A small hole of area of cross-section 2 is present near the bottom of a fully filled open tank of height 2 m. Taking , the rate of flow of water through the open hole would be nearly [2019]

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Q.
A small hole of area of cross-section 2 ${mm}^{2}$ is present near the bottom of a fully filled open tank of height 2 m. Taking $g = 10$ $m / s^{2}$ , the rate of flow of water through the open hole would be nearly [2019]

1 $6.4 \times 10^{- 6}$ $m^{3} / s$

2 $12.6 \times 10^{- 6}$ $m^{3} / s$

3 $8.9 \times 10^{- 6}$ $m^{3} / s$

4 $2.23 \times 10^{- 6}$ $m^{3} / s$

Ans.
(2)

According to Torricelli's theorem,

Velocity, $v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 2} = 6.32$ $m/s .$

From equation of continuity,

Volume of liquid flowing per second, $Q = A v$

$= 2 \times 10^{- 6} \times 6.32 = 12.6 \times 10^{- 6}$ $m^{3} / s$

Solution

Q.
A small hole of area of cross-section 2 ${mm}^{2}$ is present near the bottom of a fully filled open tank of height 2 m. Taking $g = 10$ $m / s^{2}$ , the rate of flow of water through the open hole would be nearly [2019]

1 $6.4 \times 10^{- 6}$ $m^{3} / s$

2 $12.6 \times 10^{- 6}$ $m^{3} / s$

3 $8.9 \times 10^{- 6}$ $m^{3} / s$

4 $2.23 \times 10^{- 6}$ $m^{3} / s$

Ans.
(2)

According to Torricelli's theorem,

Velocity, $v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 2} = 6.32$ $m/s .$

From equation of continuity,

Volume of liquid flowing per second, $Q = A v$

$= 2 \times 10^{- 6} \times 6.32 = 12.6 \times 10^{- 6}$ $m^{3} / s$

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Solution

Q. A small hole of area of cross-section 2 mm2 is present near the bottom of a fully filled open tank of height 2 m. Taking g=10 m/s2, the rate of flow of water through the open hole would be nearly [2019]

1 6.4×10-6 m3/s

2 12.6×10-6 m3/s

3 8.9×10-6 m3/s

4 2.23×10-6 m3/s

Ans. (2) According to Torricelli's theorem, Velocity, v=2gh=2×10×2=6.32 m/s. From equation of continuity, Volume of liquid flowing per second, Q=Av =2×10-6×6.32=12.6×10-6 m3/s

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Q.
A small hole of area of cross-section 2 ${mm}^{2}$ is present near the bottom of a fully filled open tank of height 2 m. Taking $g = 10$ $m / s^{2}$ , the rate of flow of water through the open hole would be nearly [2019]

1 $6.4 \times 10^{- 6}$ $m^{3} / s$

2 $12.6 \times 10^{- 6}$ $m^{3} / s$

3 $8.9 \times 10^{- 6}$ $m^{3} / s$

4 $2.23 \times 10^{- 6}$ $m^{3} / s$

Ans.
(2)

According to Torricelli's theorem,

Velocity, $v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 2} = 6.32$ $m/s .$

From equation of continuity,

Volume of liquid flowing per second, $Q = A v$

$= 2 \times 10^{- 6} \times 6.32 = 12.6 \times 10^{- 6}$ $m^{3} / s$