A string of area of cross-section 4 mm ^ 2 and length 0 . 5 is...

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 $\dfrac{∆ l}{l} = \dfrac{T}{AY}$

$= \dfrac{mg + \dfrac{m{v}_{b}^{2}}{l}}{AY}$ $= \dfrac{20 + \dfrac{2{\left(5\right)}^{2}}{0 . 5}}{3 \times {10}^{- 6} \times {10}^{11}} = 30 \times {10}^{- 5}$