Heater of electric kettle is made of a wire of length $L$ and diameter $d$. It takes $4$ minutes to raise the temperature of $0.5 \ kg$ water by $40\ K$. This heater is replaced by a new heater having two wires of the same material, each of length $L$ and diameter $2 d$. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by $40K$ ?
$(A)$ $4$ if wires are in parallel
$(B)$ $2$ if wires are in series
$(C)$ $1$ if wires are in series
$(D)$ $0.5$ if wires are in parallel.
IIT 2014, Advanced
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In given Kettle $R=\rho \frac{ L }{\pi\left(\frac{ d }{2}\right)^2}=\frac{4 \rho L }{\pi d ^2}$
$P=\frac{V^2}{R}$
In second Kettle $R_1=\rho \frac{L}{\pi d^2}$
$R_2=\frac{\rho L}{\pi d^2}$
So $\quad R_1=R_2=\frac{R}{4}$
If wires are in parallel equivalent resistance
$R_P=\frac{R}{8}$
then power $P_P=8 P$
so it will take 0.5 minute
If wires are in series equivalent resistance
$R_s=\frac{R}{2}$
then power $P_s=2 P$ so it will take $2$ minutes
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