JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Fluid Mechanics

The top of a water tank is open to air and its water level is maintained. It is giving out

0.74 m^{3}

water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:

(A)6.0 m
(B)4.8 m✓ Correct
(C)9.6 m
(D)2.9 m

Answer: (B)

Solution

This is a Torricelli problem. The water leaves the hole with speed

v = 2 g h

, so the volume flow rate is

\frac{d V}{d t} = A v = π r^{2} 2 g h

. Convert the rate:

0.74 m^{3}

per minute is

\frac{0.74}{60} \approx 0.0123 m^{3} / s

, and the opening area is

π (0.02)^{2} \approx 1.26 \times 1 0^{- 3} m^{2}

. Then

2 g h = \frac{0.0123}{1.26 \times 1 0 ^{- 3}} \approx 9.8 m/s

, so

2 g h \approx 96

and

h \approx \frac{96}{19.6} \approx 4.9 m

, closest to

4.8 m

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Fluid Mechanics →

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Fluid Mechanics

Solution

More JEE Main questions on Fluid Mechanics