When a rubber ball is taken to a depth of _________ m in deep sea, its volume decreases by 0.5%.
(The bulk modulus of rubber = $$9.8 \times 10^8$$ N m$$^{-2}$$, Density of sea water = $$10^3$$ kg m$$^{-3}$$, g = 9.8 m s$$^{-2}$$)

We have to find that depth in sea water where the pressure becomes high enough to compress the rubber ball so that its volume falls by 0.5 %. The datum values supplied are:

Bulk modulus of rubber $$B = 9.8 \times 10^{8}\ \text{N m}^{-2}$$,   density of sea-water $$\rho = 10^{3}\ \text{kg m}^{-3}$$,   and the acceleration due to gravity $$g = 9.8\ \text{m s}^{-2}$$.

The fractional change in volume is given as $$0.5\%$$, i.e.

$$\frac{\Delta V}{V} = -0.5\% = -\frac{0.5}{100} = -0.005.$$

For bulk deformation the defining relation is first stated:

$$B = -\frac{\Delta P}{\dfrac{\Delta V}{V}}.$$

Here $$\Delta P$$ is the increase in external pressure which produces the fractional volume change $$\dfrac{\Delta V}{V}$$. In our situation the sign merely tells the direction of change; we can work with magnitudes:

$$B = \frac{\Delta P}{\left|\dfrac{\Delta V}{V}\right|}.$$

The pressure increase at a depth $$h$$ in a fluid of density $$\rho$$ is furnished by the hydrostatic relation

$$\Delta P = \rho\,g\,h.$$

Substituting this expression for $$\Delta P$$ into the bulk-modulus formula gives

$$B \;=\; \frac{\rho\,g\,h}{\left|\dfrac{\Delta V}{V}\right|}.$$

Solving for the depth $$h$$ yields

$$h \;=\; \frac{B\,\left|\dfrac{\Delta V}{V}\right|}{\rho\,g}.$$

Next we substitute the numerical quantities step by step:

$$h = \frac{\displaystyle 9.8 \times 10^{8}\ \text{N m}^{-2} \;\times\; 0.005}{\displaystyle 10^{3}\ \text{kg m}^{-3} \;\times\; 9.8\ \text{m s}^{-2}}.$$

First multiply the numerator:

$$9.8 \times 10^{8} \times 0.005 \;=\; 9.8 \times 5 \times 10^{8} \times 10^{-3} = 49 \times 10^{5} = 4.9 \times 10^{6}.$$

Now compute the denominator:

$$10^{3} \times 9.8 \;=\; 9.8 \times 10^{3}.$$

Dividing the two magnitudes, we have

$$h = \frac{4.9 \times 10^{6}}{9.8 \times 10^{3}} = \left(\frac{4.9}{9.8}\right) \times 10^{3} = 0.5 \times 10^{3} = 5.0 \times 10^{2} \ \text{m}.$$

Therefore

$$h = 500 \ \text{m}.$$

So, the answer is $$500\ \text{m}$$.