JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Elasticity & SHM

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by

0.02%

is ____ m.
(Take density of sea water

= 1 0^{3} kgm^{- 3}

, Bulk modulus of rubber

= 9 \times 1 0^{8} Nm^{- 2}

, and

g = 10 ms^{- 2}

)

Answer: 18

Solution

Bulk modulus:

B = - \frac{Δ P}{Δ V / V}

, so

Δ P = B \cdot \frac{Δ V}{V}

.

The increase in pressure at depth

h

in the sea is

Δ P = ρ g h

. Equating:

(1 0^{3}) (10) h = 9 \times 1 0^{8} \times (\frac{0.02}{100})

1 0^{4} h = 9 \times 1 0^{8} \times 2 \times 1 0^{- 4} = 1.8 \times 1 0^{5}

h = 18 m

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer