Test Index

Waves Part 2

Section: Physics

Question : 4 of 54

Marks: +1, -0

Solution:

Concept:

Intensity from a point source varies inversely with the square of the radius.

Volume increase leads to radius increase, affecting intensity.

Explanation:

Volume of a sphere:

V = \frac{4}{3} \pi R^3

Volume increases by 8 times:

V' = 8V

8 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R')^3

Cancel common terms:

(R')^3 = 8R^3

, hence

R' = 2R

Intensity

I = \frac{P}{4\pi R^2}

, so

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

New intensity:

I' = \frac{P}{4\pi (2R)^2} = \frac{P}{4\pi \cdot 4R^2} = \frac{1}{4} \cdot \frac{P}{4\pi R^2} = \frac{I}{4}

Thus intensity decreases by 4 times.

Answer:

Option B: decrease by 4 times.

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