Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage in

A
15 minutes
B
12 minutes
C
10 minutes
D
8 minutes

Solution:

$H = \frac{V^{2}}{R} t$

Since supply voltage is same and equal amount of heat will be produced, therefore

$\begin{array}{l} \frac{R_{1}}{t_{1}} = \frac{R_{2}}{t_{2}} or \frac{R_{1}}{R_{2}} + \frac{t_{1}}{t_{2}} \\ But R \propto l \Rightarrow \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \end{array}$

By (i) and (ii),

$\frac{l_{1}}{l_{2}} = \frac{t_{1}}{t_{2}}$ …(iii)

Now $l_{2} = \frac{2}{3} l_{1} \Rightarrow \frac{l_{1}}{l_{2}} = \frac{3}{2}$

$∴ By equation (iii), \frac{3}{2} = \frac{15}{t_{2}} \Rightarrow t_{2} = 10 minutes$

Solution:

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