Q52 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
The increase in the pressure required to decrease the volume $(\Delta V)$ of water is $6.3 \times 10^7$ N/m$^2$. The percentage decrease in the volume is _____. (Bulk modulus of water $= 2.1 \times 10^9$ N/m$^2$.)
- A$2\%$
- Correct answerB$3\%$
- C$6\%$
- D$4\%$
Solution
Bulk modulus is given by $B = \dfrac{\Delta P}{-\dfrac{\Delta V}{V}}$
The fractional decrease in volume is $-\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}$
Substituting the given values:
$-\dfrac{\Delta V}{V} = \dfrac{6.3 \times 10^7}{2.1 \times 10^9} = 3 \times 10^{-2}$
The percentage decrease in volume is $\left(-\dfrac{\Delta V}{V}\right) \times 100$
Percentage decrease $= 3 \times 10^{-2} \times 100 = 3\%$
Answer: $3\%$
The fractional decrease in volume is $-\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}$
Substituting the given values:
$-\dfrac{\Delta V}{V} = \dfrac{6.3 \times 10^7}{2.1 \times 10^9} = 3 \times 10^{-2}$
The percentage decrease in volume is $\left(-\dfrac{\Delta V}{V}\right) \times 100$
Percentage decrease $= 3 \times 10^{-2} \times 100 = 3\%$
Answer: $3\%$
Hence, the correct answer is option: 'B'.