A solid cube of copper of edge 10 , cm is sub | vedclass

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(B) Given:Edge of the cube $a = 10 \, cm = 0.1 \, m = 10^{-1} \, m$.Initial volume $V = a^3 = (10^{-1} \, m)^3 = 10^{-3} \, m^3$.Pressure $P = 7 	ime

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$5 	imes 10^{-8}$

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(B) Given:Edge of the cube $a = 10 \, cm = 0.1 \, m = 10^{-1} \, m$.Initial volume $V = a^3 = (10^{-1} \, m)^3 = 10^{-3} \, m^3$.Pressure $P = 7 	imes 10^6 \, Pa$.Bulk modulus $B = 140 \, GPa = 140 	imes 10^9 \, Pa$.We know the formula for Bulk modulus is $B = \frac{P}{(-\Delta V / V)}$,where $-\Delta V$ is the contraction in volume.Rearranging for contraction in volume: $-\Delta V = \frac{P 	imes V}{B}$.Substituting the values:$-\Delta V = \frac{(7 	imes 10^6 \, Pa) 	imes (10^{-3} \, m^3)}{140 	imes 10^9 \, Pa}$.$-\Delta V = \frac{7 	imes 10^3}{140 	imes 10^9} = \frac{1}{20} 	imes 10^{-6} = 0.05 	imes 10^{-6} = 5 	imes 10^{-8} \, m^3$.Thus,the contraction in volume is $5 	imes 10^{-8} \, m^3$.

$A$ solid cube of copper of edge $10 \, cm$ is subjected to a hydraulic pressure of $7 \times 10^6 \, Pa$. If the Bulk modulus of copper is $140 \, GPa$,then the contraction in its volume will be ................ $m^3$.

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