JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2026
- Easy
- MCQ
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\(^2\).)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3\%\)
Step-by-step Solution
Detailed explanation
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: \(3\%\)
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