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A dam provides water fall for a power ho...

A dam provides water fall for a power house 300 m below. The catchment area of `10^(10)m^(2)` has annual rainfall of 0.6 m. 25%+ 15% + 10% water is lost due to different reasons. Then,

A

mass of water fall is `3xx10^(12)`kg

B

work done by water fall is `8.82xx10^(15)`J

C

work done by water fail is `8.11xx10^(10) `J

D

average power output of the station is 200 MW

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To solve the problem step by step, we will follow these calculations: ### Step 1: Calculate the volume of water collected annually The volume of water collected from the rainfall can be calculated using the formula: \[ \text{Volume} = \text{Catchment Area} \times \text{Rainfall Height} \] Given: - Catchment Area = \( 10^{10} \, m^2 \) - Annual Rainfall = \( 0.6 \, m \) Calculating the volume: \[ \text{Volume} = 10^{10} \, m^2 \times 0.6 \, m = 6 \times 10^{9} \, m^3 \] ### Step 2: Calculate the effective volume of water after losses The problem states that 25% + 15% + 10% of the water is lost. First, we need to calculate the total percentage of water lost: \[ \text{Total Loss} = 25\% + 15\% + 10\% = 50\% \] Thus, the effective volume of water that can be used is: \[ \text{Effective Volume} = \text{Volume} \times (1 - \text{Total Loss}) \] \[ \text{Effective Volume} = 6 \times 10^{9} \, m^3 \times (1 - 0.5) = 3 \times 10^{9} \, m^3 \] ### Step 3: Calculate the mass of the water To find the mass of the water, we use the formula: \[ \text{Mass} = \text{Volume} \times \text{Density} \] The density of water is approximately \( 1000 \, kg/m^3 \): \[ \text{Mass} = 3 \times 10^{9} \, m^3 \times 1000 \, kg/m^3 = 3 \times 10^{12} \, kg \] ### Step 4: Calculate the work done by the waterfall The work done (W) is given by the formula: \[ W = mgh \] where: - \( m = 3 \times 10^{12} \, kg \) - \( g = 9.8 \, m/s^2 \) (acceleration due to gravity) - \( h = 300 \, m \) (height of the waterfall) Calculating the work done: \[ W = 3 \times 10^{12} \, kg \times 9.8 \, m/s^2 \times 300 \, m \] \[ W = 3 \times 10^{12} \times 9.8 \times 300 \] \[ W = 8.82 \times 10^{15} \, J \] ### Step 5: Calculate the average power output Power (P) is defined as work done per unit time: \[ P = \frac{W}{t} \] Assuming \( t = 1 \, s \): \[ P = 8.82 \times 10^{15} \, W \] To convert this to megawatts (MW): \[ P = \frac{8.82 \times 10^{15}}{10^{6}} \, MW = 8.82 \times 10^{9} \, MW \] ### Final Answers: - Mass of water: \( 3 \times 10^{12} \, kg \) - Work done: \( 8.82 \times 10^{15} \, J \) - Average power output: \( 8.82 \times 10^{9} \, MW \)
To solve the problem step by step, we will follow these calculations: ### Step 1: Calculate the volume of water collected annually The volume of water collected from the rainfall can be calculated using the formula: \[ \text{Volume} = \text{Catchment Area} \times \text{Rainfall Height} \] Given: - Catchment Area = \( 10^{10} \, m^2 \) ...
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