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Question

There are two identical small holes of area of cross section a on the opposite sides of a tank containing a liquid of density $$ \rho $$. The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which has to be applied on the tank to keep it in equilbrium is :
71918_2b1475fb665a4c8e8ef7020248d2ce07.png


A
ghρa
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B
2ghρa
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C
2ρagh
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D
ρgha
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Solution

The correct option is D $$ 2\rho agh $$
Momentum of the mass of water coming out of the hole per sec is thrust.
Resultant thrust $$F= F_1-F_2=-\rho av_1^2 - \rho av_2^2=-\rho a \sqrt{2gh_1} - \rho a \sqrt{2gh_2} =-2a\rho g(h_1-h_2)$$
$$ \therefore$$ horizontal force that has to be applied to the tank to keep it in equilibrium is
$$2a\rho g(h_1-h_2)$$
or
$$2a\rho gh$$
137720_71918_ans_4dab16173894402ea8266ff8399ceef2.png

Physics

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