The pressure $P_{1}$ at a certain depth in river water and $P_{2}$ at the same depth in sea water are related as :
(a) $P_{1} > P_{2}$ (b) $P_{1} = P_{2}$
(c) $P_{1} < P_{2}$ (d) $P_{1} - P_{2}$ = atmospheric pressure

Q. Resistors

R_{1}, R_{2}

are connected in series. If the power dissipated by them is

P_{1}, P_{2}

then

Q. Resistors

R_{1}, R_{2}

are connected in parallel. If the power dissipated by them is

P_{1}, P_{2}

then

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A person A does 500 J of work in 10 minutes and another person B does 600 J of work in 20 minutes. Let the power delivered by A and B be P1 and P2 respectively. Then,

The correct option is B P1 > P2P1=Wt=50010×60=0.83 WP2=Wt=60020×60=0.5 W⟹P1>P2

A person A does 500 J of work in 10 minutes and another person B does 600 J of work in 20 minutes. Let the power delivered by A and B be $P_{1}$ and $P_{2}$ respectively. Then,

The correct option is B $P_{1}$ > $P_{2}$
$P_{1} = \frac{W}{t} = \frac{500}{10 \times 60} = 0.83 W$

$P_{2} = \frac{W}{t} = \frac{600}{20 \times 60} = 0.5 W$

$⟹ P_{1} > P_{2}$