A uniform wire (Young's modulus 2 × 10¹¹ Nm ⁻²) is subjected to longitudinal tensile stress of 5 × 10⁷ Nm ⁻². If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to: (a) 1.0 × 10⁻⁴ (b) 1.5 × 10⁻⁴ (c) 0.25 × 10⁻⁴ (d) 5 × 10⁻⁴

Answer: (c). Longitudinal strain (Δ L)/(L) = 5×10⁷2×10¹¹ = 2.5×10⁻⁴ V = π r² L ⇒ (Δ V)/(V) = (Δ L)/(L) - 2(Δ r)/(r) (radius decreases) 2×10⁻⁴ = 2.5×10⁻⁴ - 2(Δ r)/(r) (Δ r)/(r) = 0.25×10⁻⁴ So the answer is (c).

A uniform wire (Young's modulus 2 × 10¹¹ Nm ⁻²) is subjected to… | JEE Main 2013 (22 Apr, Shift 1)

Q12JEE Main 2013 · 22 APR · SHIFT 1Difficult

Shear Modulus & Poisson's Ratio

A uniform wire (Young's modulus $2 \times 10^{11}$ Nm $^{-2}$ ) is subjected to longitudinal tensile stress of $5 \times 10^{7}$ Nm $^{-2}$ . If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to:

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Answer: (c)

Longitudinal strain $\frac{\Delta L}{L} = \frac{5\times10^{7}}{2\times10^{11}} = 2.5\times10^{-4}$
$V = \pi r^2 L \Rightarrow \frac{\Delta V}{V} = \frac{\Delta L}{L} - 2\frac{\Delta r}{r}$ (radius decreases)
$2\times10^{-4} = 2.5\times10^{-4} - 2\frac{\Delta r}{r}$
$\frac{\Delta r}{r} = 0.25\times10^{-4}$
So the answer is (c).

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Shear Modulus & Poisson's Ratio question from JEE Main 2013 (22 Apr, Shift 1)

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