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The radius of a well is 7m. Water in it ...

The radius of a well is 7m. Water in it is at a depth of 20m and depth of water column is 10m. Work done in pumping out water completely from the well is `(g=10ms^(-2))`

A

38.5 MJ

B

38.5 MJ

C

46.2 MJ

D

385 MJ

Text Solution

AI Generated Solution

The correct Answer is:
To find the work done in pumping out water completely from a well, we can follow these steps: ### Step 1: Understand the Problem We have a well with a radius of 7 m and a water column of 10 m at a depth of 20 m. We need to calculate the work done to pump all the water out of the well. ### Step 2: Calculate the Volume of Water The volume \( V \) of water in the well can be calculated using the formula for the volume of a cylinder: \[ V = \pi r^2 h \] where \( r \) is the radius of the well and \( h \) is the height of the water column. Given: - \( r = 7 \, \text{m} \) - \( h = 10 \, \text{m} \) Substituting these values: \[ V = \pi (7^2)(10) = \pi (49)(10) = 490\pi \, \text{m}^3 \] ### Step 3: Calculate the Mass of Water The mass \( m \) of the water can be calculated using the formula: \[ m = \text{density} \times \text{volume} \] The density of water is approximately \( 1000 \, \text{kg/m}^3 \). Thus, \[ m = 1000 \times 490\pi \approx 1000 \times 1539.6 \approx 1539600 \, \text{kg} \] ### Step 4: Determine the Height to Pump Water The center of mass of the water is located at half the height of the water column from the surface of the water. Since the water column is 10 m high, the center of mass is at: \[ \text{Height from the surface} = \frac{10}{2} = 5 \, \text{m} \] The total height \( h \) that the water needs to be pumped out is: \[ h = \text{depth of the well} + \text{height of center of mass} = 20 + 5 = 25 \, \text{m} \] ### Step 5: Calculate the Work Done The work done \( W \) in pumping the water out can be calculated using the formula: \[ W = mgh \] where \( g = 10 \, \text{m/s}^2 \). Substituting the values: \[ W = 1539600 \times 10 \times 25 \] \[ W = 153960000 \, \text{J} = 153.96 \times 10^6 \, \text{J} = 153.96 \, \text{MJ} \] ### Final Answer The work done in pumping out the water completely from the well is approximately: \[ W \approx 154 \, \text{MJ} \]
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