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MHT CET 2020 (16 Oct, Shift 1) · Young's Modulus & Elongation

Young's Modulus & Elongation question from MHT CET 2020 (16 Oct, Shift 1)

Q24MHT CET 2020 · 16 OCT · SHIFT 1Medium

Young's Modulus & Elongation

Two rods of same material and volume having circular cross-section are subjected to tension $T$ . Within the elastic limit, same force is applied to both the rods. Diameter of the first rod is half of the second rod, then the extensions of first rod to second rod will be in the ratio

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Answer: (c)

Same volume: $L = \frac{V}{A}$ , so $\Delta l = \frac{FL}{AY} = \frac{FV}{A^2 Y} \propto \frac{1}{A^2} \propto \frac{1}{d^4}$
$\frac{\Delta l_1}{\Delta l_2} = \left(\frac{d_2}{d_1}\right)^4 = 2^4 = 16$
So the answer is (c) $16 : 1$ .

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