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sudhir sinha

31/10/2012 7:54pm

Reply

Admin

04/11/2012 11:46am

This is attempt to provide free education to children who are preparing for engineering entrance examination.
If you are having any doubt regarding solving questions of HC Verma or understanding any topic related to this, you may leave your comment.
we will give your answer within 24 hours.
Thank You

Reply

Tanish

01/10/2013 11:00pm

Can u please explane me the 7 th questern

Reply

govinda

28/01/2014 7:20am

sir i didnt get the way tat u how u are taking the resultant so can u explain this by taking Q3 part b

Reply

m.y

16/02/2013 3:43pm

plz expln Ques 3 how take distance i dnt undrstand

Reply

Admin

16/02/2013 6:23pm

Dear m.y,
Please see the diagram given in the question.
In part a,
the mass m is placed at point O which is the mid point of AB. Now let the side of equilateral triangle = 2a
OA = a
OB = a
OC = √(AC² - OA²) using pythagoras theorem
OC = a√3

In part b

Now here please note that length of side of equilateral triangle = a
So, height of equilateral triangle (h) = √(a² - (a/2)²) = (a√3)/2

Now we know that the mass m is lying at the centroid. And centroid is at at the distance of 2h/3 from the vertex.
so, OA = OB = OC = 2h/3 = a√3

Now you can solve question as given in the solution.

Reply

govinda

28/01/2014 6:59am

sir here h =aroot3/2 then 2h/3 will be aroot 3/3 bt u hav taken aroot 3 only y

govinda

28/01/2014 7:08am

sir i also didnt understant how the resultant is taken in this question given above so can u explain this in detail ............

ruby

16/02/2013 4:01pm

plz expln ques 5

Reply

Admin

16/02/2013 6:41pm

Dear Ruby,
In order to move due to mutual gravitational force the particles should be placed symmetrical.
There are some mistakes in the question.
Please the diagram given
AC = diameter = 2R
AB = AC = c
AC² = AB² + AC² = 2c²
AB = c√2 = 2R
c = R√2
Now you may find F(cb), F(cd), F(ca)

Taking point c to be as origin,
We resolve the force along x and y direction and find the resultant force.
resolving forces along x direction F(x) = - F(ac) cos 45 - F(cd)
Net force along y direction F(y)= F(cb) + F(ac) cos 45

Net force = √((F(x))² + (F(y))² ) acting along AC.

Net force = mv² /R
find v.

Reply

Sameer

29/12/2013 8:32pm

Sir, i dont get it how Diagonal ac and side ab are equal. Moreover, the f(cb) and f(cd) should be equal. I am unable to understand sir. Please help me out with detailed explanation.

I am unable to understand cocept of integration in Q-8

Reply

theyatin

19/03/2013 3:12pm

dear rahul,
the concept of integration id to calculate whole quantity considering an infitesimal element of whole quantity and calculating whole quantity from one point to other point.
(a real life example of integration can be when you cook rice you just take one or two pieces of cooked rice and press them to see if they are prepared or not and you dont check whole cooker of rice.)
here we ned to calculate force over half circle so wee take interation from 0 to Pi
so we consider force on one element and calculate it over whole semicircle.

Reply

ranjna

24/03/2013 1:47am

Sir can you tell me questions for short answer of questions no. 12, 6 and 7.

Reply

theyatin

24/03/2013 12:15pm

dear ranjna
what's the problem with numerical 6???
its simple formula GM/(R+h)^2
r+h is distance from center of body to the particle whose acceleration we need to claculate.
where R is radius of body(planet)

Reply

theyatin

24/03/2013 12:38pm

in numerical 7
we consider two objects at rest iniitially that means their potential energy is maximum and kinetic energy is zero
as they will move toward each other potential energy will decrease and kinetic energy will increase
as final momentum is zero so whole energy is conserved
thus P.E.+K.E.(initially)=P.E.+K.E. (at any instant)
so P.E. initial=-13.34*10^-9 as calculated in numerical
K.E. initial= 0
later when bodies are 0.5m apart then
P.E. when 0.5m apart= -13.34.*10^-9/2
K.E. of first body= 1/2*M1*v1^2
K.E. of second body=1/2*M2*V2^2
so P.E.+K.E(initial)= P.E. +K.E.1+K.E.2
as given earlier that v1=v2 thus
you can replace v1 by v2 in this equation and thus you can find the other one

Reply

ranjna

26/03/2013 2:22pm

Sir, I have asked you (questions for short answer ). I want to just discuss them.

Reply

theyatin

29/03/2013 11:45am

dear ranjna
at noon sun and earth pull the objects on the earth's surface in oppoiiste direction( as you are facing sun earth attracts the body towards i self and sun towards it self) at midnight, the sun and the earth pull in same direction(as you are facing opposite to sun thus earth and sun both attract you towards center of earth that is why you feel heavy at night time and you rbody tends to sleep also its harder to work physically at midnight)
so the answer is yes if we use spring balance it will weight more at mdnight and lesser on noon.

same thing happens when there is full moon you feel lesser gravitation and your body is lightened in a very very small way
which may g oun noticed if you dont other.
and so you feel heavier on dark nights.

Reply

theyatin

29/03/2013 11:51am

ranjna,
in short answer 7
insect inside the apple falling from tree finds that earth is moving towards it but actually appleis moving towards earth it is just relativity from earth apple is moving from apple earth is moving.
but the question is that who exerts the force needed to accelerate the earth(actually apple) with this acceleration of g.
the answer is gravitation exerts force on earth(actually apple).

Reply

Prabodh

06/12/2013 10:36pm

there is pseudo force acting between earth and apple

theyatin

29/03/2013 11:58am

dear,
in short answer 12,
if radius of erth decreases by 1% means its 99%R now or 0.99R
from g=GM/R^2
g si directly propotional to R^2
thus g is proportional to 0.99R*0.99R=0.9801R^2
so change in gravity= R^2-0.9801R^2
=0.0199 or 1.99%

Reply

manvi

26/03/2013 9:40pm

Sir can you please explain me objective 1 questions 1 2 and 3 .With reason

Reply

theyatin

29/03/2013 1:19pm

dear,
1.the acceleration of moon w.r.t. earth is given (rather than gravity of moon or acceleration of some other particle on moon)
sp i f moon is stopped it will start moving towards earth with a aceeration that is of moon W.R.T. earth that is 0.0027

Reply

theyatin

29/03/2013 1:28pm

2.
as given acceleration of earth a=Gm/R^2=10
-------------------------------- of moon a=Gm/r^2=0.0027
so relative acceleration is
Gm/(R+r)^2= Gm/ (R+R/4)^2
Gm/R^2(1+1/4)^2
=10/(5/4)^2 (as Gm/R^2=10
=6.4

Reply

theyatin

29/03/2013 1:32pm

3.
when a man goes from earth to mars there are three phases
1. when man is close to earth it feels gravity nearly equal to 10 so weight is 600N
2. when man is at half of the distance between mars and earth that is gravity of both planets becomes so weak tha man feels almost no gravity thus no weight so it tends near about zero
3. when man is near mars
its feels gravity about 4 so the mass of man tends to 240N.
so the graph goes from 600 n to nearly zero and then to 240N

Reply

manvi

28/03/2013 8:25pm

Sir please explain me questions no . 13 , 33 34 of exercise.

Reply

theyatin

29/03/2013 12:33pm

dear manvi,
in question 13
in first case where the mass m' lies in between hollow sphere and outside solid sphere.
because gravitational field is zero inside spherical shel the particle will face only gravitational force due to solid sphere and that is E=G*m*d/r^# (where d is distance of particle from surface of sphere
thus F=E*m' that is F=G*m*m'(x-r)/r^3
b)
when particle is away from sphere but still inside the shell
E=G*m/d^2
so F=E*m' that is F=G*m*m'/(x-r)^2
c)
now the particle is out of the shell too and faces gravitation due to both object shell as well as sphere so,
F for sphere as is given in b) F=G*m*m'/(x-r)^2
and E for particle outside shell is E=G*m*m'/(x-R)^2
so F for shell is F=G*m*m'/(x-R)^2
ading both we have F=G*m*m'/(x-R)^2+G*m*m'/(x-r)62

Reply

theyatin

29/03/2013 12:53pm

dear manvi
in 33
if the satelite moves with same angular velocity as of earth the satelite is geostationary stelite
thus d=(G*M*T)^1/3 / (4*pi)^1/3
from this relation u can find distance time t=24hrs
b)
from north pole to equatorial line its 1/4th of total distance (circumference of earth) if angular velocity of both objects is same then angular distance covered over same time would be same
so to cover 1/4th of distance the satelite will take 1/4th of the time of whole day that is 24/4=6hrs :)

Reply

theyatin

29/03/2013 1:13pm

in 34
let us assume true weight of MGo=10N
(where Go= actual gravity)
and the given wieght in sateliite is MG
from relation G=Go(1-2h/R) multiplying both sides by mass of object
MG=MGo(1-2h/R)
MG=10( 1- (2*42 km/6400) (as distance of geostationary satalite is 42km from earth)
=10 (1- 0.9868)
MG = 0.23 N

Reply

Agney

12/06/2013 7:27pm

Sir cant we do the Q no 34 by force method.

manvi

30/03/2013 1:11am

Sir can you please explain me questions no 35, 36 and 37.

Reply

theyatin

30/03/2013 1:16pm

dear manvi,
colattitude means the latitude which makes same angle to both sides of the circle ( here in case earth) so as the geomatry tells us that a tangent drawn from a point to circle is perpendicular to the radius.
as the radius of circle (earth) is given and the distance is given as distance of geostationary satalite.
so here we have sin as altitude over hypotnese.
so sin (phi) = Radius/ satalite distance

Reply

theyatin

01/04/2013 9:20pm

dear
the first formula given in the solution is equation of time period of orbital of a planet
thsu you can square equation and extract g out of formula and then have value of g

Reply

theyatin

01/04/2013 11:14pm

dear manvi,
in Q 37
as you know total energy of object is K.E. +P.E.
as in space P.E. is gravitation felt by the particle
and
as in space the K.E. at the maximum height is zero
so P.E. becomes total energy and as R=3600 ND HEIGHT IS 6400
so here R+H= 2R
thus in relation of energy H is replace by R
as total energy which is P.E. at maximum hieght
is equal to K.E+P.E
equating both equations we can find V^2 from K.E.=1/2 m V^2

Reply

kiran

30/03/2013 1:13am

Sir please explain me objective 2 questions no.4. And objective 1 questions no . 7 , 8 and 9 .

Reply

theyatin

30/03/2013 8:59pm

dear kiran,
as the work done is force*distance
thus for a complete revolution as the distance covered is zero thus work done is zero.
as force is due to angular motion thus angle (theta) comes into play
as so where ever cos (theta) is zero as 90' 270' the work done will be zero
so work done can not be zero at any small part of orbit but at some particular angles
hence option a and d are not applicable but b and c are.

Reply

theyatin

31/03/2013 12:18am

objective 1 q 7
according tome both options are right
as the graph in both case starts from maximum point but initially there is no graph apparently when distance is less then radius of shell as the field as well potential both are zero

q 8.
in case of solid sphere the graph in both cases starts from zero increase linearly w.r.t. distance and then after maximum it starts decreasing but never meets zero.

Reply

theyatind

31/03/2013 12:24am

dear,
in q 9
option A) because gravity is maximum on poles so there can not exist any other point where gravity is equal to g there.
b) similarly gravity of g on equator is lesser than actual gravity so there can be some point away from earth where gravity is equal to equator as moving away from earth surface the gravity decreases.
so only first is true and second one is wrong.

Reply

ranjna

30/03/2013 1:42pm

Derive an expression of gravitational field strength of a uniform disc of radius r, surface mass density £ , at an axial point situated at a distance x from the centre of the disc.
Sir please explain me this question.

Reply

theyatin

31/03/2013 12:29am

dear ranjna,
it is very difficult to drive whole derivation here as its difficult to write signs and equations can you please specify whats your problem particularly as in which step. . .

Reply

ranjna

30/03/2013 1:48pm

Sir please explain me the graph of variations of potential with distance from the centre of the shell . And also explain me the graph of variations of gravitational field due to a solid sphere with the distance from the centre.

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theyatin

31/03/2013 12:05am

dear,
consider potential on y axis and distance from center of object on x axis
for shell
the graph is zero until the distance is equal to radius of shell at the surface of shell the graph is at its maximum and afterward it starts decreasing but never zero until infinity and it varies inversely to the square of distance (1/(r^2) from center of object.
for spherical shape
at the center distance is equal to zero so the graph starts from zero and as distance from center increase the graph increases respectively as the potential is directly proportional to distance (r)
at distance equal to radius it is maximum similar to shell
also similar to shell the potential decreases w.r.t. distance as inversly to the square of distance (1/(r^2))

Reply

Sumit

30/05/2013 12:05am

plz explain Exerceises 28 qntn..

Reply

theyatin

30/05/2013 8:49pm

dear,
a) speed of ship would be same as rotation of earth.
b) T0 = mg – mw^2 R
T0 – mg = mw^2 R
c)
putting angular velocity as
(v-wR)^2/R^2
that is difference of velocity of ship w.rt. earth.

Reply

Rakesh

08/06/2013 12:35am

very helpfull site sir

Reply

Rakesh

08/06/2013 12:39am

sir can you explain me Q5 of short answer

Reply

theyatin

08/06/2013 1:22pm

dear,
moon revolves around sun too as earth revolves.
as the distance of moon from earth is very small as compared to sun,
thus force between earth and moon is more effective.

Reply

Rakesh

08/06/2013 1:36pm

then why the other planets revolve around the sun even they are at a very large distance from the sun

Reply

rakesh

08/06/2013 1:53pm

sir this is the question
even there are other planets close to earth but they revolve around the sun and not the earth?

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theyatin

08/06/2013 10:13pm

dear,
distance as well as gravitational force matters.
in case of other planets they are far away from earth so as to get affected from earth's gravitational force.
as gravitational force of sun is large enough to make planets at large distance revolve around it.

Reply

faizan

21/06/2013 1:06pm

sir,i need d explaination of Q2

Reply

theyatin

07/12/2013 10:46pm

dear,
first we need to calculate gravitational force vetween o and a
then o and b
then o and c
then o and d
now we need to find resulatant between a and d
we find F(of) that is using pythagorus theorem
similarly resultant between b and c
now we need to find resultant between F(of) and F(oe)\
ths is sqrt( Fof)^ -+(Foe )^2+ 2 X Fof X Foe (cos 180)
hence you have your answer

Reply

faizan

21/06/2013 1:07pm

sir,i need d explaination of Q2

Reply

theyatin

07/12/2013 10:49pm

dear,
first we need to calculate gravitational force vetween o and a
then o and b
then o and c
then o and d
now we need to find resulatant between a and d
we find F(of) that is using pythagorus theorem
similarly resultant between b and c
now we need to find resultant between F(of) and F(oe)\
ths is sqrt( Fof)^ -+(Foe )^2+ 2 X Fof X Foe (cos 180)
hence you have your answer

Reply

Aymanzoor

28/10/2013 10:53pm

Sir
what is the formula for integration used in q 9
regarding [(x^2 + d^2)]^ -3/2

Reply

Satya Prakash Nandy

19/11/2013 9:34pm

Question 2

Reply

theyatin

07/12/2013 10:50pm

dear,
first we need to calculate gravitational force vetween o and a
then o and b
then o and c
then o and d
now we need to find resulatant between a and d
we find F(of) that is using pythagorus theorem
similarly resultant between b and c
now we need to find resultant between F(of) and F(oe)\
ths is sqrt( Fof)^ -+(Foe )^2+ 2 X Fof X Foe (cos 180)
hence you have your answer

Reply

jatin

24/11/2013 8:43pm

in Q8 y df3=df ?

Reply

theyatin

07/12/2013 11:01pm

dear,
df3 is wrongly typed
df'=2 df
coz of considering two elements at same angles simuntaneuosly on both sides of normal.

Reply

jatin

24/11/2013 8:50pm

2df*

Reply

akriti

08/12/2013 7:48pm

If I want a full solutions of the problems of hc verma how can i get that

Reply

theyatin

11/12/2013 10:15am

here.

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SHIVANI SHARMA

17/12/2013 11:42pm

SIR PLEASE EXPLAIN ME IN QUESTION NO.7 OF SHORT ANSWER TYPE,SHOULD NOT THE INSECT EXERT THE FORCE ON THE EARTH TO ACCELERATE THE EARTH? ALSO IF THERE WAS NO APPLE AND ONLY INSECT WAS FALLING,THEN ALSO THE INSECT IS EXERTING THE FORCE?
SO I THINK THAT THE INSECT SHOULD EXERT THE FORCE ON EARTH. AM I RIGHT?

Reply

Alok Kumar Sharma

18/12/2013 11:05am

You are partly correct. Acctually, the insect as well as the apple exert force on the earth, however this force is not enough to accelerate the earth by such a large amount as the insect experiences. The reality is that the insect is watching the earth from an accelerated frame of reference, hence, the from "Insect's frame fo reference" there is a pseudo force acting on the earth, which is responsible for the earth's acceleration, as seen by the insect.

Reply

theyatin

20/12/2013 8:51pm

very good dear i appreciate your effort.
keep going help each other..

theyatin

20/12/2013 8:49pm

dear,
everything exert some force on things surrounding it.
amount of force exerted by the object depends upon its mass.
and effect of this force on the other particle depends upon the mass of other particle.

Reply

SHIVANI SHARMA

17/12/2013 11:48pm

SIR PLEASE EXPLAIN ME IN Q.6 OF SHORT ANSWER TYPE ,THE WEIGHT OF THE OBJECT AT THE 2 PLACES WOULD BE DIFFERENT BUT WE SAY THAT THE SPRING BALANCE NOTE THE SAME READINGS AS IT IS RELATED ONLY TO THE UPWARD FORCE ON THE SPRING?

ALSO SIR WHAT IS THE ANSWER OF Q.4 OF SHORT ANSWER TYPE ?

Reply

theyatin

20/12/2013 8:44pm

dear,
the difference is actually very small so as to be measured by a spring balance
so it displays similar weight.

Reply

shivani sharma

23/12/2013 7:20pm

sir,I have a very big confusion that in questions of 2 stars coming close to each other and are going to collide,why we conserve the total energy of the system and why not of only one star?

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theyatin

28/12/2013 1:12am

dear,
because each star exerts force on other so we cant consider only one start to conserve energy.

Reply

aditya

09/01/2014 7:33am

SIR,how to solve que13

Reply

theyatin

15/01/2014 11:00am

dear,
in first case where the mass m' lies in between hollow sphere and outside solid sphere.
because gravitational field is zero inside spherical shel the particle will face only gravitational force due to solid sphere and that is E=G*m*d/r^# (where d is distance of particle from surface of sphere
thus F=E*m' that is F=G*m*m'(x-r)/r^3
b)
when particle is away from sphere but still inside the shell
E=G*m/d^2
so F=E*m' that is F=G*m*m'/(x-r)^2
c)
now the particle is out of the shell too and faces gravitation due to both object shell as well as sphere so,
F for sphere as is given in b) F=G*m*m'/(x-r)^2
and E for particle outside shell is E=G*m*m'/(x-R)^2
so F for shell is F=G*m*m'/(x-R)^2
adding both we have F=G*m*m'/(x-R)^2+G*m*m'/(x-r)62

Reply

anjini

19/01/2014 6:19pm

Sir in objective 2, question 1 all options are given as correct.Can you please explain why option (B) is correct?how can potential be zero when field intensity is non-zero?

theyatin

21/01/2014 11:30am

dear anjali plz ask in new comment i cant answer here

govinda

28/01/2014 7:34pm

sir explain in detail of short question 4,objective1-1,2and ex-5 please do this

dear,
total work done= work done due to distance a +work done due to distance 2a

Reply

angel

05/02/2014 2:08pm

sir can u explain hcv ch gravitation objective 1 Q3 to13
as well as objective2

Reply

theyatin

06/02/2014 7:09pm

dear,
please ask questions one by one
and you can ask 3 question max per day so visit regularly.

Reply

Saraswati

20/04/2014 6:57pm

Can we solve 7 and 8 problem without using integration

Reply

chirag jain

26/04/2014 10:18am

sir can u plz explain solutions of gravitation exercise Q-7,8&9?

Reply

ER.MIRJAIK

04/05/2014 4:24pm

In Queshtion.no.7 we consider two objects at rest iniitially that means their potential energy is maximum and kinetic energy is zero
as they will move toward each other potential energy will decrease and kinetic energy will increase
as final momentum is zero so whole energy is conserved
thus P.E.+K.E.(initially)=P.E.+K.E. (at any instant)
so P.E. initial=-13.34*10^-9 as calculated in numerical
K.E. initial= 0
later when bodies are 0.5m apart then
P.E. when 0.5m apart= -13.34.*10^-9/2
K.E. of first body= 1/2*M1*v1^2
K.E. of second body=1/2*M2*V2^2
so P.E.+K.E(initial)= P.E. +K.E.1+K.E.2
as given earlier that v1=v2 thus
you can replace v1 by v2 in this equation and thus you can find the other one

Reply

er. mirjaik

04/05/2014 4:28pm

In ques.no.8 we use the concept of integration id to calculate whole quantity considering an infitesimal element of whole quantity and calculating whole quantity from one point to other point.
(a real life example of integration can be when you cook rice you just take one or two pieces of cooked rice and press them to see if they are prepared or not and you dont check whole cooker of rice.)
here we ned to calculate force over half circle so wee take interation from 0 to Pi
so we consider force on one element and calculate it over whole semicircle.

Reply

mahesh

04/05/2014 7:18pm

sir pls explain Q.21

Mahesh

04/05/2014 7:23pm

sir pls explain Q.21
Explain it sir

Reply

ER.MIRJAIK

10/05/2014 9:34am

DEAR MAHESH
first field in the given region is to be represented along the vector component then from the resolution of the forces we have to get angle subtended along the field line as from the statement it is clear that field direction and displacement vector are orthogonal to each other when particle move along the line hence situation of no work done

Reply

niti n

08/05/2014 10:25am

sir, cud u exp me d Q8 via taking dL nd nt by d(theta).....AT D EARLIEST!

Reply

ER.MIRJAIK

10/05/2014 9:17am

DEAR NITI
the concept of integration id to calculate whole quantity considering an infitesimal element of whole quantity and calculating whole quantity from one point to other point.
(a real life example of integration can be when you cook rice you just take one or two pieces of cooked rice and press them to see if they are prepared or not and you dont check whole cooker of rice.)
here we ned to calculate force over half circle so wee take interation from 0 to Pi
so we consider force on one element and calculate it over whole semicircle.

Reply

Adi

05/06/2014 9:36pm

Sir can you please explain Questions 2 and 3, with formulae used clearly given. Thanks sir

Reply

Sagar j.

19/06/2014 9:42pm

sir can u plz explain excersise 4 question
i dont understand tht inclusion of negative sign in Force cb part
thank you

Reply

admin

23/06/2014 9:45am

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