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A solid cube is subjected to pressure of $5 \times 10^{5} {N/m}^{2}$ . Each side of cube is shortened by 1% then volumetric strain and bulk modulus of the cube are

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0.03 $, 5 \times 10^{5} {N/m}^{2}$

0.01 $, 1.67 \times 10^{- 7} {N/m}^{2}$

3 $,1.67 \times 10^{- 7} {N/m}^{2}$

0.03 $, 1.67 \times 10^{7} {N/m}^{2}$

answer is B.

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Detailed Solution

$\frac{Δ V}{V} \times 100 = 3 \frac{Δ l}{l} \times 100$

$\frac{Δ V}{V} \times 100 = 3 (1 %)$

$\frac{Δ V}{V} = \frac{3}{100} = 0.03$

$K = \frac{P}{\frac{- Δ V}{V}} = \frac{5 \times 10^{5}}{0.03}$

$= \frac{5}{3} \times 10^{7}$

$= 1.67 \times 10^{7} N / m^{2}$

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