Problems from **H C Verma**‘s **Concepts of Physics** is considered a must work out assignment by most of the **IIT **aspirants.

Here you can find the solutions to the problems chapterwise. The downloads are based on the old edition of HC Verma’s Concepts of Physics. In case the book is revised, the answers and solutions may not match. Students will have to use your senses to find out. We cannot reproduce the questions for reasons known to all.

If you donot have a copy of HC Verma’s concepts of Physics, **Buy it here at the lowest price.**

The solutions can be downloaded in pdf format by clicking on the links below.

really really useful site i have ever visited! i like it very much

rit

check problem 52 chapter 3 ans.given is wrong

it is really a useful site for weak students.

refer irodov

irodov is a little tougher but ya we can clear jee with this book

so i think you are really weak.

very helpful site

In volume 1, chapter 3, problem no 31, why is the intial velocity of the elevator taken as zero?

because the elevator starts its motion at the moment when the coin is dropped. we consider the motion of elevator when coin is dropped.therefore u of both elevator and coin is taken zero.

arey yaar isme to pta ni kahan kahan se formula's utha k lga rkhe hain………really……..

BECAUSE IT JUST STARTS FROM REST .

please read the question once again. coin is dropped when elevator just starts.

therfore u=0.

initial velocity of elevator is taken as 0 because elevator is starting from rest!

bcauz in problem they have said that A PERSON IN ELEVATOR DROPS A COIN AT THE MOMENT THE ELEVATOR START… in problem they consider this condition hence the initial velocity(u)is zero……

@Siri, Read the question very carefully, It says "a person in the elevator drops the coin at the moment the elevator starts." So u = 0(a body always start from rest… )

Can u help me with Laws Of motion Problem no.28.. In the solutions the equation for the third object is taken as

[t/2 – m3g – m3(a + a') = 0 ]

Why?

Shouldn't it be [m3g – t/2 – m3(a + a') = 0]

Please elaborate…

check…it has relative acceleration a2,which is acting in upward diection……it,'s easy if you apply contraint relation…….

you can take is anyway…. it's just the matter of direction… in the end if u chose the wrong direction… u will get a -ve sign with ur answer….

acceleration of block is a+a' . force is m(a+a') and is taken on left.

i hav such problems in ths books which are best examples for concept build up.

and i want to have my clear concepts on those problem.this website helped me so much in this situation.

thanks

sollutions are presented in an excelent way

bt sum ppl say doing this book only will not help u crack jee…………..:(((

bt i am not among those ppl…

i really liked this book. it not only clears the doubt of students but also generates interest of student towards the physics

do this book once then go for resnick nd halliday. but don't read theroy of resnick hallidayas its theory is above jee level. problems are enjoyable

Are you bloody crazy? Resnick and Halliday has a very, VERY lucid theory, definitely not above JEE level.

You must be crazy.

incredibly good site…….really helpful…concepts of physics – hc verma !!

thanks.!

wonderfulll

thanks,

its nice …………..

REALLY HELPFUL THANX A LOT………………

I can't open it?

Every link here is opening. If you can't probably you may not be having the adobe reader or any other compatible software. You can download adobe reader for free from http://www.adobe.com

its so good…that u r so idiot and u will not able to beat me in entrance exams

thanks a lot

SOLUTIONS TO HC VERMA ARE REALLY,, USE FUL FOR EVERY ENTRANCE EXAMS

thanx a ton!! :D

ch 5 ques. 33, phy. volume 1 m not able to eliminate r,t and a……help me out…

Mg – T =Ma —— A

T – Rsin* = M'a ———–B

Rcos* = mg —-c

Rsin* – ma = 0 ———D

D in B

T – ma = M'a

=> T = (m+M')a ———-E

E in A

Mg = (m+M'+M)a ———F

c / D

tan* = a/g => a = gtan* ———G

G in F

Mg = (m+M'+M) gtan*

M(1- tan*) = M'+m

M= ((m+M')tan*)/ (1- tan*)

M = (m+M') / (cot* – 1 )

ideal book for iit-jee

ch 6, q 22, b part.. 12 newton is not shown on lower block…y so?

when 12 N is applied on lower block…..there are two forces in horizontal direction..so…..12N-friction force=ma………(note……in friction force..the friction is exerted by 2kg mass so…m=2kg)…but in ma ……m=4kg

srry silly mistake of mine…..pls dont bother about it..

q-27…..ch 7..vol. 1… y normal is taken towards center????

the normal acts along the radial direction……………that is because the person is pushed towards the walls.hence the walls provide a force in the opposite direction….towards the center……….(actually its a cylinder…..)…..

it's wonderful.

kmhg

its a wonderful site

excellent…..!very helpful!

accurate(2 quite n xtent) n well organised..!

tnx!

its not accurate.some questions are wrong….i mean ans are given in a wrong fashion

anybody can explain Q 14 CHAPTER 9 VOL 1 PLS PLS PLS PLS

see uday,

first , in these type of problems , there is a shift in the centre of mass of the body– and whenever these type of problems comes, we generally use the formula —

dispalcement = CM1 – CM2

now in this situation , consider , the ballon as the origin (0,0) and the monkey's position to be (L,0) or (0,L) {whatever you please} as the monkey is away from the origin by a distance L (the length of the rope)…

so first of all , let us find the CM (centre of mass) when monkey is not moving i.e at rest —

so CM1 = (M1X1 + M2X2 / M1 + M2) = ml / M + m

WHERE M1 – mass of monkey

M2 – mass of ballon

X1 – L

X2 – 0

Now as soon as monkey starts moving towards the ballon ,, and finally reaches there ..,,, so X1 = 0 as new position of monkey becomes (0,0)

so ,, CM2 = 0

by applying the above mentioned formula ,, we grt the answer as —–

ml / M + m

where the symbols have been put according to the symbols used in the question,,,,,

if then also any problem comes,,, then please ask me ==

DHAWAL HARKAWAT

a person brings mass of 1 kg from infinity to a pt A .initially the mass was at rest but moves at a speed of 2m/s as it reaches A .the work done by person on the mass is -3 J .the potential at A is..?

I don think I could hv solved all d qstns on my own. Thanks a lot…

one who solved both the books completely i can write and give he will crack iitjee!!!

very-very usefull and helpful

THANKS ALOT THT WAS REALLY VERY HELPFUL….

its nice

i have a doubt in chapter 9 .3rd sum

if am wrong please notify me

what i think is :the mass i distributed along the length but,

the solution given is link http://plustwophysics.com/wp-content/uploads/2010…

according to the solution the entire mass is concentrated in the point L/10,L/2.————->just look for this m(L/2+L/10)

am i correct ?

Luk vivek’

wt u r sayin is correct!!

bt as d rods r unifrm each of d rod has its COM at L/2.

(u cn asume d mass 2 b concntratd at COM)

& then u cn easily gt d positn of COM of whole systm!!

gr88

when a light beam passes through the glass slab first it reduceses it velocity and the get it again why this happends

why

VELOCITY DEPEND ON REFRACTIVE INDEX GLASS TO AIR REF. INDEX DECRSES SO VELOCITY INCREASES

take it this way…. u r runnin at ur max speed on a non crowded track… and u r runnin at ur maximum speed on a crowded track… obviously max speed on non crowded track > max speed on crowded track…. this is because density of the crowded place is more than density of non crowded place…. same happens in the case of light…. when it goes to a denser medium its max speed i.e 3.8 x 10 raise to power 8 decreases….

ap ko achhi tarh se soive karna chahiye

can we also find the solutions to the objective type questions?if anyone knows…..pls let me know.

gud solutions but i need shorter methods to solve d questions……..

tnx

in chapter 7 prob no 26 hw is v cos theta/2=ucostheta/cos(theta/2)

aray yar solutions ka pecha mat karo concepts ka pecha karo solutions jak mar kay tumharay dimag mai ayai gay

Chap 3, Ojective 1, Question 13. The answer is given as (b). I think it should be (d). Is this correct?

in vol 1,chapter 5,Q.no:22 – why is the acceleration of mass m1 taken downward when its mass is less than that of m2…….it should accelerate upward in such a case ???please explain

mass m2 is only acting downwards & not m1…..

is it possible to get solutions for d c pandey?

i like this site.its very helpful

hwcum in prob 21 there is no tension shown..how the force (T) is taken in upward direction………….and then how is it equated to tension acting on lower block…….lot of mistakes have been done…………just they are trying to bring the answer……………..no concept

volume 1, chapter 9, Q 37….in the solution….what is l and a ????

volume 1, chapter 9, Q 37…in the solution given…what is l and a ????

best book on physics for IIT-JEE preparation………………………………….

hey why don’t u try resnick haliday’s fundamental physics

it is one of the finest book of haliday of pittsburg universty..why dnt u try shaumns series..

Hey can anyone help me wid this question

Q>42 chp 3 volume one

cudnt get it how the writer to the solutions derived the formula….

help plzzz!

ITS VERY VERY HELPFUL…

THANKYOU

in chap 12 q 1 the angle taken out as phi/6 but when magnitude of acc is taken out then in shm formula the angle is taken as phi/3 why????

GIVE THE SOLUTIONS OF I.E IRODOV

The solutions to Irodov are available at http://irodovsolutions.blogspot.com/http://plustwophysics.com/get-complete-solutions-to-irodov-problems/

Done completely…

hc verma is very nice

Baaaaa……….

hc verma is very nice

thnxxxxx……….

……….this site helped me whn I needed a guidance at most

H.C.Verma, the best book of theory for IIT-JEE preparation….& secnd 2 Irodov as whole……..

I think a faster way to solve problem no 60 chapt9 will be to conserve energgy and momentum along X-axis = 0. Momentum along Y – Axis shall not be conserved since an external reactin force N is being applied on the system.

in chapter 10 Q no. 40 mei m1f*r3g 1 and 3 is in subscript… what is it

in chapter 10 Q no. 68 mei rod ka moment of inertia about an end mr2 le kar solve kiya hai… magar ml2/3 karne par kuch aur ata hai….2 is in superscript

PLS ASK VERY SHORT ANSWER QUESTIONS TO ME BCOZ THEY ARE VERY GOOD TO SOLVE THAN EXERCISE

Irodov is d best book for IIT-JEE……

I think the answer to problem 28 in chapter 5 is wrong. In the solution it is written :

T = 1g – 1a2 = 0 …(i) from fig (2)

which is impossible because the block moves up with acceleration a1.

Let me rephrase that. The answer given is correct (of course) but the method of solving I find to be wrong.

Sir, I want the solution of sir H. C. Verma but I dont know how to get this. Can u plz give me ur mobile no.to talk in detail. Plz…

i want derivations for formulas

they are vry hlpful bt where cn i gt d explanations 4 objective 1 n 2

can ne1 help in chapter 13 question 5 part b

where can i download hv verma volume2?

Solutions are available here. Better buy the book if you are asking about the original

Concepts of PhysicsbyH C Vermachp-10,ques 25,b)-wats wrong in using work done=torque X radian angle(teta)

Is it right t0 say that centrifugal f0rce is a reacti0n t0 centripetal f0rce.In s0me b00ks it is given the same.But acti0n and reacti0n acts 0n tw0 different b0dies in c0ntact and centripetal and centrifugal f0rces act 0n same b0dy

can anyone tell me that chapter 5 Que 7 how come the force f and force F are in the direction of the block

provides a good conceptual grip…for preparation of various exams.

i love this book……….its just awesome…………………….hats off to you verma sir….

can anyone explian me d solution for chaptr 15 24th sum hw dd thy gt T’

gr8 gr8 gr8 gr8 gr8 links and SITE…..

i just have 1 question : what is the difference between the triangle addition of vector law and the parallelogram addition of vector

it’s very good . dis is for all weak n intelligent students . thnxxxxxxxxxxx 2 h. c. verma

fantastic

see suyash………….

in traingle addition of vector law :-

if the two vectors AB & AC are strating from the same pt.A their resultant will be in

and the resultant of BC.

but in parellelogram law of vector addition:-

the resultant of AB & AC will be orginating from

from pt.A making some angle with AB and AC

ya! its really very helpful site

THANKS TO PUBLISHER OF WEBSITE IT VERY USEFULL SITE

thank you

very help full information

for quest 42..chptr 3 of vol1..is der any odr mthd wich dsnt include eq of trajectory?

IT HELPS A LOT TO GET THE IDEA!!!!!!! TO SOLVE THE PROBLMS….!!!! REDUCES THE STRESS…#!!!$#!!!$#!

in chapter 6(friction) in question 29.i dont understood what is written there in the solution.

chapter 6 problm 21 2 mistakes

1)at 3 last step it should b U-1 not u+1

2)T means max force where is tension because if no tension the block will nt experience = force from both sides and wl accelerate.please verify eating my head off.

very very useful

@Dimple Chouhan

Yep there is a much better way:

Use the vertical distance to calculate time (here initial velocity is zero, since it has only horizontal velocity)

Then use the time to find the horizontal velocity using the minimum horizontal distance (40cm)

relly helpful 1!!!!!

really nice work .. i had been looking 4 dis !!

really helpful site…but stdnt shd try der bst to solve dese prblms….in case very tough que…1 shd rfr dis st…

very helpful site for weak students!

very true but we should try it first then look out for solutions

this book is really the best book for IITJEE preparation!! :)

site is wonderful for iit aspirants

c’mon guys its just HC VERMA………..nd u ppl. need help for this silly book………have u ever seen the standard of he book……….its shit………..try out irodov or krotov……..nd u will cum to kno where u ppl stand………..if u guys r preparing for IIT………just forget thinking that u wud ever clear the entrance with this buk…………..rest is upto u…………..n’jpoy….!!!!

hahahha this book is enough guru dont boast :P

hmmmm…..really…erodov nd all….r vry tuf !

:(

can anyone tell me from where i can download arihant dc pandey optics solutions…

Very helpful thankyou

THANK YOU SO MUCH. REALLY HELPS A LOT.

can i get answers

for short answer

this is an amazing site i have ever visited………..Thank you^1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 Very much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

wow guys i think myself lucky enough to find this site thank you very very……………………….. much

someone please help me solve part1-3rd chapter-49 pilot problem

thank you… i am greatly benefited!!!

is the spotted line elevator’s floor?ifyes,then what is the darken line over which the man is standing?plz help me.

THIS PROBLEM IS FROM CHAPTER-3 NO-31.

yes…you are right…the eqn for the third body should be as u put it…they just made a mistake there…even i was gonna ask that

can any1 tell me what is initial velocity??

Hello,

I am a student, and i was facing trouble in some questions of HCV, reached to this site, really awesome explained answers, thanks.

can anyone tell me that is there any site for asit das gupta’s problem plus book?

this book is not sufficient for iit..

it can help you clear the cut off

it is very useful for me,,,,thanx,…………

hey guys can any one help me out with : chapter 10

exercises question 19 and 20?

hc verma part1 ch-5 laws of motion

question number 31

can u tell me the answer briefly

specially why there is different acceleration of the body

its really a wonderful n hlpful site…. all d solutions r sooo clear n specified…i luv it…

excellent work!!!!! its really helpful…!!thnxx

hc verma is being very helpful! :) i’ve nicknamed the sums in it…. “worms”. :D

really a nce site…good 1… :D

thnqu ! ;)

if u dont get any problem’s ans go for key

but if u r not aware of the method pls do not watch the key,u will not gain any thing

but not the method of approach

Really….. It’s very helpfull

Is HC verma is enough to crack iit jee pleaseeeeeeee tellllllllllll

sir plz capter viz ouestion phy subj start 1998 to 2011 end bord exam 9423844255 cell no

12 th phy subj bord papers question 1998 to 2011 plzzzzzzzzzzzzzzzzzzzzzzzzzzzz sir

Really helped me to complte my notes….. :)

THANX A LOT!!!! :D :D

IT IS REALLY VERY HELPFUL FOR IIT ENTRANCE STUDY………

sir this is good fr getting the answers if the solution is explained in concept wise irs better to under stand easily and get the concept soon so please arrange the solutions in concept wise

how to find the direction of static friction

Static friction is a self adjusting force and its direction is so as to prevent the

relative motionthis is a nyc site……..especially for those who are preparing fr exams seriously…if a bit more explanation is given…then it wud have been better..bt still nt bad

– realy very Help Full ! =)

I also like iT !

I.E Irodov ka bhi solution mil sakta hai kya..???

The solutions are already posted. Please try a search at this site

it is very useful information over there

would u please tell me the short answer type’s solution(ques. no. 2,some mechanical properties of solids)

post the question please

what about solutions to QUESTIONS FOR SHORT ANSWERS and OBJECTIVES plzzzzzzzzzzzzzzzzz reply

no difference and where can i get solutions to objectives and que for short answers

the solutions given on this site is very complicated….

plzz do it in easiar way…!

i hav seen some known ans. of some question which u hav done it in a very complicated way!

You are always welcome to post alternate methods. Come on! Help the community.

hey its not complicate… then u should refer resnick hallyday 8th edition to clear ur concepts…it will help u to solve the problem….

its not about the concept VIDYALANKAR

ok fine….take chapter no.9 Question no.42

we can directly put the distance travelled along the incline as 0 and solve it…

the answer to this question hardly completes in 4 to 5 line and u have done all that complicated maths calculation….H C Verma is a great book and we should answer it in a very simple way (logically)

but never mind!….some answers are done in a very good manner…!

it’s the most useful site i have ever visited.and now i feel that i don’t need any more tutions.this site is the best tutor

in chapter no.6 of part 1 how the question no.18 is solved.

i mean i don’t understand what is P.

plz help me out

Thanks a lot…..Its really helpful to do the problems.

thanku ……….for you have solution of h.c.verma it is really helpful..

The answers haven’t been explained properly…Not the ideal solutions..

can anyone answer this question..plz mail it to my email address nainam.007don@gmail.com

my question is:

a rod of length L is placed along x-axis at leftmost end its linear mass density is lemda1(^1)

and at right most end it is lemda 2 (^2)

as we move from left to right linear mass density increases LINEARLY from lemda 1 to lemda 2….find the COM of this rod?

I have problem in Chap-5 , Ques 28.

I made equations as follows –

for mass , m1 –>

T – m1g = m1a1

for mass , m2 –>

T/2 – m2g = m2(a1 – a2 )

for mass , m3–>

m3g – T/2 = m3(a1 + a2)

but in the solution only my 2nd equation is correct..

mass m1 is going up so it should be T – m1g = m1a1 .. , right ??

and m3 is going down….. but in sol they have marked its acc. upwards… why so ???

very useful site,,but also want hcv short answer type questions…..

Sir in chap 16 ques no 14. Can we take average velocity between t1 and t2

ans is coming by this method also

v1=v(root)T1/T

v2=v(root)T2/T

GR8 Site this!! n GR8 Stuff!!

i m working for the JEE on my own at home..

So tis was of gr8.gr8.gr8 use 2 me ..i almost did more than half f t chaps f specially physics frm tis…….SO TNKS 2 tis site.!!!

Really helpful site for solutions

Can anyone give some info on the “ISEET” exam this year

thanks a lot.:)

really through it we can easily solve all the question without any problem………….i like it……..

Please explain the problem-68,in chapter-10

sir how to downlaod solutions?and it very good site for solutions

Click on the links (Or right click and choose “save link as” option

really nice site …easy solutions..

really gr8 for all students…..

gr8 book….but i find it a little diff 2 follow d calculus part in a few problems

its good &helpful some solutions r confusing must be made more descriptive

its really good .it will help me in clearing my doubts and i will be able to finish book in quick time

really nice site. All the solutions are given in organised form,easy to understand…

Thank you very much. best site i have ever visited

really useful website ……thank you……

Most useful and very informative for inquisitive students

nice solutions

thnxxxx a lot…

http://sites.google.com/site/physicsclassroomonline/h-c-verma-concepts-of-physics-chapter-wise-solutions पर पूरा सोलुशन हैं

can anyone plz explain this Q. deeply????:-

volume 1 -chapter no 3 – Q21—(pg 52)

<<<>>>

very useful…………. give me the link for .ZIP file

its helpfull

see tne conversion its wrong in chapter no 15 part 1 hcv question no 15 please explain me..

In chapter 6 Q.no-28 Your solution says that Tension force is acting on the mass of block M in the vertical direction and you have written a equilibrium condition for the forces on block M in the vertical direction Where as you haven’t mentioned any tension in the vertical direction on block M in the free body diagram-2. How could you come to such a conclusion with out mentioning in the body diagram?

i didn’t understand the 2nd equation of chap 10 , ques 34 pls help .

I didn’t understood that why the acceleration of block of mass 2M in ch-5 Q-31 is taken as a/2

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