Question

A small hole of area of cross-section 2 mm² of height 2 m. Taking g = 10 m/s², the rate of water through the open hole would be nearly
(1) 12.6 × 10⁻⁶ m³/s
(2) 8.9 × 10⁻⁶ m³/s
(3) 2.23 × 10⁻⁶ m³/s
(4) 6.4 × 10⁻⁶ m³/s

Answer 1

The rate of water through the open hole would be nearly $\bold{12.6\times 10^{-6}m^{3}/s}$

step by step explanation:

Rate of flow of liquid can be calculated by the following formula.

                  $\bold{Q=au=a\sqrt{2gh}}$...........................(1)

From the given,

area of cross-section = $2mm^{2}=2\times 10^{-6}m^{2}$

g = $10m/s^{2}$

h = 2

Substitute the given values in the equation (1)

                                    $\Rightarrow 2\times 10^{-6}m^{2}\times \sqrt{2\times 10\times 2m/s }$

                                    $\Rightarrow 12.56\times 10^{-6}m^{3}/s$

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

Therefore, The rate of water through the open hole would be nearly $\bold{12.6\times 10^{-6}m^{3}/s}$.

Hence, correct answer is 1