Question:

The Young's modulus of brass and steel are respectively $ 1.0\times 10^{11}Nm^{-2} $ and $ 2.0\times 10^{11}Nm^{-2}. $ .A brass wire and a steel wire of the same length are extended by $1 \,mm$ each under the same force. If radii of brass and steel wires are $R_B$ and $R_S$ respectively, then

Updated On: Apr 7, 2024
  • $R_s = \sqrt{2} R_B$
  • $R_S = \frac{R_{B}}{\sqrt{2}} $
  • $R_s = 4R_B$
  • $R_s = \frac{R_{B}}{2} $
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The Correct Option is B

Solution and Explanation

Increase in length, $\Delta L=\frac{F L}{Y \cdot A}=\frac{F L}{Y \cdot \pi R^{2}}$ As $F, L$ and $\Delta L$ are same hence, $Y \cdot R^{2}= a$ constant $\therefore 2.0 \times 10^{11} R_{S}^{2}=1.0 \times 10^{11} R_{B}^{2}$ $\Rightarrow R_{S}=\frac{R_{B}}{\sqrt{2}}$
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