A uniform heating wire of resistance 36 , Ome | vedclass

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(A) In the first case,the power dissipation is given by $P_1 = \frac{V^2}{R} = \frac{(240)^2}{36} \, W$.When the wire is cut into two equal halves,the

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$4$

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(A) In the first case,the power dissipation is given by $P_1 = \frac{V^2}{R} = \frac{(240)^2}{36} \, W$.When the wire is cut into two equal halves,the resistance of each half becomes $R' = \frac{R}{2} = \frac{36}{2} = 18 \, \Omega$.In the second case,a potential difference of $240 \, V$ is applied across each half separately. The power dissipated in each half is $P_{half} = \frac{V^2}{R'} = \frac{(240)^2}{18} \, W$.The total power dissipation in the second case is $P_2 = P_{half} + P_{half} = 2 	imes \frac{(240)^2}{18} = \frac{(240)^2}{9} \, W$.The ratio of power dissipation is $\frac{P_1}{P_2} = \frac{(240)^2 / 36}{(240)^2 / 9} = \frac{9}{36} = \frac{1}{4}$.Comparing this to $1:x$,we find $x = 4$.

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