A uniform heavy rod of weight 10 kg ms−2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 1011 Nm−2. Neglecting the lateral contraction, find the elongation of rod due to its own weight. (a) 2 × 10−9 m (b) 5 × 10−8 m (c) 4 × 10−8 m (d) 5 × 10−10 m

Answer: (d). A rod hanging under its own weight stretches by Δ l = WL/2AY (average tension is W/2). Δ l = 10 × 0.202 × 100 × 10⁻⁴ × 2 × 10¹¹ = 24 × 10⁹ Δ l = 5 × 10⁻¹⁰ m, so the answer is (d).

A uniform heavy rod of weight 10 kg ms−2, cross-sectional area 100… | JEE Main 2021 (31 Aug, Shift 1)

Q19JEE Main 2021 · 31 AUG · SHIFT 1Medium

Wire Combinations & Special Cases

A uniform heavy rod of weight 10 kg ms−2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 $\times$ 1011 Nm−2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.

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Answer: (d)

A rod hanging under its own weight stretches by $\Delta l = \frac{WL}{2AY}$ (average tension is $W/2$ ).
$\Delta l = \frac{10 \times 0.20}{2 \times 100 \times 10^{-4} \times 2 \times 10^{11}} = \frac{2}{4 \times 10^{9}}$
$\Delta l = 5 \times 10^{-10}\text{ m}$ , so the answer is (d).

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