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Question: A small hole of an area of cross-section \(2\,\,m{m^2}\) is present near the bottom of a fully fille...

A small hole of an area of cross-section 2  mm22\,\,m{m^2} is present near the bottom of a fully filled open tank of height 2  m2\,\,m. Taking g=10  ms−2g = 10\,\,m{s^{ - 2}}, the rate of flow of water through the open hole would be nearly;
A) 12.6×10−6  m3 s−112.6 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}}
B) 8.9×10−6  m3 s−18.9 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}}
C) 2.23×10−6  m3 s−12.23 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}}
D) 6.4×10−6  m3 s−16.4 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}}

Explanation

Solution

Hint:- The given above problem can be solved using the formula that is derived from the rate of flow of the liquid due the acceleration due to gravity with respect to the area of cross-section through which the liquid has to flows from with a certain velocity

Useful formula:
The rate of flow of liquid is given by;
Q=auQ = au
Where, QQ denotes the rate of flow of liquid through the open hole, aa denotes the cross-sectional area of the hole, uu denotes the velocity with which the liquid flows.

Complete step by step solution:
The data given in the problem is;
The area of the cross section of the hole is, a=2  mm2a = 2\,\,m{m^2}.
The height of the tank filed with the liquid is, h=2  mh = 2\,\,m.
The value of the gravity is, g=10  ms−2g = 10\,\,m{s^{ - 2}}
The rate of flow of liquid through the hole is given by;
Q=au  ..........(1)Q = au\,\,..........\left( 1 \right)
Since we know that the,
u=2ghu = \sqrt {2gh}
Where, ggis the acceleration due to gravity, hh denotes the height of the tank that is filled with the liquid.
Substitute the value of the height of the tank, h=2  mh = 2\,\,m and the value of the acceleration due to gravity, g=10  ms−2g = 10\,\,m{s^{ - 2}} in the above formula;
u=2×10×2 u=40 u=6.324  ms−1  u = \sqrt {2 \times 10 \times 2} \\\ u = \sqrt {40} \\\ u = 6.324\,\,m{s^{ - 1}} \\\

Substitute the values of the area if the cross-section, a=2  mm2a = 2\,\,m{m^2} and the value of the velocity through which the liquid flows is, u=6.324  ms−1u = 6.324\,\,m{s^{ - 1}} in the equation (1);
Q=au  ..........(1)Q = au\,\,..........\left( 1 \right)
Q=2×10−6  m2×6.324  ms−1Q = 2 \times {10^{ - 6}}\,\,{m^2} \times 6.324\,\,m{s^{ - 1}}
On equating the equation, we get;
Q=12.6×10−6  m3 s−1Q = 12.6 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}}

Therefore, the rate of flow liquid through a cross- sectional area of a hole is Q=12.6×10−6  m3 s−1Q = 12.6 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}}.

Hence the option (A), Q=12.6×10−6  m3 s−1Q = 12.6 \times {10^{ - 6}}\,\,{m^3}\,{s^{ - 1}} is the correct answer.

Note: The rate of flow of the liquid changes drastically when the area of cross-section through which it flows increases or decreases in its area. The height of the liquid from which it flows can also change the rate of flow of the liquid as it affects the velocity with which the liquid flows.