A pump of $200 W$ power is lifting $2 k g$ water from an average depth of $10 m$ in one second. Velocity of water delivered by the pump is :
$(g = 10 m / s^{2})$

10 m / s

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2 m / s

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4 m / s

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1 m / s

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Solution

The correct option is A $10 m / s$
acceleration due to gravity $g = 10 m / s e c^{2}$
height $h = 10 m e t e r s$
mass $= 2 k g$
Potential energy $= m g h = 2 \times 10 \times 10 = 200 J$
as power $= \frac{w o r k d o n e}{t i m e}$
when power of motor $= 200 w$
$200 = 200 / t$
$\Rightarrow t = 1 s e c$
here displacement of water= height $= 10 m$
time = 1sec
Hence,
Velocity $V = \frac{D i s p l a c e m e n t}{T i m e}$
$V = \frac{10}{1}$
$V = 10 m / s e c$

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A pump of 200W power is lifting 2kg water from an average depth of 10m in one second. Velocity of water delivered by the pump is :(g=10m/s2)

The correct option is A 10m/sacceleration due to gravity g=10m/sec2height h=10 metersmass =2kgPotential energy =mgh=2×10×10=200Jas power =work donetimewhen power of motor =200w200=200/t⇒t=1sechere displacement of water= height =10mtime = 1secHence,Velocity V=DisplacementTime V=101 V=10m/sec

A pump of $200 W$ power is lifting $2 k g$ water from an average depth of $10 m$ in one second. Velocity of water delivered by the pump is :
$(g = 10 m / s^{2})$