Practice portal › Mechanical Properties of Fluids › Torricelli's Law and Efflux
Asked in JEE Main 27th July 2nd Shift 2021 · Efflux speed and range
Given: water depth H=12 m, hole at depth h below the surface, so the hole is H-h above the ground.
Torricelli gives the horizontal launch speed v=√2gh, and the jet falls H-h in time t=√(2(H-h))/g.
Range R=vt=2√h(H-h).
The product h(H-h) is largest when the two factors are equal, at h=H/2.
h=(12)/2=6 m, and the maximum range is then R=H=12 m.
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