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

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

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

$\begin{array}{l}\frac{{\mathrm{R}}_1}{{\mathrm{t}}_1}=\frac{{\mathrm{R}}_2}{{\mathrm{t}}_2}\mathrm{or}\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}+\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}\\ \mathrm{But}\mathrm{R}\propto \mathrm{l}\Rightarrow \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}\end{array}$

By (i) and (ii),

$\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$              …(iii)

Now  ${\mathrm{l}}_2=\frac{2}{3}{\mathrm{l}}_1\Rightarrow \frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{3}{2}$

$\therefore \mathrm{By}\mathrm{equation}\left(\mathrm{iii}\right),\frac{3}{2}=\frac{15}{{\mathrm{t}}_2}\Rightarrow {\mathrm{t}}_2=10\mathrm{minutes}$