Q. The rate of flow of water in a capillary tube of length $l$ and radius r is V. The rate of flow in another capillary tube of length $2l$ and radius $2r$ for same pressure difference would be
A) $16\ V$
B) $9\ V$
C) $8\ V$
D) $2\ V$
Explanation
(c) Rate of flow of water through a capillary tube is
$V = \frac{\pi Rr^{4}}{8\eta l}$
As $P,\eta$ remain the same
$\therefore\frac{V^{'}}{V} = \frac{(2r)^{4}}{(r)^{4}} \times \frac{(l)}{(2l)} = \frac{16}{2} = 8 \Rightarrow V^{'} = 8V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.