Q41 (2026)
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
- $\text{increase by } 8 \text{ times}$
- $\text{increase by } 64 \text{ times}$
- $\text{decrease by } 8 \text{ times}$
- $\text{decrease by } 4 \text{ times}$
### Related Formula
$$I = \frac{P}{4\pi R^2}$$
$$V = \frac{4}{3}\pi R^3$$
### Core Logic
Since volume $V \propto R^3$, if volume is increased by 8 times:
$$V \rightarrow 8V \implies R^3 \rightarrow 8R^3 \implies R \rightarrow 2R$$
### Step 1: Intensity Relation
Since intensity $I \propto \frac{1}{R^2}$:
If $R \rightarrow 2R$, the area $A = 4\pi R^2$ increases by $2^2 = 4$ times ($A \rightarrow 4A$).
Therefore, intensity becomes $I \rightarrow \frac{I_0}{4}$.
### Pattern Recognition
Volume ratio cubed root dictates radius multiplier. The square of that radius multiplier is the area multiplier, which is the exact inverse of the intensity multiplier.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
