Mechanical Properties of Solids

JEE Main level, single correct, medium

85 of 118

Question 85 of 118

The increase in the pressure required to decrease the volume (ΔV)(\Delta V) of water is 6.3×1076.3 \times 10^7 N/m2^2. The percentage decrease in the volume is _____. (Bulk modulus of water =2.1×109= 2.1 \times 10^9 N/m2^2.)

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Answer: (B) 3%3\%

Bulk modulus is given by B=ΔP−ΔVVB = \dfrac{\Delta P}{-\dfrac{\Delta V}{V}}

The fractional decrease in volume is −ΔVV=ΔPB-\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}

Substituting the given values:

−ΔVV=6.3×1072.1×109=3×10−2-\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 (−ΔVV)×100\left(-\dfrac{\Delta V}{V}\right) \times 100

Percentage decrease =3×10−2×100=3%= 3 \times 10^{-2} \times 100 = 3\%

Answer: 3%3\%

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