Question

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

Answer 1

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Answer 2

The rate  of flow of water through the open hole is $12.6\times10^-^6m^3/s$.

Explanation:

Given the crossectional area of the small hole, $A =2mm^2=2\times10^-^6m^2$.

The height of the tank, h = 2 m.

The rate of the flow of the water is given as

Q = A v

Here, v is the velocity of the water.

Velocity is given as,

$v=\sqrt{2gh}$

g is the acceleration due to gravity.

substitute the values, we get velocity,

$v=\sqrt{2\times10m/s^2\times2m}$

v = 6.3 m/s.

thefore,

$Q=2\times10^-^6m^2\times6.3m/s$

$Q=12.6\times10^-^6m^3/s$.

Thus, the rate  of flow of water through the open hole is $12.6\times10^-^6m^3/s$.

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