Q. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.1%? [Take : density of sea water $= 1000kgm^{- 3}$, Bulk modulus of rubber $= 9 \times 10^{8}Nm^{- 2}$; acceleration due to gravity $= 10\ ms^{- 2}$]
A) $9\ m$
B) $18\ m$
C) $180\ m$
D) $90\ m$
Explanation
(d) $K = \frac{p}{- \frac{\Delta V}{V}} \Rightarrow K = \frac{h\rho g}{0.1 \times 10^{- 2}}$
$\Rightarrow h = \frac{K \times 0.1 \times 10^{- 2}}{\rho g} = \frac{9 \times 10^{8} \times 10^{3}}{10^{3} \times 10} = 90\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.