Q: 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 ω. The force exerted by the liquid at the other end is

(a)(Mω^{2} L)/2

(b)Mω^{2} L

(c) (Mω^{2} L)/4

(d) (Mω^{2} L^{2})/2

Ans: (a)

Sol: Taking element of length dx at a distance x from the axis of rotation .

Mass of element dx is $\large m = \frac{M}{L}dx$

For rotation , this element require centripetal force .

$\large dF = m \omega^2 x$

$\large dF = (\frac{M}{L}dx) \omega^2 x$

$\large F = \int_{0}^{L}dF $

$\large F = \int_{0}^{L} \frac{M}{L} \omega^2 x dx $

$\large F = \frac{M}{L} \omega^2 \int_{0}^{L} x dx $

$\large F = \frac{M}{L} \omega^2 (\frac{L^2}{2})$

$\large F = \frac{M\omega^2 L}{2} $

The force exerted by the liquid at the other end is

$\large F = \frac{M\omega^2 L}{2} $