To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%.

Answers (1)

B = Bulk modulus = 9.8 \times 108 N/m^{2}

p = density of water = 1000 kg/m3

percentage change in volume = \frac{\Delta V}{V} \times 100 = 0.1

let h be the depth to which the rubber ball is being taken = hpg

p = B \times \frac{\Delta V}{V}

h \times 9.8 \times 10^3 = 9.8 \times 10^8 \times \frac{1}{1000}

hence, h=100 m

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