A solid sphere is taken to a depth of 1500 m in an ocean. If the...

The increase in hydrostatic pressure at a depth $h$ is given by:
$\Delta P = \rho_w g h$
$\Delta P = 10^3 \times 10 \times 1500 = 1.5 \times 10^7 \text{ N m}^{-2}$

The fractional change in volume is related to the bulk modulus $B$ by:
$\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}$
$\dfrac{\Delta V}{V} = \dfrac{1.5 \times 10^7}{5 \times 10^9} = 3 \times 10^{-3}$

Since the mass of the sphere remains constant, the density $\rho = \dfrac{M}{V}$.
For small changes, the fractional increase in density is approximately equal to the fractional decrease in volume:
$\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta V}{V} = 3 \times 10^{-3}$

The percentage increase in density is:
$\dfrac{\Delta \rho}{\rho} \times 100\% = 3 \times 10^{-3} \times 100\% = 0.3\%$

Comparing this with $x \times 10^{-1} \%$, we get:
$x \times 10^{-1} = 0.3 \implies x = 3$

Answer: $3$