Mechanical Properties Of Solids Question 8
Question 8 - 2024 (31 Jan Shift 1)
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02 \%$ is $m$.
(Take density of sea water $=10^{3} kgm^{-3}$, Bulk modulus of rubber $=9 \times 10^{8} Nm^{-2}$, and $g=10 ms^{-2}$ )
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Answer: (18)
Solution:
$\beta=\frac{-\Delta P}{\frac{\Delta V}{V}}$
$\Delta P=-\beta \frac{\Delta V}{V}$
$\rho g h=-\beta \frac{\Delta V}{V}$
$10^{3} \times 10 \times h=-9 \times 10^{8} \times\left(-\frac{0.02}{100}\right)$
$\Rightarrow h=18 m$
