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A tube of length 50 \mathrm{~cm} is filled completely with an incompressible liquid of mass 250 \mathrm{~g}  and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity \mathrm{x \sqrt{F}\: \: \mathrm{rad} \: \: \mathrm{s}^{-1}}. If \mathrm{F} be the force exerted by the liquid at the other end then the value of \mathrm{x} will be____________.

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{F=\int d F =\int d n x \omega^2 }

    \mathrm{=\int\left(\frac{M}{L} d x\right) x \omega^2 }

    \mathrm{=\frac{M \omega^2}{L} \int_0^L x d x }

\mathrm{F =\frac{M L \omega^2}{2} }

\mathrm{F =\frac{(0.25) \times(0.5) \times\left(x^2 F\right)}{2} }

\mathrm{{16}=x^2}

\mathrm{x=4}

The value of x is 4

Posted by

Gautam harsolia

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