A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

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one fourth

halved

doubled

four times

answer is C.

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Detailed Solution

$\begin{array}{l} H = \frac{V^{2} t}{R} \Rightarrow \frac{H_{Half}}{H_{Full}} = (\frac{R_{Full}}{R_{Half}}) = \frac{R}{R / 2} = 2 \\ \Rightarrow H_{Half} = 2 \times H_{Full} \end{array}$

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