AN electric motor of power $100 W$ is used to drive the stirrer in a water bath . If $50 %$ of the energy supplied to the motor is spent in stirrering the water , work done on the water in $1 m i n$ is

3000 J

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300 J

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30 J

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3 J

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Solution

The correct option is A $3000 J$
work done = $\frac{50}{100} \times P \times t$
$= \frac{1}{2} \times 100 \times 60$
$= 3000 J$

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AN electric motor of power 100W is used to drive the stirrer in a water bath . If 50% of the energy supplied to the motor is spent in stirrering the water , work done on the water in 1min is

The correct option is A 3000Jwork done = 50100×P×t=12×100×60=3000J

AN electric motor of power $100 W$ is used to drive the stirrer in a water bath . If $50 %$ of the energy supplied to the motor is spent in stirrering the water , work done on the water in $1 m i n$ is

The correct option is A $3000 J$
work done = $\frac{50}{100} \times P \times t$
$= \frac{1}{2} \times 100 \times 60$
$= 3000 J$