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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 \(x=......... .\)

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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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