Question

A circular well of diameter \( 2 \, \text{m} \) has water up to the ground level. If the bottom of the well is at a depth of \( 14 \, \text{m} \), the time taken in seconds to empty the well using a \( 1.4 \, \text{kW} \) motor is (Acceleration due to gravity = \( 10 \, \text{m/s}^2 \))

• A. 1860
• B. 2200
• C. 2660
• D. 3300

Hint:

When calculating the time to pump water out of a well, consider the average height of the water column and use the formula for work done against gravity.

Answer 1

The Correct Option is B

Step 1: Known Information.
Diameter of the well: \( d = 2 \, \text{m} \)
Radius of the well: \( r = \frac{d}{2} = \frac{2}{2} = 1 \, \text{m} \)
Depth of the well: \( h = 14 \, \text{m} \)
Power of the motor: \( P = 1.4 \, \text{kW} = 1400 \, \text{W} \)
Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)
Density of water: \( \rho = 1000 \, \text{kg/m}^3 \)
Step 2: Volume of Water in the Well.
The volume \( V \) of water is: $$ V = \pi r^2 h = \pi (1)^2 (14) = 14\pi \, \text{m}^3 $$ Step 3: Mass of Water. The mass \( m \) of the water is: $$ m = \rho V = 1000 \cdot 14\pi = 14000\pi \, \text{kg} $$ Step 4: Work Done to Pump the Water Out.
The work \( W \) required to lift the water is: $$ W = mgh_{\text{avg}} $$ where \( h_{\text{avg}} \) is the average height through which the water is lifted. For a uniform distribution, \( h_{\text{avg}} = \frac{h}{2} = \frac{14}{2} = 7 \, \text{m} \). Thus: $$ W = (14000\pi)(10)(7) = 980000\pi \, \text{J} $$ Step 5: Time to Empty the Well.
The time \( t \) is given by: $$ t = \frac{W}{P} = \frac{980000\pi}{1400} $$ Substitute \( \pi \approx 3.1416 \): $$ t \approx \frac{980000 \cdot 3.1416}{1400} \approx \frac{3077480}{1400} \approx 2200 \, \text{s} $$ Final Answer: \( \boxed{2200} \)