal profit = Capital Employed × Normal rate of return
Capital employed = Asset – Liabilities
= (10,000 + 20,000 + 31,500 + 30,000+ 20,000) – 11,500
= 1,00,000
Normal profit = 1,00,000 × 10/100 = 10,000
Superprofit =40,000 – 10,000 = 30,000
Goodwill = 30,000 × 2 = 60,000
New partner’s share of Goodwill – 60,000 × 1/3
= 20,000
Sacrifacina ratio
Manu = 3/5 – 1/3 = 9 – 5/15 = 4/15
Raju = 2/5 – 1/3 = 6 – 5/15 = 1/15
Sacrificing ratio = 4:1
Manu = 3/5 – 1/3 = 9 – 5/15 = 4/15
Manu’s Sacrifice = 20,000 × 4/5 = 16,000
Raju’s Sacrifice = 20,000 × 1/5 = 4,000 [Hint: New ratio 1:1:1].

Question 10.
Observe the following table.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 81
Furniture was sold at Rs. 2700 on the date of admission. You are required:

Revaluation A/c
Capital Accounts
Balance sheet of the new firm

Answer:
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 82
Capital A/c

Cash A/c

Balance Sheet

Question 11.
A and B are partners in a firm sharing profits in the ratio of 2:1 ‘C’ is admitted into the firm with 1/4 share in Profits. He will bring in Rs. 30,000 as capital and capital of A and B are to be adjusted in the profit sharing ratio. The balance sheet of A and B as on 31/03/2017 (before c’s admission) was as under.
Balance sheet of A and B as on 31/03/2017
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 86
The terms of agreement are as follows:

‘C’ will bring in Rs. 12,000ashisshareofgoodwill.
Building was valued at Rs. 45,000 and Machinery at Rs. 23,000
A provision for bad debts is to be created @ 6% on debtors
The capital accounts of A and B are to be adjusted by opening current accounts.

Prepare necessary ledger accounts and new balance of the firm.
Answer:
Revaluation A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner - 87
Capital A/c

Balance Sheet

Working Note:
1. New Profit Sharing ratio C’s share of Profit = 1 /4

2. New capital of A and B on the basis of c’s capital
Total capital of the new firm = 30,000 × \(\frac{4}{1}\) = 1,20,000
As new capital = 1,20,000 × \(\frac{2}{4}\) = 60,000
The Existing capital of A = 63,680 Excess (A) = 3,680
B’s new capital = 1,20,000 × \(\frac{1}{4}\) = 30,000
The existing capital of B = 38,840
Excess (B) – 8,840
The current accounts can be opened and the amount to be withdrawn by A and B will be transferred to their respective current accounts.

Kerala SSLC Chemistry Model Question Paper 3 English Medium

April 13, 2021March 13, 2020 by Prasanna

Students can Download Kerala SSLC Chemistry Model Question Paper 3 English Medium Pdf, Kerala SSLC Chemistry Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Chemistry Model Question Paper 3 English Medium

General Instructions:

The first 15 minutes is the cool off time. You may use the time to read and plan your answers.
Answer the questions only after reading the instructions and questions thoroughly.
Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
Five questions are given in each section. Answer any four from each section.
Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

(Answer any 4 questions from 1 to 5. Each question carries 1 score) (4 × 1 = 4)

Question 1.
Which one of the following electron configuration is not correct?
a) 1s²2s²3s²
b) 1s²2s²
c) 1s²2s²2p⁶
d) 1s²2s²2p⁶3s¹
Answer:
a) 1s²2s²3s²

Question 2.
The volume of 5 mol ammonia (NH30 at STP will be …….
a) 5L
b) 22.4L
c) 17L
d) 112L
Answer:
1121

Question 3.
The metal that cannot liberate hydrogen when react with dilute hydrochloric acid is …….
a) Copper
b) Magnesium
c) Zinc
d) Aluminium
Answer:
Copper

Question 4.
To which category the compound CH₃-COO-CH₃ belongs to?
a) Carboxylic acid
b)Ethene
c) alcohol
d) Ester
Answer:
Ester

Question 5.
Which of the following substance is used to identify a sulphate salt?
a) Hydrochloric acid
b) Silver nitrate
c) Barium chloride
d) Liquor Ammonia
Answer:
Barium chloride

Section-B

(Answer any 4 questions from 6 to 10. Each question carries 2 scores)(4 × 2 = 8)

Question 6.
From the statements given below select those applicable to ‘s’ block elements.
a) high electronegativity
b) high ipnisation energy
c) loses electrons in chemical reactions
d) forms acidic oxides
Answer:
Loses electrons in chemical reactions.

Question 7.
Complete the blanks in the table given below.
Kerala SSLC Chemistry Model Question Paper 3 English Medium - 1
Answer:
a) 17 × 3 = 51 g
b) 22.4 L
c) 10 × 6.022 × 10²³
d) 22.4 × 5 = 112L

Question 8.
An iron nail is immersed in copper sulphate solution.
a) What will be the observation after some time?
b) Which of the following reactions include in this change?
Kerala SSLC Chemistry Model Question Paper 3 English Medium - 2
Answer:
a) Iron nail will be coated with copper
b) Fe → Fe²⁺ + 2e^–
Cu²⁺ + 2e^– → Cu

Question 9.
Match those given in the columns A, B and C suitably.

A	B	C
Blast furnance	Manufacture of Aluminium	Reduction using CO
Hall-Heroult	Concentration process of the ore	Electrolysis
Calcination	Manufacture of iron	Calamine

Answer:
Blast Furnace – Production of Iron – Reduction using CO
Hall – Heroult process – Production of Aluminium – Electrolysis
Calcination-Concentration of ore-Calamine

Question 10.
Find out a pair of isomers from those given below. Also write what type of isomers they are?
a) CH₃-CH₂-CH₂-OH
b) CH₃-CH₂-CH₂-CH₃
c) CH₃-O-CH₂-CH₃
d) CH₃-CH₂-CH₂-CH₂-OH
Answer:
CH₃-CH₂-CH₂-OH and CH₃-O-CH₂-CH₃ Funtional group isomers

Section – C

(Answer any 4 questions from 11 to 15. Each question carries 3 scores) (4 × 3 = 12)

Question 11.
The subshell electron configuration of two ions are given below. (Symbols are not real)
X²⁺ = 1s²2s²2p⁶
Y^1- = 1 s²2s²2p⁶
a) Write the subshell configuration of the elements X, and Y.
b) Write the chemical formula of the compound formed by the elements X and Y.
Answer:
a) X – 1s²2s² 2p⁶3s²
Y – 1s²2s²2p⁵
b) XY₂

Question 12.
The atomic mass of oxygen is 16. Match those in – the following table suitably.

A	B	C
16 g Oxygen	1GMM	6.022 × 10 ²³ molecules
160 g Oxygen	1 GAM	6.022 × 10²³ atoms
32 g Oxygen	5 mole	112 L volume at STP

Answer:
16g Oxygen – 1 GAM – 6.022 × 10²³ atoms
16g Oxygen – 5 mol – 112 L volume at STP
32g Oxygen – 1 GMM – 6.022 × 10²³ molecules

Question 13.
Write the products obtained in the following circumstances.
a) Molten NaCl is electrolysed.
b) Aqueous solution of NaCl is electrolysed
Answer:
a) Molten NaCI
At anode – Chlorine
At cathode – Sodium

b) Aqueous solution of NaCI
At anode-Chlorine
At cathode-Hydrogen.

Question 14.
The different stages involved in the concentration of the ore Of aluminium is given as a flow chart. Complete the missing columns.
Answer:
A – Sodium aluminate – NaAlO₂
B – Aluminium hydroxide-AI(OH)₃
C – Alumina -Al₂O₃

Question 15.
a) Complete the following equations by writing the formula/name of the products.
Kerala SSLC Chemistry Model Question Paper 3 English Medium - 3
b) Write balanced equation for the combustion of the fuel propane'(C₃H₈).
Answer:

Section – D

(Answer any 4 questions from 16 to 20. Each question carries 4 scores)(4 × 4 = 16)

Question 16.
Analyse the structural formula of a Compound given below.
Kerala SSLC Chemistry Model Question Paper 3 English Medium - 5
a) Write the IUPAC name of this compound.
b) Write the structural formula of a possible isomer of this compound.
c) Which type of isomers these compounds are?
Answer:
a) 2,3-dimethyl butane

c) Chain isomers

Question 17.
The alloys Nichrome and Alnico contains same components.
a) Which are the component elements in these alloys?
b) Write any one use of each alloy.
c) Why does these alloys differ in properties eventhough they contain same components?
Answer:
a. Al, Ni, Cr, Fe
b. Alnico – used to omake magnets Nichrome – used to makeJieating coil of heating appliances.
c. The ratio of components is different

Question 18.
The equation of a reversible reaction is given below.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g) + Heat
What is the influence of the following factors?
a) Increases pressure
b) Increases the concentration of N₂
c) Ammonia (NH₃) is continuously removed from the sytem.
d) A catalyst is used.
Answer:
a) Forward reaction increases
b) Forward reaction increases
c) Forward reaction increases
d) Both forward and backward reaction increases and equillibrium state is attained at an early stage.

Question 19.
Write the observation and the reason for it in the following experiments.
a) Barium chloride (BaCl₂) is added to the solution of a sulphate salt.
b) Cone. H₂SO₄ is added to cane sugar crystals.
c) Cone. H₂SO₄ is added to water taken in a beaker.
d) Cone. H₂SO₄ is poured in a dry paper.
Answer:
a. A white precipitate is formed. Sulphate salts reacts with Barium chloride to form white Barium sulphate.
b. Sugar changes to sugar charcol. Con. H₂SO₄ absorbs hydrogen and oxygen present in sugar in the ratio as that of water. (2:1)
c. Heat is liberated. Dissolution of H₂SO₄ in water is exothermic!
d. Paper get charred. Dehydration property of H₂SO₄

Question 20.
a) How ethanol is manufactured on a large scale?
b) Write any two uses of ethanol
c) What is power alcohol?
d) What is methylated spirit?
Answer:
a) Ethanol is manufactured by the fermentation of ‘ sugar solution.
b) As a solvent
As a preservative
As beverage
c) Power alcohol is a mixture of absolute alcohol and petrol. It is used as a fuel for motor vehicles.
d) Ethanol added with poisonous substances is called methylated spirit.

Kerala SSLC Maths Model Question Paper 4 English Medium

April 13, 2021March 13, 2020 by Prasanna

Students can Download Kerala SSLC Maths Model Question Paper 4 English Medium, Kerala SSLC Maths Model Question Papers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala SSLC Maths Model Question Paper 4 English Medium

Time: 2½ Hours
Total Score: 80 Marks

Instructions

Read each question carefully before writing the answer.
Give explanations wherever necessary.
First 15 minutes is Cool-off time. You may use the time to read the questions and plan your answers.
No need to simplify irrationals like √2, √3, π etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
In quadrilateral ABCD, ∠A = 100°, ∠B = 110°, ∠C = 50°
a) Where is the position of B?
b) Where is the position of D?
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 1
a) Inside the circle
b) ∠D = 360 – 270 = 90°
On the circle.

Question 2.
In an arithmetic sequence, the 6th term is 14 and 14th term is 6.
a) Find a common difference.
b) Find the 20th term.
Answer:
a) d = \(\frac{6-14}{14-6}\) = \(\frac{-8}{8}\) = 1
b) t₂₀= t₁₄ + 6d = 6 + 6 × -1 = 6 – 6 = 0

Question 3.
First, five odd numbers which are multiples of five are given
a) Find mean.
b) Find the median.
Answer:
5, 15, 25, 35, 45
\(\frac{5+15+25+35+45}{5}=\frac{125}{5}=25\)
b) Median = 25

Question 4.
In a box, there are pieces of paper in which the digits 1.5 are written. If we take a paper without looking.
a) Probability of being odd is.
b) What is the probability of being prime?
Answer:
1, 2, 3, 4, 5
a) 3/5
b) 3/5

Answer any 5 questions from 5 to 11. Each question carries 3 marks each. (5 × 3 = 15)

Question 5.
Consider the arithmetic sequence \(\frac{17}{7}, \frac{31}{7}, \frac{45}{7}, \ldots \ldots\)
a) Write the algebric form of this sequence.
b) Find wheather.natural numbers are a term of this sequence.
Answer:
17, 31, 45, ……….
a) 14n + 3/7
b) No
Reason = \(\frac{14 n+3}{7}=2 n+3 / 7\)

Question 6.
When a circle of radius 13 unit and mid-point (4, 5) is drawn, the point A and B cuts the x-axis.
a) Find the coordinates of A and B
b) Find the length of AB.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 2
a) (x – 4)² + 5² = 13²
(x – 4)² = 169 – 25
(x – 4)² = 144
x – 4 = 12, x – 4 = -12
x = 8, x = -8
A (8, 0), B (-8, 0)
b) AB = 16

Question 7.
a) Draw a circle of radius 3 cm
b) Draw a triangle with angles 55° and 75°, and the abovecirle as the incircle of the triangle.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 3

Question 8.
A chord AB is drawn 4 cm away from the centre of a circle. Chord makes an angle of 120° at the centre.
a) Draw a rough figure.
b) Find the radius of the circle.
c) Find the length of AB.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 4

Question 9.
In polynomial x² – 3x
a) What should be added with this polynomial for (x – 2) to be a factor.
b) If (x – 1) is a factor of x² – kx + 2. What is the value of k?
Answer:
a) P(2) = 2² – 3 × 2 = 4 – 6 = -2
P(2) = 0
(To make P(2) = 0 add 2)
b) x² – kx + 2 if x – 1 is a factor
P(1) = 0
⇒ 1 – k ×1 + 2 = 0
⇒ 3 – k = 0
⇒ k = 3

Question 10.
In the picture, a rhombus is drawn joining the mid-points of a rectangle and another rectangle is drawn joining the mid-points of the rhombus and it is shaded. If we put a dot on the figure without looking at what is the probability that it is inside the shaded part.
Kerala SSLC Maths Model Question Paper 4 English Medium 5
Answer:

PQ = ½ AB
SR = ½ AB
PS = ½ CD
QR = ½ CD
Area of PQSR = PQ × QR
= ½ AB × ½ CD
= \(\frac { 1 }{ 4 }\) (AB × CD)
= \(\frac { 1 }{ 4 }\) × (area of large rectangle)
Probability = \(\frac { 1 }{ 4 }\) or \(\frac{4}{16}=\frac{1}{4}\)
Kerala SSLC Maths Model Question Paper 4 English Medium 7

Question 11.
If the first term of the arithmetic sequence is x and the common difference is 3.
a) What is the next term?
b) Write arithmetic sequences in which the product of the first two terms is 28, and the common difference is 3.
Answer:
First term = x
a) Second term = x + 3
b) x (x + 3) = 28
x² + 3x = 28
x² + 3x – 28 = 0
(x + 7) (x – 4) = 0
x = -7, 4
Sequence = -7, -4, -3 … or 4, 7, 10

Answer any 7 questions from 12 to 21. Each question carries 4 score. (7 × 4 = 28)

Question 12.
2x + 1, 4x + 2, 6x + 3
a) If this is an arithmetic sequence what is the common difference?
b) What is the 10th term?
c) If the 10th term is 70. Find the value of x.
Answer:
a) d = 4x + 2 – (2x + 1) = 2x + 1
b) t₁₀ = 2x + 1 + 9 (2x + 1) = 20x + 10
c) 20x + 10 = 70
20x = 60
x = 60/20 = 3

Question 13.
In the figure ABCD is a cyclic quadrilateral PQ, R are the tangents of points at A, C.
∠BAQ = 50°, ∠ABD = 60°, ∠BCR = 30°.
Find the following angles.
Kerala SSLC Maths Model Question Paper 4 English Medium 8
a) ∠ADB
b) ∠CDB
c) ∠BCD
d) ∠DCS
Answer:
a) ∠ADB = 50°
b) ∠CDB = 30°
c) ∠BCD = 110°
d) ∠DCS = ∠DBC = 40°

Question 14.
The distance from the centre to a point outside the circle of radius is 3 units larger than 2 times its radius. The length of the tangent is 1 unit less than the distance between the centre and external point.
a) If the radius is x. Find the distance between the centre and external point and length at a tangent.
b) Find its length using a second-degree equation.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 9
a) 2x + 3
2x + 3 – 1 = 2x + 2
b) (2x + 3)² = (2x + 2)² + x²
⇒ 4x² + 12x + 9 = 4x² + 8x + 4 + x²
⇒ x² + 8x – 12x + 4 – 9 = 0
⇒ x² – 4x -5 = 0
⇒ (x – 5)(x + 1) = 0
⇒ x = 5, -1
∴ r = 5, d = 13, l = 12

Question 15.
The angle between lateral edge and base diagonal of a square pyramid is 30°. If the length of lateral edge is 20 cm.
a) Find the height of the square pyramid.
b) Find the length of diagonal.
c) Find the length of base edge.
d) Find slant height.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 10

Question 16.
If x² – 1 is a factor of P(x) = ax³ + bx² + cx + d
a) Write two first degree factors of P(x)
b) Prove a + c = b + d
Answer:
a) (x + 1), (x – 1)
b) P(1) = 0, P(-1) = 0
If P(-1) = 0 P(-1) = 0
⇒ – a × (-1)³ + b × (-1)² + c × -1 + d = 0
⇒ -a + b – c + d = 0
⇒ a + c = b + d

Question 17.
Kerala SSLC Maths Model Question Paper 4 English Medium 11
The centre of a regular hexagon is the origin with sides 4 cm.
Write the co-ordinates of all its vertices.
Answer:

x = 2, y = 2√3
E = (-2, 2√3)
D = (2, 2√3)
C = (4, 0)
B = (2, -2√3)
A = (-2, -2√3)
F = (-4, 0)

Question 18.
The radius of a vessel of cone shape is 6 cm and height 18 cm. It is filled with water. The radius of another cone-shaped vessel is 27 cm and height 24 cm.
a) What is the volume of the first cone?
b) Find the volume of the larger cone.
c) If we pour the water from the small cone to the larger one by holding it upside down. What will be the height of the water level?
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 13

Question 19.
The number of students in the 10th standard is arranged in a table according to their height.

Height (cm)	No. of Students
130 – 135	2
135 – 140	5
140 – 145	8
145 – 150	10
150 – 155	7
155 – 160	6
160 – 165	3

a) Find the height of the student in 16th place, if all are arranged according to their heights in order.
b) Which student’s height comes as a median.
c) Find the median height.
Answer:
Less than 135 – 2
Less than 140 – 7
Less than 145 – 15
Less than 150 – 25
Less than 155 – 32
Less than 160 – 38
Less than 165 – 41
N = 41
N/2 = 20.5
Kerala SSLC Maths Model Question Paper 4 English Medium 15
b) 21st
c) 145.25 + 5 × 0.5 = 145.25 + 2.5 = 147.75

Question 20.
a) Draw a rectangle with sides 3 cm, 5 cm.
b) Draw a square with one sides √15 cm and the same area that of the rectangle.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 16
Question 21.
The line joining A(1, 3), B(5, 2) and C(1, 1), D (5, 4) intersects at P.
a) Write the equation of the line AB.
b) Write the equation of the line CD
c) Find the coordinates of P.
Answer:

Write any 5 questions from 22 to 28. Each question carries 5 marks. (5 × 5 = 25)

Question 22.
In the figure a model of aframe which hangs a weight is given BC = 5 unit, BE = 2 unit, AE = 6 unit. ∠BDC = 35°
Kerala SSLC Maths Model Question Paper 4 English Medium 18
a) Find the length of AB.
b) Find the length of CD
c) Find the length of ED
Answer:

Question 23.
Consider the following arithmetic sequences
8, 14, 20, 26, …………
34, 38, 42, 46, ………….
a) Write the common difference of the above two sequences.
b) Write the algebraic form of two sequences.
c) Check whether is there any common term at the same position for these two series.
Answer:
a) 6, 4
b) 6n + 2
2n + 30
c) 6n + 2 = 4n + 30
2n = 28
n = 14
(14th term will be equal)
6 × 14 + 2 = 84 + 2 = 86
Equal term = 86

Question 24.
There are two taps for a tank. One is to fill and the other is to empty. The second tap takes 5 minutes more to empty the full tanks than the first tap to fill the tank. If two were open it took 1 hr to fill the tank.
a) If the first tap takes x minutes to fill the tank. What is the time taken to empty the tank by the second tap?
b) From a quadratic equation by taking the volume fill in the tank in a minute,
c) Find the time taken by the two taps.
Answer:
a) x + 5
Kerala SSLC Maths Model Question Paper 4 English Medium 20

Question 25.
A solid hemisphere is attached at one end of a solid core. Height of the cone is equal to the diameter of the hemisphere. This solid figure is melted and recast into a sphere of radius equal to \(\frac{1}{3}^{\mathrm{rd}}\) of the diameter of the hemisphere.
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a) If the radius of the hemisphere is ‘r’. What is the height of the cone?
b) Find the volume of the solid shape.
c) What is the radius of the sphere?
d) Find the volume of the sphere.
e) How many such spheres can be made from the solid shape?
Answer:
a) 2r
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Question 26.
Two lines 2x – 3y + 7 = 0, 3x + 2y – 9 = 0 intersects at P.
a) Find the co-ordinates of P.
b) Find the equation of the line which passes through P and has a slope ½.
c) Prove that the two lines 2x – 3y + 7 = 0 and 3x + 2y – 9 = 0 are perpendicular.
Answer:
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Question 27.
CQ and CS are tangent from C to the outer circle of ∆ABC. I is the centre of the incircle and J is the centre of the outer circle.
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a) Prove that the perimeter of ∆ABC is equal to CQ + CS.
b) If CP = 5 cm, CQ = 8 cm, IP = 2 cm. Find the radius of the outer circle.
Answer:
Perimeter of ∆ABC = CB + BR + AR + AC
= (CB + BQ) + (AS + AC)
= CQ + CS

Question 28.
a) Draw a square of side 5 cm.
b) Draw the incircle and circumcircle of the square.
c) Find the ratio of the incircle and circumcircle of the square.
Answer:
Kerala SSLC Maths Model Question Paper 4 English Medium 26

Read the mathematical concept carefully and answer the following question. (6 × 1 = 6)

Question 29.
We can draw an incircle and circumcircle to a triangle. What about in the case of quadrilaterals….?. Can we draw both to a quadrilateral…?
No. we cannot. If we can draw a circumcircle to a quadrilateral. It is known as a cyclic quadrilateral. What is its property…?. The sum of opposite angles of a cyclic quadrilateral will be 180°. What about if we can draw an incircle to a quadrilateral…?. The sum of opposite sides of that quadrilateral will be equal. If these two properties come together for a quadrilateral then we can draw both circumcircle and incircle. These type of quadrilaterals are known as “Bicentric quadrilaterals”. The area of a “Bicentric quadrilaterals” is given by the formula √abcd where a, b, c, d are the sides of the quadrilateral.
a) Can we draw both circumcircle and incircle to a rectangle? (1)
b) Write two examples of “Bicentric quadrilateral. (1)
c) If one angle of a “Bicentric quadrilateral” is 110°, find its opposite angle. (1)
d) If the sum of two opposite sides of a “Bicentric quadrilateral” is 15 cm. Find its perimeter. (1)
e) If the four sides of a “Bicentric quadrilateral” is 10 cm, 7 cm, 12 cm, 15 cm. Find its area. (2)
Answer:
a) No
b) Square, Rhombus
c) 70° (180 – 110)
d) 2 × 15 = 30 cm
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