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(B) The intensity of radiation $I$ at a distance $r$ from a source of power $P$ is given by $I = \frac{P}{4 \pi r^2}$.Also,the intensity is related to

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$3$

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(B) The intensity of radiation $I$ at a distance $r$ from a source of power $P$ is given by $I = \frac{P}{4 \pi r^2}$.Also,the intensity is related to the electric field amplitude $E$ by the relation $I = \frac{1}{2} c \epsilon_0 E^2$.For the $100 \, W$ bulb: $\frac{1}{2} c \epsilon_0 E^2 = \frac{100}{4 \pi (3)^2}$.For the $60 \, W$ bulb: $\frac{1}{2} c \epsilon_0 (\sqrt{\frac{x}{5}} E)^2 = \frac{60}{4 \pi (3)^2}$.Dividing the second equation by the first,we get: $(\sqrt{\frac{x}{5}})^2 = \frac{60}{100}$.$\frac{x}{5} = \frac{6}{10} = \frac{3}{5}$.Therefore,$x = 3$.

The electric field intensity produced by the radiation coming from a $100 \, W$ bulb at a distance of $3 \, m$ is $E$. The electric field intensity produced by the radiation coming from a $60 \, W$ bulb at the same distance is $\sqrt{\frac{x}{5}} E$. The value of $x$ is ......... .

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