IIT JEE , AIEEE Forum - Physics , Mechanics - Doubt in NEWTONS LAWS OF MOTION

Pitchu

SIR
I 'm now solving HC Verma Physics.......hv some doubts in Newtons laws ..........Pg No.79 problem nos. 7, 8, 12, 16, 20 21 &22. Has it been solved in this forum by anyone.....if so can u pls give me some links?..........pls help..........

Pitchu

anyone pls reply..............................waiting .....

deepak_agarwal

hello pitchu....u need to tell me the questions..hc verma may not be the standard book for all of ppl here..infact i never touched that book..so write the questions...we will definitely solve it for u

aditya_arora04

Hi pitchu !!!

First of all let me tell you that Pitchu is a very cool an cute name !!!!

Now, lets get back to business !!! For the questions i have not been able to make free body diagrams. Sorry, but i will try to explain with the best of my ability.

Q. 7) Here, consider m_a. Here, forces acting on it are :-

1. Force F₂ exerted by the experimenter.

2. Force F exerted by block m_b.

As the block accelerates, F₂ - F = m_aa [ where a=acceleration ]

Now, consider m_b . Here, forces are :-

1. Force F exerted by block m_b . This is because Force

exerted by both blocks on each other is 0 due to

action-reaction pair.

As, this block accelerates with same acceleration,

F = m_ba

Solve both the equations i.e

F₂ - m_ba = m_aa

F₂ = a ( m_a+m_b )

instead of a you can put F / m_b [ Eq. 2 ]

So, you get = F₂ = F ( 1 + m_a/m_b)

Q. 9) Look,

Here,

use v² = u² + 2as

v = 0 [ as drops come to rest ]

u = 30 m/s [ Velocity of drops before hitting the head ]

s = 0.001 m [ Distance covered by each drop ]

So,

By solving this we get the retardation[as drops retard],

a = 450000 m/s²fsdfsf

So, force provided by each drop = ma

= 4*10^-6* 450000 N

= 1.8 N

Q.12) Look,

Here,

Let the force exerted by each friend be F. This is equal to the tension

in each string. Let that tension be T

Now, at any height, the ropes make an equal angle p with the

vertical.

Wen you break the components for the tensions by making

free body diagram for the person in the "khadda", you get

2Tcosp = mg

As, the friends pull him slowly the acceleration can be assumed to be

0.

T = mg/2cosp

Now, mg is constant

As, the person comes up, p increases.

If, p increases cosp decreases.

So, T increases.

i.e first part proved.

At a height h, [ please see the diagram ]

cosp = h / [h^2 + (d/2)^2 ]^2

T = mg / 2cosp

Solve it and you get the answer as given in the book.

aditya_arora04

Q.16) Here,

Look,

For 3 kg block,

3g_{effective 1} - T = 3a

For 1.5 kg block.

T - 3g_{effective 2} = 1.5a

Here, g_{effective 1} = g + g/10 [ Block goes down g' = g + g/10]

g_{effective 2} = g- g/10 [ Block goes up g' = g- g/10 ]

Solve to get values of T.

Tension in spring balance's string = 2*Tension in string = 2T

Reading of balance = 2T/10

Solve to get the answer !!!!

Q. 20) Here,

Forces acting on the block :-

Bouyant force = B = Vpg [ p =density of water ]

Here, V,p,g remain constant in all cases. Volume,V is constant

Weight of body = mg

Now,

In first case,

B - mg = mg / 6

B = 7mg / 6 ------------------- Eq. 1

In second case, buoyant force remains same.

For second case,

m'g - B = m'g/6 ------------------------ Eq. 2

here, m' = m + k( mass of sand )

Solve both to get k = 2m / 5

Q. 21 Here,

F = v x A

If it continues to move undeflected with constant velocity,[ velocity of projection ] F should balance mg

So, F = mg

vAsinp = mg

Here, p is angle between A and v = pie/2

this is because, F is upward, velocity is perpendicular to bth F and A .

so, v = mg / A

Q.22) After above six, i am sure you can solve the last one !!!!!

Hope this helps !!!

If you can't solve Q.22, please tell me !!!!

INDIAN_ARMY19890

hi all

the q-22 of HCverma pg no-80

sol- .6g-T=.6a........(1)

T-.3g=.3a.......(2)

solving 1and 2 we get a=3.3metre per second square

(a) distance move can be calculated using eq second eq of motion=6.5m

(b) tension can be calculated by puting value of a in eq 1 or 2

(c) force=2T

Pitchu

Thks for liking my name........Hi Aditya (I too liked yr name)

Thanks yaar for solving them. u r rated.

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