Water is flowing in a river at $2 m s^{- 1}$ . The river is $50 m$ wide and has an average depth of $5 m$ . The power available from the current in the river is (Density of water $= 1000 k g m^{- 3})$ .

0.5 M W

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1 M W

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1.5 M W

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2 M W

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Solution

The correct option is B $1 M W$
Given data,
speed of water= $= v = 2 m / s$
area of flow $= 5 m \times 50 m = 250 m^{2}$
Density of water $ρ = 1000 k g / m^{3}$
Form of the continuity equation the volume of water flowing per second $v = A v$
Hence, mass of water $m = v ρ = A v ρ$
$K . E = \frac{1}{2} m v^{2} = \frac{1}{2} A v ρ v^{2}$
$= \frac{1}{2} A ρ v^{3}$
$ρ = \frac{1}{2} A ρ v^{3} = \frac{1}{2} \times 250 \times (2)^{3} \times 1000$
$ρ = 10^{6} W = 1 M W$

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Water is flowing in a river at 2 ms−1. The river is 50 m wide and has an average depth of 5 m. The power available from the current in the river is (Density of water =1000 kg m−3).

The correct option is B 1 MWGiven data,speed of water= =v=2m/sarea of flow =5m×50m=250m2Density of water ρ=1000kg/m3Form of the continuity equation the volume of water flowing per second v=AvHence, mass of water m=vρ=AvρK.E=12mv2=12Avρv2=12Aρv3ρ=12Aρv3=12×250×(2)3×1000ρ=106W=1MW

Water is flowing in a river at $2 m s^{- 1}$ . The river is $50 m$ wide and has an average depth of $5 m$ . The power available from the current in the river is (Density of water $= 1000 k g m^{- 3})$ .