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A machine which is 75 percent efficient, uses 12 joules of energy in lifting up a $1 \mathrm{~kg}$ mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in $\mathrm{ms}^{-1}$)
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The correct answer is:
$\sqrt{18}$
Potential energy of a
$\begin{aligned}
& \text { body }=75 \% \text { of } 12 J \\
& \text { mgh }=9 J \Rightarrow h=\frac{9}{1 \times 10}=0.9 \mathrm{~m}
\end{aligned}$
Now when this mass allow to fall then it acquire velocity
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 0.9}=\sqrt{18} \mathrm{~m} / \mathrm{s} .$
$\begin{aligned}
& \text { body }=75 \% \text { of } 12 J \\
& \text { mgh }=9 J \Rightarrow h=\frac{9}{1 \times 10}=0.9 \mathrm{~m}
\end{aligned}$
Now when this mass allow to fall then it acquire velocity
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 0.9}=\sqrt{18} \mathrm{~m} / \mathrm{s} .$
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