If the average depth of an ocean is $4000 \ m$ and the bulk modulus of water is $2 \times 10^9 \ N m^{-2}$,then the fractional compression $\frac{\Delta V}{V}$ of water at the bottom of the ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is . . . . . . (Given,$g=10 \ m s^{-2}, \rho=1000 \ kg m^{-3}$)

If the compressibility of water is $\sigma$ per unit atmospheric pressure,then the decrease in volume $V$ due to $P$ atmospheric pressure will be

The density of a metal at normal pressure is $\rho$. When an additional pressure $P$ is applied,its density becomes $\rho'$. If its bulk modulus is $B$,then the ratio $\frac{\rho'}{\rho}$ is:

The approximate depth of an ocean is $2700 \, m$. The compressibility of water is $45.4 \times 10^{-11} \, Pa^{-1}$ and the density of water is $10^3 \, kg/m^3$. What fractional compression of water will be obtained at the bottom of the ocean?

Compute the bulk modulus of water from the following data: Initial volume $= 100.0 \; L$,Pressure increase $= 100.0 \; atm$ $(1 \; atm = 1.013 \times 10^{5} \; Pa)$,Final volume $= 100.5 \; L$. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

$A$ bottle has an opening of radius $a$ and length $b$. $A$ cork of length $b$ and radius $(a + \Delta a)$,where $(\Delta a << a)$,is compressed to fit into the opening completely (see figure). If the bulk modulus of the cork is $B$ and the frictional coefficient between the bottle and the cork is $\mu$,then the force needed to push the cork into the bottle is: