At what speed the velocity head of water is equal to the pressure head of 40cm of mercury?
(A) $ 10.32m/s $
(B) $ 10.23m/s $
(C) $ 10.52m/s $
(D) $ 10.54m/s $

Hint
We have to equate the formula for the velocity head of water to the pressure head of mercury according to the formula, $ \dfrac{1}{2}{\rho _w}{v^2} = {\rho _m}gh $ and then substitute the values to find the velocity of the stream of water.
In this solution we will be using the following formula,
 $\Rightarrow {\text{velocity head}} = \dfrac{1}{2}{\rho _w}{v^2} $
where $ {\rho _w} $ is the density of water and $ v $ is the velocity of water
 $\Rightarrow {\text{Pressure head}} = {\rho _m}gh $
where $ {\rho _m} $ is the density of mercury, $ g $ is the acceleration due to gravity and $ h $ is the height of mercury.

Complete step by step answer
Let us consider the velocity of the water stream as $ v $. So the velocity head of the water stream will be given by the formula,
 $\Rightarrow {\text{velocity head}} = \dfrac{1}{2}{\rho _w}{v^2} $
Now the density of water denoted by $ {\rho _w} $ has a value of $ {\rho _w} = {10^3}kg/{m^3} $
So by substituting the value we get,
 $\Rightarrow {\text{velocity head}} = \dfrac{1}{2} \times {10^3}{v^2} = \dfrac{{{{10}^3}}}{2}{v^2} $
Now the pressure head of the mercury column is given by the formula,
 $\Rightarrow {\text{Pressure head}} = {\rho _m}gh $
The density of mercury is denoted by $ {\rho _m} $ and has a value of $ {\rho _m} = 13.6 \times {10^3}kg/{m^3} $
The height of the mercury column as given in the question is $ h = 40cm = 0.4m $
The acceleration due to gravity has a value of $ g = 9.8m/{s^2} $
By substituting these values, we get the value of the pressure head as,
 $\Rightarrow {\text{Pressure head}} = 13.6 \times {10^3} \times 9.8 \times 0.4 $
On doing the multiplication above, we get the value of pressure head as,
 $\Rightarrow {\text{Pressure head}} = 53312kg/m{s^2} $
Now in the question it is said that the velocity head of water is equal to the pressure head of mercury. So by equating the above obtained values of the velocity head and the pressure head, we get
 $\Rightarrow \dfrac{{{{10}^3}}}{2}{v^2} = 53312 $
On taking the numeric values on the LHS to the RHS we get
 $\Rightarrow {v^2} = 53312 \times \dfrac{2}{{{{10}^3}}} $
On calculating this we get a value of
 $\Rightarrow {v^2} = 106.624 $
So by doing square root on both the sides of the equation we get the velocity of water as,
 $\Rightarrow v = \sqrt {106.624} = 10.32m/s $
Therefore, the correct answer will be option (A).

Note
In the fluid mechanics, the energy per unit weight of a fluid is termed as head. Therefore, the velocity head refers to the amount of kinetic energy per unit weight of fluid, which when provided to that certain fluid will rise to a specific height. Similarly the pressure energy per unit weight of the fluid is termed as pressure head.