JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Elasticity & SHM

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A cube of metal is subjected to a hydrostatic pressure 4 GPa. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal,

B = 8 \times 1 0^{10}

Pa)

(A)0.6
(B)5
(C)1.67✓ Correct
(D)20

Answer: (C)

Solution

The bulk modulus is

B = - \frac{Δ P}{Δ V / V}

, so the fractional volume change is

\frac{Δ V}{V} = \frac{Δ P}{B} = \frac{4 \times 1 0 ^{9}}{8 \times 1 0 ^{10}} = \frac{1}{20}

. For a cube

V = ℓ^{3}

, so

\frac{Δ V}{V} = 3 \frac{Δ ℓ}{ℓ}

, giving

\frac{Δ ℓ}{ℓ} = \frac{1}{60}

. As a percentage the side changes by about

1.67%

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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