JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 28)
A certain pressure '$$\mathrm{P}$$' is applied to 1 litre of water and 2 litre of a liquid separately. Water gets compressed to $$0.01 \%$$ whereas the liquid gets compressed to $$0.03 \%$$. The ratio of Bulk modulus of water to that of the liquid is $$\frac{3}{x}$$. The value of $$x$$ is ____________.
Answer
1
Explanation
Given, Volume of water, $V_1=1$ litre
Volume of liquid, $V_2=2$ litre,
Pressure $=P$
$$ \begin{aligned} & \left(\frac{\Delta V}{V} \times 100\right)_\text { water }=0.01 \% ;\\\\ & \left(\frac{\Delta V}{V} \times 100\right)_\text { liquid }=0.03 \% \\\\ & \text { Bulk modulus, } B=\frac{-P V}{\Delta V} ; \\\\ & \frac{B_{\text {water }}}{B_{\text {liquid }}}=\frac{\left(\frac{\Delta V}{V}\right)_\text { liquid }}{\left(\frac{\Delta V}{V}\right)_\text { water }}=\frac{\frac{0.03}{100}}{\frac{0.01}{100}}=3 \end{aligned} $$
On comparing the given value with $\frac{3}{x}$, we get $x=1$.
Volume of liquid, $V_2=2$ litre,
Pressure $=P$
$$ \begin{aligned} & \left(\frac{\Delta V}{V} \times 100\right)_\text { water }=0.01 \% ;\\\\ & \left(\frac{\Delta V}{V} \times 100\right)_\text { liquid }=0.03 \% \\\\ & \text { Bulk modulus, } B=\frac{-P V}{\Delta V} ; \\\\ & \frac{B_{\text {water }}}{B_{\text {liquid }}}=\frac{\left(\frac{\Delta V}{V}\right)_\text { liquid }}{\left(\frac{\Delta V}{V}\right)_\text { water }}=\frac{\frac{0.03}{100}}{\frac{0.01}{100}}=3 \end{aligned} $$
On comparing the given value with $\frac{3}{x}$, we get $x=1$.
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