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 […

Question

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 [NEET 2019]

Accepted Answer

Correct answer: D. (d) The rate of water flow moving with velocity v through an area a is given by Rate (R )= Area (a ) Velocity (v ) Given, area of cross-section of hole, a=2 mm^2=2 10^-6 m^2 height of tank, h=2 m As the velocity of liquid flow is given as v= 2 gh ∴R= av=a 2 gh Substituting the given values, we get, R=2 10^-6 2 10 2=2 10^-6 6.32=12.64 10^-6 m^3/s ≃12.6 10^-6 m^3/s

$A small hole of area of cross-section 2 mm^{2} is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m / s^{2}, the rate of flow of water through the open hole would be nearly [NEET 2019]$

Solution

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