A horse drinks water from a cubical container of side $1 m$ with water filled upto $1 m$ . The level of the stomach of horse is at $2 m$ from the ground. Assume that all the water drunk by the horse is at a level of $2$ m from the ground, then minimum work done by the horse in drinking the entire water of the container is:
(Take $ρ_{w a t e r} = 1000 k g / m^{3}$ and $g = 10 m / s^{2}$ )

10 k J

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15 k J

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20 k J

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Zero

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Solution

The correct option is B $15 k J$
The mass of water is $m = 1 \times 10^{3} k g$
Work done by horse is the increase in potential energy.
The increase in potential energy of water is $= m g h = (1 \times 10^{3}) (10) (1.5) = 15 k J$

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