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

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0.6

1.67

answer is B.

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Detailed Solution

$\begin{array}{l} B u l k M o d u l u s = B = \frac{- Δ P}{\frac{Δ V}{V}} \\ \Rightarrow - \frac{Δ V}{V} = \frac{Δ P}{B} \\ \Rightarrow - 3 \frac{Δ ℓ}{ℓ} = \frac{Δ P}{B} [S i n c e, v o l u m e o f c u b e = V = ℓ^{3}] \\ \Rightarrow |\frac{Δ ℓ}{ℓ}| = \frac{1}{3} \frac{Δ P}{B} = \frac{1}{3} \times \frac{4 \times 10^{9}}{8 \times 10^{10}} = \frac{1}{60} \\ \Rightarrow |\frac{Δ ℓ}{ℓ}| \times 100 = \frac{1}{60} \times 100 = 1.67 % \end{array}$