$A$ rubber ball is taken into the deep sea such that its volume decreases by $x \%$. The bulk modulus of rubber is $K$ and the density of sea water is $\rho$. The depth $h$ to which the rubber ball is taken is proportional to $(g = \text{acceleration due to gravity})$:
- A$\frac{Kx}{\rho g}$
- B$\frac{\rho g}{Kx}$
- C$\frac{K}{x \rho g}$
- D$\frac{x \rho g}{K}$
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$A$ swimming pool has a depth of $22 \ m$ and area $700 \ m^2$. Calculate the fractional change $\frac{\Delta V}{V}$ of water at the bottom of the swimming pool. Given that the bulk modulus of water is $2.2 \times 10^9 \ N \ m^{-2}$,$g = 10 \ m \ s^{-2}$,and the density of water is $1000 \ kg \ m^{-3}$.
When a pressure of $100$ atmosphere is applied on a spherical ball,its volume reduces by $0.01\%$. The bulk modulus of the material of the ball in $dyne/cm^2$ is:
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View SolutionThe compressibility of water is $5 \times 10^{-10} \ m^2/N$. $A$ pressure of $15 \times 10^6 \ Pa$ is applied on $100 \ ml$ volume of water. The change in the volume of water is:
The wrong statement in the following is
The density of a metal at normal pressure $P$ is $\varrho$. When it is subjected to an excess pressure $p$,the density becomes $\varrho^{\prime}$. If $K$ is the bulk modulus of the metal,then the ratio $\frac{\varrho^{\prime}}{\varrho}$ is
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