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(D) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$, which implies $V \propto R^3$.If the volume $V$ increases by $8$ times, then $R^3$ i

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decrease by $4$ times

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(D) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$, which implies $V \propto R^3$.If the volume $V$ increases by $8$ times, then $R^3$ increases by $8$ times, so the radius $R$ increases by $2$ times $(R \rightarrow 2R)$.The surface area of the sphere is $A = 4 \pi R^2$, which implies $A \propto R^2$.Since $R$ increases by $2$ times, the area $A$ increases by $2^2 = 4$ times $(A \rightarrow 4A)$.Intensity $I$ is defined as power per unit area, $I = \frac{P}{A}$.Since the power $P$ of the point source remains constant, $I \propto \frac{1}{A}$.Therefore, if the area $A$ increases by $4$ times, the intensity $I$ will decrease by $4$ times $(I \rightarrow \frac{I}{4})$.

$A$ point source is kept at the center of a spherically enclosed detector. If the volume of the detector is increased by $8$ times, the intensity will

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