Question

Heater of an electric kettle is made of a wire of length $L$ and diameter $d.$ It takes 4 minutes to raise the temperature of $0.5\text{kg}$ water by $40\text{K}.$ This heater is replaced by a new heater having two wires of the same material, each of length $L$ and diameter $2d.$ 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 $40\text{K}$?

• 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

Answer 1

Answer: (B), (D)

$0.5,$ if wires are in parallel

Heat required to increase the temperature of water by $40\text{K}$
$\Delta Q=mc\Delta T=\left(\frac{V^2}{R}\right)t$

from the above we can deduce that $t\propto R$
Here $t=4\text{min}$
$\text{Where}R=\frac{\rho L}{\left(\frac{\pi d^2}{4}\right)}$
The resistance of wire of length $L$ and diameter $2d.$
$R_1=\frac{pL}{\pi (2d{)}^2/4}=\frac{R}{4}$
$\text{If resistance are in series,}{R}_{eq}=\frac{R}{4}+\frac{R}{4}=\frac{R}{2}$
Time taken to rise the temperature by same amout will be $\frac{t}{2}=2\text{min}$

$\text{If resistances are in parallel,}{R}_{eq}=\frac{\frac{R}{4}\times \frac{R}{4}}{\frac{R}{4}+\frac{R}{4}}=\frac{R}{8}$
Time taken to rise the temperature by same amout will be $\frac{t}{8}=0.5\text{min}$