JEE PYQ: Electromagnetic Waves Question 4
Question 4 - 2021 (17 Mar 2021 Shift 2)
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 60 W bulb at the same distance is $\sqrt{\frac{x}{5}}E$. Where the value of $x$ is ______.
Show Answer
Answer: 3
Solution
$c \epsilon_0 E^2 = \frac{100}{4\pi \times 3^2}$, $c \epsilon_0 \left(\sqrt{\frac{x}{5}} E\right)^2 = \frac{60}{4\pi \times 3^2}$. $\frac{x}{5} = \frac{3}{5}$, so $x = 3$.
BETA
