Question:
The intensity at spherical surface due to an isotropic point source placed at its center is $I_0$. If its volume is increased by $8$ times, what will be intensity at the spherical surface?
The intensity at spherical surface due to an isotropic point source placed at its center is $I_0$. If its volume is increased by $8$ times, what will be intensity at the spherical surface?
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For isotropic sources, intensity varies inversely as square of distance.
Updated On: Feb 2, 2026
- Increase by $128$ times
- Increase by $8$ times
- Decrease by $4$ times
- Decrease by $8$ times
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The Correct Option is C
Solution and Explanation
Step 1: Relation between volume and radius.
\[ V \propto R^3 \Rightarrow 8V \Rightarrow R' = 2R \]
Step 2: Relation between intensity and radius.
\[ I \propto \dfrac{1}{R^2} \]
Step 3: Finding new intensity.
\[ I' = \dfrac{I_0}{(2)^2} = \dfrac{I_0}{4} \]
\[ V \propto R^3 \Rightarrow 8V \Rightarrow R' = 2R \]
Step 2: Relation between intensity and radius.
\[ I \propto \dfrac{1}{R^2} \]
Step 3: Finding new intensity.
\[ I' = \dfrac{I_0}{(2)^2} = \dfrac{I_0}{4} \]
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