JEE Main 2025 Jan 24 Shift 2, Physics Q24: Properties of Fluids

JEE Main2025Jan 24, Shift 2Physics

The increase in pressure required to decrease the volume of a water sample by 0.2% is $P \times 1 0^{5} Nm^{- 2}$ . Bulk modulus of water is $2.15 \times 1 0^{9} Nm^{- 2}$ . The value of $P$ is ______.

Your answer

Solution

The bulk modulus is defined by

$B = - \frac{Δ P}{Δ V / V}$

For a compression, $Δ V / V = - 0.2% = - 2 \times 1 0^{- 3}$ , so $- Δ V / V = + 2 \times 1 0^{- 3}$ .

$Δ P = B \cdot \frac{Δ V}{V} = (2.15 \times 1 0^{9}) (2 \times 1 0^{- 3}) = 4.3 \times 1 0^{6} N/m^{2}$

Express as $P \times 1 0^{5}$ :

$Δ P = 4.3 \times 1 0^{6} = 43 \times 1 0^{5} N/m^{2}$

So $P = 43$ .

Answer: 43

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JEE Main 2025 Jan 24 Shift 2, Physics Q24: Properties of Fluids

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