Question

The Young's modulus of brass and steel are $10 \times 10^{10} \, \text{Nm}^{-2}$ and $20 \times 10^{10} \, \text{Nm}^{-2}$, respectively. A brass wire and steel wire of the same length are extended by 1 mm under the same force. If the radii of the brass and steel wires are $R_B$ and $R_S$ respectively, then

• A. $R_S = \frac{R_B}{4}$
• B. $R_S = \frac{R_B}{2}$
• C. $R_S = \sqrt{2} R_B$
• D. $R_S = \frac{R_B}{\sqrt{2}}$

Hint:

When the Young's modulus differs, the radius required to maintain the same elongation under the same force can be calculated using the relationship between Young's modulus and radius.

Answer 1

The Correct Option is D

For a wire under tension, the elongation $\Delta L$ is given by: \[ \Delta L = \frac{F L}{A Y} \] where: $F$ is the force, $L$ is the length of the wire, $A$ is the cross-sectional area, and $Y$ is the Young's modulus. Since the elongation is the same for both wires, we can equate the elongations: \[ \frac{F L}{A_B Y_B} = \frac{F L}{A_S Y_S} \] The cross-sectional area $A = \pi r^2$, so: \[ \frac{r_B^2}{Y_B} = \frac{r_S^2}{Y_S} \] Thus, we find: \[ r_S = \frac{r_B}{\sqrt{2}} \]