JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 1)
What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10$$-$$3 Pa s, respectively. (Density of water : 103 kg/m3)
Steady flow to unsteady flow
Unsteady to steady flow
Remains turbulent flow
Remains steady flow
Explanation
The nature of flow is determined by reynolds no
$$R = {{\rho VD} \over \eta }$$
If R < 1000 $$ \to $$ flow is steady
1000 < R < 2000 $$ \to $$ flow becomes unsteady
R > 2000 $$ \to $$ flow is turbulent
$${R_1} = {{4 \times {{10}^3} \times 0.18 \times {{10}^{ - 3}}} \over {60 \times \pi \times {{10}^{ - 2}} \times {{10}^{ - 3}}}} = {{4 \times {{10}^5} \times 0.18} \over {60\pi }}$$
$$ = 0.0038 \times {10^5} = 380$$
$${R_2} = {{0.48} \over {0.18}} \times 380 = 1018$$
$$R = {{\rho VD} \over \eta }$$
If R < 1000 $$ \to $$ flow is steady
1000 < R < 2000 $$ \to $$ flow becomes unsteady
R > 2000 $$ \to $$ flow is turbulent
$${R_1} = {{4 \times {{10}^3} \times 0.18 \times {{10}^{ - 3}}} \over {60 \times \pi \times {{10}^{ - 2}} \times {{10}^{ - 3}}}} = {{4 \times {{10}^5} \times 0.18} \over {60\pi }}$$
$$ = 0.0038 \times {10^5} = 380$$
$${R_2} = {{0.48} \over {0.18}} \times 380 = 1018$$
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