HC Verma Physics is like a bible for student who are aspiring for IIT-JEE, AIPMT, UG NEET or any engineering or medical entrance. All the questions in the book is of high level, which requires all basics to applied.
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A rod of length 1m rest on a smooth horizontal base.When heated the length becomes 1.01 then find the strain doveloped.

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Aman Kumar

09/04/2014 10:44pm

Solve this question:
An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and length l. Left part of the cylinder contains 1 mole of helium and right part contains 0.5 mole each of helium and oxygen. If initial pressure of gas in each part is P°. Calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.

please tell us about the specific problem or topic that you are not able to understand. We will you detail explanation about that.

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bhavya

12/01/2013 8:49am

how did the specific heat capacity of water come in q-2???

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Admin

13/01/2013 3:26am

Dear Bhayva,
You need to learn the specific heat capacity of water. There will be many questions where you will not be given Specific heat of water but still they are using it.

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bhavya

12/01/2013 10:13am

actually, if u refer example given on page 43 u would see that they added the water equivalent and the mass.if we do the same way we and take specific heat as pre described 4180,we get the answer as 449.3 which is the answer at the back of the book..........just wanted to know why have we added it...........................

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Admin

13/01/2013 3:35am

Dear Bhavya,
Here do not get confused with the word water equivalent.
You can understand in this way that the calorimeter is also made up of some material. The heat which is given to the water, some heat will also be taken up by the calorimeter or we can say that some heat will be lost to the calorimeter material. so, we see that how much heat is taken by calorimeter mterial and calculate the equivalent mass of water that would take the same heat to change its temperature as that of the calorimeter.
That's why we are adding it with the given mass of water.

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bhavya

13/01/2013 6:53am

but water equivalent is explained on pg.40 of the book........which i have not understood

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Admin

14/01/2013 3:41am

Dear Bhaivya,
Here i am going to explain you about the water equivalent.
First of all the question arises from where this water equivalent is coming.
Now, see that there is calorimeter which is vessel made up of copper. Inside the calorimeter we have kept some cold water.
Now when this calorimeter comes in contact with some hot body then there will be transferring of heat energy, the whole system will try to attain equilibrium temperature in which the temperature of hot body will decrease and the temperature of cold body or the calorimeter containing water will increase. so here the heat from the hot body is entering the calorimeter. Let us assume that the temperature of colt water is increasing from T1 to T2. This means that the temperature of the container or the calorimeter in which the water is kept will also increase from T1 to T2. so, here we need to realize that some heat is used is increasing the temperature of the container.
Now I am going to explain that from where water equivalent is coming.

For this you need to understand the heat capacity.
Heat capacity of the material is the amount of heat required or released to change the temperature of the material of mass m by 1˚. so, the calorimeter is made of the copper whose temperature is increased from T1 to T2.
so, Total heat use by the container = ms(T2 - T1)
so, water equivalent is the mass of the water having same heat capacity as that of the container.
mˊc(T2 -T1) = ms(T2 - T1)
c = specific heat capacity of water
mˊ = ms/c
where mˊ = water equivalent
Now the question arises that why we are adding it with the given mass of water. This is so because we know that
Heat lost by hot body = Heat gained by calorimeter
= Heat gained by water filled in container + heat gained by container material
heat lost = mc(T2-T1) + heat gained by container ................eq(1)
where m = mass of water
c = specific heat
In above discussion i have shown you that Heat gained by container = Heat gained by water equivalent
SO, eq(1) changes to
Heat lost = mc(T2-T1) + mˊc(T2-T1) = (m+mˊ)c(T2-T1)
where m = mass of water
mˊ = water equivalent

Reply

Satyam

17/01/2013 1:06am

PLESE HELP IN SOLVING THIS PROBLEM?????
A Cooper Cube Of mass 200g slides down a rough incline plane of inclination 37° at a constant speed.Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in temp. of the block as it slides down 60cm. Specific heat capacity of cooper = 420J/kg-K

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Admin

17/01/2013 2:16am

Dear Satyam,
Here a cube is moving on the rough incline plane. so, we will get two components of weight. These are -
1. mg cos Ɵ perpendicular to the inclined plane.
2. mg sin Ɵ parallel to the inclined plane.
Here in this question it is given that the surface is rough, so friction force is acting on the body.
Also it is given that the body is moving with constant velocity. This means that acceleration is zero or the force acting along the inclined plane is zero. This means that friction force is balancing mg sin Ɵ.
friction force = mg sin Ɵ
As cube slides down 60cm. Then work done by the friction force or we can say loss of mechanical energy = friction force x distance
work done = mg sin Ɵ x 0.60 J
where m = 0.2 kg
Ɵ = 37°
In the question it is given that loss of mechanical energy goes into the copper as thermal energy.
so, Thermal energy = Loss of mechanical Energy
Thermal Energy = mcT
where c = heat capacity of cooper
T = increase in temperature
so,
mcT = mg sin Ɵ x 0.60
T = (g sin Ɵ x 0.60)/c

Dear Ashish,
If you are facing problem in solving any question, then please tell us.
we will give you full explanation about that.

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ruthwik

07/02/2013 7:59am

how to do q-4 (b)??

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Admin

08/02/2013 4:12am

Dear Ruthwik,
In the first part we see that the water and ice attains equilibruim.
The heat released by the water will be used by the ice to change into water but the heat released by water is not adequate to change the whole ice into water. Let us assume mass m of the ice changes into water.
so,
m x latent heat of fusion = Heat released by water
Now calculate m

answer of ques 15 in h.c. verma is 3000 and by ur method it is coming 333.33 J.....is this the correct solution?

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Admin

16/02/2013 4:47am

Dear Akash,
It may be possible that answer given in the book.
Still i am telling you how we solved this question. Please go to page no 148 in HC verma part i.
There you will see the formula for loss of kinetic energy in in elastic collision.
And we know that loss kinetic energy = thermal energy developed.
You may check yourself.

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ritesh kumar

22/02/2013 1:27am

please calorimetry ka pura solution de

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priyanka

28/03/2013 7:12am

pura solution to diya hua hai. ab kya chahiye?

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manvi

03/03/2013 7:00am

Can you explains me questions for short answer, question no. 2 , 4, 5, 6 and 10 .

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manvi

03/03/2013 7:05am

Can you please explain me questions for short answer questions no. 2 , 4, 10.

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Admin

04/03/2013 3:24am

This service will start from 13th march. Please wait till then.

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Abhishek

26/08/2013 9:44am

cannot find..????????

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rana vishwajeet

29/03/2013 1:28am

it is very helpful site

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rana vishwajeet

29/03/2013 1:35am

please explain me question no.4

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rana vishwajeet

30/03/2013 7:19pm

please explain me question no. 5

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theyatin

03/05/2013 8:51am

dear,
as we know the water is cooling down so it is transferring its heat to atmosphere hence we need to find total heat tranfer and then rate of flow of heat secondly.
and as rate of flow of heat is = Q/T
thus T=total heat /rate of flow of heat

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abhi

02/05/2013 4:08am

please explain 4Q PART A why the tempertature chages to 0 degrees of 200 ml

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theyatin

03/05/2013 8:43am

dear,
as the minimum temperature attained can be freezing point that is 0 when energy absorbed by ice to melt is lesser than the energy of ice thus whole cube will not melt.

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himank

30/05/2013 12:38am

i m new to this site. i am unable to get full answers of calorimetry. please help. plus i dont know if the comments i submit are able to reach to u

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theyatin

30/05/2013 8:04am

dear,
we try to answer al questions asked here.
but in case we miss your question you are free to ask any time again.

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himank

30/05/2013 12:45am

finally got it. thanks a lot

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karan randhawa

27/08/2013 9:42am

please help me sove the first problem

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theyatin

02/10/2013 8:02am

dear,
as heat lost by one object = heat gain by other object.
now heat gain by water is
heat of iron+heat of aluminum
that is
mass of iron* change in temp * specific heat of iron + mass of aluminum * change in temp * specific heat of aluminum

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shabd

01/10/2013 9:24am

sir, Is heat a conserved quantity?

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theyatin

02/10/2013 8:15am

dear,
no heat is not conserved its transient energy.

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shabd

01/10/2013 9:58pm

sir, after solving the que.6,i got the temperature 85.5*C but its different from given ans...
Plz give solution

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theyatin

02/10/2013 8:08am

dear,
calculations in given solution is right plz check if you have done any mistake.

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shabd

01/10/2013 11:30pm

sir, plz give the solution of Q.16 of exercise.

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theyatin

02/10/2013 8:37am

dear,
the question is that we calculate mechanical energy that later changes into 40% of heat
so first we get mechanical energy then we gets its 40% and then we have change in temperature
from change in heat=specific heat *mass*change in temperature
so we can find change in temperature

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shabd

02/10/2013 8:18pm

plz, show how i get mechanical energy

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theyatin

03/10/2013 9:53pm

dear,
mechanical energy=K.E.+P.E.
here we only have P.E.
so we can find P.E. in each case
thus we will have mechanical loss by subtracting it.

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shabd

02/10/2013 8:20pm

plz give the solution of Q.18

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theyatin

04/10/2013 8:18am

dear,
loss in energy=potential energy
thus =mgh
now heat gained by both matters
mass of block*specific heat of block* change in temperature +
mass of water* specific heat of water* change in temperature= 4.707
thus change in temperature *( Mb * Cb + Mw *Cw)=4.707

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Sid trivedi

15/10/2013 6:31am

how can
I get formulas of physics...

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theyatin

16/10/2013 10:23am

there are handbooks for formulas of physics in market you can purchase them
or you better study them in books and make notes.

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Ashutosh

08/11/2013 4:00am

in ch-25 objective-I Q-1,
specific heat (s)=q/m(dt)
where q=heat ; m=mass ; dt=change in temp
hence the answer should be (a),(b),(c)

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theyatin

08/11/2013 6:01am

dear,
this is change in heat. . .
specific heat means specifically for a particular material hence it does depends upon material.

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manas

06/12/2013 10:24pm

Thanks for these precious solutions and better too.

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sumanta

15/12/2013 12:25am

i want to know the true meaning of water equivalent

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theyatin

17/12/2013 8:11am

dear,
please share the whole line or paragraph you have read water equivalent only then i can explain true meaning

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Rahul

19/12/2013 8:47am

where i Get the solution of exersise of chapter 25

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theyatin

22/12/2013 8:08am

here you have solutions of exercise

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shubhu

13/01/2014 10:53am

In q7 what is the temperature at which thermal equillobrium is reached

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theyatin

14/01/2014 9:51pm

dear it is not required for this instance.
as well we dont have all quantities to find equilibrium temp

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theyatin

20/01/2014 10:52pm

dear, ok
let us say the equilibrium temperature is 40' now what is use of it?
where do you think you would use it when we have calculated it.

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Nikhil Sharma

21/01/2014 6:42am

in Q16 where did mass m go

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shubhu

22/01/2014 5:30am

Dear mass get cancelled all over . Don't see solution before solving it on your own dear!!!!

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theyatin

23/01/2014 9:14am

dear you can ask again if you want to..

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sumit

09/02/2014 10:53am

sir please tell me ques 1of chapter 25 i used final temp as Q but i got wrong

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theyatin

10/02/2014 9:54am

dear where did you use temp as Q
Q is sign of heat you cant use it in place of temp

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shiva

11/02/2014 1:01am

A calorimeter contains some ice 2.1kg of heat is required to increase the temp of the calorimeter and its contents from 270K and 69.72kg of heat is required from 272K to 274K.Find the mass

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swikriti

12/02/2014 4:56pm

Inside an oven,there is high temp.we can put hand in it unless we touch anything.but since air inside is also of high temp, why is not hand burnt,plz provide me answer

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theyatin

14/02/2014 7:47am

dear,
heat is transfered through medium if medium is densed it will transfer more heat,
and if its gas it will transfer lesser heat.
hence when you touch hot solid at same temp you get more heat on your hand
but at same temp of air you get lesser heat.

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shubhu

15/02/2014 9:27am

Sir , i have a question . Force between two identical charges placed at a distance R in vacuum is F now a slab of dielectric constant 4 is inserted between two charges if the thickness of slab is R/2 then the force will become between th two charge ==
Sorry for puting it here!!

Reply

sham

05/04/2014 5:09am

Reply

Aman Kumar

09/04/2014 10:42pm

Solve this question:
An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and length l. Left part of the cylinder contains 1 mole of helium and right part contains 0.5 mole each of helium and oxygen. If initial pressure of gas in each part is P°. Calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.

Reply

Aman Kumar

09/04/2014 10:42pm

Solve this question:
An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and length l. Left part of the cylinder contains 1 mole of helium and right part contains 0.5 mole each of helium and oxygen. If initial pressure of gas in each part is P°. Calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.

Reply

Aman Kumar

09/04/2014 10:42pm

Solve this question:
An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and length l. Left part of the cylinder contains 1 mole of helium and right part contains 0.5 mole each of helium and oxygen. If initial pressure of gas in each part is P°. Calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.

Reply

Aman Kumar

09/04/2014 10:44pm

Solve this question:
An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and length l. Left part of the cylinder contains 1 mole of helium and right part contains 0.5 mole each of helium and oxygen. If initial pressure of gas in each part is P°. Calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.

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roamy

24/04/2014 9:58am

sir in ques 4 of chp 29 how we come to know that whole ice wont melt

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Archit Wadalkar

28/04/2014 8:14pm

Solution of Q15 is incorrect.
BY WORK ENERGY THEOREM,
Change in KE=Work done.
so W=1/2*(m1m2/m1+m2)*(u1-u2)2
=1/2*(20*10/30)*30*30
=3000J

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devansh

30/06/2014 11:30am

Answer of 15th question is not matching with the answer given in HC Verma......you have given 333.3 but in the book it is given 3000

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devansh

30/06/2014 11:39am

Sir, in question 13th you have given a different answer while HC Verma has given a different one........your answer is 4030.33 , in HC Verma it is 4000......plus help.....

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rishabh jain

17/10/2014 10:43pm

I m sorry sir...most of ur solutions r incorrect.....for example solution of question number 15...kindly correct ur solution....u can take my help if required...

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ASHIF GAZI

31/10/2014 4:56am

listen rishabh..when they have applied conservation of momentum-there they have made a big mistake.they have not taken opposite sign of the other ball...by taking opposite sign the final velocity comes out to be 10m/s...then apply rest of the steps given in solution....ur ans will be 3000....thank u

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Parth Sarthi Sharma

22/11/2014 8:11am

many solutions are wrong... like ques 15 and 18 trust me...

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mekha

23/11/2014 7:05am

How to do ques 3

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mekha

23/11/2014 7:11am

I am not able to understand the way u solved sir..can u help me ??

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Anirudh

10/12/2014 12:20am

I like the solution which will help me in studies ! (y)

Thank you So much! ~ Team _my college Bag_

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Mansha Kalra

28/12/2014 2:03am

Hello sir
This is with ref to Q7
I would like to know how you found the amount of steam that was converted to water.
I mean, why did you divide the heat used by ice by the latent heat of fusion of ice?

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sneha

03/01/2015 3:48am

would u please explain me question no.13 of HC Verma of calorimetry chapter

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sneha

03/01/2015 3:57am

would u please explain me question no.13 of HC Verma of calorimetry chapter

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vikas

14/01/2015 2:32am

answer of Q-15 of calorimetry has been given 3000J in my book , but here, it is 333.33J , which one is correct ???

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Harshit Shroff

26/01/2015 10:37am

Can u pls elaborate the heat gain in question number: 1!!!

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ak

06/03/2015 3:19am

why the temp is take 293 nd 373 in heat gain in question1