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Reply

GB

23/12/2012 2:51pm

In 28th sum, why is the acceleration of m1 block a2 instead of being a1?

Reply

Admin

23/12/2012 5:48pm

Yes, there is a printing mistake in the solutions. you just take the acceleration of m1 as a1 and solve the question. you will get correct answer.

Reply

prakhar

08/01/2013 6:55pm

in problem twenty eight. from fig 4 the equation for m3 is T/2-3g-3(a1+a2)=0. Why not it is 3g-T/2=3(a1+a2). Because think like this way. m3 will go downwards when released therefore equation should be

3g-T/2=3(a1+a2)

please clear my doubt

Reply

Admin

09/01/2013 4:30pm

Dear Prakhar,
Yes there are some mistakes in the above solution. so I am writing the full solution of question 28.
first we are solving for m1
acceleration of m1 = a1
solving the equation we get
T = g+a1 .....(eq 1)
solving for mass 2(m2),
it is having two accelerations,
a1 = acceleration acting downwards
a2 = acceleration acting upwards
net acceleration = a2-a1
we take a2-a1 because we assume that m2 is moving upwards.
solving the equation for m2, we get
T/2 - 2g - 2(a2-a1) = 0
T/2 = 2g+2(a2-a1) ......eq(2)

solving for mass 3(m3)
it is also having two accelerations. these are a1 and a2 acting in downward direction.

so, net acceleration = a1+a2
solving the equation for m3, we get
T/2 = 3g - 3(a1+a2) ....eq(3)

Now solving equation 1 and 2 by eliminating T, we get

5a1 - 4a2 = 3g ..............eq(4)

again solving equation 2 and 3, we get

5a2+a1 = g . ........eq(5)

now solving 4 and 5, we get
we get
a1 = acceleration for a1 = 19g/29
a2 = 2g/29
acceleration for a2 = a2 - a1 = (2g-19g)/29 = -17g/29
this means that m2 is accelerating downwards with acceleration 17g/29.
acceleration for m3 = a1+a2 = (2g+19g)/29 = 21g/29 (downwards)

Reply

kashyap

25/08/2013 8:56pm

Sir,
Instead of that we can write constraint relations(equations) and solve the problem more easily and in a better way.I tried it in that way and got the answer more easily.

kashyap

25/08/2013 9:04pm

sir,can we do like that in that method or not pls suggest??????

manju yadav

01/02/2013 10:34pm

Q:25 plz shw the diagram with cmponents and ful explnatn i m cnfused

Reply

Admin

02/02/2013 3:51pm

Dear Manju,
In this question there are two blocks on an inclined plane. one on the left side of the pulley and another on the right side of the pulley. Now see that the mass of the both blocks are same but block on the left side of the pulley is at the greater angle ( Ɵ1)than the block on the right of the pulley (Ɵ2).
The sides of the inclined plane triangle are in the ration 3:4:5. so, this means that the triangle given in the figure is right angle traingle which is making 90° at the vertex opposite to 5 m length.
Ɵ1 = angle between 3 m and 5 m line.
Ɵ2 = angle between 4 m and 5 m line.
Now see the figure-1 given in the solutions.
so, sin Ɵ1 = 4/5
sin Ɵ2 = 3/5
Now we are making free body diagram of the bock on left side of the pulley.
For this please see the figure 2 given in the solutions.
There are three forces acting on it. These are -
1. mg = 1g = g = This force is acting in the downward direction.
2. T = tension = This is the tension force due to the string and it acting along the string.
3. R = Reaction force = This force is acting due to reaction exerted by the base on which block is moving.
Now we resolve mg along the plane and we get
mg sin Ɵ1 = g sin Ɵ1
Now calculating the net force acting on the block
Net force = g sin Ɵ1 - T
Net force = ma = a
where a = acceleration with which block is moving
we get,
a = g sin Ɵ1 - T ..........................eq 1

Now let us see the free body diagram of the second block which is on the right side of the pulley. For this please see figure 3 given in the question.
There are also three forces acting on it. These are
1. g sin Ɵ2 = This force is acting along the plane on which block is lying.
2. T = tension in the string
3. R2 = reaction force on the block due to the base.
Net force = ma = a = T - g sin Ɵ2 ................eq 2
Here T is greater because the block is moving upward.
Now solve these two equations and get the answer.

Reply

manvi

12/02/2013 7:23pm

Two identical billiards balls strike a rigid wall with the same speed but at different angles and get reflected without any change in speed . What is the direction of the force on the wall due to each ball if in 1st case the ball is normal to a vertical wall and in 2nd case it is inclined at an angle 30 to the normal to the vertical wall and get reflected with the same angles?

Reply

Admin

13/02/2013 5:39pm

Dear Manvi,
In this question we need to find out the direction of force on the wall. We will solve this question with the newtons second law which says that force = (mv2 - mv1) / t
force is equal to rate of change of momentum. so, direction of the force depends upon the direction of change of momentum.
In the first case when the ball is moving normal to the wall. so,
initial momentum = mv
final momentum after reflecting = - mv
change of momentum in normal direction = mv - (-mv) = 2mv
So, force is acting normal to the wall.

case 2 - when ball was moving at angle 30 to the normal.
Now we will resolve the velocity in the vertical and horizontal direction. velocity normal to wall before collision= v cos 30
velocity parallel to wall before collision = v sin 30
Velocity normal to wall after collision = - v cos 30
velocity parallel to wall after collision = v sin 30

change of momentum normal to the wall =
= mv cos 30 - (- mv cos 30) = 2 mv cos 30

change in momentum parallel to wall = mv sin 30 - mv sin 30 = 0

so, we get change in momentum in the direction normal to wall. so, here also force acting on the wall will be in the direction normal to wall. But the difference is that the magnitude is smaller as compared to above case.

Reply

manvi

13/02/2013 7:52am

Force is not always in the direction of motion. Depending on the situation , f may be along v , opposite to v , , normal to v or may take some other angle with v. In every case it is parallel to acceleration.can you please give me example of each.

Reply

Admin

13/02/2013 6:15pm

Dear Manvi,
Yes the force is not always parallel to the velocity. Then in this case we resolve first resolve the velocity in the direction of force and in the
normal direction of force. Velocity which is acting parallel to force will change and velocity which is perpendicular to the force will not change. You may see the example of projectile motion where velocity in the horizontal direction remains constant and velocity in vertical in changing due to acceleration g.

Reply

manvi

14/02/2013 1:45am

Thanks.

manvi

14/02/2013 3:20am

A stone of mass 0.25 tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?

Reply

Admin

14/02/2013 5:35pm

Dear Manvi,
This question is just based upon the concept of centrifugal force.
We know that centrifugal force = mrw²
where m = mass
r = radius
w = angular speed
w = 2πrN/60
where N = Rotational speed = 40 rev/sec
Now we know that centrifugal force acts along the radius and away from the center. Here string is acting as radius. so, centrifugal force is acting as tension in the string.
Tension = mrw² = mr(2πrN/60)²
where N = 40 rev/ sec

For second part again use
Tension = mrw²
Put tension = 200 and
calculate N(rotational speed) and w(angular velocity)

Reply

manvi

14/02/2013 3:25am

A batsman deflects a ball by an angle of 45 without changing its initial speed which is equal to 54 km/hr. What is the impulse imparted to the ball? (mass of ball is 0.15 kg)

Reply

Admin

14/02/2013 5:52pm

Dear Manvi,
Let us assume that the ball is moving in the positive x direction with 54 km/hr.
The ball is deflected through 45˚ through x direction.
Now resolving velocity
Final velocity in x direction = 54 cos 45˚
Final velocity in y direction = 54 sin 45˚

Initial velocity in x direction = 54 km/hr
Final velocity in y direction = 0 km/hr

Now Impulse = change in momentum
Impulse in x direction = change in momentum in x direction
Impulse in y direction = change in momentum in y direction

change in momentum in x direction = mv - mv cos 45˚
where v = 54 km/hr
m = .15 kg

Impulse in x direction (I(x)) = mv - mv cos 45˚
Impulse in y direction(I(y)) = mu - mv sin 45˚ = - mv sin 45˚
here u = initial velocity of ball in y direction = 0

Now we get two impulse I(x) in x direction and I(y) in y direction.
Net impulse = √((I(x))² +(I(y))² )
Now you can find the net impulse.

Reply

manvi

15/02/2013 9:13am

I am unable to understand the concept of tension can you please explain me. For example 5.4 and 5.3 on pg. No. 68 and 69.

Reply

Admin

15/02/2013 5:43pm

Dear Manvi,
Tension if just a type of force. For example - there is a string which is attached to the wall and you are pulling it with the some force and also you not moving backward. So, now imagine that there should be some force which is balancing your pulling force. This force is in the string which acting opposite to you and it is called tension in the string. Do not get confused with the electromagnetic force that is given on page 68.
This force is same as that of spring force that acts when we pull the spring. Same as here when you pull the string then the reaction force will develop in the string that will oppose your applied force to maintain the equilibrium. So, it is just a type of reaction force in the string. Also Tension has some maximum value above which the if we apply the force then the string will break.

Reply

manvi

16/02/2013 10:04am

How can we get answer for question for short answer , objective 1 and objective 2. By download from above i am not getting answer.

Reply

Admin

16/02/2013 6:25pm

Dear Manvi,
We haven't uploaded the solution for short question answer. You may ask if you are facing problem in understand any question.

Reply

manvi

17/02/2013 3:35pm

Plz explain me questions no. 7 ,30 and 31.

Reply

Admin

18/02/2013 6:33pm

Dear Manvi,
Explanation for question 7 -
In this question the experimenter has applied force F on block A.
Now see the diagram in the solutions.
Free body diagram for A. There are three force acting on A in the horizontal direction. These are -
1. f = This is the applied by the experimenter.
2. F = This is the reaction force by block B and A and is acting towards the left direction.
3. ma = This is the pseudo force. As the block A is accelerating towards right direction then pseudo force will act towards left.

Now you can form the equation as given in the solution.

Free body diagram for Block B -
There are two force acting on block B in horizontal direction. These are -
F = This is the reaction force by A ob B and is acting in the right direction.
m(B)a = This is the pseudo force and is acting towards left.
Though this we can form another equation.

Reply

Admin

18/02/2013 6:47pm

Dear Manvi,
In this question let the blocks of one Kg are moving with acceleration a.
Let us see the free body diagram of the blocks as shown in the solution.
T = tension in the string
Free body diagram for horizontally placed block.
1. T = tension in the string
2. ma = a = pseudo acting towards right

free body diagram for vertically suspended block.
1. T = tension upward
2. mg = downward
3. ma = a = pseudo force acting upward.

Now form two equations and solve the answer.

Reply

Admin

18/02/2013 7:03pm

Dear Manvi,
Here also we need to understand the free body diagram of the two blocks
a = acceleration of block M.
1. For block 2M - This block will move with acceleration a/2 in the right direction because if M moves by x distance down then 2M will move by x/2 distance.
There are two forces acting in the horizontal direction. These are -
a. T = tension towards right.
b. (2M)a/2 - This is pseudo force acing in right direction.

for block M -
1. mg = downward
2. T = tension upward
3. ma = pseudo force upward.

Now you form equations an solve part a.

part b = You can also find

Part c= for this part see the diagram given in the solution.
R = force by clamp on pulley. = Resultant force of tension on pulley as shown in diagram.

Reply

Rajat

20/02/2013 4:34pm

How to we deal with the questions based on throwing a particle from a moving bus? And its various cases like when bus is moving down the incline with acceleration or constant velocity..what is the method to determine in which direction we must throw the particle to land back on the bus in above cases?

Reply

Admin

20/02/2013 5:39pm

Dear Rajat,

Whenever you are solving any question which an object is moving with some angle then solve the question resolving the components in the x an y direction and solve them differently and at last use the result of one equation into another.
For example if a bus is moving down an inclined plane with some angle then there will be motion in x as well as y direction . Now calculate them and solve them.

Reply

Admin

23/02/2013 2:27pm

The doubt solving service for HC verma has been disabled for few days. We are very happy with the faith the students have shown in our service. We are hiring more and more teachers and upgrading our features like audio video techniques to solve your doubts.

Reply

Kushagra

27/03/2013 10:21pm

Q12

Reply

theyatin

28/03/2013 11:04am

a) at any depth let the ropes make angle θ with the vertical
From the free body diagram
F cos θ + F cos θ – mg = 0
(here F is force applied to each rope and then force applied to the weight would be at some angle θ depending on position of the object) so for both ropes each Fcosθ is added and as weight of the body is in downward thus its subtracted...
mg=2F cos(theta)
As the man moves up. θ increases i.e. cos decreases. Thus F
increases.
( it means more force is needed to pull the man up as man moves up)
F=mg/2 cos(theta)

2) now we need to calculate force applied to a man for any depth H from ground level.
H can be any depth where width of trench is D
so considering force applied by one rope lets assume angle of object is θ from horizon.
thus from width D we have D/2 as height of the triangle and H as base of triangle thus apllying pythagoras theoram we have hypotnese of this triangle
as we know cos θ= base/hypotnese
thus in solution we can find cosθ
now from previous equation as we have
F=mg/2cosθ
put value of cos in this equation we have equation for force acting on body at any depth H

Reply

Kushagra

28/03/2013 8:22pm

thanx

kushagra

28/03/2013 8:23pm

Q.29 plz....

Reply

theyatin

28/03/2013 10:42pm

to solve this question we need to devide whole system into two systems as the one system of pulley is of m2 and m3 and other system is m1 and (m2+m3)
now,
m1 should be at rest.
T – m1g = 0 ( as there is no acceleration as well gravity because both cancel out because m1 is at rest so acceleration and gravity both are equal and opposite in direction)
T = m1g …(i)
T/2 – 2g – 2a1 = 0 ( tension on rope because of pulley of m2+m3)
T – 4g – 4a1 = 0 …(ii) ( multiplying whole equation by 2)
T/2 – 3g – 3a1 =0 (tension on rope of pulley due to weight m2 or m3)
T = 6g – 6a1 …(iii) ( multiplying whole equation by 2)
From eqn (ii) & (iii) we get
3T – 12g = 12g – 2T
T = 24g/5= 4.8kg.
Putting the value of T eqn (i) we get, T=m1*g=4.8g ( here g is gravity)
m1 = 4.8kg

Reply

Kushagra

28/03/2013 11:30pm

Thank you

Kushagra

28/03/2013 11:11pm

m not able to get the Q 31.... why acceleration is half in the case of 2M..

Reply

theyatin

29/03/2013 11:08am

dear kushagra
here as you can see in image the mass 2m is joined to A through B
that is there are two pully as the mass M is attached to A and the mass 2M is attached to B that is again attached to A thus the rope is doubled thus if mass M moves a distance suppose D then mass 2M will move only D/2. so acceleration is also half as much of M.

Reply

KUSHAGRA

28/03/2013 11:26pm

in Q31 only.why tension is twice of 2m...?
.

Reply

Kushagra

28/03/2013 11:27pm

can't i write as 2Ma=T for Q.31 only..

Reply

theyatin

29/03/2013 11:25am

no dear you can only write it as 2M*(a/2)-2T=0 => MT-2T=0
as acceletration is half and tension on rope of B is twice as much because of twice of mass as M.

Reply

Kushagra

29/03/2013 2:00pm

THanks

Kushagra

29/03/2013 3:47pm

Q39,33plz..........

Reply

theyatin

29/03/2013 9:07pm

dear kushagra,
if only mass M' moves and m doesn't then acceleration of m should be equal to acceleration of M' (just like water in bucket which doesn't fall even if you rotate it upward downward,because angular acceleration of water eliminates gravity)

T + Ma – Mg = 0 ( tension on mass M)
...(i) (From FBD – 1)

T – M'a – R sin θ = 0 (tension on mass M')
...(ii) (From FBD -2)

R sin θ – ma = 0 (equation for acceleration on mass m)
...(iii) (From FBD -3)

R cos θ– mg =0 (equation for acceleration due to gravity on m)
...(iv) (From FBD -4)
now subtracting ...2 from ....1
we can eleminate T ...5
R sin θ = ma
R cos θ= mg
dividing these equation and then putting value of g in ....5
we have M=(M'+m)/(cot θ-1)

Reply

theyatin

29/03/2013 9:31pm

dear,
now 39
1. Given, Mass of man = 60 kg.
Let R' = apparent weight of man in this case.
Now, R' + T – 60g = 0 [Tension From FBD of man]
T = 60g – R'
T – R' – 30g = 0 [Tension from FDB of box]
60g – R' = R' – 30g = 0 [ equating both equations]

R' = 15g The weight shown by the machine is 15kg.

2.
as in above case there was no motion so thus no acceleration but in this case the man must move so as to maintain his original weight.
From the FBD of the box
let R be its weight on the machine
T – R – 30g – 30a = 0 (as the mass of box is 30)

T – 60g – 30g – 30a = 0 (as the weight of man is 60g on the machine)

T – 30a -90g

T = 30a – 900 (as gravity is 10)
...(1)

From the FBD of the man

T + R – 60g – 60a = 0

T – 60a = 0 [as R = 60g]

T = 60a
...(2)

From eqn (i) and eqn (ii) we get T = 2T – 1800 T = 1800N.
So, he should exert 1800 N force on the rope to get correct reading.

Reply

pankaj

03/04/2013 10:41pm

dear admin.
In explanation of part (a ) of Q. 34, i.n fbd 2 , why is tension taken in the same direction of 4g.

Reply

theyatin

04/04/2013 2:05pm

dear,
first the tension for 5 kg is determined.
and then tension for one side of horizontal weight is determined that is why tension is taken as half of its value. (t/2)
and
because the tension is in same direction as of acceleration of horizontal weights.

Reply

Shubhankar

04/04/2013 11:54am

please help me in understanding problem number 30 chapter Newton's laws of motion. From HC Verma

Reply

theyatin

04/04/2013 2:10pm

dear,
the motion of both objects is in different direction thus
firstly we consider
T+1a=1g (as here object is moving downward thus gravity is also considered
thus T-1a=0 (object is moving horizontal)

Reply

Shubhankar

04/04/2013 3:07pm

So the tension would be same throughout the string? The part of string attached to the mass which is on the table would have different tension?

Reply

theyatin

04/04/2013 8:25pm

dear shubh
yes the tension throughout the string is constant or it is considered constant depending upon uniformity of the string.
but there is no possibility that string would have more tension where it is bound friction is different phenomenon but tension is constant throughout

T + 0.5a – 0.5g = 0 .... 1
T1 – 0.05a – 0.05g = 0 ...2
T1 + 0.1a – T + 0.05g = 0....3 (tension due to 500g=T1 is in same directio to the motion of 100g of weight and the tension T due to 50gm is in oppositte direction to the motion of 100gm so it has negative sign)
above 3 eq. for tension are respectively for wight 50gm 500gm and 100gm
taking tension on a side from first two equations we have
T = 0.05g + 0.05a ....4
T1 = 0.5g – 0.5a ......5
respectively and putting values of T and T1 in ...3 we can solve it further to get a while g=10

Reply

Shubhankar

04/04/2013 9:05pm

thank you for your support theyatin.

Reply

Tanya Singh Rathore

05/04/2013 2:44pm

sir,
can you please explain me Q 13?? I'm not getting it.

Reply

theyatin

06/04/2013 9:33pm

dear tanya,
its as simple as that
mass of A=0.5kg
gravity=10
accelration=2
as lift is moving downward so accelration will be subtracted so
0.5 (10-2)= m a = 4N

Reply

ayushi

14/04/2013 9:49pm

please exp ques no. 16

Reply

theyatin

14/04/2013 10:00pm

dear,
as you can see from free body diagram we have two equations
first we multiply with appropriate constant and then substract or add both equations to extract T from b
now we put value of T in any of equation to find acceleration as we know value of g that is gravity. ..

sir can u explain the free body diagram in q.no 16 in detail

Darshita

21/04/2013 11:20am

dear sir,
what are the explanations to questions for short answers 4,6 and 7 from nlm in concepts of physics (hc verma).
thank you.

Reply

theyatin

21/04/2013 9:33pm

dear
if no force is applied a particle can undergo only linear motion. . .
but in case we see from other frame of reference answer may differ,
imagine you are on marry go round which is rotating, you roll a ball outward in straight line according to your view the ball is under linear motion. . .
but from a man sitting on ground the ball may have a curved motion. . .

Reply

theyatin

21/04/2013 9:37pm

dear,
in Q6
imagine you are in car and you through a ball upward according to you the ball undergo a straight up and down motion. . .
but from a man standing at ground the motion is parabolic.
hence actual motion of ball is straight up and down and there can be no effect on motion of ball thus there is no such frame but for ideal motion of ball we need to consider such inertial frame of reference.

Reply

theyatin

21/04/2013 9:42pm

dear,
Q7
no it is not neccessary as the object is far away from any external force. . .

Reply

Darshita

22/04/2013 11:07am

thank you:)

Darshita

23/04/2013 12:23pm

Sir, can you please help me in the following questions:
Q. A string passing over a light frictionless pulley carries at its ends two variable unequal masses where sum of the masses (m) is constant. If the breking tension of the string is 15/32 of the weight of the sum of masses (a) what is the least acceleration (b) what is the least value of the greater mass
Q. a spring force constant k is cut into two pieces such that one piece is double the length of the other .then the long piece will have what force constant?
Q. A physicist hanged a cylinder shaped container of base area 100 cm square to a spring. he slowly poured water into the container and found that the surface of water remained at the same level with respect to ground. Find the spring constant k of the spring. Take density of water as 1000kg/metre cube.
thank you

Reply

Shubham

23/04/2013 7:58pm

Plz explain Q5 why T-ma=0

Reply

theyatin

24/04/2013 5:00pm

it is tension applied ob b
that is total force on B is T - ma=0 where acceleration is in opposite dirction from static momentum of B.

Reply

Shubham

27/04/2013 1:23am

Didn't Get you..Plz explain properly and in detail

Shubham

27/04/2013 1:23am

Can't Understand.Explain in Detail Please

Shubham

27/04/2013 1:23am

Can't Understand. Explain in Detail Please

Shubham

27/04/2013 1:25am

Reply

Shubham

27/04/2013 1:24am

Can't get you plz explain in detail

Reply

theyatin

28/04/2013 10:22am

Dear Shubham,
please see the free body diagram for both block A and B.
figure 2 is for block A
fig 3 is for block B

now assuming that both block are moving with constant acceleration a.
for block A
there are two horizontal forces
these are
1. F = acting toward right
2. T = acting towards left
net force = F - T = ma

for block B,
There is only one acting in horizontal direction.
1. T = acting towards right

so, we get net force = T
ma = T
Now you can solve the question

Reply

AKHIL

05/05/2013 11:47pm

please help me with quest no q 33 )

Reply

theyatin

09/05/2013 12:24am

dear,
if only mass M' moves and m doesn't then acceleration of m should be equal to acceleration of M' (just like water in bucket which doesn't fall even if you rotate it upward downward,because angular acceleration of water eliminates gravity)

T + Ma – Mg = 0 ( tension on mass M)
...(i) (From FBD – 1)

T – M'a – R sin θ = 0 (tension on mass M')
...(ii) (From FBD -2)

R sin θ – ma = 0 (equation for acceleration on mass m)
...(iii) (From FBD -3)

R cos θ– mg =0 (equation for acceleration due to gravity on m)
...(iv) (From FBD -4)
now subtracting ...2 from ....1
we can eleminate T ...5
R sin θ = ma
R cos θ= mg
dividing these equation and then putting value of g in ....5
we have M=(M'+m)/(cot θ-1)

ask me again if you have any doubt. . .

Reply

ankit

15/05/2013 2:02pm

in question no.10 why have you shown kx=f

Reply

theyatin

16/05/2013 11:10am

dear,
as F=kX
here f is force and k is coefficient of elasticity,
X is displacement this formula is to assign motion of SHM in spring system.

Reply

Rajat

15/05/2013 5:52pm

In Q.35, why have we taken two different tension values, T and T1 ?
Can't we just assume the tension "T" is same throughout the system ?

Reply

theyatin

16/05/2013 11:36am

dear,
because of in system there are two different kind of tensions
one is due to hanging weight and other is due to dragging weight in middle.

Reply

Vishal

02/06/2013 7:21pm

Can u plz explain Q.7 properly with a FBD on blocks

Reply

theyatin

02/06/2013 8:30pm

dear,
as it is not possible to draw FBD here in this comment box.
as FBD are given in the solution.
Now see the diagram in the solutions.
Free body diagram for A. There are three force acting on A in the horizontal direction. These are -
1. f = This is the applied by the experimenter.
2. F = This is the reaction force by block B and A and is acting towards the left direction.
3. ma = This is the pseudo force. As the block A is accelerating towards right direction then pseudo force will act towards left.

Now you can form the equation as given in the solution.

Free body diagram for Block B -
There are two force acting on block B in horizontal direction. These are -
F = This is the reaction force by A ob B and is acting in the right direction.
m(B)a = This is the pseudo force and is acting towards left.
Though this we can form another equation.

Reply

Ragib Khan

13/06/2013 4:24pm

plz help me in Q 16---y there is a plus sign in second equation before 3a...and a minus sign in thefirst equation..before 1.5a...explain in detail plzzz

Reply

Admin

13/06/2013 7:31pm

Dear Ragib,
Yatin sir is not well. Please wait for 1-2 days, he will answer your question.

Reply

theyatin

16/06/2013 1:56pm

dear,
you can ask only 3 question per day.

Reply

theyatin

16/06/2013 2:00pm

dear,
you can ask only 3 question per day.

Reply

Ragib

16/06/2013 10:56am

could u plz exppan objective 1 questions 5,6,7 on pg 77

Reply

theyatin

16/06/2013 2:10pm

dear,
Q5
the answer is inertia inertia pushes you forward,
there is no particular physical force which pushes you forward but your own momentum that is equivalent to car. but when momentum of car decreases you feel forward push.
Q6
no actual inertial exist so as we need to suppose this frame to understand certain problems to understand exact motions of bodies.
Q7
if there is no object to compare motion of the observed object's motion thus it is not necessary that the frame would be inertial.

Reply

Ragib

17/06/2013 10:29am

sir i was asking the objective questions 5,6 & 7 which are on page 77

Reply

theyatin

17/06/2013 7:12pm

dear,
i have answered these questions.

Reply

ragib

17/06/2013 8:41pm

dear sir but u have answered the short question answwrs......here i am talking about the objective questions which have four options....y cant u understand???

Reply

theyatin

19/06/2013 9:31pm

sorry dear.
my fault. . .

Reply

theyatin

19/06/2013 9:39pm

dear,
5.force exerted by plane on the box is horizontal component or x component of the weight so it is given as
mg cos theta.
6.here horizontal component of the force must equalize weight of the box to stop it from slipping down.
so Fcos theta=mg
thus F=mg/cos theta
7. in case the man does not apply any force downward the man will be standing on the place he is. in case he applies force downward like walking or jumping it will fly upward tangentially.

Reply

aka

21/06/2013 12:00pm

can you please explain why the acceleration of block m1 will be downawards

Reply

aka

21/06/2013 12:02pm

can you please explain why the acceleration of block m1 will be downwards in ques no.26

Reply

ragib

21/06/2013 11:16pm

dear sir.....now Q11 of objective I whivh has only one correct answer and Q2 and Q5 of objevtive II which has more than one correct answers in pg-78

Reply

theyatin

22/06/2013 3:48pm

dear,
Q11
as the block slides down due to gravity but when cable breaks the wedge freely falls that means no gravity so the block remains at the top of the wedge until it strikes ground.

Reply

theyatin

22/06/2013 3:53pm

dear,
Q2 objective 2
as in case particle is at rest.
either the particle is at rest completely so as no force is applied at it.
or
the whole frame is in motion as well as the particle so as there is a force acting on particle as the particle would be moving with same speed as of frame.
hence c). & d)

Reply

theyatin

22/06/2013 3:59pm

dear,
Q5 obj 2
as the forces would be equal before time started as block is at rest so all forces would be equal.
but there would be no force F2 after force F2 is withdrawn.
hence only a) justifies these statements.

Reply

Aritro

29/06/2013 1:23pm

could u plz explain me Q.38 and 39

Reply

theyatin

01/07/2013 9:08pm

dear,
Q38. Suppose A move upward with acceleration a, such that in the tail of A maximum tension 30N produced.
T – 5g – 30 – 5a = 0 ...(i)
30 – 2g – 2a = 0 ...(ii)
30 – 20 – 2a = 0 a = 5 m/s^2
so from ......1
T = 50 + 30 +(5 × 5) = 105 N (max)
So, A can apply a maximum force of 105 N in the rope to carry the monkey B with it.
now
For minimum force there is no acceleration of monkey ‘A’ and B. that is, a = 0
Now equation (ii) is T'1 – 2g = 0
T'1 = 20 N (wt. of monkey B) (gravity =10)

Equation (i) is T-5g-T'1=0
T – 5g – 20 = 0 [As T'1 = 20 N]
T = 5g + 20 = 50 + 20 = 70 N.
The monkey A should apply force between 70 N and 105 N to carry the monkey B with it.

Reply

theyatin

01/07/2013 9:09pm

dear,
Q39
1. Given, Mass of man = 60 kg.
Let R' = apparent weight of man in this case.
Now, R' + T – 60g = 0 [Tension From FBD of man]
T = 60g – R'
T – R' – 30g = 0 [Tension from FDB of box]
60g – R' = R' – 30g = 0 [ equating both equations]

R' = 15g The weight shown by the machine is 15kg.

2.
as in above case there was no motion so thus no acceleration but in this case the man must move so as to maintain his original weight.
From the FBD of the box
let R be its weight on the machine
T – R – 30g – 30a = 0 (as the mass of box is 30)

T – 60g – 30g – 30a = 0 (as the weight of man is 60g on the machine)

T – 30a -90g

T = 30a – 900 (as gravity is 10)
...(1)

From the FBD of the man

T + R – 60g – 60a = 0

T – 60a = 0 [as R = 60g]

T = 60a
...(2)

From eqn (i) and eqn (ii) we get T = 2T – 1800
T = 1800N.
So, he should exert 1800 N force on the rope to get correct reading.

Reply

Sreedhar

03/07/2013 10:50am

Is the answer for Objective Question number 8 correct? If so can you explain the answer?
This question is about three charged particles and belongs to the chapter Newton's Laws of Motion

Reply

theyatin

03/07/2013 9:42pm

dear,
which objective you are asking for objective 1 or objective 2??

Reply

Sreedhar

05/07/2013 12:56pm

This is for Objective 1

theyatin

05/07/2013 10:01pm

dear,
as the rods are rigid so there is no force that has been exerted by other particles on the particle at A.
the rods cancels all the force acting on each particle.
ur answer may be right in case particles are kept freely.

Reply

Kycor

07/07/2013 7:52pm

In an inclined plane , suppose we have a mass 'm' (Inclination='A') .
Then , Is the normal force N=mgCos(A) or N=mg/(Cos(A)) ?

Reply

theyatin

07/07/2013 10:39pm

dear,
depends upon the motion too but generally the normal of force may be
mg/cos theta.

Reply

rohan

09/07/2013 11:18am

can somebody explain me question no 25 in a proper manner please??? :(

Reply

theyatin

09/07/2013 1:09pm

dear,
as it has been shown in the figure shorter side has more angle so if masses are same then acceleration would be in the direction of shorter side.
first FDB diagram shows motion along shorter side.
second FDB diagram shows motion along longer side.
by comparing acceleration on both sides we can have acceleration.

Reply

Kycor

09/07/2013 8:30pm

Sir , as to my last question "In an inclined plane , suppose we have a mass 'm' (Inclination='A') .
Then , Is the normal force N=mgCos(A) or N=mg/(Cos(A)) ? "

If it is given that the bodies are in equilibrium , then what is the case ? It's a puzzling scenario sir

Reply

Rohan

10/07/2013 10:39am

Sir,
in question 28 and 29 why the tension in block 2 is taken as t/2??

Reply

theyatin

10/07/2013 2:51pm

dear,
as T is assumed to be tension of the sting of system m1 and m2+m3
so now tension of system m2 and m3 would be half.
as other half of the tension is to balance m1

Sir/Ma'am
in question no 16, I fail to understand the concept after you got acceleration. (I guess i lack in my basics)
After getting T, you apparently multiplied T by 2 to get 2T, which, you say, is shown in the spring. Why is that so? Are you adding up the 2 Tensions in the pulley to get the net force?
Thankyou

Oh, and, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct? Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Reply

theyatin

13/07/2013 11:30am

dear,
we ddnt multiply to get 2T we multiplied to get 3g so that we can subtract it from other equation so as to remove one variable so as to find other one.
hence we get T=6a

I'm sorry, but you misunderstand, sir.
I was actually referring to the end part of the question, where, after getting the values of T, you multiplied it by T and then divided it by g to get the final answer. However, I got this doubt cleared from some other source.
Rather, I'd like to know why, in the FBD, is g/10 acting in the downward direction for both the blocks. Could you please explain the concept step by step, without using pseudo force, and by explaining how and with respect to what have you taken a and how is g-10 related to a?
Thank You again.

theyatin

16/07/2013 10:41pm

dear,
physical significance of psuedo forces is to solve these kind of problems. so i cant suggest any simple way to solve this numerical without psuedo forces.

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Reply

theyatin

13/07/2013 12:21pm

dear,
this link is not working properly right now i will try later and will answer your doubts. . .

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Oh, and, in question 16, why is g-10 acting in the downward direction for both the blocks? I found this solution on the Internet...is this correct?
LINK: http://www.scribd.com/doc/52121841/Solutions-to-Concepts-of-Physics-by-HC-Verma-Chapter-5
Why/Why Not? And if it is correct, could you explain to me this method too? Thank You very much for your help.

Reply

Vedabit Saha

14/07/2013 10:45am

I'm sorry, but you misunderstand, sir.
I was actually referring to the end part of the question, where, after getting the values of T, you multiplied it by T and then divided it by g to get the final answer. However, I got this doubt cleared from some other source.
Rather, I'd like to know why, in the FBD, is g/10 acting in the downward direction for both the blocks. Could you please explain the concept step by step, without using pseudo force, and by explaining how and with respect to what have you taken a and how is g-10 related to a?
Thank You again.

Reply

theyatin

14/07/2013 10:23pm

dear, it is given in the numerical acceleration of elevator is g/10 . . .

Reply

Vedabit Saha

16/07/2013 7:41am

uhhm...you didn't understand, i guess :)
i sort of figured it out though, but could you perhaps solve this question without using pseudo force?

dear,
in Q12
a) at any depth let the ropes make angle θ with the vertical
From the free body diagram
F cos θ + F cos θ – mg = 0
(here F is force applied to each rope and then force applied to the weight would be at some angle θ depending on position of the object) so for both ropes each Fcosθ is added and as weight of the body is in downward thus its subtracted...
mg=2F cos(theta)
As the man moves up. θ increases i.e. cos decreases. Thus F
increases.
( it means more force is needed to pull the man up as man moves up)
F=mg/2 cos(theta)

2) now we need to calculate force applied to a man for any depth H from ground level.
H can be any depth where width of trench is D
so considering force applied by one rope lets assume angle of object is θ from horizon.
thus from width D we have D/2 as height of the triangle and H as base of triangle thus applying Pythagoras theorem we have hypotnese of this triangle
as we know cos θ= base/hypotenuse
thus in solution we can find cosθ
now from previous equation as we have
F=mg/2cosθ
put value of cos in this equation we have equation for force acting on body at any depth H

Reply

Vedabit Saha

18/07/2013 11:33pm

Could you please solve question 31 using constraint?

Reply

theyatin

22/07/2013 8:43pm

dear,
it is not easy to solve whole question here as notations are not easy to type here as well read.

Reply

Shadab

21/07/2013 11:24am

Sir
In Q39(b) we found out the tension in the string. How is it equal to the force applied by the man?

Reply

theyatin

22/07/2013 8:51pm

dear,
if tension is equal to force applied by man only then the system would be stable.

dear,
in Q 34 a).
first we need to find tension due to system of weight 5kg.
then we need to find tension due to system of weight 4 kg. while tension of first system is T then tension of system 2 is only half of tension.
b).
similarly in this part
first we need to find tension due to system of weight 5kg.
then we need to find tension due to system of weight 4 kg. while tension of second system is T then tension of system 1 is only half of tension.
c).
similarly in this part
first we need to find tension due to system of weight 1 kg.
then we need to find tension due to system of weight 2 kg. while tension of first system is T then tension of system 2 is only half of tension.

Reply

brajinder

25/07/2013 8:38am

explain ques no 16 sir

Reply

theyatin

01/08/2013 12:04am

dear,
as you can see from free body diagram we have two equations
first we multiply with appropriate constant and then subtract or add both equations to extract T from b
now we put value of T in any of equation to find acceleration as we know value of g that is gravity. ..

Reply

sejal

26/07/2013 9:33pm

in ques 34(a) how acceleration is 2a??

Reply

theyatin

01/08/2013 12:03am

dear,
as you can see in diagram as the for a single moment of 5kg weight the 4kg weight has to move twice due to double and single wire assigned to each block
now as the motion is twice thus acceleration would be twice too

Reply

Gunjan Sharma

27/07/2013 8:01pm

SIR PLS EXPLAIN ME Q8 OF Qs FOR SHORT ANSWER of NEWTON's LoM

Reply

Gunjan Sharma

27/07/2013 8:05pm

PLZ MAKE ME UNDERSTAND THE CONCEPT OF ATWOOD MACHINE THANKYOU

Reply

theyatin

01/08/2013 12:24am

dear,
The ideal Atwood Machine consists of two objects of mass m1 and m2, connected by an inextensible massless string over an ideal massless pulley. [1]
When m1 = m2, the machine is in neutral equilibrium regardless of the position of the weights.
When m1 ≠ m2 both masses experience uniform acceleration.

Reply

GUNJAN SHARMA

29/07/2013 2:42pm

I NEED YOUR REPLY AS SOON AS POSSIBLE PLZ

Reply

LIVIA DEY

30/07/2013 11:29pm

I COULDN'T UNDERSTAND Q.11 OF CHAPTER5

Reply

theyatin

01/08/2013 12:20am

dear,
as the small block has no force applied on it so it would not move
and the acceleration of big block can be calculated by F=ma so we have 2ms^-2
so we have distance initial velocity and acceleration we can find time taken
using S=ut+1/2at ^2

dear,
first we need to find tension on both masses
we get two equations one for 1.5kg aceelration upward thus negative.
and one for 3kg acceleration is downward thus positive
as acceleration is given g/10 upward
multiplying equation 1 and subtracting we get
T=6a
putting it back
we can find now a. thus tension T

sir, in question 34 a why acceleration is taken as 2a please explain me

Reply

theyatin

12/08/2013 8:06pm

dear,
as the for the unit motion of 5kg, 4kg mass mvoes twice
hence for unit acceleratin of 5kg mass , he 4kg mass moves twice.
it is becuase of two pulley attached to 4kg and single pully attached to 5kg

Reply

Rameeza

11/08/2013 10:10am

Explain me q no 7 , the free body diagram...

Reply

theyatin

12/08/2013 8:18pm

dear,
the only difference between both cases of A and B is that when force of contact is considered in between two boxes and the force applied by experimenter is only on first box so on second box it is zero hence f=0
thus two equations are
F+ma-f=0
and F-ma=0

in q.no 28 for the mass m2 , the relative string acceleration is a1-a2 or a2-a1.... How shall we differentiate between two accelerations which are differing in magnitu

in q.no 28 for the mass m2 , the relative string acceleration is a1-a2 or a2-a1.... How shall we differentiate whether to take a1-a2 or a2-a1 pls clarify my doubt...........

Reply

theyatin

12/08/2013 8:25pm

dear.
as we dont know the magnitude or value of a1 or a2
so now we can only consider a1-a2
as it will not effect any equation whether a1>a2 or a1<a2

in the example problems of newtons law of motion ..q.no 9 the relative accelartions of m2 is given as a0-a.whereas
in q.no 28 for the mass m2 , the relative string acceleration is a1-a2 . How shall we differentiate whether to take a1-a2 or a2-a1 pls clarify my doubt...........

i am not getting 3,6,8,12,18,24 problems can you please explain

Reply

theyatin

12/08/2013 8:33pm

dear,
in Q3
we can use V^2-U^2=2AS
as A= V^2-U^2 / 2S

Reply

theyatin

12/08/2013 8:39pm

dear,Q6
as acceleration is rate of change of velocity that is 15-0/3
now when velocity is constant acceleration is zero
now again velocity is decreasing w.r.t. time as now acceleration is negative or in opposite direction. that is 0-15/3

Reply

theyatin

12/08/2013 8:43pm

dear.
Q8 is similar as Q6
you can use v^2-u^2=2as

Reply

ishika

12/08/2013 5:47pm

in ques 29, if we take tension in the thread connecting m2 and m3 as T and therefore tension in thread of m1 as 2T, the answer coming is different from when we use tensions as T/2 and T respectively.
can u please explain why is this happening? and also give me the solution using T and2T, if possible.

Reply

theyatin

12/08/2013 8:52pm

dear, Q29
to solve this question we need to divide whole system into two systems as the one system of pulley is of m2 and m3 and other system is m1 and (m2+m3)
now,
m1 should be at rest.
T – m1g = 0 ( as there is no acceleration as well gravity because both cancel out because m1 is at rest so acceleration and gravity both are equal and opposite in direction)
T = m1g …(i)
T/2 – 2g – 2a1 = 0 ( tension on rope because of pulley of m2+m3)
T – 4g – 4a1 = 0 …(ii) ( multiplying whole equation by 2)
T/2 – 3g – 3a1 =0 (tension on rope of pulley due to weight m2 or m3)
T = 6g – 6a1 …(iii) ( multiplying whole equation by 2)
From eqn (ii) & (iii) we get
3T – 12g = 12g – 2T
T = 24g/5= 4.8kg.
Putting the value of T eqn (i) we get, T=m1*g=4.8g ( here g is gravity)
m1 = 4.8kg .

Reply

ishika

12/08/2013 5:51pm

can u please thoroughly explain ques 21 and 31?

Reply

theyatin

12/08/2013 8:58pm

dear,
Q31
Here also we need to understand the free body diagram of the two blocks
a = acceleration of block M.
1. For block 2M - This block will move with acceleration a/2 in the right direction because if M moves by x distance down then 2M will move by x/2 distance.
There are two forces acting in the horizontal direction. These are -
a. T = tension towards right.
b. (2M)a/2 - This is pseudo force acing in right direction.

for block M -
1. mg = downward
2. T = tension upward
3. ma = pseudo force upward.

Now you form equations an solve part a.
b )= You can also find

c)= for this part see the diagram given in the solution.
R = force by clamp on pulley. = Resultant force of tension on pulley as shown in diagram.

Reply

ishika

14/08/2013 8:22am

thanku sir. can u please explain ques 21 as well?

Reply

theyatin

15/08/2013 2:45pm

dear,
there are two forces acting on particle F and weight.
AS DIRECTION IS NOT CHANGED THUS TOTAL FORCE IS ZERO
as vector product results in perpendicular to axes of each variable.
u A sin theta = mg
thus we can find u
as sin theta is in denominator thus u will be minimum when
sin theta is maximum. that is 1.
thus u=mg/A

Reply

eklavya

15/08/2013 9:44am

to travel from eqn of accln time to velocity time what do we have to do (diff/int)? and why

Reply

thyatin

15/08/2013 2:51pm

dear,
we derivate distance w.r.t. time to calculate velocity
and derivate velocity w.r.t. to time to calculate acceleration.
hence now we integrate acceleration w.r.t. time to get velocity
and integrate velocity w.r.t time to calculate distance. . .

Reply

Abc

15/08/2013 6:20pm

Sir Can you do exercise question 16 without pseodo force

Reply

Abc

15/08/2013 6:24pm

Or please explain the concept ofhow to use pseudo force in these type of questions to solve them

Reply

kashyap

25/08/2013 11:39am

A person is moving in a lift.
when we see that person from the ground,it is called inertial frame of reference and if we see that person standing beside him it is called non-inertial frame of reference .Suppose we see that person from another lift then how to do those type of given sums.

Reply

theyatin

27/08/2013 3:17pm

dear,
in this case it is called frame of reference problem.
means you have a frame as reference and other frame is moving w.r.t. to that frame. . .

Reply

kashyap

25/08/2013 9:16pm

the solution given for the question 34(b) is wrong.pls correct it.it is not T=8a,it should be T=4a.

Reply

ping

26/08/2013 1:09pm

sir, could you please explain objective 1, ques 12 and 13.

Reply

theyatin

27/08/2013 3:39pm

dear,
Q13
as in both case relative gravity in both cases is same so is the actual distance between coin and floor.
when lift is still g=g
when its moving up uniform speed so the acceleration is also zero
hence g+0=g
in case moving downward
g-0=g

Reply

gopal

26/08/2013 8:01pm

sir can you pls explain the answer of Q-10 of objective 1. at page no-78

Reply

theyatin

27/08/2013 3:32pm

dear,
while the speed of both particles is constant
and mass of A is greater than mass of B
thus momentum of A would be greater that momentum of B
hence more momentum leads to higher distance covered by A.

dear,
when balloon is moving downward the friction offered by air is positive
hence buoyant force=mg-F of air
hence buoyant force is less than weight so balloon dips
now if balloon rises
the air restant is negative
so bouyant force
B=mg+Fa
hence balloon rises
calculating mass from both equations
we have the diference of mass that is to be reduced of box so that it rises

Reply

alisha

29/08/2013 11:28am

explain ques 15

Reply

theyatin

30/08/2013 9:56pm

dear,
15. When the elevator is accelerating upwards, maximum weight will be recorded.
as acceleration is added with gravity
R=m(g+a)=72 * 9.9
now
when acceleration is decreasing
R=m(g-a)=60*9.9
now calculating mass from both equations
we have m=66
now from equation 1
we can calculate acceleration that is a

Reply

ping

29/08/2013 9:07pm

sir, could you please explain that in Q 31 why does the body having 2M mass have tension 2T ?

Reply

theyatin

30/08/2013 9:58pm

dear,
when T is tension for mass m
then 2T would be for mass 2m

Reply

Alisha

08/09/2013 11:24am

well its easy.
analyze the forces on the pulley. REMEMBER: in these questions the pulley is assumed to be weightless
so you have 2 threads pulling the pulley with tension T each. so total force on the rightside is 2T.
on the left side you have another thread exerting force(tension) of (suppose) T1.
Now we know the pulley is accelerating toward s right with some acc. (suppose) a m/s^2 towards right
so equating forces you get
2T - T1 = ma
but pulley is weightless. so m =0
2T-T1=0
=> 2T=T1
so the body having mass 2M is having a tension 2T
Hope this helps...

Reply

Alisha

08/09/2013 11:24am

well its easy.
analyze the forces on the pulley. REMEMBER: in these questions the pulley is assumed to be weightless
so you have 2 threads pulling the pulley with tension T each. so total force on the rightside is 2T.
on the left side you have another thread exerting force(tension) of (suppose) T1.
Now we know the pulley is accelerating toward s right with some acc. (suppose) a m/s^2 towards right
so equating forces you get
2T - T1 = ma
but pulley is weightless. so m =0
2T-T1=0
=> 2T=T1
so the body having mass 2M is having a tension 2T
Hope this helps...

Reply

saloni

31/08/2013 9:31pm

in question number 5 ....why a force ma is acting on block b in the left direction

Reply

theyatin

01/09/2013 9:15pm

dear,
because it is the force applied by A to B. . .

Reply

paras

02/09/2013 8:36pm

i need answers for objective worksheet.. can anyone help me?

Reply

alisha

04/09/2013 10:01pm

Can u pls explain Q10 and Q.17 of SHORT ANSWERS...

Reply

theyatin

05/09/2013 9:37pm

dear,
Q10
a) when projectile is at its maximum hieght.
b)when breaks ae applied the velocity is opposite to the direction of acceleration.
c) in case of projectile moving parallel to horizon velocity is perpendicular to acceleration due to gravity.

Reply

Alisha

08/09/2013 11:14am

umm.. thanx but these are not the questions i asked..
I wanted the answers to q10 and q17 of short answers.
10. if you see a snake in front of you, you stop suddenly. What force causes this deceleration?
17. a plumb bob hangs from the ceiling of a train compartment. the train moves with acceleration a.......(check in book............................. Can a person in the compartment tell by looking at the bob if the train is accelerated on a incline plane or horizontal?

theyatin

11/09/2013 8:58pm

dear
10. if you see a snake in front of you, you stop suddenly. What force causes this deceleration?
frictional force causes deceleration
you instantaneously lean back and force your steps to stop moving forward and friction between floor and feet causes stop.

Reply

theyatin

11/09/2013 9:02pm

dear,
17. a plumb bob hangs from the ceiling of a train compartment. the train moves with acceleration a.......(check in book............................. Can a person in the compartment tell by looking at the bob if the train is accelerated on a incline plane or horizontal?

as the train accelerates the bob moves backward so as to maintain its original motion
as train decelerates the bob moves forward so as to maintain its motion.

my teachers are telling that solve of no 39 is wrong the ans. cant be 1800N pls help me

Reply

Alisha

08/09/2013 11:08am

no @arko
the solution is correct it IS 1800 N
You should analyze the forces on the man and then on lift this is the answer that you will get...
on the man there is a tension T upwards his weight is acting downwards, a normal N from weight box and he is accelerating upwards( draw a diagram it will become clear)
so you get
T+N- 60g=60a
weight box measures N and you are given N = 60g
=> T=60a
now analyzing forces on lift
there is a force 2T acting on the lift upwards and its weight(along with the weight of the man) acting downwards.
so you get
2T-(60+30)g= (60+30)a
=> 2T -90g=90a
we know T=60a
so put it in the eqn you get
120a-90g=90a
solve to get a = 30m/s^2
=>T=60a=60*30=1800N
So he should apply a force of 1800N
Hope this helps :D

Reply

Alisha

08/09/2013 11:08am

no @arko
the solution is correct it IS 1800 N
You should analyze the forces on the man and then on lift this is the answer that you will get...
on the man there is a tension T upwards his weight is acting downwards, a normal N from weight box and he is accelerating upwards( draw a diagram it will become clear)
so you get
T+N- 60g=60a
weight box measures N and you are given N = 60g
=> T=60a
now analyzing forces on lift
there is a force 2T acting on the lift upwards and its weight(along with the weight of the man) acting downwards.
so you get
2T-(60+30)g= (60+30)a
=> 2T -90g=90a
we know T=60a
so put it in the eqn you get
120a-90g=90a
solve to get a = 30m/s^2
=>T=60a=60*30=1800N
So he should apply a force of 1800N
Hope this helps :D

Reply

Alisha

08/09/2013 11:08am

no @arko
the solution is correct it IS 1800 N
You should analyze the forces on the man and then on lift this is the answer that you will get...
on the man there is a tension T upwards his weight is acting downwards, a normal N from weight box and he is accelerating upwards( draw a diagram it will become clear)
so you get
T+N- 60g=60a
weight box measures N and you are given N = 60g
=> T=60a
now analyzing forces on lift
there is a force 2T acting on the lift upwards and its weight(along with the weight of the man) acting downwards.
so you get
2T-(60+30)g= (60+30)a
=> 2T -90g=90a
we know T=60a
so put it in the eqn you get
120a-90g=90a
solve to get a = 30m/s^2
=>T=60a=60*30=1800N
So he should apply a force of 1800N
Hope this helps :D

Reply

desai

08/09/2013 4:54pm

explain Q28

Reply

theyatin

11/09/2013 9:20pm

dear,
Let the block m+1+ moves upward with acceleration a, and the two blocks m2 an m3 have relative acceleration a2 due to the difference of weight between them. So, the actual acceleration at the blocks m1, m2 and m3 will be a1.
(a1-a2) is acceleration of m2
and (a1+a2) is of m3
now we can calculate tension in each case
we can get (a1-a2)
and (a1+a2)
now calculating a1 ,a2 and a3 w.r.t. to g.

Reply

swapnum

14/09/2013 9:04am

Worked out example 7 .....plz explain with alternative ...thank u

Reply

shikha

20/09/2013 11:31pm

sir,a vehicle moving on a road with an accelerationa=20m/s^ .The frictional coefficient between the block of mass(m)and the vehicle so thatblock does not fall forward.

Reply

theyatin

22/09/2013 11:35pm

dear,
if this block is offered friction equal to force of the object then it may fall forward.
thus frictional force must be equal to the force of the object.

Reply

shikha

20/09/2013 11:36pm

sir, a weight mg is suspended from the middle of the rope whose ends are at the same level.The rope is no longer horizontal. the minimum tension required to completely straighten the rope.

Reply

the yatin

22/09/2013 11:06pm

dear,
mg sin theta may be the tension required to pull it upward on each side.
so minimum tension required is 2mg sin theta

Reply

shikha

20/09/2013 11:42pm

sir, arocket of mass 5700 kg ejects mass at a constant rate of15kg/s with constant speed of 12km/s.The acceleration of the rocket 1minute after the blast is .

Reply

theyatin

22/09/2013 11:12pm

dear,
from v=u+at
we have 0=12000-a 60 thus a=12000/60=200ms^-2

Reply

shikha

23/09/2013 9:56am

sir,but the ans is 27.5m/s^2

pragya

20/09/2013 11:49pm

sir,a projectile is fired with velocity u at an angle theta with horizontal. at the highest point of its trajectory it splits up into 3 segments of massesm,m and 2m. firstpart falls vertically downward with0 initial velocity and 2nd part returns via same path to the point of projection. the velocity of third part of mass 2m just after explosion is.

Reply

theyatin

22/09/2013 11:21pm

dear,
force is constant thus F=4m*a
while the first part of mass m falls straight down thus its force is zero
the force of second part is of force similar to the part of original force
thus final force would be F'=3m*a
thus 3 part of force is directly proportional to 2 part of masses of original particle.
thus final force would be = 3/2u

Reply

pragya

20/09/2013 11:53pm

sir,
a ball of mass 9 kg explodes into 2 pieces of masses 3 kg and 6 kg. the velocity of mass 3 kg is 16 m/s. the kinetic energy of mass 6 kg is

Reply

theyatin

22/09/2013 11:27pm

dear,
nass is in ration of 1:2
thus kinetic energy would be also in ratio of 1:2
so as K.E.= 1/2 m v^2= 1/2*3*16*16=384
so K.e. of larger part would be double as of smaller hence 768

Reply

pragya

20/09/2013 11:57pm

sir,
a uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support then the tension in the rope at the distance l from rigid support is

Reply

theyatin

22/09/2013 11:31pm

dear,
the tension on the rope at distance l from top would be due to weight of rope of length L-l
mass of whole rope of length L is M then mass of length L-l would be
M(L-l)/L
thus weight would be M(L-l)/L * g

Reply

shivendra

22/09/2013 1:00am

sir how to deal with problems when one body is moving on another..????

Reply

theyatin

22/09/2013 11:32pm

dear,
in that case you need to find relative motion of the bodies.

Reply

shikha

23/09/2013 10:10am

sir, a metallic chain 1m long lies on a horizontal surface of a table . The chain starts sliding on the table if 25cm (or more of it)hangs over the edge of a table .The correct vaue of the coefficient of friction between the table and the chain is

Reply

theyatin

24/09/2013 12:19am

dear,
force of friction is F=u m g
where force to balance is weight of the chain hanging down the table
so if both forces are equal only when the chain would lie on table otherwise it would slide down
hence
F=u m1 g = m2 g (whee m1 is weight of 25cm chain and m2 is weight of 75cm of chain)
so um1=m2
or u=m2/m1
if chain is uniform then ratio of masses would be directly proportional to ratio of its lengths
hence m2/m1=L2/L1=75/25=3
thus u=3

Reply

shikha

23/09/2013 10:15am

sir ,a boy is sitting on the horizontal platform of a joy wheel at a distance of 5 m from the center. The wheel begins to rotate and when the angular speed exceeds 1rad/s the boy just slips. The coefficient of friction between the boy and the wheel is

Reply

theyatin

24/09/2013 12:22am

dear,
force of angular rotation is F=mw^2 r
that must be equal to force of friction so as if speed exceed the boys lips away
thus
F=u m g = m w^2 r
or u g= w^2 r
os u= w^2 *r /g
you have values of w and r=5 and g you can have your answer.

Reply

shikha

23/09/2013 10:20am

sir, if a pushing force makes an angle alpha with the horizontal is pplied on a block of mass m placed on horizontal table and angle of friction is bita then minimum magnitude of force required to move the block is

Reply

theyatin

24/09/2013 12:31am

dear,
as force applied is making angle alpha (a) with horizontal then
effective force would be horizontal component of the force that is
Fcos a
now frictional force is making angle beta (b) to the horizontal then again effective frictional force wold be horizontal component of the force that is F'cos b
now force applied must be greater than force offered by frictional force so as to move the box

Reply

pragya

23/09/2013 10:24am

sir,a horizontal force 10 N is necessary to just hold a block stationary against a wall.the coefficient of frction between the block and the wall is 0.2, the weight of the block is

Reply

theyatin

24/09/2013 12:35am

dear,
frictional force is equal to force applied that is why stationary is held against wall
so 10=umg
given that u=0.2
so 10/0.2=mg
that is 50N=mg

Reply

pragya

25/09/2013 6:13pm

sir ,please check out i too got 50Nas answer but the ans is 20Nwhere is the problem.

pragya

23/09/2013 10:30am

sir, a 40kg slab rests on a frictionless floor . a 10 kgblock restson the top of the slab .the coefficient of friction between the block and the slab is 0.40.The10kg block is acted upon by a horizontal force of 100N.then accleration of the slab is

Reply

theyatin

24/09/2013 12:41am

dear,
the force acting on block is 100N
then fricitonal force acting on it would be
Ff=100*0.4
that is F= 40N
now this is force acting on slab
so from F=ma
40N=40*a
a=1

sir, in the vol.1 chapt 5 question no. 28 ,initially we took covension that m1 is going upward while other two downward . However while drawing individual FBD we considered m1 to be accelerating downward. Why are we taking two differnet assumptions ?

Reply

theyatin

01/10/2013 10:50pm

dear,
in fdb
that a1 is not associated to m1 but to the whole system at left hand side

Reply

Anish

30/09/2013 8:28am

in question no. 42 why aren't we taking psuedo force into account as acceleration of body is given as a=10 (downward direction ) not a =12-10=2 (upward direction ) ?

Reply

theyatin

01/10/2013 10:58pm

dear,
in your way the displacement would only be relative to elevator we need to calculate actual displacement

Reply

ashu

11/10/2013 3:51pm

In ques no. 6th. why can't we find acceleration by dv/dt and then calculate the force ??

Reply

theyatin

12/10/2013 11:33pm

dear,
which equation is there to derivate w.r.t time??

Reply

ashu

22/10/2013 11:57am

plzz reply fast..

ashu

17/10/2013 10:40am

there is no such equation given..only the graph given is of velocity versus time....

Reply

pranita

12/10/2013 1:59pm

sir,
plz explain Q.16 and Q.35

Reply

theyatin

12/10/2013 11:39pm

dear,
Q16
first we need to find tension on both masses
we get two equations one for 1.5kg acceleration upward thus negative.
and one for 3kg acceleration is downward thus positive
as acceleration is given g/10 upward
multiplying equation 1 and subtracting we get
T=6a
putting it back
we can find now a. thus tension T

T + 0.5a – 0.5g = 0 .... 1
T1 – 0.05a – 0.05g = 0 ...2
T1 + 0.1a – T + 0.05g = 0....3 (tension due to 500g=T1 is in same directio to the motion of 100g of weight and the tension T due to 50gm is in oppositte direction to the motion of 100gm so it has negative sign)
above 3 eq. for tension are respectively for wight 50gm 500gm and 100gm
taking tension on a side from first two equations we have
T = 0.05g + 0.05a ....4
T1 = 0.5g – 0.5a ......5
respectively and putting values of T and T1 in ...3 we can solve it further to get a while g=10

please could you explain the whole of question 34 ...
it is mind boggling
please sir ,

Reply

theyatin

13/10/2013 10:49pm

dear,
in Q 34 a).
first we need to find tension due to system of weight 5kg.
then we need to find tension due to system of weight 4 kg. while tension of first system is T then tension of system 2 is only half of tension.
b).
similarly in this part
first we need to find tension due to system of weight 5kg.
then we need to find tension due to system of weight 4 kg. while tension of second system is T then tension of system 1 is only half of tension.
c).
similarly in this part
first we need to find tension due to system of weight 1 kg.
then we need to find tension due to system of weight 2 kg. while tension of first system is T then tension of system 2 is only half of tension

Reply

priyansha

17/10/2013 11:35am

sir,
in ques. no. 7th
F- f = mass of A block * acc. ..........(1)
F = mass of B block * acc. ............(2)

are these two eqn. correct ??

Reply

theyatin

24/10/2013 12:10pm

dear,
it is correct interpretation.

Reply

priyansha

28/10/2013 11:34am

but my answer is coming
2F - mass of block A * acc. -mass of block B * acc.

rebekah

18/10/2013 12:35am

in question 32 how will we conclude that acceleration of M is 2a and that of 2M is a?

Reply

theyatin

24/10/2013 12:13pm

dear,
when mass M moves a distance 2d then mass 2M moves d because of the pulley connected to 2M rope is wrapped thus is of twice length
hence acceleration of M is twice as of acceleration of 2M

Reply

manish

21/10/2013 1:01am

I am very confused in banking of roads and and motion on a level curved road

Reply

theyatn

24/10/2013 12:16pm

dear,
we bank road just to balance the inertial force on the vehicle so that it doesn't get off the road while taking a turn
more acute the turn would be higher we need the banking.

Reply

priyansha

22/10/2013 11:56am

sir,
in ques. no. 7th
F- f = mass of A block * acc. ..........(1)
F = mass of B block * acc. ............(2)

are these two eqn. correct ??

Reply

priyansha

22/10/2013 11:57am

sir,
in ques. no. 7th
F- f = mass of A block * acc. ..........(1)
F = mass of B block * acc. ............(2)

are these two eqn. correct ??

Reply

Ishan

22/10/2013 3:01pm

can u explain solution for ques 8 objective 1 , newton laws of motion on pg no. 77

Reply

theyatin

24/10/2013 12:19pm

dear,
as all the charges are at equal distances and of equal magnitude thus
al the three charges would cancel out forces at each of the vertex thus at A magnitude of force is zero

Reply

Pranav

24/10/2013 3:37pm

Sir i can't solve any diagram question of newtons law

Reply

theyatin

26/10/2013 11:14pm

dear,
you can ask a particular question and i will tell you how to do it

Reply

tushar

25/10/2013 11:53am

q 32 pls.

Reply

theyatin

26/10/2013 11:26pm

dear,
we need to balance forces in case of 2M and in case of M
2Ma + Mg sin theta – T = 0

2T + 2Ma – 2Mg = 0
when mass M moves a distance 2d then mass 2M moves d because of the pulley connected to 2M rope is wrapped thus is of twice length
hence acceleration of M is twice as of acceleration of 2M
now solving and getting value of a

Reply

priyansha

28/10/2013 11:36am

in ques. no 7th, my answer is coming :
2F - mass of block A * acc. -mass of block B * acc.

Reply

theyatin

28/10/2013 11:01pm

dear,
please tell me how do you obtain this answer??

Reply

ashu

28/10/2013 11:39am

In ques no. 6th. why can't we find acceleration by dv/dt and then calculate the force ?? there is no equation given..only the graph given is of velocity versus time....

Reply

theyatin

28/10/2013 11:03pm

dear,
you can not derivate a constant
you can derivate it only if it has variables in equation.

Reply

priyansha

01/11/2013 9:26am

sir,
in ques. no. 7th
F- f = mass of A block * acc. ..........(1)
F = mass of B block * acc. ............(2)

but my answer is coming 2F - mass of block A * acc. -mass of block B * acc.

Reply

theyatin

11/11/2013 9:51pm

dear,
we need to compare not adding both of them. . .

Reply

pranjal

10/11/2013 10:34pm

sir i thiink there is a problem in ques no 28....please see......and can u please explain that can we take any arbitary direction of acc. of the block m2 and m3.....and why have we taken a2+a1 and a1-a2....please explain sir....these r sum basics i need to clear

Reply

theyatin

11/11/2013 10:05pm

dear,
no you cant take acceleration in any arbitrary direction
in pulley its upward iff the body is moving upward
and downward if body is moving downward

and in system where two bodies are attached to a string
the system it self is having an acceleration a1 so if the acceleration of either of the mass is along with the direction of a1 then it would add
a1+a2
and in case of acceleration of one body of the system is against the acceleration of system then a1-a2

Reply

pranjal

12/11/2013 12:13am

sir please explain in ques no 39 that why is is neccessary to have acc. to get his original wieght....and wether the acc. be upward or downward

Reply

Sidharth

22/11/2013 10:10pm

I have a question. In questions like pendulum at an angle @ or object on an incline of angle, sometimes we make the cos & sin components of mg and sometimes of tension or normal. How exactly do I get to know whose components do I have to make in which question?

Reply

shivani sharma

18/12/2013 8:09pm

there is no hard and fast rule ,we can break any force in any direction but we break the forces in that direction where there are maximum no. of forces.

Reply

shabd

06/12/2013 8:28am

sir, plz explain me Q10 of objective1, why will A go higher than B?

Reply

theyatin

20/12/2013 7:57pm

dear,
while we throw them upward with same velocity lets say v
if Ma>Mb
then kinetic energy of a must be greater than b
hence K.Ea>K.E.b
so if same amount of force is exerted on each ball thus ball a will go higher than b.
as kinetic energy is measure of motion of an object.

Reply

shivani sharma

18/12/2013 8:10pm

sir,I also have a similar doubt.Why A will go higher than B(in Q.10 of objective-1).

Reply

theyatin

20/12/2013 7:58pm

dear,
while we throw them upward with same velocity lets say v
if Ma>Mb
then kinetic energy of a must be greater than b
hence K.Ea>K.E.b
so if same amount of force is exerted on each ball thus ball a will go higher than b.
as kinetic energy is measure of motion of an object.

Reply

shivani sharma

20/12/2013 3:37pm

sir,please give me the reply

Reply

theyatin

20/12/2013 7:58pm

done that. . .

Reply

shivani sharma

21/12/2013 12:59am

Thankyou sir...

Aniket

01/01/2014 11:56pm

sir can u please tell me how the tension in the string to which another pulley is attached comes to be T/2 in problem 28?

Reply

siddharth

02/01/2014 7:50pm

give me solution of q no 33 39 40 of ch 5

Reply

theyatin

03/01/2014 9:54pm

dear
Q33
if only mass M' moves and m doesn't then acceleration of m should be equal to acceleration of M' (just like water in bucket which doesn't fall even if you rotate it upward downward,because angular acceleration of water eliminates gravity)

T + Ma – Mg = 0 ( tension on mass M)
...(i) (From FBD – 1)

T – M'a – R sin θ = 0 (tension on mass M')
...(ii) (From FBD -2)

R sin θ – ma = 0 (equation for acceleration on mass m)
...(iii) (From FBD -3)

R cos θ– mg =0 (equation for acceleration due to gravity on m)
...(iv) (From FBD -4)
now subtracting ...2 from ....1
we can eleminate T ...5
R sin θ = ma
R cos θ= mg
dividing these equation and then putting value of g in ....5
we have M=(M'+m)/(cot θ-1)

Reply

theyatin

03/01/2014 9:56pm

dear,
Q39
1. Given, Mass of man = 60 kg.
Let R' = apparent weight of man in this case.
Now, R' + T – 60g = 0 [Tension From FBD of man]
T = 60g – R'
T – R' – 30g = 0 [Tension from FDB of box]
60g – R' = R' – 30g = 0 [ equating both equations]

R' = 15g The weight shown by the machine is 15kg.

2.
as in above case there was no motion so thus no acceleration but in this case the man must move so as to maintain his original weight.
From the FBD of the box
let R be its weight on the machine
T – R – 30g – 30a = 0 (as the mass of box is 30)

T – 60g – 30g – 30a = 0 (as the weight of man is 60g on the machine)

T – 30a -90g

T = 30a – 900 (as gravity is 10)
...(1)

From the FBD of the man

T + R – 60g – 60a = 0

T – 60a = 0 [as R = 60g]

T = 60a
...(2)

From eqn (i) and eqn (ii) we get T = 2T – 1800 T = 1800N.
So, he should exert 1800 N force on the rope to get correct reading.

Reply

theyatin

03/01/2014 10:14pm

dear,
Q40
force F=ma=mgsin@
so a=gsin@
hence using motion formula s=ut+1/2 a t ^2
s=l
u=0
we can find t
(there is typing mistake its not g^2 its t^2)

Reply

Anupam

09/01/2014 9:00pm

Please explain the free body diagram of Q.NO 33.

Reply

theyatin

03/02/2014 7:58pm

dear,
if only mass M' moves and m doesn't then acceleration of m should be equal to acceleration of M' (just like water in bucket which doesn't fall even if you rotate it upward downward,because angular acceleration of water eliminates gravity)

T + Ma – Mg = 0 ( tension on mass M)
...(i) (From FBD – 1)

T – M'a – R sin θ = 0 (tension on mass M')
...(ii) (From FBD -2)

R sin θ – ma = 0 (equation for acceleration on mass m)
...(iii) (From FBD -3)

R cos θ– mg =0 (equation for acceleration due to gravity on m)
...(iv) (From FBD -4)
now subtracting ...2 from ....1
we can eliminate T ...5
R sin θ = ma
R cos θ= mg
dividing these equation and then putting value of g in ....5
we have M=(M'+m)/(cot θ-1)

Reply

debasish nayak

25/01/2014 8:22pm

Q.31 (c )

Reply

theyatin

03/02/2014 7:57pm

dear,
Part c
for this part see the diagram given in the solution.
R = force by clamp on pulley. = Resultant force of tension on pulley as shown in diagram.

Reply

shahil Akhtar

30/01/2014 1:19pm

plz help me q.no.38

Reply

theyatin

03/02/2014 7:59pm

dear,
Q38. Suppose A move upward with acceleration a, such that in the tail of A maximum tension 30N produced.
T – 5g – 30 – 5a = 0 ...(i)
30 – 2g – 2a = 0 ...(ii)
30 – 20 – 2a = 0 a = 5 m/s^2
so from ......1
T = 50 + 30 +(5 × 5) = 105 N (max)
So, A can apply a maximum force of 105 N in the rope to carry the monkey B with it.
now
For minimum force there is no acceleration of monkey ‘A’ and B. that is, a = 0
Now equation (ii) is T'1 – 2g = 0
T'1 = 20 N (wt. of monkey B) (gravity =10)

Equation (i) is T-5g-T'1=0
T – 5g – 20 = 0 [As T'1 = 20 N]
T = 5g + 20 = 50 + 20 = 70 N.
The monkey A should apply force between 70 N and 105 N to carry the monkey B with it.

Reply

Mihir

23/02/2014 12:07pm

Sir, please explaine me 39 th c.
I think that the 1 kg block should go upword.

Reply

Mihir

23/02/2014 12:15pm

I m sorry plz explaine 32 instade of 39

Reply

theyatin

26/02/2014 12:14pm

dear,
there is no mass of 1kg in 32nd q.

niti n

01/04/2014 6:42pm

could u give me d ans of q 2 of short answer question.

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niti n

03/04/2014 8:33pm

y dont u reply me quickly!!!!!! it has been 3 days since I posted my Q and ur site tells dat u will reply under 24hrs ,but still u dint reply, i dont know wat is d reason ,maybe due to poor mngmnt or u people dont check d posed Qs regularly . UTTER , ABSOLUTE DISSAPPOINTMENT. I HOPE U RESPOND AT D EARLIEST.