If a brass sphere of radius 36 cm is submerg | vedclass

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Question

(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V/V}$.Given: $B = 60 \ GPa = 60 	imes 10^9 \ Pa$, $\Delta P = 10^7 \ Pa$, $r = 36

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$2 	imes 10^{-3} \ cm$

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(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V/V}$.Given: $B = 60 \ GPa = 60 	imes 10^9 \ Pa$, $\Delta P = 10^7 \ Pa$, $r = 36 \ cm = 0.36 \ m$.The volume of a sphere is $V = \frac{4}{3}\pi r^3$, so the fractional change in volume is $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.Substituting this into the Bulk modulus formula: $B = -\frac{\Delta P}{3 \Delta r / r}$.Rearranging for $\Delta r$: $\Delta r = -\frac{\Delta P \cdot r}{3B}$.Taking the magnitude: $|\Delta r| = \frac{10^7 	imes 0.36}{3 	imes 60 	imes 10^9}$.$|\Delta r| = \frac{3.6 	imes 10^6}{180 	imes 10^9} = \frac{3.6}{180} 	imes 10^{-3} = 0.02 	imes 10^{-3} \ m = 2 	imes 10^{-5} \ m$.Converting to cm: $2 	imes 10^{-5} 	imes 10^2 \ cm = 2 	imes 10^{-3} \ cm$.

If a brass sphere of radius $36 \ cm$ is submerged in a lake at a depth where the pressure is $10^7 \ Pa$, then the change in the radius of the sphere is $($Bulk modulus of brass $= 60 \ GPa)$.

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