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\,Kg{m^{ - 3}}$ ).
(A) $0.5\,MW$
(B) $1\,MW$
(C) $1.5\,MW$
(D) $2\,MW$

Hint: The kinetic energy is the power available. Apply the continuity equation in the formula of the mass and derive its relation. Find the area of the river from the given width and the depth. Substitute these in the formula of the kinetic energy, simplifying it we get the value of the kinetic energy.
Formula used:
(1) The continuity equation is given by
$V = Av$
Where $V$ is the volume of the river per second, $A$ is the area of the river and $v$ is the velocity of the water in the river.
(2) The kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$
Where $K$ is the kinetic energy of the water and $m$ is the mass of the water flow.

Complete step-by-step solution:
It is given that the
Speed of the water in a river, $v = 2\,m{s^{ - 1}}$
Width of the river, $b = 50\,m$
Depth of the river, $d = 5\,m$
Density of water, $\rho = 1000\,Kg{m^{ - 3}}$
From the width and the depth of the river, let us calculate the area of it.
$A = wd$
Substituting the known values, we get
$A = 50 \times 5 = 250\,{m^2}$
It is known that mass is obtained by the product of volume and density.
$m = V\rho $
Substituting the continuity equation in the above formula, $m = Av\rho $ .
Let us take the formula of the kinetic energy,
$K = \dfrac{1}{2}m{v^2}$
Substitute the relation of mass in it.
$K = \dfrac{1}{2}A{v^3}\rho $
Substituting the area, velocity, and density in it.
$K = \dfrac{1}{2} \times 250 \times 1000 \times {2^3}$
By performing various arithmetic operations, we get
$K = {10^6}\,W = 1\,MW$
Hence the kinetic energy of the water flow in the river is obtained as $1\,MW$.
Thus the option (B) is correct.

Note: The kinetic energy mainly depends on the mass of the water flows and also the velocity of the water. If the velocity of the water is less, then the kinetic energy will also be less. This energy is used to run the turbine in order to derive the current in the hydroelectric power projects.