Mechanical Properties of Solids

JEE Main level, numerical answer, medium

86 of 118

Question 86 of 118

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 cm30.8 \mathrm{~cm}^3. If the bulk modulus of the liquid is 2 GPa , the initial volume of the liquid was _______ litre. (\left(\right. Take 1 atm=105 Pa)\left.1 \mathrm{~atm}=10^5 \mathrm{~Pa}\right)

Show answer and solution

Answer: 4

Given, Initial pressure of liquid (Pi)=1 atm\left(P_i\right)=1 \mathrm{~atm}
Final pressure of liquid (Pf)=5 atm\left(\mathrm{P}_{\mathrm{f}}\right)=5 \mathrm{~atm}
Change in pressure (dP)=Pf−Pi=4 atm(\mathrm{dP})=\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}=4 \mathrm{~atm}
=4×105 Pa=4 \times 10^5 \mathrm{~Pa}
Change in volume (dV)=−0.8 cm3(\mathrm{dV})=-0.8 \mathrm{~cm}^3
Bulk modulus (B)=2×109 Pa(B)=2 \times 10^9 \mathrm{~Pa}
Now, B=−dP(dV/V)⇒V=−B(dVdP)B=\frac{-d P}{(d V / V)} \Rightarrow V=-B\left(\frac{d V}{d P}\right)
$\begin{aligned}
\Rightarrow & V=-2 \times 10^9 \times \frac{\left(-0.8 \times 10^{-6}\right)}{4 \times 10^5} \\
& =4 \times 10^{-3} \mathrm{~m}^3=4 \text { litre }
\end{aligned}$

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Elasticity: stress, strain and Young’s modulus simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Mechanical Properties of Solids (Bulk Modulus), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.