Q175 / 200 · Manipal (MET) 2013 · Physics
The rate of flow of glycerine of density $1.25 \times 10^3 \mathrm{kgm}^{-3}$ through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length $10 \mathrm{Nm}^{-2}$ is
- Correct answerA$6.93 \times 10^{-4} \mathrm{~m}^3 \mathrm{~s}^{-1}$
- B$7.8 \times 10^{-4} \mathrm{~m}^3 \mathrm{~s}^{-1}$
- C$10.4 \times 10^{-5} \mathrm{~m}^3 \mathrm{~s}^{-1}$
- D$14.5 \times 10^{-5} \mathrm{~m}^3 \mathrm{~s}^{-1}$
Solution
From Bernoulli's theorem
$p_1+\frac{1}{2} \rho v_1^2=p_2+\frac{1}{\alpha} \rho v_2^2$
$\therefore \quad p_1-p_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)$
$\therefore \quad 10=\frac{1}{2} \times 1.25 \times 10^3\left(v_2^2-v_1^2\right)$
$\therefore \quad v_2^2-v_1^2=\frac{10 \times 2}{1.25 \times 10^3}=16 \times 10^{-3}$ ....(i)
Also from equation of continuity
$\begin{aligned} & =A_1 v_1=A_2 v_2 \\ & \pi r_1^2 v_1=\pi r_2^2 v_2\end{aligned}$
$\therefore \quad \frac{v_1}{v_2}=\left[\frac{r_2}{r_1}\right]^2=\frac{0.04}{0.1}=0.4$
$v_1=0.4 v_2$ ....(ii)
Substituting this value in Eq. (i)
$\begin{gathered}v_2^2-\left(0.4 v_2\right)^2=16 \times 10^{-3} \\ v_2=1.38 \times 10^{-1}=0.138 \mathrm{~ms}^{-1}\end{gathered}$
Rate of flow of glycrine $v=A_2 v_2$
$\begin{aligned} & =\pi r_2^2 v_2 \\ & =6.93 \times 10^{-4} \mathrm{~m}^3 \mathrm{~s}^{-1}\end{aligned}$
Hence, the correct answer is option: 'A'.