(i) (a) Define work and state the mathematical expression for wor… - Vidyadip

Question

$(i) (a)$ Define work and state the mathematical expression for work.
$(b)$ State $\text{CGS}$ and $SI$ unit of work.
$(ii)$ An electric pump is $60 \%$ efficient and is rated $2 HP$ . Calculate the maximum amount of water it can lift through a height of $5 m$ in $40 s .\ [$Take $g =10 ms^{-2}$ and $1 HP =750 W$ ]
$(iii)$ In a hydroelectric power station, $1000 \ kg$ water is allowed to drop through a height of $100 m$ in $1s$ . If the conversion of potential energy to electric energy is $60 \%$, calculate the power output in $kW ($take $g =10 m / s ^2).$

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Answer

$(i) (a)$ Work is said to be done only when the force applied on a body causes a displacement of the body in the direction of force or any resolved component of the force.
Mathematically, work is the product of force and the displacement in the direction of force.
$W=\overrightarrow{F} \cdot \overrightarrow{S}=FS \cos \theta$
$(b) \ \text{CGS}$ unit of work $- \text{erg}$
$SI$ unit of work $- $joule
$(ii)$ Energy supplied to the motor $= P \times t$
$ 2 HP \times 40 s=2 \times 750 W \times 40 s=1500 W \times 40 s=6 \times 10^4 J$
Useful work done by the motor $=6 \times 10^4 J \times \frac{60}{100}=36000 J$
Useful work done in lifting water $=m \ \text{mgh}$
$=m \times 10 ms^{-2} \times 5 m$
$\Rightarrow m \times 50 m^2 s^{-2}=36000 J$
$\therefore m=\frac{36000 J}{50 m^2 s^{-2}}=720 \ kg$
$(iii) \ P.E$. of water $= \text{mgh} =1000 \ kg \times 10 ms^{-2} \times 100 m=10^6 J$
Useful $P.E. =10^6 J \times \frac{60}{100}=6 \times 10^5 J$
$\therefore$ Electrical power output $=\frac{ W }{t}=\frac{6 \times 10^5 J}{1 s}$
$=6 \times 10^5 W$
$=6 \times 10^2 kW=600 kW$.

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