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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E₁of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= $\frac{V}{50 + R}$ = $\frac{10}{50 + R}$

Potential difference across wire of potentiometer

IR'= $\frac{10 R'}{50 + R}$

Since with 50 ohm resistor, null point is not obtained it is possible when

$\frac{10 * R'}{50 + R} < 8$

10R' $< 400 + 8 R'$

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

$\frac{10 * R'}{50 + R} < 8$

2R'>40

R'>40

$\frac{10 * \frac{3}{4} R'}{10 + R'} < 8$

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4^th segment of

...more

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= $\frac{V}{50 + R}$ = $\frac{10}{50 + R}$

Potential difference across wire of potentiometer

IR'= $\frac{10 R'}{50 + R}$

Since with 50 ohm resistor, null point is not obtained it is possible when

$\frac{10 * R'}{50 + R} < 8$

10R' $< 400 + 8 R'$

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

$\frac{10 * R'}{50 + R} < 8$

2R'>40

R'>40

$\frac{10 * \frac{3}{4} R'}{10 + R'} < 8$

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4^th segment of the potentiometer, therefore potential drop across 400cm of wire >8V

Potential gradient K $* 400 c m > 8 V$

K*4m>8V

K>2V/m

Similarly potential drop across 300 cm wire <8V

k*3m<8V, K<2*2/3V/m

2*2/3V/m>k>2V/m

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

A system goes from P to Q by two different paths in the P-V diagram as shown in Fig. 12.8. Heat given to the system in path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

For path1

Heat Q₁= 1000J

Work done =W₁

For path 2

Work done W₂= W₁-100

As change in internal energy is same

dU=Q₁-W₁=Q₂-W₂

1000-W₁=Q₂-W₁+100

Q₂= 1000-100= 900J

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

A room AC runs for 5 a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – power consumption in a day i.e in 5 = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{ρ l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{ρ l}{A}$ l²= 1.7 $*$ 10^{-8
$* \frac{10}{π * 10^{- 6}} * 81$}j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000} *$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000} *$ 100= 0.32%

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Can a system be heated and its temperature remains constant?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Yes this is possible when the entire heat supplied to the system is utilised in expansion.

So its working against the surroundings.

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider one mole of perfect gas in a cylinder of unit cross section with a piston attached . A spring (spring constant k) is attached (unstretched length L ) to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from V_o to V₁.

(a) What is the initial pressure of the system?

(b) What is the final pressure of the system?

(c) Using the first law of thermodynamics, write down a relation between Q, Pa, V, Vo and k.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) Initially the piston is in equilibrium P_i=P_a

(b) On supplying heat , the gas expands from V_o to V_i

so increase in volume of the gas =V_i-V_o

as the piston is of unit cross sectional area hence extension in the spring

x= $\frac{V_{i} - V_{0}}{a r e a} = V_{i} - V_{0}$

force exerted by the spring on the piston= F= kx= K(V_i- V_o)

hence final pressure =P_f =P_a +kx

= P_a+K $*$ ( $V_{i} - V_{0}$ )

(c) From first law of thermodynamics

dQ=dU+dW

dU=C_v(T-T_o) = C_v(T-T_o)

T= $\frac{P_{f} V_{1}}{R} = [\frac{P_{a} + K (V_{i} - V_{0})}{R}] \frac{V_{1}}{R}$

Work done by the gas =pdV+ increase in PE of the spring

= P_a(V₁-V_o) + $\frac{1}{2} k$ x²

dQ=dU+dW

= C_v(T-T_o)+P_a(V-V_o)+ $\frac{1}{2} k$ x²

= C_v(T-T_o)+P_a(v-V_o)+1/2 ( $V_{i} - V_{0}$ )²

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New answer posted

10 months ago

Physics physics ncert exemplar solution class 12th chapter one

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – power consumption in a day i.e in 5 = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{ρ l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{ρ l}{A}$ l²= 1.7 $*$ 10^{-8
$* \frac{10}{π * 10^{- 6}} * 81$}j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000} *$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000} *$ 100= 0.32%

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider that an ideal gas (n moles) is expanding in a process given by P = f ( V ), which passes through a point (P_o, V₀ ). Show that the gas is absorbing heat at (P₀, V_o ) if the slope of the curve P = f (V ) is larger than the slope of the adiabat passing through (P_o, V₀ ).

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Slope of the curve = f(V) , where V is the volume

Slope of P = f(V) curve at ((P_o, V₀ )= f(V_o)

Slope of adiabatic at (P_o, V₀ )= k(-Y)V_o^-1-Y =-YP_o/V_o

Now heat absorbed in the process P= f(V)

dQ=dU+dW= nC_vdT+pdV

pV=nRT

T= pV/nR

T= $\frac{1}{n R} [f (V) + V f^{'} (V) d V]$

$\frac{d Q}{d V} =$ nC_{v
$\frac{d T}{d V} + p \frac{d V}{d V} = n C_{V} \frac{d T}{d V} + P$}

$\frac{C_{V}}{R} [f (V_{o} + V_{0} f^{'} (V) + f (V)]$

After solving we get

${[\frac{d Q}{d V}]}_{v = v o} =$ $[\frac{1}{Y - 1} + 1] f (V_{o}) + \frac{V_{o} f^{' V_{o}}}{Y - 1}$

= $\frac{Y}{Y - 1} P_{o} + \frac{V_{o}}{Y - 1} f^{'} (V_{o})$

Heat is absorbed where dQ/dV>0 when gas expands

Hence YP_o+V_of'(V_o)>0 or f'(V_o)>(-Y $\frac{P_{o}}{V_{o}}$ )

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)

AB : constant volume

BC : constant pressure

CD : adiabatic

DA : constant pressure

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) For process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

= nC_vdT= nC_v(T_B-T_A)

= $\frac{3}{2} R (T_{B} - T_{A})$

= $\frac{3}{2} (R T_{B} - R T_{A}) = \frac{3}{2} (P_{B} V_{B} - P_{A} V_{A})$

Heat exchanged = $\frac{3}{2} (P_{B} V_{B} - P_{A} V_{A})$

(b) For process BC , p =constant

dQ= dU+dW = $\frac{3}{2} R (T_{C} - T_{B}) + P_{B} (V_{C} - V_{B})$

heat exchanged = $\frac{5}{2} P_{B} (V_{C} {- V}_{A})$

(c) For process CD , because CD is adiabatic , dQ= heat exchanged =0

(d) DA involves compression of gas from V_D to V_A at constant pressure P_A

heat transferred as similar way as BC₁

hence dQ = $\frac{5}{2}$ P_A(V_A-V_D)

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in Fig .

A to B : volume constant B to C : adiabatic C to D : volume constant D to A : adiabatic V_c=V_D=2V_A=2V_B

(a) In which part of the cycle heat is supplied to the engine from outside?

(b) In which part of the cycle heat is being given to the surrounding by the engine?

(c) What is the work done by the engine in one cycle? Write your answer in term of P_A, P_B, V_A

(d) What is the efficiency of the engine? [γ =5/3 for the gas], (C_v= 3/2R for one mole)

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) For the process AB

dV=0 and dW=0

dQ=dU+dW=dU

dQ=dU= change in internal energy , so heat utilised is equal to change in internal energy.

Since p= $\frac{n R}{V} T,$ in adiabatic temperature is directly proportional to pressure. So heat is supplied to the system in process AB.

(b) For the process CD volume is constant but the pressure decreases, hence temperature also decreases . so heat is also given to the surroundings.

(c) W_AB= $\int_{A}^{B} p d V = 0$ , W_CD= $\int_{V c}^{V d} p d V = 0$

W_BC= $\int_{V B}^{V c} p d V = k \int_{V B}^{V c} \frac{d V}{V^{Y}} = \frac{k}{1 - Y} [V^{1 - Y}]$

= $\frac{1}{1 - Y}$ [pV]= $\frac{P_{c} V_{c} - P_{B} V_{B}}{1 - Y}$

W_DA= $\frac{P_{A} V_{A} - P_{D} V_{D}}{1 - Y}$

B and C lies on adiabatic curve BC

P_BV_B^Y= P_CV_C^Y

P_C = P_B( $\frac{V_{B}}{V_{C}}$ )^Y = P_B( $\frac{1}{2}$ )^Y= 2^-YP_B

Total work done by the engine in one cycle ABCD

...more

This is a long answer type question as classified in NCERT Exemplar

(a) For the process AB

dV=0 and dW=0

dQ=dU+dW=dU

dQ=dU= change in internal energy , so heat utilised is equal to change in internal energy.

Since p= $\frac{n R}{V} T,$ in adiabatic temperature is directly proportional to pressure. So heat is supplied to the system in process AB.

(b) For the process CD volume is constant but the pressure decreases, hence temperature also decreases . so heat is also given to the surroundings.

(c) W_AB= $\int_{A}^{B} p d V = 0$ , W_CD= $\int_{V c}^{V d} p d V = 0$

W_BC= $\int_{V B}^{V c} p d V = k \int_{V B}^{V c} \frac{d V}{V^{Y}} = \frac{k}{1 - Y} [V^{1 - Y}]$

= $\frac{1}{1 - Y}$ [pV]= $\frac{P_{c} V_{c} - P_{B} V_{B}}{1 - Y}$

W_DA= $\frac{P_{A} V_{A} - P_{D} V_{D}}{1 - Y}$

B and C lies on adiabatic curve BC

P_BV_B^Y= P_CV_C^Y

P_C = P_B( $\frac{V_{B}}{V_{C}}$ )^Y = P_B( $\frac{1}{2}$ )^Y= 2^-YP_B

Total work done by the engine in one cycle ABCDA

W= W_AB+W_BC+W_CD+W_DA= W_BC+W_DA

= $\frac{P_{C} V_{C} - P_{B} V_{B}}{1 - Y} + \frac{P_{A} V_{A} - P_{D} V_{D}}{1 - Y}$

W = $\frac{1}{1 - Y} [2^{- Y} P_{B} 2 V_{B} - P_{B} V_{B} + P_{A} P_{A} - 2^{- Y} P_{B} 2 V_{B}]$

W = $\frac{1}{1 - Y} (2^{1 - Y} - 1) (P_{B} - P_{A}) V_{A}$

W= $\frac{3}{2} [1 - (\frac{1}{2})$ ^2/3]( $P_{B} - P_{A}$ )V_A

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New answer posted

10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig.

(a) Find the work done when the gas is taken from state 1 to state 2.

(b) What is the ratio of temperature T₁/T₂, if V₂ = 2V₁?

(c) Given the internal energy for one mole of gas at temperature T is (3/2) RT, find the heat supplied to the gas when it is taken from state 1 to 2, with V₂ = 2V₁.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

pV^1/2= constant

P=k/ $\sqrt[]{V}$

Work done from 1 to 2

W= $\int_{V 1}^{V 2} p d V = K \int_{V 1}^{V 2} \frac{d V}{\sqrt[]{V}} = k {[\frac{\sqrt[]{V}}{\frac{1}{2}}]}_{V 1}^{V 2} = 2 K (\sqrt[]{V_{2}} - \sqrt[]{V_{1}})$

from ideal equation = pV=nRT

T= pV/nR= $\frac{p \sqrt[]{V} \sqrt[]{V}}{n R}$

T= $\frac{K \sqrt[]{V}}{n R}$

T₁= $\frac{K \sqrt[]{V 1}}{n R}$ , T₁= $\frac{K \sqrt[]{V 2}}{n R}$

$\frac{T_{1}}{T_{2}} = \frac{\frac{k \sqrt[]{V_{1}}}{n R}}{\frac{k \sqrt[]{V_{2}}}{n R}} = {\sqrt[]{\frac{V_{1}}{V_{2}}}}^{}$ = $\sqrt[]{\frac{V_{1}}{2 V_{1}}} = \frac{1}{\sqrt[]{2}}$

U= $\frac{3}{2} R T$

$? U = U_{2} - U_{1} = \frac{3}{2} R (T_{1} - T_{2})$

= $\frac{3}{2}$ RT₁( $\sqrt[]{V} - 1$ )

$? W$ =2p₁V₁^1/2( $\sqrt[]{V_{2}} - \sqrt[]{V_{2}}$ )

= 2p₁V₁^1/2(2 $\sqrt[]{V_{1}} - \sqrt[]{V_{1}}$ )

= 2p₁V₁( $\sqrt[]{2} - 1$ )= 2RT₁( $\sqrt[]{2} - 1$ )

$? Q$ = $? U + ? W$

= $\frac{3}{2}$ RT₁( $\sqrt[]{2} - 1$ )+ 2RT₁( $\sqrt[]{2} - 1$ )

= $\frac{7}{2} R T_{1} (\sqrt[]{2} - 1)$

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