Question

A large tank open to atmosphere at top and filled with water, develops a small hole in the side at a point \(20 \, \text{m}\) below the water level. If the rate of flow of water from the hole is \(3\times 10^{-3}\,\text{m}^3/\text{min}\), then the area of hole is \((g=10\,\text{m s}^{-2})\):

• A. \(4\,\text{mm}^2\)
• B. \(1.5\,\text{mm}^2\)
• C. \(2.5\,\text{mm}^2\)
• D. \(2\,\text{mm}^2\)

Hint:

For water flowing out of a small hole in a tank, first use Torricelli's theorem \(v=\sqrt{2gh}\), then use \(Q=Av\) to find the area of the hole.

Answer 1

The Correct Option is C

Step 1: Use Torricelli's theorem.
Velocity of efflux from a hole at depth \(h\) is \[ v=\sqrt{2gh} \] Given, \[ g=10\,\text{m s}^{-2},\quad h=20\,\text{m} \] Therefore, \[ v=\sqrt{2\times 10\times 20} \] \[ v=\sqrt{400} \] \[ v=20\,\text{m s}^{-1} \]

Step 2: Convert rate of flow into SI unit.
Given rate of flow is \[ Q=3\times 10^{-3}\,\text{m}^3/\text{min} \] Since, \[ 1\,\text{min}=60\,\text{s} \] So, \[ Q=\frac{3\times 10^{-3}}{60}\,\text{m}^3/\text{s} \] \[ Q=5\times 10^{-5}\,\text{m}^3/\text{s} \]

Step 3: Use discharge formula.
Rate of flow is given by \[ Q=Av \] Therefore, \[ A=\frac{Q}{v} \] Substituting the values, \[ A=\frac{5\times 10^{-5}}{20} \] \[ A=2.5\times 10^{-6}\,\text{m}^2 \]

Step 4: Convert area into \(\text{mm}^2\).
Since, \[ 1\,\text{m}^2=10^6\,\text{mm}^2 \] Therefore, \[ A=2.5\times 10^{-6}\times 10^6\,\text{mm}^2 \] \[ A=2.5\,\text{mm}^2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{2.5\,\text{mm}^2} \]