Question

A uniform chain of length is L = (3/2)m and mass M is placed on a smooth spherical surface of radius R =
(2\7) m with one end A held by thread PA. then just after PA is burnt, the chain starts with acceleration 'a'
then:
08
_
0:
(A) Tension at point C = 1 (L - TR) g
(B) Tension at point A is zero
(C) Tension at point B
08
MR ()
(D) Tension will increase from A to B if a>g​

Answer 1

Option B - Tension at point A is zero.   (R = 2/pi m )


Here are the calculations:


Mass of this elemental length = λ(Rdθ)

Consider the tangential forces

(T + dT) + λRg sinθdθ-T = λRadθ

Or  dT = λR(a – g sinθ)dθ

To solve the equation integrate it with 0 to π/2.

ʃ dT = λRʃ(a – g sinθ)dθ

Here θ=0 means at top point A, tension is zero because the thread is burnt and tension is assumed to be T1 at θ=π/2.

Therefore T1 = λR(a π/2 – g)……..(1)

Consider top l m length and lower 0.5 m length.

0.5λg - T1 = 0.5 λa……….(2)

λR(a π/2 – g) = 0.5λg - 0.5 λa

From equations 1 and 2 we get

 a= (4+π/3π)g = 7.58 ms-2


Thus, option B is correct.


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