Question 13 Marks
To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1\%$.
(The bulk modulus of rubber is $9.8 \times 10^8N m^{-2}$, and the density of sea water is $10^3kg m^{-3}$.)
(The bulk modulus of rubber is $9.8 \times 10^8N m^{-2}$, and the density of sea water is $10^3kg m^{-3}$.)
Answer
View full question & answer→According to the problem, Bulk modulus of rubber (B) = $9.8 \times 10^8N/ m^2$
Density of sea water (p) = $10^3kg/ m^3$ Percentage decrease in volume$\frac{\Delta\text{V}}{\text{V}}=0.1\%=\frac{0.1}{100}=10^{-3}$
$\rho = 10^3kg m^{-3}, h = ?$
Let the rubber ball be taken up to depth h.$\therefore$ Change in pressure $\text{(P)}=\text{h}\rho\text{g}$
we know, $\text{B}=\frac{\Delta\text{P}}{(\Delta\text{V/V})}\Rightarrow\Delta\text{P}=\text{B}\times\frac{\Delta\text{V}}{\text{V}}$$\Rightarrow\Delta \text{P}=9.8\times10^8\times10^{-3}=9.8\times10^5\text{Nm}^{-2}$
Also, $\Delta\text{P}=\rho\text{gh}$$\text{h}=\frac{\Delta\text{P}}{\rho\text{g}}=\frac{9.8\times10^5}{10^3\times9.8}\Rightarrow\text{h}=10^2\text{m}=100\text{m}$
Density of sea water (p) = $10^3kg/ m^3$ Percentage decrease in volume$\frac{\Delta\text{V}}{\text{V}}=0.1\%=\frac{0.1}{100}=10^{-3}$
$\rho = 10^3kg m^{-3}, h = ?$
Let the rubber ball be taken up to depth h.$\therefore$ Change in pressure $\text{(P)}=\text{h}\rho\text{g}$
we know, $\text{B}=\frac{\Delta\text{P}}{(\Delta\text{V/V})}\Rightarrow\Delta\text{P}=\text{B}\times\frac{\Delta\text{V}}{\text{V}}$$\Rightarrow\Delta \text{P}=9.8\times10^8\times10^{-3}=9.8\times10^5\text{Nm}^{-2}$
Also, $\Delta\text{P}=\rho\text{gh}$$\text{h}=\frac{\Delta\text{P}}{\rho\text{g}}=\frac{9.8\times10^5}{10^3\times9.8}\Rightarrow\text{h}=10^2\text{m}=100\text{m}$