JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 22)
The increase in pressure required to decrease the volume of a water sample by $0.2 \%$ is $\mathrm{P} \times 10^5 \mathrm{Nm}^{-2}$. Bulk modulus of water is $2.15 \times 10^9 \mathrm{Nm}^{-2}$. The value of P is _________ .
Answer
43
Explanation
$$\begin{aligned}
&\text { Since bulk modulus is given as }\\
&\begin{aligned}
& \mathrm{B}=\frac{-\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{~V}}{\mathrm{~V}}\right)} \\
& 2.15 \times 10^9=\frac{-\Delta \mathrm{P}}{-\left(\frac{0.2}{100}\right)} \\
& \Delta \mathrm{P}=2.15 \times 10^9 \times 2 \times 10^{-3} \\
& =4.3 \times 10^6=43 \times 10^5 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\end{aligned}$$
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