Question 260: Properties of Matter

Question

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 $$=10^3 \mathrm{kgm}^{-3}$$, Bulk modulus of rubber $$=9 	imes 10^8 \mathrm{~Nm}^{-2}$$, and $$g=10 \mathrm{~ms}^{-2}$$)

Accepted Answer

$$\begin{aligned}
& \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \
& \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \
& ho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \
& 10^3 	imes 10 	imes \mathrm{h}=-9 	imes 10^8 	imes\left(-\frac{0.02}{100}ight) \
& \Rightarrow \mathrm{h}=18 \mathrm{~m}
\end{aligned}$$

Properties of Matter

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02 \% is _______ m. (Take density of sea water =10^3 \mathrm{kgm}^{-3}, Bulk modulus of rubber =9 \times 10^8 \mathrm{~Nm}^{-2}, and g=10 \mathrm{~ms}^{-2}$)

Integer / Numeric Answer Type