Question 45 of 118

A string of area of cross-section $4{mm}^{2}$ and length $0.5$ is connected with a rigid body of mass $2kg$. The body is rotated in a vertical circular path of radius $0.5m$. The body acquires a speed of $5m{s}^{-1}$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _____ $\times {10}^{-5}$. (Use Young's modulus ${10}^{11}N{m}^{-2}$ and $g=10m{s}^{-2}$)

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Answer: 30

Given here, $A=4{mm}^{2}$, $l=0.5m$, $m=2kg$ and ${v}_{b}=5m{s}^{-1}$.
Tension at the bottom of the circular motion is given by $T=mg+\dfrac{m{v}_{b}^{2}}{R}$

Using Young's modulus, we have
Strain produced in the string when the body is at the bottom of the circle is $\dfrac{∆l}{l}=\dfrac{T}{AY}$
$=\dfrac{mg+\dfrac{m{v}_{b}^{2}}{l}}{AY}$ $=\dfrac{20+\dfrac{2{\left(5\right)}^{2}}{0.5}}{3\times {10}^{-6}\times {10}^{11}}=30\times {10}^{-5}$