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Question

Physics Question on elastic moduli

The Young's modulus of brass and steel are respectively 1.0×1011Nm21.0\times 10^{11}Nm^{-2} and 2.0×1011Nm2.2.0\times 10^{11}Nm^{-2}. .A brass wire and a steel wire of the same length are extended by 1mm1 \,mm each under the same force. If radii of brass and steel wires are RBR_B and RSR_S respectively, then

A

Rs=2RBR_s = \sqrt{2} R_B

B

RS=RB2R_S = \frac{R_{B}}{\sqrt{2}}

C

Rs=4RBR_s = 4R_B

D

Rs=RB2R_s = \frac{R_{B}}{2}

Answer

RS=RB2R_S = \frac{R_{B}}{\sqrt{2}}

Explanation

Solution

Increase in length, ΔL=FLYA=FLYπR2\Delta L=\frac{F L}{Y \cdot A}=\frac{F L}{Y \cdot \pi R^{2}} As F,LF, L and ΔL\Delta L are same hence, YR2=aY \cdot R^{2}= a constant 2.0×1011RS2=1.0×1011RB2\therefore 2.0 \times 10^{11} R_{S}^{2}=1.0 \times 10^{11} R_{B}^{2} RS=RB2\Rightarrow R_{S}=\frac{R_{B}}{\sqrt{2}}