JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2025
- Medium
- Numerical
A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to change of volume of \(0.8 \mathrm{~cm}^3\). If the bulk modulus of the liquid is 2 GPa , the initial volume of the liquid was _______ litre. \(\left(\right.\) Take \(\left.1 \mathrm{~atm}=10^5 \mathrm{~Pa}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
Given, Initial pressure of liquid \(\left(P_i\right)=1 \mathrm{~atm}\)
Final pressure of liquid \(\left(\mathrm{P}_{\mathrm{f}}\right)=5 \mathrm{~atm}\)
Change in pressure \((\mathrm{dP})=\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}=4 \mathrm{~atm}\)
\(=4 \times 10^5 \mathrm{~Pa}\)
Change in volume \((\mathrm{dV})=-0.8 \mathrm{~cm}^3\)
Bulk modulus \((B)=2 \times 10^9 \mathrm{~Pa}\)
Now, \(B=\frac{-d P}{(d V / V)} \Rightarrow V=-B\left(\frac{d V}{d P}\right)\)
\(\begin{aligned}
\Rightarrow & V=-2 \times 10^9 \times \frac{\left(-0.8 \times 10^{-6}\right)}{4 \times 10^5} \\
& =4 \times 10^{-3} \mathrm{~m}^3=4 \text { litre }
\end{aligned}\)
Final pressure of liquid \(\left(\mathrm{P}_{\mathrm{f}}\right)=5 \mathrm{~atm}\)
Change in pressure \((\mathrm{dP})=\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}=4 \mathrm{~atm}\)
\(=4 \times 10^5 \mathrm{~Pa}\)
Change in volume \((\mathrm{dV})=-0.8 \mathrm{~cm}^3\)
Bulk modulus \((B)=2 \times 10^9 \mathrm{~Pa}\)
Now, \(B=\frac{-d P}{(d V / V)} \Rightarrow V=-B\left(\frac{d V}{d P}\right)\)
\(\begin{aligned}
\Rightarrow & V=-2 \times 10^9 \times \frac{\left(-0.8 \times 10^{-6}\right)}{4 \times 10^5} \\
& =4 \times 10^{-3} \mathrm{~m}^3=4 \text { litre }
\end{aligned}\)
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