Question

If the water falls from a dam into a turbine wheel $19.6\;m$ below, then the velocity of water at the turbine is (take $g=9.8m/s^{2}$)

& A.19.6m/s \\\ & B.39.0m/s \\\ & C.98.8m/s \\\ & D.9.8m/s \\\ \end{aligned}$$

Answer 1

We know that the water stored in a dam is used to produce electricity using a hydropower plant. Clearly, the water stored in the dam experiences two types of energy namely, the potential energy and the kinetic energy. Using this we can solve the question.

Formula used: $K.E+P.E=0$

Complete step by step answer:
We know that the hydropower plant uses the potential energy of water stored in the dam to rotate a turbine. It converts the energy of the falling water into electrical energy. Initially, the kinetic energy is converted into mechanical energy by a turbine and then the mechanical energy is converted to electrical energy by a generator. Water that is pumped uphill into reservoirs during periods of low demand is then released for a generation when the demand is high.
Clearly, the water falling from a height $h$ of the dam experiences potential energy given as $-mgh$, where the negative sign indicates the fall of water. Also, the flowing water experiences a kinetic energy given as$\dfrac{1}{2}mv^{2}$ where $v$ is the velocity of the flowing water.
We know from the law of conservation of energy that the energy of the water is conserved. Then we can say that, $-mgh+\dfrac{mv^{2}}{2}=0$
$\implies mgh=\dfrac{mv^{2}}{2}$
$\implies v=\sqrt{2gh}$
Given the height from which the water falls as $h=19.6\;m$ and$g=9.8m/s^{2}$. Then substituting the values, we get, $v=\sqrt{2\times 9.8\times 19.6}=19.6m/s$

So, the correct answer is “Option A”.

Note: We know that potential energy is the energy gained by an object when it is thrown up to some height. Here since the water falls from some height, we have a negative sign. Also note that, we don’t need the mass of the water to calculate the velocity, as it gets cancelled.