The efficiency of a bulb of power $60 \ W$ is $16 \%$. The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of $2 \ m$ from the bulb is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}\right)$ (in $Vm^{-1}$)

  • A
    $24$
  • B
    $16$
  • C
    $9$
  • D
    $12$

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