The bulk modulus of rubber is 9.1 times 10^8 | vedclass

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(B) The bulk modulus $K$ is defined as $K = \frac{\Delta P}{\Delta V/V}$.Here, the change in pressure $\Delta P$ at depth $h$ is given by $\Delta P =

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$91$

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(B) The bulk modulus $K$ is defined as $K = \frac{\Delta P}{\Delta V/V}$.Here, the change in pressure $\Delta P$ at depth $h$ is given by $\Delta P = ho gh$, where $ho$ is the density of water $(10^3 \, kg/m^3)$ and $g$ is the acceleration due to gravity $(10 \, m/s^2)$.The fractional change in volume is $\frac{\Delta V}{V} = 0.1 \, \% = \frac{0.1}{100} = 10^{-3}$.Substituting the values into the formula:$9.1 	imes 10^8 = \frac{10^3 	imes 10 	imes h}{10^{-3}}$$9.1 	imes 10^8 = \frac{10^4 	imes h}{10^{-3}}$$9.1 	imes 10^8 = 10^7 	imes h$$h = \frac{9.1 	imes 10^8}{10^7} = 9.1 	imes 10 = 91 \, m$.

The bulk modulus of rubber is $9.1 \times 10^8 \, N/m^2$. To what depth $h$ (in $m$) must a rubber ball be taken in a lake so that its volume is decreased by $0.1 \, \%$?

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