Question

12. The velocity of flow of a liquid through a pipe
at a point of radius r is v. At a point where the
radius is 2r, the mass of the liquid flowing out
per second is (d = density of the liquid)
Tur?vd
1) 2tr vd 2) 4Tr vd 3) Trở vd 4)​

Answer 1

Your given option is not clear. But i try my best.

Given:

The velocity of flow of a liquid through a pipe  at a point of radius r is v. At second point where the  radius is 2r. Density of the liquid is d.

To Find:

What is the mass of the liquid flowing out  per second?

Solution:

During flow of liquid mass flow rate is constant at every cross- section of flow:

Let us take first point in pipe:

$\mathbf{\textrm{Velocity at point 1 }= V_{1}= v \ \ \ \frac{m}{s}}$

$\mathbf{\textrm{Radius at point 1 }= R_{1}= r \ \ \ (m)}$

$\mathbf{\textrm{Area at point 1 }= A_{1}= \pi \times r^{2} \ \ \ (m^{2})}$

Let us take second point in pipe:

$\mathbf{\textrm{Velocity at point 2}= V_{2}= Unknown }$

$\mathbf{\textrm{Radius at point 2 }= R_{2}= 2r \ \ \ (m)}$

$\mathbf{\textrm{Area at point 2 }= A_{2}= \pi \times (2r)^{2} = 4\times \pi \times r^{2} \ \ \ (m^{2})}$

Since density is constant , means density at point 1 and point 2 is same.

Value of density = d

Mass flow rate at point 2 = Mass flow rate at point 1

$\mathbf{\rho _{2}\times A_{2}\times V_{2}=\rho _{1}\times A_{1}\times V_{1}}$

On putting respective value in above equation:

$\mathbf{d\times 4\times \pi \times r^{2}\times V_{2}=d\times \pi \times r^{2}\times v}$

On simplify:

$\mathbf{Velocity\ at \ point \ 2 = V_{2}= \dfrac{v}{4}}$

Let us calculate mass flow rate at point 2:

$\mathbf{Mass\ flow\ rate\ at\ point\ 2= \rho _{2}\times A_{2}\times V_{2}}$

$\mathbf{Mass\ flow\ rate\ at\ point\ 2= d\times 4\times \pi \times r^{2}\times \dfrac{v}{4}}$

$\mathbf{Mass\ flow\ rate\ at\ point\ 2= \pi \times d\times v\times r^{2} }$

Means mass flow rate of liquid is πdvr² kg per second through pipe.