A given quantity of water boils in an electric kettle in 12 min- The length of the heating element in the kettle is - If the same quantity of water is to boil in 10 min on the same mains the length of an identical heating element is -

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Standard XII

Physics

Parallel and Series Combination of Capacitors

A given quant...

Question

A given quantity of water boils in an electric kettle in $12 m i n$ . The length of the heating element in the kettle is $ι$ . If the same quantity of water is to boil in $10 m i n$ on the same mains the length of an identical heating element is _______.

ι

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\frac{ι}{2}

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\frac{3 ι}{4}

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\frac{5 ι}{6}

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Solution

The correct option is D $\frac{5 ι}{6}$
Amount of heat required to boil a given quantity of water in a constant
$P = \frac{W}{t}$ or $W = H = P \times t = \frac{V^{2}}{R} \times t = \frac{V^{2} t_{1}}{R_{1}} = \frac{V^{2} t_{2}}{R_{2}}$
$\Rightarrow \frac{t_{2}}{t_{1}} = \frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}}$
or $\frac{10}{12} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = \frac{10}{12} \times l_{1}$
$l_{2} = \frac{5}{6} l$

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A given quantity of water boils in an electric kettle in 12min. The length of the heating element in the kettle is ι. If the same quantity of water is to boil in 10min on the same mains the length of an identical heating element is _______.

The correct option is D 5ι6Amount of heat required to boil a given quantity of water in a constantP=Wt or W=H=P×t=V2R×t=V2t1R1=V2t2R2⇒t2t1=R2R1=l2l1or 1012=l2l1⇒l2=1012×l1l2=56l

A given quantity of water boils in an electric kettle in $12 m i n$ . The length of the heating element in the kettle is $ι$ . If the same quantity of water is to boil in $10 m i n$ on the same mains the length of an identical heating element is _______.