
and $\quad \frac{Q}{P}=\frac{405}{5}$ .......$(2)$
Solving $S^{2}=400 \times 405$
$\Rightarrow \quad S=402.5\, \Omega$
Statement $1 :$ The possibility of an electric bulb fusing is higher at the time of switching $ON.$
Statement $2:$ Resistance of an electric bulb when it is not lit up is much smaller than when it is lit up.
