Question

A block of mass 2 kg is tied to one end of a 2 m long metal wire of 1.0 mm² 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:

• A. \( 1.0 \times 10^{11} \, \text{Nm}^{-2} \)
• B. \( 1.2 \times 10^{11} \, \text{Nm}^{-2} \)
• C. \( 2.0 \times 10^{11} \, \text{Nm}^{-2} \)
• D. \( 0.2 \times 10^{11} \, \text{Nm}^{-2} \) \bigskip

Hint:

Remember the tension is zero at the highest point in the vertical circle.

Answer 1

The Correct Option is B

We are given that the mass \(m = 2 \, \text{kg}\), the length of the wire \(L = 2 \, \text{m}\), the area of cross-section \(A = 1.0 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2\), and the maximum elongation in the wire is 2 mm. The tension is zero at the highest point in the vertical circle. Step 1: Calculate the force causing the elongation At the highest point, the centripetal force is provided by the weight of the block, which is \(F = mg\). The tension \(T\) at the highest point is zero, so the maximum elongation is caused by the force \(mg\). We use Hooke's law to calculate the elongation: \[ F = Y \cdot \frac{\Delta L}{L} \cdot A \] where: - \(Y\) is Young’s modulus, - \(\Delta L = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\), - \(L = 2 \, \text{m}\), - \(A = 1 \times 10^{-6} \, \text{m}^2\). Substituting the known values: \[ mg = Y \cdot \frac{2 \times 10^{-3}}{2} \cdot 1 \times 10^{-6} \] \[ 2 \times 10 \, \text{N} = Y \cdot 10^{-9} \] \[ Y = \frac{2 \times 10}{10^{-9}} = 1.2 \times 10^{11} \, \text{Nm}^{-2} \] Thus, the Young’s modulus is \( 1.2 \times 10^{11} \, \text{Nm}^{-2} \).