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Q.
A $500\, W$ heater is designed to operate at $200\, V$ potential difference. If it is connected across $160 \,V$ line, the heat it will produce in $20$ minutes is
Case-I Resistance of heater when it operates at $200\, V$ and power, $P_{1}=500\, W$
$\because$ Power, $P=\frac{V^{2}}{R} $
$\Rightarrow R_{1}=\frac{V_{1}^{2}}{P_{1}}$
$R_{1}=\frac{(200)^{2}}{500} =\frac{200 \times 200}{500} $
$R_{1} =80 \,\Omega$ Case - II Now, it connected across $160 \,V$ line the heat it will produce in $20$ minute is
$V_{2} =160 \,V$
$\therefore P_{2} =\frac{V_{2}^{2}}{R}=\frac{(160)^{2}}{R} $
$=\frac{(160)^{2}}{80}=\frac{160 \times 160}{80}=2 \times 160 \,\,\,(\because R=80 \,\Omega) $
$P_{2} =320\, W$
$\therefore $ Now, Heat produce is $20$ minutes
$d Q=P_{2} \times $ time $ =320 \times 20 \times 60 $
$d Q =384 \,kJ$