Q. A small hole of area of cross-section 2 $mm^2$ is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 $m/s^2$, the rate of flow of water through the open hole would be nearly :
A) $12.6 \times 10^{-6}$ $m^3/s$
B) $8.9 \times 10^{-6}$ $m^3/s$
C) $2.23 \times 10^{-6}$ $m^3/s$
D) $6.4 \times 10^{-6}$ $m^3/s$
Explanation
[Solution] $v = \sqrt{2gh} = \sqrt{40} \approx 6.32$ m/s; $Q = Av = 2 \times 10^{-6} \times 6.32 = 12.6 \times 10^{-6}$ $m^3/s$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.