Equalising levels between two identical tanks via a small orifice Two identical cylindrical vessels (cross-sectional area A each) are interconnected by an orifice of area a and coefficient of discharge C_d. If the initial difference in liquid levels.

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:Interconnected-tank transients are common in laboratory hydraulics and process engineering. When two identical tanks are connected by a small orifice, the level difference decays in time, and the equalisation time can be derived from unsteady continuity combined with the orifice equation. Given Data / Assumptions: Two tanks, each of area A, connected by an orifice of area a with discharge coefficient C_d. Head loss confined to the orifice; inviscid elsewhere; free discharge between tanks. Initial level difference Δh(0) = H; final equalised difference Δh(T) = 0. Concept / Approach: Flow through the orifice is Q = C_d a √(2 g Δh). Because one tank empties while the other fills, the rate of change of difference is doubled: d(Δh)/dt = −(2Q)/A. This yields a first-order, separable differential equation in √Δh, leading to a closed-form time to reach Δh = 0 from H. Step-by-Step Solution: Q = C_d a √(2 g Δh).A d(h_1)/dt = −Q, A d(h_2)/dt = +Q ⇒ d(Δh)/dt = −(2 Q)/A.d(Δh)/dt = −(2 C_d a/A) √(2 g) √(Δh).Integrate: ∫ dΔh/√(Δh) = −(2 C_d a/A) √(2 g) ∫ dt.2 √(Δh) = −(2 C_d a/A) √(2 g) t + 2 √H.Set Δh = 0 at t = T ⇒ T = (A / (C_d a √(2 g))) √H. Verification / Alternative check: Dimensional analysis: [T] = [A]/([a] √(g) √[H]) → seconds, consistent. Why Other Options Are Wrong: (b) and (c) have incorrect factors of 2; (d) is dimensionally inconsistent; (e) omits √2 in the orifice velocity. Common Pitfalls: Forgetting the factor of 2 because both tanks change level; misapplying Torricelli’s law without the discharge coefficient. Final Answer: T = (A / (C_d a √(2 g))) * √H

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