$\Rightarrow \text { In 't' time , } \Delta Q =\left(\frac{ V ^{2}}{ R }\right) t$
Given that, (for same source, $v=$ same)
$Q _{0}=\frac{ v ^{2}}{ R _{1}} \times 20=\frac{ V ^{2}}{ R _{2}} \times 60 \ldots$ $(i)$
$\Rightarrow R _{2}=3 R _{1} \ldots.$ $(ii)$
If they are connected in parallel then
$\operatorname{Req}=\frac{ R _{2} R _{1}}{ R _{1}+ R _{2}}=\frac{3 R _{1} \cdot R _{1}}{3 R _{1}+ R _{1}}=\left(\frac{3 R _{1}}{4}\right)$
To produce same heat, using equation ...$(i)$
$Q _{0}=\frac{ V ^{2}}{ R _{1}} \times 20=\frac{ v ^{2}}{\left(\frac{3 R _{1}}{4}\right)} \times t$
$t =\frac{3 \times 20}{4}=15 \,min$
(Given resistivities of iron and copper-nickel alloy wire are $12 \;\mu \Omega {cm}$ and $51\; \mu \Omega {cm}$ respectively) (in ${m}$)



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.
