Question:
A cylindrical tank having large diameter is filled with water to a height \(H\). A hole of cross-sectional area \(5\ \text{cm}^2\) in the tank allows water to drain out. If the water drains out at the rate of
\[
2\times10^{-3}\ \text{m}^3\text{s}^{-1},
\]
then the value of \(H\) is
\[
(\text{acceleration due to gravity }=10\ \text{m s}^{-2})
\]
A cylindrical tank having large diameter is filled with water to a height \(H\). A hole of cross-sectional area \(5\ \text{cm}^2\) in the tank allows water to drain out. If the water drains out at the rate of
\[
2\times10^{-3}\ \text{m}^3\text{s}^{-1},
\]
then the value of \(H\) is
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]
Show Hint
For a large tank with a small outlet,
\[
v=\sqrt{2gH}
\]
(Torricelli's theorem).
Also,
\[
Q=Av,
\]
where \(Q\) is the volume flow rate, \(A\) is the area of the hole, and \(v\) is the speed of efflux.
Updated On: Jun 26, 2026
- \(80\ \text{cm}\)
- \(120\ \text{cm}\)
- \(60\ \text{cm}\)
- \(90\ \text{cm}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the equation of continuity.
The rate of flow of water is \[ Q=Av, \] where \[ Q=2\times10^{-3}\ \text{m}^3\text{s}^{-1} \] and the area of the hole is \[ A=5\ \text{cm}^2. \] Converting into SI units, \[ A=5\times10^{-4}\ \text{m}^2. \] Therefore, \[ v=\frac{Q}{A}. \] \[ v=\frac{2\times10^{-3}}{5\times10^{-4}}. \] \[ v=4\ \text{m s}^{-1}. \]
Step 2: Apply Torricelli's theorem.
For a tank of large cross-sectional area, \[ v=\sqrt{2gH}. \] Substituting \[ v=4\ \text{m s}^{-1} \] and \[ g=10\ \text{m s}^{-2}, \] we get \[ 4=\sqrt{20H}. \] Squaring both sides, \[ 16=20H. \] \[ H=\frac{16}{20}. \] \[ H=0.8\ \text{m}. \]
Step 3: Convert into centimetres.
\[ H=0.8\times100. \] \[ H=80\ \text{cm}. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{H=80\ \text{cm}} \] Hence, the correct option is \[ \boxed{(1)} \]
The rate of flow of water is \[ Q=Av, \] where \[ Q=2\times10^{-3}\ \text{m}^3\text{s}^{-1} \] and the area of the hole is \[ A=5\ \text{cm}^2. \] Converting into SI units, \[ A=5\times10^{-4}\ \text{m}^2. \] Therefore, \[ v=\frac{Q}{A}. \] \[ v=\frac{2\times10^{-3}}{5\times10^{-4}}. \] \[ v=4\ \text{m s}^{-1}. \]
Step 2: Apply Torricelli's theorem.
For a tank of large cross-sectional area, \[ v=\sqrt{2gH}. \] Substituting \[ v=4\ \text{m s}^{-1} \] and \[ g=10\ \text{m s}^{-2}, \] we get \[ 4=\sqrt{20H}. \] Squaring both sides, \[ 16=20H. \] \[ H=\frac{16}{20}. \] \[ H=0.8\ \text{m}. \]
Step 3: Convert into centimetres.
\[ H=0.8\times100. \] \[ H=80\ \text{cm}. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{H=80\ \text{cm}} \] Hence, the correct option is \[ \boxed{(1)} \]
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