A vertical tank, open at the top, is filled with a liquid and rests on a smooth horizontal surface.$A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole, where $N$ is a large number. Neglect mass of the tank itself. The initial acceleration of the tank is
Diffcult
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We know that force exerted by fluid coming out on the container is $\rho A v^{2}$
$\mathrm{v}=$ velocity of fluid
$v=\sqrt{2 g \frac{H}{2}}$
$A=$ area of the hole
Acceleration of the tank
$=\frac{\rho \mathrm{Av}^{2}}{\rho(\mathrm{NAH})}$
$=\frac{\rho(\mathrm{AgH})}{\rho \mathrm{NAH}}=\frac{\mathrm{g}}{\mathrm{N}}$
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