A small hole of an area of cross-section $2 ext{ mm}^2$ is present near the bottom of a fully filled open ta

Question

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

Accepted Answer

1. **Torricelli's Law:** The speed of efflux ($v$) of a fluid flowing out of an open tank from a hole at a depth $h$ below the free surface is given by the formula derived from Bernoulli's theorem: $v = \sqrt{2gh}$ .
2. **Rate of Flow:** The rate of flow (or volume flux, $Q$) is defined as the product of the cross-sectional area ($A$) of the hole and the velocity ($v$) of the fluid emerging from it: $Q = A 	imes v$ .
3. **Given Data:**
   - Area, $A = 2 	ext{ mm}^2 = 2 	imes 10^{-6} 	ext{ m}^2$.
   - Height, $h = 2 	ext{ m}$.
   - Acceleration due to gravity, $g = 10 	ext{ m/s}^2$.
4. **Calculation:**
   - Calculate velocity: $v = \sqrt{2 	imes 10 	imes 2} = \sqrt{40} \approx 6.32 	ext{ m/s}$.
   - Calculate flow rate: $Q = (2 	imes 10^{-6}) 	imes 6.32 = 12.64 	imes 10^{-6} 	ext{ m}^3/	ext{s}$.
   - Rounding to one decimal place gives $12.6 	imes 10^{-6} 	ext{ m}^3/	ext{s}$.

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