physics ncert solutions class 11th
Get insights from 952 questions on physics ncert solutions class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about physics ncert solutions class 11th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
4 months agoContributor-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/
Where n = N/V
n=number of aeroplanes/volume
= -3
T=
T= =
New answer posted
4 months agoContributor-Level 10
This is a long answer type question as classified in NCERT Exemplar

New answer posted
4 months agoContributor-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
New answer posted
4 months agoContributor-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
New answer posted
4 months agoContributor-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.
New answer posted
4 months agoContributor-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
New answer posted
4 months agoContributor-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
New answer posted
4 months agoContributor-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.
New answer posted
4 months agoContributor-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 =
New answer posted
4 months agoContributor-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= ( )yPo= 2yPo
Hence ratio of final pressure =
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 686k Reviews
- 1800k Answers




