A block of mass 2 kg is tied to one end of a 2 m long metal wire of $1.0 \mathrm{~mm}^2$ area of cross-section and rotated in a vertical circle such that the tension in the wire is zero at the highest point. If the maximum elongation in the wire is 2 mm , the Young's modulus of the metal is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

• A

  $1.0 \times 10^{11} \mathrm{Nm}^{-2}$
• Correct answer
  B

  $1.2 \times 10^{11} \mathrm{Nm}^{-2}$
• C

  $2.0 \times 10^{11} \mathrm{Nm}^{-2}$
• D

  $0.2 \times 10^{11} \mathrm{Nm}^{-2}$

Solution

$\mathrm{M}=2 \mathrm{~kg}, \mathrm{l}=2 \mathrm{~m}, \mathrm{~A}=1 \mathrm{~mm}^2$
Velocity of mass at bottom, for tension in wire at top is zero, $\begin{aligned} & \mathrm{v}=\sqrt{5 \mathrm{gl}}=\sqrt{5 \times 10 \times 2}=10 \mathrm{~m} / \mathrm{s} \\ & \mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mv}^2}{1}=20+\frac{2 \times 100}{2}=120 \mathrm{~N} \\ & \therefore(\delta 1)_{\max }=\frac{\mathrm{T}_{\max } \cdot 1}{\mathrm{AY}} \Rightarrow \mathrm{Y}=\frac{120 \times 2}{1 \times 10^{-6} \times 2 \times 10^{-3}} \\ & =1.2 \times 10^{11} \mathrm{Nm}^{-2}\end{aligned}$

Hence, the correct answer is option: 'B'.