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Question: A stone of mass m tied to one end of a thread of length l. The diameter of the thread is d and it is...

A stone of mass m tied to one end of a thread of length l. The diameter of the thread is d and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle θ with the vertical. Find the increase in length of the wire. Youngs modulus of the wire is Y.

A

4mglπd2Ycos⁡θ\frac{4mgl}{\pi d^{2}Y\cos\theta}

B

4mglπd2Ysin⁡θ\frac{4mgl}{\pi d^{2}Y\sin\theta}

C

4mglπd2Y\frac{4mgl}{\pi d^{2}Y}

D

4mglπd2Ysec⁡θ\frac{4mgl}{\pi d^{2}Y\sec\theta}

Answer

4mglπd2Ycos⁡θ\frac{4mgl}{\pi d^{2}Y\cos\theta}

Explanation

Solution

T cos θ = mg.

Y = TLAΔLorΔL=TLAY=mgLAYcos⁡θ=4mglπd2Ycos⁡θ\frac{TL}{A\Delta L}or\Delta L = \frac{TL}{AY} = \frac{mgL}{AY\cos\theta} = \frac{4mgl}{\pi d^{2}Y\cos\theta}