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Physics-

General

Easy

Question

A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $omega$ . The force exerted by the liquid at the other end is

$\frac{M L ω^{2}}{2}$
$M L ω^{2}$
$\frac{M L ω^{2}}{4}$
$\frac{M L^{2} ω^{2}}{2}$

The correct answer is: $fraction numerator M L omega to the power of 2 end exponent over denominator 2 end fraction$

$d M equals open parentheses fraction numerator M over denominator L end fraction close parentheses d x$
Force on $´ d M ´$ mass is
$d F equals open parentheses d M close parentheses omega to the power of 2 end exponent x$

By integration we can get the force exerted by whole liquid
$rightwards double arrow F equals not stretchy integral subscript 0 end subscript superscript L end superscript fraction numerator M over denominator L end fraction omega to the power of 2 end exponent x blank d x equals fraction numerator 1 over denominator 2 end fraction M omega to the power of 2 end exponent L$

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