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An open-topped vertical tank filled with liquid sits on a frictionless horizontal floor. A small hole is made at the middle of one vertical wall. The tank's cross-sectional area is N times that of the hole, where N is large. Ignoring the tank's own mass, what is the initial acceleration of the tank?

1. g/(2N)
2. g/sqrt(2N)
3. g/N
4. g/(2*sqrt(N))

Correct answer: g/N

Solution

Let the total liquid height be H. The hole is at the centre of the side, so the head above the hole is H/2 and the efflux speed is v = sqrt(2*g*(H/2)) = sqrt(g*H). The reaction thrust on the tank equals the rate of momentum carried out: F = rho*a*v² = rho*a*(g*H). The mass of liquid in the tank is m = rho*A*H, where A = N*a. The initial acceleration is F/m = (rho*a*g*H)/(rho*A*H) = g*(a/A) = g/N.

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