JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2021
- Hard
- Numerical
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 \(60\, W\) at the same distance is \(\sqrt{\frac{x}{5}} E\). Where the value of \(x=......... .\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(c \in_{0} E ^{2}=\frac{100}{4 \pi \times 3^{2}}\)
\(c \epsilon_{0}\left(\sqrt{\frac{ x }{5}} E \right)^{2}=\frac{60}{4 \pi \times 3^{2}}\)
\(\Rightarrow \frac{ x }{5}=\frac{3}{5}\)
\(\Rightarrow x =3\)
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