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Question: At what speed, the velocity head of the water is equal to the pressure head of 40 cm of mercury? \...

At what speed, the velocity head of the water is equal to the pressure head of 40 cm of mercury?
A.2.8ms1 B.10.32ms1 C.5.6ms1 D.8.4ms1 \begin{aligned} & A.2.8m{{s}^{-1}} \\\ & B.10.32m{{s}^{-1}} \\\ & C.5.6m{{s}^{-1}} \\\ & D.8.4m{{s}^{-1}} \\\ \end{aligned}

Explanation

Solution

The velocity head of water is given by the equation,
v=12ρwV2v=\dfrac{1}{2}{{\rho }_{w}}{{V}^{2}}
Where ρw{{\rho }_{w}}the density of water,VVthe velocity of water and also vvis the velocity of water head.v=12×1000×V2 V=10.32ms1 P=13600×10×40×102 P=54400 ρm=13600Kgm3 \begin{aligned} & v=\dfrac{1}{2}\times 1000\times {{V}^{2}} \\\ & V=10.32m{{s}^{-1}} \\\ & P=13600\times 10\times 40\times {{10}^{-2}} \\\ & P=54400 \\\ & {{\rho }_{m}}=13600Kg{{m}^{-3}} \\\ \end{aligned}
The pressure head of mercury will be given as
P=ρmghP={{\rho }_{m}}gh
Where PPthe pressure head of the mercury, ρm{{\rho }_{m}}the density of mercury, ggis the acceleration due to gravity and hhis denoted as the height.
These two equations are being used up here in order to solve this question.

Complete step-by-step answer:
First of all let us look at the values of density of water as well as mercury in general.
The density of water is given as,
ρw=1000Kgm3{{\rho }_{w}}=1000Kg{{m}^{-3}}
And also the density of mercury is written as,
ρm=13600Kgm3{{\rho }_{m}}=13600Kg{{m}^{-3}}
It is mentioned the velocity head of the water is equal to the pressure head of mercury.
That is, the velocity head of water is given as,
v=12ρwV2v=\dfrac{1}{2}{{\rho }_{w}}{{V}^{2}}
Substituting the values will give,
v=12×1000×V2v=\dfrac{1}{2}\times 1000\times {{V}^{2}}
Thus the velocity will be,
v=500V2v=500{{V}^{2}}
And the pressure head of mercury can be written in the form of an equation,
P=ρmghP={{\rho }_{m}}gh
Substituting the values in it,
P=13600×10×40×102P=13600\times 10\times 40\times {{10}^{-2}}
Simplifying this will give,
P=54400P=54400
Now let us make a comparison in between the both equations,
Therefore we can write that,
54400=500V254400=500{{V}^{2}}
After necessary rearrangements, we can write that
V2=5445{{V}^{2}}=\dfrac{544}{5}
That is,
V=10.32ms1V=10.32m{{s}^{-1}}
Therefore the correct answer is obtained.

So, the correct answer is “Option B”.

Note: Mercury is basically an element. It is a liquid at room temperature and also it is very heavy. It will not sink in water, as it is denser. But heavy solid bodies such as iron cannonballs, will float in a pool of the silvery metal.