Chapter wise solutions to H C Verma’s Concepts of Physics Part 1

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.

  1. CHAPTER 1
  2. CHAPTER 2
  3. CHAPTER 3
  4. CHAPTER 4
  5. CHAPTER 5
  6. CHAPTER 6
  7. CHAPTER 7
  8. CHAPTER 8
  9. CHAPTER 9
  10. CHAPTER 10
  11. CHAPTER 11
  12. CHAPTER 12
  13. CHAPTER 13
  14. CHAPTER 14
  15. CHAPTER 15
  16. CHAPTER 16
  17. CHAPTER 17
  18. CHAPTER 18
  19. CHAPTER 19
  20. CHAPTER 20
  21. CHAPTER 21
  22. CHAPTER 22


218 thoughts on “Chapter wise solutions to H C Verma’s Concepts of Physics Part 1

    1. 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.

    2. 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……

  1. @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 ]

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

    Please elaborate…

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

    2. 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….

  2. 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.

      1. 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

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

    1. 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 )

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

    1. 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…..)…..

    1. 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 ==


  3. 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..?

    1. 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!!

    1. 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….

  4. in vol 1,chapter 5, – 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

  5. hwcum in prob 21 there is no tension 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…………… concept

  6. 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????

  7. 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.

  8. 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

  9. 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.

  10. 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

  11. 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

  12. 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

  13. 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.

  14. @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)

  15. 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………… is upto u…………..n’jpoy….!!!!

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

  17. 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

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

  19. 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

  20. 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

  21. this 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 still nt bad

  22. 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!

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

  23. 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…!

  24. 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

  25. can anyone answer this question..plz mail it to my email address

    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?

  26. 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 ???

  27. Sir in chap 16 ques no 14. Can we take average velocity between t1 and t2
    ans is coming by this method also

  28. 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.!!!

  29. In chapter 6 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?

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