JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Elasticity & SHM

A uniform heavy rod of weight

10 kg ms^{- 2}

, cross-sectional area

100 cm^{2} f

and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is

2 \times 1 0^{11} Nm^{- 2}

. Neglecting the lateral contraction, find the elongation of rod due to its own weight:

Answer: (B)

Solution

A rod stretched under its own weight elongates as though its whole weight acted at its centre of mass, giving

Δ L = \frac{W L}{2 A Y}

. With

W = 10

L = 0.20

A = 100 c m^{2} = 1.0 \times 1 0^{- 2} m^{2}

and

Y = 2 \times 1 0^{11} N m^{- 2}

Δ L = \frac{10 \times 0.20}{2 \times ( 1.0 \times 1 0 ^{- 2} ) \times ( 2 \times 1 0 ^{11} )} = 5 \times 1 0^{- 10} m .

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option