JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 22)
Water from a pipe is coming at a rate of
100 litres 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/m3,
coefficient of viscosity of water = 1mPas)
106
104
103
102
Explanation
Flow rate of water (Q) = 100 lit/min
= $${{100 \times {{10}^{ - 3}}} \over {60}} = {5 \over 3} \times {10^{ - 3}}{m^3}$$
$$ \therefore $$ Velocity of flow (v)
= $${Q \over A} = {{5 \times {{10}^{ - 3}}} \over {3 \times \pi \times {{(5 \times {{10}^{ - 2}})}^2}}}$$
$$ = {{10} \over {15\pi }} = {2 \over {3\pi }}\,m/s$$
= 0.2 m/s
$$ \therefore $$ Reynold number (Re) = $${{Dv\rho } \over \eta }$$
$$ = {{\left( {10 \times {{10}^{ - 2}}} \right) \times {2 \over {3\pi }} \times 1000} \over 1} = 2 \times {10^4}$$
$$ \therefore $$ Order of Re = 104
= $${{100 \times {{10}^{ - 3}}} \over {60}} = {5 \over 3} \times {10^{ - 3}}{m^3}$$
$$ \therefore $$ Velocity of flow (v)
= $${Q \over A} = {{5 \times {{10}^{ - 3}}} \over {3 \times \pi \times {{(5 \times {{10}^{ - 2}})}^2}}}$$
$$ = {{10} \over {15\pi }} = {2 \over {3\pi }}\,m/s$$
= 0.2 m/s
$$ \therefore $$ Reynold number (Re) = $${{Dv\rho } \over \eta }$$
$$ = {{\left( {10 \times {{10}^{ - 2}}} \right) \times {2 \over {3\pi }} \times 1000} \over 1} = 2 \times {10^4}$$
$$ \therefore $$ Order of Re = 104
Comments (0)
