Mechanical Properties of Solids

JEE Main level, single correct, medium

99 of 118

Question 99 of 118

A cube of metal is subjected to a hydrostatic pressure 4GPa4GPa. The percentage change in the length of the side of the cube is close to : (Given bulk modulus of metal, B=8×1010PaB=8\times {10}^{10}Pa)

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Answer: (D) 1.671.67

B=−∆P∆VVB=-\dfrac{∆P}{\dfrac{∆V}{V}}
∆VV=−∆PB\dfrac{∆V}{V}=-\dfrac{∆P}{B}
Magnitude is given by,
∣∆VV∣=∆PB\left|\dfrac{∆V}{V}\right|=\dfrac{∆P}{B}
Volume in terms of length can be represented as,
3×∣∆LL∣=∆PB3\times \left|\dfrac{∆L}{L}\right|=\dfrac{∆P}{B}
Therefore,
∣∆LL∣=13×∆PB\left|\dfrac{∆L}{L}\right|=\dfrac{1}{3}\times \dfrac{∆P}{B}
Substituting the values we get,
∣∆LL∣=13×4×1098×1010=160\left|\dfrac{∆L}{L}\right|=\dfrac{1}{3}\times \dfrac{4\times {10}^{9}}{8\times {10}^{10}}=\dfrac{1}{60}
Percantage change is,
160×100=53=1.67\dfrac{1}{60}\times 100=\dfrac{5}{3}=1.67

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