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(C) Let the height of the cylinder be $h$ and the radius be $r$.The volume of the cylinder is $V_{cyl} = \pi r^2 h$ and the volume of the hemisphere i

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(C) Let the height of the cylinder be $h$ and the radius be $r$.The volume of the cylinder is $V_{cyl} = \pi r^2 h$ and the volume of the hemisphere is $V_{hemi} = \frac{2}{3} \pi r^3$.The total volume of the system is $V_1 = \pi r^2 h + \frac{2}{3} \pi r^3$.When the height of the cylinder is doubled $(h 	o 2h)$,the new volume $V_2$ is:$V_2 = \pi r^2 (2h) + \frac{2}{3} \pi r^3 = 2 \pi r^2 h + \frac{2}{3} \pi r^3$.Given that the volume increases by $50\,\%$,we have $V_2 = 1.5 V_1 = \frac{3}{2} V_1$.$\frac{2 \pi r^2 h + \frac{2}{3} \pi r^3}{\pi r^2 h + \frac{2}{3} \pi r^3} = \frac{3}{2}$.Cross-multiplying gives $4 \pi r^2 h + \frac{4}{3} \pi r^3 = 3 \pi r^2 h + 2 \pi r^3$.$\pi r^2 h = \frac{2}{3} \pi r^3$,which implies $h = \frac{2}{3} r$.Now,if the radii are doubled $(r 	o 2r)$ while keeping the height $h$ fixed,the new volume $V'_2$ is:$V'_2 = \pi (2r)^2 h + \frac{2}{3} \pi (2r)^3 = 4 \pi r^2 h + \frac{16}{3} \pi r^3$.Substituting $h = \frac{2}{3} r$ into the expression for $V'_2$:$V'_2 = 4 \pi r^2 (\frac{2}{3} r) + \frac{16}{3} \pi r^3 = \frac{8}{3} \pi r^3 + \frac{16}{3} \pi r^3 = \frac{24}{3} \pi r^3 = 8 \pi r^3$.The original volume $V_1$ in terms of $r$ is:$V_1 = \pi r^2 (\frac{2}{3} r) + \frac{2}{3} \pi r^3 = \frac{2}{3} \pi r^3 + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi r^3$.The ratio $\frac{V'_2}{V_1} = \frac{8 \pi r^3}{\frac{4}{3} \pi r^3} = 8 	imes \frac{3}{4} = 6$.The increase in volume is $V'_2 - V_1 = 6 V_1 - V_1 = 5 V_1$.Percentage increase = $\frac{5 V_1}{V_1} 	imes 100\,\% = 500\,\%$.

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