If a rubber ball is taken at the depth of 200 m in a pool. Its volume decreases by 0.1%. If the density of the water is 1 x 10³ kg / m³ and g = 10 m/s² , then the volume elasticity in N/m² will be

The correct option is 2 x 10⁹.

To find the volume elasticity (also known as the bulk modulus, $B$) of the rubber ball, we use the definition of bulk modulus:
$B=\frac{\Delta P}{\left(\frac{-\Delta V}{V}\right)}$
where:
- $\Delta P$ is the change in pressure (increase in pressure due to depth),
- $\frac{-\Delta V}{V}$ is the fractional change in volume (volume strain).

Step 1: Calculate the change in pressure ($\Delta P$)
The increase in pressure at a depth $h$ in a liquid of density $\rho$ under gravity $g$ is given by:
$\Delta P=\rho \cdot g\cdot h$

Given data:
- Depth ($h$) = 200 m
- Density of water ($\rho$) = 1 × 10³ kg/m³
- Acceleration due to gravity ($g$) = 10 m/s²

Substituting these values:
$\Delta P=(1\times {10}^3)\cdot 10\cdot 200$
$\Delta P=2\times {10}^6{\text{N/m}}^2$

Step 2: Determine the volume strain ($\frac{-\Delta V}{V}$)
The volume decreases by 0.1%. Therefore, the magnitude of the fractional change in volume is:
$\frac{\Delta V}{V}=0.1\%=\frac{0.1}{100}={10}^{-3}$

Step 3: Calculate the volume elasticity ($B$)
Substitute the values of $\Delta P$ and volume strain into the bulk modulus formula:
$B=\frac{2\times {10}^6}{{10}^{-3}}$
$B=2\times {10}^6\times {10}^3$
$B=2\times {10}^9{\text{N/m}}^2$

Thus, the volume elasticity is 2 × 10⁹ N/m².