JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Elasticity & SHM

A uniform heavy rod of mass 20 kg, Cross sectional area

0.4 m^{2}

and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is

x \times 1 0^{- 9} m

. The value of x is ______.
(Given: Young's modulus

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

and

g = 10 ms^{- 2}

)

Answer: 25

Solution

For a rod hanging under its own weight, the elongation is given by

Δ ℓ = \frac{m g ℓ}{2 A Y}

, where the factor of 2 accounts for the varying tension along the rod (zero at the bottom, maximum at the top).

Substituting the given values:

Δ ℓ = \frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 1 0 ^{11}} = \frac{4000}{1.6 \times 1 0 ^{11}} = 25 \times 1 0^{- 9} m

Therefore,

x = 25

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Integer answer