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Physics

Fluid-Mechanics

Easy

Question

A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side ‘L’ at a depth ‘4y’ from the top and the other is a circular hole of radius 'R' at a depth 'y’ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, ‘R’ is equal to:

$\frac{L}{\sqrt{2 π}}$
$2 π L$
$\sqrt{\frac{2}{π}} \cdot L$
$\frac{L}{2 π}$

The correct answer is: $square root of 2 over pi end root times L$

$table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell v subscript 1 equals square root of 8 g y end root text and end text v subscript 2 equals square root of 2 g left parenthesis y right parenthesis end root end cell row cell Q subscript 1 equals v subscript 1 S subscript 1 equals square root of 8 g y end root L squared semicolon Q subscript 2 equals v subscript 2 S subscript 2 end cell row cell equals square root of 2 g y end root pi R squared times Q subscript 1 equals Q subscript 2 end cell row cell therefore space of 1em R equals square root of 2 over pi end root times L end cell end table$

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