JEE PYQ: Mechanical Properties Of Fluids Question 23
Question 23 - 2019 (08 Apr 2019 Shift 1)
Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of: (density of water = 1000 kg/m$^3$, coefficient of viscosity of water = 1 mPa s)
(1) $10^3$
(2) $10^4$
(3) $10^2$
(4) $10^6$
Show Answer
Answer: (2) $10^4$
Solution
Rate of flow of water ($V$) = 100 lit/min $= \frac{100 \times 10^{-3}}{60} = \frac{5}{3} \times 10^{-3}$ m$^3$/s $\therefore$ Velocity of flow of water ($v$): $v = \frac{V}{A} = \frac{5 \times 10^{-3}}{3 \times \pi \times (5 \times 10^{-2})^2} = \frac{10}{15\pi} = \frac{2}{3\pi}$ m/s $= 0.2$ m/s $\therefore$ Reynold number ($N_R$) $= \frac{Dv\rho}{\eta}$ $= \frac{(10 \times 10^{-2}) \times \frac{2}{3\pi} \times 1000}{1} = 2 \times 10^4$ Order of $N_R = 10^4$
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