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:
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0:
(A) Tension at point C = 1 (L - TR) g
(B) Tension at point A is zero
(C) Tension at point B
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MR ()
(D) Tension will increase from A to B if a>g
Answers
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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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