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

Author:  Published Date: 28 Feb 2010 Comments:  Comments

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

SEE PART TWO FOR MORE

Author: physicsfundamentals

Comments

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

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

    4. please read the question once again. coin is dropped when elevator just starts.
      therfore u=0.

    6. 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……

  5. @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…

    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….

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

      2. 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.

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

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

    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 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

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

    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 ==

      DHAWAL HARKAWAT

  28. 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..?

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

  34. 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 ?

    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!!

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

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

    3. 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….

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

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

  44. 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

  47. 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

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

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

  51. 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!

  53. 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????

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

  59. 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.

  61. 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

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

  64. 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.

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

    1. Solutions are available here. Better buy the book if you are asking about the original Concepts of Physics by H C Verma

  72. 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

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

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

  77. 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

  79. 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

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

  86. 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.

  88. @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)

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

  96. 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….!!!!

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

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

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

  107. 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

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

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

  113. hey guys can any one help me out with : chapter 10
    exercises question 19 and 20?

  114. 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

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

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

  119. 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

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

  127. 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

  129. 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 better..bt still nt bad

  130. – realy very Help Full ! =)
    I also like iT !
    I.E Irodov ka bhi solution mil sakta hai kya..???

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

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

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

  135. 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!

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

  136. 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…!

  137. 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

  138. 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

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

  142. 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?

  143. 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 ???

  145. 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

  146. 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.!!!

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

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

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

    volume 1 -chapter no 3 – Q21—(pg 52)
    <<<>>>

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

  169. 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?

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