Question

One end of a steel rod is clamped to the roof and the other end is attached to a mass of \(1000\ \text{kg}\) as shown in the figure. The length of the rod is \(50\ \text{cm}\) and its cross-sectional area is \(1000\ \text{mm}^2\). The change in the length of the rod due to the weight of the mass is
\[ (\text{Young's modulus of steel }=2\times10^{11}\ \text{N m}^{-2}) \] \[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]

• A. \(0.025\ \text{mm}\)
• B. \(0.10\ \text{mm}\)
• C. \(0.050\ \text{mm}\)
• D. \(0.075\ \text{mm}\)

Hint:

For elongation of a wire or rod under a load, \[ \Delta L=\frac{FL}{AY}, \] where \(F\) is the applied force, \(L\) is the original length, \(A\) is the cross-sectional area, and \(Y\) is Young's modulus.

Answer 1

The Correct Option is A

Step 1: Apply Young's modulus definition.
\( Y = \frac{FL}{A\,\Delta L} \Rightarrow \Delta L = \frac{FL}{AY} \).

Step 2: Substitute values.
\( F = 1000\times10 = 10^4\text{ N},\; L = 0.5\text{ m},\; A = 1000\times10^{-6}\text{ m}^2 = 10^{-3}\text{ m}^2 \).
\( \Delta L = \frac{10^4 \times 0.5}{10^{-3}\times 2\times10^{11}} = \frac{5000}{2\times10^8} = 2.5\times10^{-5}\text{ m} = 0.025\text{ mm} \)

\[ \boxed{\Delta L = 0.025\text{ mm}} \]