Water boils in the electric kettle in 1.5 minutes after switching on. If the length of 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

0.8 minutes

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1 minutes

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1.2 minutes

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1.5 minutes

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Solution

The correct option is B 1 minutes
Resistance, $R = \frac{ρ l}{A}$ or $R \propto l$
If initial resistance is $R_{i}$ , final resistance $R_{f} = (2 / 3) R_{i}$
Heat produced, $H = \frac{V^{2}}{R} t$
Here, $H, V$ is constant so, $R \propto t$
Thus, $\frac{R_{f}}{R_{i}} = \frac{t_{f}}{t_{i}}$
$\Rightarrow t_{f} = \frac{R_{f}}{R_{i}} t_{i} = \frac{(2 / 3) R_{i}}{R_{i}} \times 1.5 = 1 m i n$

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Water boils in the electric kettle in 1.5 minutes after switching on. If the length of 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

The correct option is B 1 minutesResistance, R=ρlA or R∝lIf initial resistance is Ri, final resistance Rf=(2/3)RiHeat produced, H=V2RtHere, H,V is constant so, R∝tThus, RfRi=tfti⇒tf=RfRiti=(2/3)RiRi×1.5=1 min