Answer

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**Hint:**You could note down all the quantities that are given in the question. Then, you could recall the relation that is apt for solving the given question which here would be the continuity equation. You could now substitute the value accordingly and thereby find the answer that is the velocity of water in pipe D.

**Formula used:**

Continuity equation,

${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}$

**Complete answer:**

In the question, we are given 3 pipes that are fitted together. If you look at the figure, you could clearly see that the pipe GB is fitted to the other two pipes C and D. The area of cross section at G is given as,

$A=24{{m}^{2}}$

Then we are given the values of velocities of water at G and C respectively as,

${{v}_{G}}=10m{{s}^{-1}}$

${{v}_{C}}=6m{{s}^{-1}}$

We are supposed to find the velocity of water at D.

For that we could use the continuity equation. This equation if you may recall could be given by,

${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}$

Applying this here,

${{A}_{G}}{{v}_{G}}={{A}_{C}}{{v}_{C}}+{{A}_{D}}{{v}_{D}}$

$\Rightarrow A\left( 10 \right)=\dfrac{A}{2}\left( 6 \right)+\dfrac{A}{3}\left( v \right)$

$\Rightarrow 10=\dfrac{6}{2}+\dfrac{v}{3}$

$\Rightarrow \dfrac{v}{3}=7$

$\therefore v=21m{{s}^{-1}}$

Therefore, we found that the velocity of water in the pipe D would be 21m/s.

**Hence, option A is found to be the answer.**

**Note:**

The continuity equation, in general, is known to describe the transport. As per the continuity equation, for a steady state process, the rate of mass entering the system is found to be equal to the mass leaving the system. We could clearly see that the law of conservation of mass is being applied to the fluid flow.

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