JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 19)
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 $$x \sqrt{F} \,\mathrm{rad} \,\mathrm{s}^{-1}$$. If $$\mathrm{F}$$ be the force exerted by the liquid at the other end then the value of $$x$$ will be __________.
Answer
4
Explanation
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$\mathrm{F}=\int(\mathrm{dm}) \omega^2 \mathrm{x}$
$=\int_0^{\mathrm{L}}\left(\frac{\mathrm{m}}{\mathrm{L}} \mathrm{dx}\right) \omega^2 \mathrm{x}$
$=\frac{\mathrm{m}}{\mathrm{L}} \omega^2 \frac{\mathrm{L}^2}{2}$
$=\frac{\mathrm{m} \omega^2 \mathrm{~L}}{2}$
$\omega=\sqrt{\frac{2}{\mathrm{~mL}}} \sqrt{\mathrm{F}}$
$=\sqrt{\frac{2}{0.25 \times 0.5}} \sqrt{\mathrm{F}}$
$=\sqrt{16} \sqrt{\mathrm{F}}$
$=4 \sqrt{\mathrm{F}}$
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