Question 85: Waves

Question

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

Accepted Answer

The volume ( $V$ ) of a sphere of radius $R$ is given by the formula :
$$ V=\frac{4}{3} \pi R^3 $$ .......(i)
According to the problem, the volume increases by 8 times ( $\mathrm{V}^{\prime}=8 \mathrm{~V}$ ) :
$$ 8 V=\frac{4}{3} \pi\left(R^{\prime}ight)^3 \ldots $$(ii)
$$\Rightarrow $$ $$ 8 	imes \frac{4}{3} \pi R^3=\frac{4}{3} \pi\left(R^{\prime}ight)^3 \Rightarrow R^{\prime}=2 R $$
The intensity (I) of a point source (of power P) is the power spread over the surface area of the sphere. The formula for intensity is :
$$ I=\frac{P}{A}=\frac{P}{4 \pi R^2} \Rightarrow I \propto \frac{1}{R^2} $$
Using the new radius $R^{\prime}=2 R$ :
$$ \frac{\mathrm{I}^{\prime}}{\mathrm{I}}=\frac{\left(1 / \mathrm{R}^{\prime 2}ight)}{\left(\frac{1}{\mathrm{R}^2}ight)}=\frac{\mathrm{R}^2}{\mathrm{R}^{\prime 2}}=\frac{\mathrm{R}^2}{4 \mathrm{R}^2}=\frac{1}{4} $$
$$\Rightarrow $$ $$ \mathrm{I}^{\prime}=\frac{\mathrm{I}}{4} $$
Therefore, when the volume of the spherical detector increases by 8 times, the intensity decreases by 4 times. Hence, the correct option is (B).

Answer

increase by 64 times

Answer

decrease by 4 times

Answer

decrease by 8 times

Answer

increase by 8 times

Waves

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

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