The increase in pressure required to decrease the volume of a water sample byy 0.2%, is , Bulk modulus of water is . The value of P is __________. [2025]

Question

The increase in pressure required to decrease the volume of a water sample by 0.2% is $P \times 10^{5} {Nm}^{- 2}$ . Bulk modulus of water is $2.15 \times 10^{9} {Nm}^{- 2}$ . The value of P is __________. [2025]

Smart Achievers · Accepted Answer

(43)

Since bulk modulus is given as

$B = \frac{- △ P}{(\frac{△ V}{V})}$

$2.15 \times 10^{9} = \frac{- △ P}{- (\frac{0.2}{100})}$

$△ P = 2.15 \times 10^{9} \times 2 \times 10^{- 3} = 4.3 \times 10^{6} = 43 \times 10^{5} N / m^{2}$

Solution

Q.
The increase in pressure required to decrease the volume of a water sample by 0.2% is $P \times 10^{5} {Nm}^{- 2}$ . Bulk modulus of water is $2.15 \times 10^{9} {Nm}^{- 2}$ . The value of P is __________. [2025]

Ans.
(43)

Since bulk modulus is given as

$B = \frac{- △ P}{(\frac{△ V}{V})}$

$2.15 \times 10^{9} = \frac{- △ P}{- (\frac{0.2}{100})}$

$△ P = 2.15 \times 10^{9} \times 2 \times 10^{- 3} = 4.3 \times 10^{6} = 43 \times 10^{5} N / m^{2}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The increase in pressure required to decrease the volume of a water sample by 0.2% is P×105 Nm–2. Bulk modulus of water is 2.15×109 Nm–2. The value of P is __________. [2025]

Ans. (43) Since bulk modulus is given as B=–△P(△VV) 2.15×109=–△P–(0.2100) △P=2.15×109×2×10–3=4.3×106=43×105 N/m2

Want Us to Email you About Special Offers & Updates?

Q.
The increase in pressure required to decrease the volume of a water sample by 0.2% is $P \times 10^{5} {Nm}^{- 2}$ . Bulk modulus of water is $2.15 \times 10^{9} {Nm}^{- 2}$ . The value of P is __________. [2025]

Ans.
(43)

Since bulk modulus is given as

$B = \frac{- △ P}{(\frac{△ V}{V})}$

$2.15 \times 10^{9} = \frac{- △ P}{- (\frac{0.2}{100})}$

$△ P = 2.15 \times 10^{9} \times 2 \times 10^{- 3} = 4.3 \times 10^{6} = 43 \times 10^{5} N / m^{2}$