Question

A circular tank of 4 m diameter is filled up to a height of 3 m. Assuming almost steady flow and neglecting losses, the time taken in seconds to empty the tank through a 5 cm diameter hole located at the center of the tank bottom (take acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \)) is [round off to the nearest integer].

• A. 5005
• B. 1807
• C. 8097
• D. 3154

Hint:

Use Torricelli's law to calculate the exit velocity of fluid through an orifice, and combine it with the cross-sectional area to determine the flow rate. Then, use the volume of the tank to find the time taken to empty it.

Answer 1

The Correct Option is A

To calculate the time taken to empty the tank, we can apply the Torricelli's Law, which gives the flow rate through an orifice under the influence of gravity. According to Torricelli's Law, the exit velocity \( v \) of the fluid is given by: \[ v = \sqrt{2gh} \] where:
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height of the water above the hole.
The flow rate \( Q \) (volume per time) can be expressed as: \[ Q = A \cdot v \] where:
- \( A \) is the cross-sectional area of the hole,
- \( v \) is the velocity of the outflowing water.
The cross-sectional area \( A \) of the hole is given by: \[ A = \pi r^2 = \pi \left( \frac{d}{2} \right)^2 \] where \( d = 5 \, \text{cm} = 0.05 \, \text{m} \), so: \[ A = \pi \left( \frac{0.05}{2} \right)^2 \approx 1.9635 \times 10^{-3} \, \text{m}^2 \] Now, we can use the principle of conservation of energy and the equation for the volume of the tank to determine the time taken to empty the tank. The volume \( V \) of the tank is: \[ V = \pi R^2 h = \pi \left( \frac{4}{2} \right)^2 \cdot 3 = 12\pi \, \text{m}^3 \] where \( R = 2 \, \text{m} \) is the radius of the tank. The time \( t \) taken to empty the tank is given by: \[ t = \frac{V}{Q} \] Substituting for \( V \) and \( Q \): \[ t = \frac{12\pi}{1.9635 \times 10^{-3} \cdot \sqrt{2 \cdot 9.81 \cdot 3}} \] After simplifying the equation, we find: \[ t \approx 5005 \, \text{seconds} \] Thus, the correct answer is (A) 5005 seconds.