To what depth must a rubber ball be taken in | vedclass

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(A) The percentage decrease in volume is given by $\frac{\Delta V}{V} 	imes 100 = 0.1$. Thus,the fractional change in volume is $\frac{\Delta V}{V} =

Vedclass · Accepted Answer

$100$

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(A) The percentage decrease in volume is given by $\frac{\Delta V}{V} 	imes 100 = 0.1$. Thus,the fractional change in volume is $\frac{\Delta V}{V} = \frac{0.1}{100} = 0.001$. The pressure exerted by the sea water at a depth $h$ is $P = h ho g$. The bulk modulus $B$ is defined as $B = \frac{P}{\Delta V / V}$. Rearranging the formula to solve for depth $h$,we get $h = \frac{B 	imes (\Delta V / V)}{ho g}$. Substituting the given values: $B = 9.8 	imes 10^8 \, N/m^2$,$\frac{\Delta V}{V} = 0.001$,$ho = 10^3 \, kg/m^3$,and $g = 9.8 \, m/s^2$. $h = \frac{9.8 	imes 10^8 	imes 0.001}{10^3 	imes 9.8} = \frac{9.8 	imes 10^5}{9.8 	imes 10^3} = 10^2 = 100 \, m$. Therefore,the required depth is $100 \, m$.

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 $9.8 \times 10^8 \, N/m^2$ and the density of sea water is $10^3 \, kg/m^3$.) (in $, m$)

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