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

Solution

Physics sir · 4 steps

Volume tells you the radius, and the radius tells you the area the same power spreads over.

1. Intensity from power. A point source sends its power P out equally in all directions. At the detector wall, a sphere of radius r, that power is spread over an area 4π r².

   I = P/(4π r²) ∝ 1/r²

2. Volume to radius. Volume V = 4/3π r³, so r ∝ V^(1/3). If V becomes 8V, then r becomes 8^(1/3)r = 2r.
3. Radius to intensity. With the radius doubled, the area becomes 2² = 4 times larger. The same power over 4 times the area gives I' = I/4.
4. So. The intensity decreases by 4 times, which is option (b).

The idea to keep. Intensity from a point source goes as 1/r². When a question gives volume or area instead of distance, first convert it to the radius, then square.

Answer from the NTA final answer key. Solution by PrepWiser.