A point source is kept at the center of a spherically enclosed de… — Solution | JEE Physics

MCQEasyJEE 2026Electromagnetic Spectrum & Applications

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

Answer

Correct answer:C

Standard Method

Given: A point source is at the center of a spherical detector, and the detector volume is increased by $8$ times.

Find: How the intensity changes.

For a spherical detector,

V \propto r^3

If the volume becomes $8V$ , then

r_{\text{new}}^3 = 8r^3

So,

r_{\text{new}} = 2r

For a point source, intensity follows the inverse square law:

I \propto \frac{1}{r^2}

Therefore,

I_{\text{new}} \propto \frac{1}{(2r)^2} = \frac{1}{4r^2}

Thus,

I_{\text{new}} = \frac{I}{4}

Therefore, the intensity decreases by $4$ times. The correct option is C.

Using a direct inverse relation with volume and assuming $I \propto \frac{1}{V}$ . This is wrong because intensity depends on distance from the point source, not directly on enclosed volume. First relate volume to radius using $V \propto r^3$ , then apply $I \propto \frac{1}{r^2}$ .
Assuming that increasing volume by $8$ times makes radius $8r$ . This is wrong because radius scales as the cube root of volume. Since $8 = 2^3$ , the new radius is $2r$ , not $8r$ .
Applying the inverse square law incorrectly and concluding the intensity decreases by $8$ times. This is wrong because the radius only doubles, so intensity becomes $\frac{1}{2^2} = \frac{1}{4}$ of the original value.

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