Question

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

(Take, density of sea water $= \text{1}\text{0}^{3}\text{ kg }\text{m}^{- 3},$ bulk modulus of rubber$= 9 \times \text{1}\text{0}^{8}\text{ N }\text{m}^{- 2}\text{, }g = 10\text{ m }\text{s}^{- 2})$

• A. 9 m
• B. 18 m
• C. 90 m
• D. 180 m

Answer 1

Answer: (C)

90 m

Answer 2

: Let h be the depth at which the rubber ball be taken. Then

$P = h\rho g$ …(i)

By definition of bulk modulus,

$B = - \frac{P}{\Delta V/V}$

The negative sign shows that with increase in pressure, a decrease in volume occurs.

$\therefore P = B\frac{\Delta V}{V}$

Using (i)

$h\rho g = \mathbf{B}\frac{\Delta V}{V}orh = \frac{B}{\rho g}\frac{\Delta V}{V}$

Substituting the given values we get

$h = \frac{9 \times 10^{8}Nm^{- 2}}{10^{3}kgm^{- 3} \times 10ms^{- 2}}\left( \frac{0.1}{100} \right) = 90m$