JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 18)
Two tubes of radii r1 and r2, and lengths l1 and l2 , respectively, are connected in series
and a liquid flows through each of them in stream line conditions. P1 and P2 are pressure differences across the two tubes.
If P2 is 4P1 and l2 is $${{{1_1}} \over 4}$$, then the radius r2 will be equal to :
If P2 is 4P1 and l2 is $${{{1_1}} \over 4}$$, then the radius r2 will be equal to :
r1
2r1
4r1
$${{{r_1}} \over 2}$$
Explanation
We know,
Rate of flow of liquid through narrow tube,
$${{dv} \over {dt}}$$ = $${{\pi {{\Pr }^4}} \over {8\eta l}}$$
Both tubes are connected in series so rate of flow of liquid is same.
$$\therefore\,\,\,$$ $${{\pi {P_1}r_1^4} \over {8\eta {l_1}}}$$ = $${{\pi {P_2}r_2^4} \over {8\eta {l_2}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{{P_1}{r_1}^4} \over {{l_1}}}$$ = $${{{P_2}{r_2}^4} \over {{l_2}}}$$
Given,
P2 = 4P1 and $${l_2}$$ = $${{{l_1}} \over 4}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{{P_1}{r_1}^4} \over {{l_1}}}$$ = $${{4{P_1}{r_2}^4} \over {{{{l_1}} \over 4}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $${r_2}^4$$ = $${{{r_1}^4} \over {16}}$$
$$ \Rightarrow $$$$\,\,\,$$ r2 = $${{{r_1}} \over 2}$$
Rate of flow of liquid through narrow tube,
$${{dv} \over {dt}}$$ = $${{\pi {{\Pr }^4}} \over {8\eta l}}$$
Both tubes are connected in series so rate of flow of liquid is same.
$$\therefore\,\,\,$$ $${{\pi {P_1}r_1^4} \over {8\eta {l_1}}}$$ = $${{\pi {P_2}r_2^4} \over {8\eta {l_2}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{{P_1}{r_1}^4} \over {{l_1}}}$$ = $${{{P_2}{r_2}^4} \over {{l_2}}}$$
Given,
P2 = 4P1 and $${l_2}$$ = $${{{l_1}} \over 4}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{{P_1}{r_1}^4} \over {{l_1}}}$$ = $${{4{P_1}{r_2}^4} \over {{{{l_1}} \over 4}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $${r_2}^4$$ = $${{{r_1}^4} \over {16}}$$
$$ \Rightarrow $$$$\,\,\,$$ r2 = $${{{r_1}} \over 2}$$
Comments (0)
