JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 14)
An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and
minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum
and the maximum velocities of fluid in this pipe is :
$${3 \over 4}$$
$${9 \over {16}}$$
$${{\sqrt 3 } \over 2}$$
$${{81} \over {256}}$$
Explanation
Using equation of continuity
A1V1 = A2V2
$$ \Rightarrow $$ $${{{V_1}} \over {{V_2}}} = {{{A_2}} \over {{A_1}}}$$ = $${\left( {{{4.8} \over {6.4}}} \right)^2}$$ = $${9 \over {16}}$$
A1V1 = A2V2
$$ \Rightarrow $$ $${{{V_1}} \over {{V_2}}} = {{{A_2}} \over {{A_1}}}$$ = $${\left( {{{4.8} \over {6.4}}} \right)^2}$$ = $${9 \over {16}}$$
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