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A regular cube of side 0.1 m is subjected to a pressure of 15 atm. Then its volume decreases by $2.5 m m^{3}$ . Then bulk modulus of cube is $(1 a t m = 0.1 \times 10^{6} P a)$

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300 GPa

450 GPa

600 GPa

None

answer is C.

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Detailed Solution

$S = 0.1 m$

$P = 15 a t m$

$= 15 \times 0.1 \times 10^{6} P a$

$∆ V = 2.5 \times 10^{- 6} m^{3}$

$B = ?$

$B = - \frac{ρ}{\frac{∆ V}{V}}$

$B = \frac{15 \times 0.1 \times 10^{6} \times {(0.1)}^{3}}{2.5 \times 10^{- 6}}$ $B = 0.6 \times 10^{12} \times 10^{- 3}$

$= 6 \times 10^{11}$

$= 600 G p a$

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