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Question: The diagram shows a venturimeter through which water is flowing. The speed of water \(X\) is \(2\,cm...

The diagram shows a venturimeter through which water is flowing. The speed of water XX is 2cms12\,cm{s^{ - 1}}. The speed of water at YY ( taking g=10ms2g = 10\,m{s^2} ) is

A. 23cms123\,cm{s^{ - 1}}
B. 32cms132\,cm{s^{ - 1}}
C. 101cms1101\,cm{s^{ - 1}}
D. 1024cms11024\,cm{s^{ - 1}}

Explanation

Solution

Here we have to use the formula of the venturimeter. The formula shows the pressure difference between XX that is before narrowing of the pipe and YY that is after narrowing of the pipe. After that using the pressure difference formula of fluid we can solve the equation to find the value of speed of water at YY.

Complete step by step answer:
As per the problem we have a venturimeter through which water is flowing. The speed of water X is 2cms12\,cm{s^{ - 1}}. Now we need to calculate the speed of water at Y.
We know venture effect is represented as,
pxpy=ρ2(vy2vx2)(1)p_x - p_y = \dfrac{\rho }{2}\left( {v{y^2} - {v_x}^2} \right) \ldots \ldots \left( 1 \right)
Where, Pressure at position X before the narrowing of the pipe is pxpx, Pressure at position Y after the narrowing of the pipe is pyp_y, Density of fluid travelling in the pipe is ρ\rho , Velocity of the fluid at position Y is vyv_y and Velocity of the fluid at position X is vxv_x.

The pressure difference is also represented as,
pxpy=Δhρg(2)p_x - p_y = \Delta h\rho g \ldots \ldots \left( 2 \right)
Where,
Δh\Delta h is the change in height due to change in before and after narrowing of the pipe.
Now equation equation (1)\left( 1 \right) and (2)\left( 2 \right) we will get,
ρ2(vy2vx2)=Δhρg\dfrac{\rho }{2}\left( {{v_y}^2 - v_{x^2}} \right) = \Delta h\rho g
Now cancelling the common terms we will get,
12(vy2vx2)=Δhg\dfrac{1}{2}\left( {{v_y}^2 - {v_x}^2} \right) = \Delta hg

We know,
Δh=5.1mm=0.51cm\Delta h = 5.1mm = 0.51cm
vx=2cms1\Rightarrow v_x = 2cm{s^{ - 1}}
g=10ms1=1000cms1\Rightarrow g = 10\,m{s^{ - 1}} = 1000\,cm{s^{ - 1}}
Now putting the given values in the above equation we will get,
12(vy222)=0.51×1000\dfrac{1}{2}\left( {{v_y}^2 - {2^2}} \right) = 0.51 \times 1000
Now on rearranging and solving further solving we will get,
vy2=(510×2)+2=1024{v_y}^2 = \left( {510 \times 2} \right) + 2 = 1024
Hence we can say the speed of water at Y is 32cms132\,cm{s^{ - 1}}.

Therefore the correct option is (B)\left( B \right).

Note: Here we converted all the units of length in centimeter as our option is given in centimeter. Remember that a venturimeter is a device which is used to measure the rate of flow of fluid flowing through a pipe. The principle of venturimeter is that when a fluid flows through it, it’s acceleration in the convergent section and decelerates in the divergent section which results in drop in static pressure.