physics ncert solutions class 11th

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4 months ago

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Payal Gupta

Contributor-Level 10

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

Time require to avoid the collision T= l/v where l = mean free path =1/ 2 π d 2 n

Where n = N/V

n=number of aeroplanes/volume

= 10 20 * 20 * 1.5 = 0.0167 k m -3

T= 1 V 2 π d 2 N * 1 v

T= 1 2 * 3.14 * 20 2 * 0.0167 * 10 - 6 * 150 = 10 6 1776.25 * 2.505

= 10 6 4449.5 = 224.74 h 225 h

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

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

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

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

(a) The moon has small gravitational force and hence, the escape velocity is small .

As the moon in the proximity of the earth as seen from the sun, the moonhas the same amount of heat per unit area as that of the earth.

The air molecules have large range of speeds . even though the rms speed of the air molcules is smaller than the escape velocity on the moon, a significant number of molecules have speed grater than the escape velocity.

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At g

...more

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) Q1= W+Q2

W=Q1-Q2>0

Q1>Q2>0

We can also write Q21

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) the given process is a cyclic process i.e returns to the original state 1

Hence change in internal energy dU =0

dQ= dU+dW=0+dW

hence total heat supplied is converted to work done by the gas which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . it is not possible to return to the same state without being heat supplied hence 3 to 1 cannot be adiabatic.

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4 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Change in internal energy for process A to B

dU=nCvdT=nCv (dT)=nCv (TB-TA)

work done from A to B  = area under the PV curve which is maximum for path I

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) For isothermal dT= 0 so T=constant

For an ideal gas dU = change in internal energy = nCvdT=0

From first law of thermodynamics dQ= dU+dW

dQ= dW

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b), (d) When the rod is hammered the external work is done on the rod which increases its temperature.

Heat is transferred to the gas in the small container by big reservoir at temperature T2

As the weight is added to the cylinder arrangement in the form of external pressure so it cannot reversed.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Let us assume that T1, T2T3

According to questions there is no loss of heat in the surroundings

Heat lost by M3 = heat gained by M1+ heat gained by M2

M3s (T3-T)= M1s (T-T1)+M2s (T-T2)

T [M1+M2+M3]= M3T3+M1T1+M2T2

T = M 1 T 1 + M 2 T 2 + M 3 T 3 M 1 + M 2 + M 3

New answer posted

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) Container A is isothermal and container B is adiabatic

For isothermal process P1V1=P2V2

Po (2Vo)= P2 (Vo)

P2= 2Po

for adiabatic process

P1V1y= P2V2y

Po (2Vo)y=P2 (Vo)y

P2= ( 2 V o V o )yPo= 2yPo

Hence ratio of final pressure = 2 γ P o 2 P o = 2 γ - 1

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