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4m^(3) of water is to be pumped to a he...

`4m^(3)` of water is to be pumped to a height of 20 m and forced into a reservoir at a pressure of `2xx10^(5)Nm^(-2)`. The work done by the motor is (external pressure `= 10^(5)Nm^(-2)`)

A

`8xx10^(5)J`

B

`16xx10^(5)J`

C

`12xx10^(5)J`

D

`32xx10^(5)J`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of calculating the work done by the motor in pumping water to a height and forcing it into a reservoir, we can break down the solution into clear steps. ### Step 1: Identify the given data - Volume of water, \( V = 4 \, m^3 \) - Height to which water is pumped, \( h = 20 \, m \) - Pressure in the reservoir, \( P_{\text{reservoir}} = 2 \times 10^5 \, N/m^2 \) - External pressure, \( P_{\text{external}} = 1 \times 10^5 \, N/m^2 \) - Density of water, \( \rho = 10^3 \, kg/m^3 \) - Acceleration due to gravity, \( g = 10 \, m/s^2 \) ### Step 2: Calculate the work done to lift the water The work done to lift the water to a height \( h \) is equal to the change in potential energy, which can be calculated using the formula: \[ W_{\text{lift}} = mgh \] Where: - \( m \) is the mass of the water, calculated as \( m = \rho V \) Calculating the mass: \[ m = \rho V = 10^3 \, kg/m^3 \times 4 \, m^3 = 4000 \, kg \] Now, substituting the values into the potential energy formula: \[ W_{\text{lift}} = 4000 \, kg \times 10 \, m/s^2 \times 20 \, m = 800000 \, J = 8 \times 10^5 \, J \] ### Step 3: Calculate the work done against external pressure The work done against the external pressure is given by: \[ W_{\text{pressure}} = (P_{\text{reservoir}} - P_{\text{external}}) \times V \] Calculating the change in pressure: \[ \Delta P = P_{\text{reservoir}} - P_{\text{external}} = (2 \times 10^5 \, N/m^2) - (1 \times 10^5 \, N/m^2) = 1 \times 10^5 \, N/m^2 \] Now substituting into the work done against external pressure formula: \[ W_{\text{pressure}} = (1 \times 10^5 \, N/m^2) \times (4 \, m^3) = 400000 \, J = 4 \times 10^5 \, J \] ### Step 4: Calculate total work done by the motor The total work done by the motor is the sum of the work done to lift the water and the work done against external pressure: \[ W_{\text{total}} = W_{\text{lift}} + W_{\text{pressure}} = (8 \times 10^5 \, J) + (4 \times 10^5 \, J) = 12 \times 10^5 \, J \] ### Final Answer The work done by the motor is: \[ W_{\text{total}} = 12 \times 10^5 \, J \]
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