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

# 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

1. A

15 minutes

2. B

12 minutes

3. C

10 minutes

4. D

8 minutes

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### Solution:

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

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

$\begin{array}{l}\frac{{\mathrm{R}}_{1}}{{\mathrm{t}}_{1}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{R}}_{2}}{{\mathrm{t}}_{2}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{R}}_{1}}{{\mathrm{R}}_{2}}+\frac{{\mathrm{t}}_{1}}{{\mathrm{t}}_{2}}\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{R}\text{\hspace{0.17em}}\propto \text{\hspace{0.17em}}\mathrm{l}\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}}\frac{{\mathrm{R}}_{1}}{{\mathrm{R}}_{2}}=\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}\end{array}$

By (i) and (ii),

$\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{t}}_{1}}{{\mathrm{t}}_{2}}$              …(iii)

Now  ${\mathrm{l}}_{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}{\mathrm{l}}_{1}\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}\frac{3}{2}$

$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{By}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{equation}\text{\hspace{0.17em}}\left(\mathrm{iii}\right),\text{\hspace{0.17em}\hspace{0.17em}}\frac{3}{2}\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}\frac{15}{{\mathrm{t}}_{2}}\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}_{2}\text{\hspace{0.17em}}=10\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{minutes}$

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