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A liquid emerges from a hole at depth h below the free surface of a large open tank. The exit speed (Torricelli's law) is:
Av = g h
Bv = √(g h)
Cv = √(2 g h)
Dv = 2 g h
Answer & Solution
Correct answer: C. v = √(2 g h)
1. Apply Bernoulli between the free surface (slow, P = P_atm) and the hole (fast, P = P_atm).
2. ρ g h = ½ ρ v² ⇒ v = √(2 g h).
3. Same as free-fall from height h — Torricelli's law.
_Source: NCERT Class 11 Physics Ch 9 §9.6 Torricelli's Law_
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