A rubber ball is taken into the deep sea such | vedclass

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Question

(A) The bulk modulus $K$ is defined as $K = -\frac{\Delta p}{\Delta V / V}$.Given that the volume decreases by $x \%$,we have $\frac{\Delta V}{V} = \f

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$\frac{Kx}{\rho g}$

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(A) The bulk modulus $K$ is defined as $K = -\frac{\Delta p}{\Delta V / V}$.Given that the volume decreases by $x \%$,we have $\frac{\Delta V}{V} = \frac{x}{100}$.The change in pressure $\Delta p$ at a depth $h$ in the sea is given by $\Delta p = \rho g h$.Substituting these into the bulk modulus formula: $K = \frac{\rho g h}{x / 100}$.Rearranging for depth $h$: $h = \frac{K \cdot x}{100 \cdot \rho \cdot g}$.Since $100$ is a constant,the depth $h$ is proportional to $\frac{Kx}{\rho g}$.

$A$ 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 $(g = \text{acceleration due to gravity})$:

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