Question

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 $1\times {10}^{3}kg/m^3$and$g=10m/s^2$ then the volume elasticity in $N/m^2$ will be

• A. ${10}^8$
• B. $2\times {10}^8$
• C. ${10}^9$
• D. $2\times {10}^9$

Answer 1

The correct option is D $2\times {10}^9$

Pressure change on the rubber ball at the depth of $200m$,

$P=200\times 1\times {10}^3\times 10$

$2\times {10}^{6}N/m^2.$

$\therefore$ fractional change in volume,

$\frac{△V}{V}=\frac{0.1}{100}$

$\therefore$ Volume elasticity$\left(K\right)=\frac{p}{\frac{△V}{V}}$

$=\frac{2\times {10}^6}{\frac{0.1}{100}}$

$=2\times {10}^{9}N/m^2$.