The level of water in tank is 5 m high. A hole of area 1 cm2 is made at the bottom of the tank. The rate of leakage of water from the hole is (g=10 ms−2)

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10−3 m3s−1

10−4 m3s−1

10 m3s−1

10−2 m3s−1

answer is A.

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Detailed Solution

v=2gh Since volume flowing per second is Q=Av=A2gh=(1×10−4)×2×10×5 ⇒ Q=10−3 m3s−1