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(B) The power $P$ required to pump water is the sum of the power required to lift the water to height $h$ and the power required to provide kinetic en

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) The power $P$ required to pump water is the sum of the power required to lift the water to height $h$ and the power required to provide kinetic energy to the water.However, in this context, we calculate the power required to lift the water against gravity at a steady rate.Power $P = \frac{	ext{Work}}{	ext{Time}} = \frac{mgh}{t}$.From the equation of continuity, the mass flow rate is $\frac{m}{t} = A v ho$.Substituting this into the power equation: $P = (A v ho) g h$.Given values: $r = 1 \ cm = 10^{-2} \ m$, $v = 10 \ m \ s^{-1}$, $ho = 1000 \ kg \ m^{-3}$, $g = 10 \ m \ s^{-2}$, $h = 3 \ m$.Area $A = \pi r^2 = \pi (10^{-2})^2 = \pi 	imes 10^{-4} \ m^2$.Calculating power: $P = (\pi 	imes 10^{-4}) 	imes 10 	imes 1000 	imes 10 	imes 3$.$P = \pi 	imes 10^{-4} 	imes 10^5 	imes 3 = 30\pi \ W$.

Column - $I$	Column - $II$
$(a)$ Velocity head	$(i)$ $\frac{P}{\rho g}$
$(b)$ Pressure head	$(ii)$ $h$
$(c)$ Potential head	$(iii)$ $\frac{v^2}{2g}$

Water is pumped steadily out of a flooded basement at a speed of $10 \ m \ s^{-1}$ through a hose (tube) of radius $1 \ cm$, passing through a window $3 \ m$ above the water level. The power of the pump is (Assume $g = 10 \ m \ s^{-2}$, density of water $= 1000 \ kg \ m^{-3}$): (in $\pi \ W$)

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Different heads are in Column - $I$ and their formulas are given in Column - $II$. Match them appropriately.
Column - $I$ Column - $II$
$(a)$ Velocity head $(i)$ $\frac{P}{\rho g}$
$(b)$ Pressure head $(ii)$ $h$
$(c)$ Potential head $(iii)$ $\frac{v^2}{2g}$

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