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8.Mechanical Properties of Solids
hard
A string of area of cross-section $4\,mm ^{2}$ and length $0.5$ is connected with a rigid body of mass $2\,kg$. The body is rotated in a vertical circular path of radius $0.5\,m$. The body acquires a speed of $5\,m / s$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is $\ldots . . \times 10^{-5}$. (Use Young's modulus $10^{11}\,N / m ^{2}$ and $g =10\,m / s ^{2}$ )
A
$29$
B
$300$
C
$30$
D
$303$
(JEE MAIN-2022)
Solution

Strain $=$ F/AY
$=\frac{ mg +\frac{ mv ^{2}}{ R }}{ AY }$
$=\frac{20+\frac{2(5)^{2}}{0.5}}{3 \times 10^{-6} \times 10^{11}}=30 \times 10^{-5}$
Standard 11
Physics