Answer

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**Hint:**From the given value of the internal diameter of the hemispherical tank, we can determine its internal radius. The internal volume of the tank can be found by using the formula for the volume of a hemisphere, which is given by $V=\dfrac{2}{3}\pi {{r}^{3}}$ where $r$ is the internal radius. The initial volume of water in the tank is given to be equal to $50$ kilo litres, which must be converted to meter cubes by using the relation \[1l={{10}^{-3}}{{m}^{3}}\]. Finally, the required volume of the water pumped into the tank will be equal to the difference between the internal volume of the hemisphere and the initial volume of water in the tank.

**Complete step by step solution:**

In the above question, the initial volume of the water in the tank is given to be equal to $50$ kilo litres, so that we have

$\begin{align}

& \Rightarrow {{V}_{i}}=50kl \\

& \Rightarrow {{V}_{i}}=50\times {{10}^{3}}l \\

\end{align}$

Now, we know that \[1l={{10}^{-3}}{{m}^{3}}\]. So the initial volume of water becomes

\[\begin{align}

& \Rightarrow {{V}_{i}}=50\times {{10}^{3}}\times {{10}^{-3}}{{m}^{3}} \\

& \Rightarrow {{V}_{i}}=50{{m}^{3}}.......(i) \\

\end{align}\]

Now, according to the above question, water is pumped into the tank to fill to its capacity. So the final volume of water in the tank will be equal to the internal volume of the hemisphere. So we have to determine the internal volume of the tank.

The internal diameter of the hemispherical tank is given to be equal to $14m$. Let us write it as

$\Rightarrow D=14m........(ii)$

We know that the radius is equal to the half of the diameter. So the internal radius of the tank is

$\Rightarrow r=\dfrac{D}{2}$

Substituting (i) in the above, we get

$\begin{align}

& \Rightarrow r=\dfrac{14}{2}m \\

& \Rightarrow r=7m........(iii) \\

\end{align}$

Now, we know that the volume of a hemisphere is given by the formula

$\Rightarrow V=\dfrac{2}{3}\pi {{r}^{3}}$

Putting (iii) in the above equation, we get the internal volume of the hemisphere as

$\Rightarrow V=\dfrac{2}{3}\pi \times {{7}^{3}}$

Putting $\pi =\dfrac{22}{7}$ in the above equation, we get

$\begin{align}

& \Rightarrow V=\dfrac{2}{3}\times \dfrac{22}{7}\times 7\times 7\times 7 \\

& \Rightarrow V=\dfrac{2}{3}\times 22\times 7\times 7 \\

& \Rightarrow V=\dfrac{44\times 49}{3} \\

& \Rightarrow V=718.67{{m}^{3}} \\

\end{align}$

According to the question, this is the final volume of water in the tank, that is,

$\Rightarrow {{V}_{f}}=718.67{{m}^{3}}.......(iv)$

Therefore, the volume of water pumped into the tank is given by

$\Rightarrow {{V}_{p}}={{V}_{f}}-{{V}_{i}}$

Putting (i) and (iv) we get

$\begin{align}

& \Rightarrow {{V}_{p}}=718.67{{m}^{3}}-50{{m}^{3}} \\

& \Rightarrow {{V}_{p}}=668.67{{m}^{3}} \\

\end{align}$

**Hence, the correct answer is option (d).**

**Note:**Do not worry about the external diameter of the hemispherical tank. This is because the water occupies the internal volume of a tank and not the external. Also, do not forget to convert the unit of the volume, which is given in litres, to cubic metres. For this conversion, we must be familiar with the basic units of measurements and their conversions.

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