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
A.
$$\frac{{ML{\omega ^2}}}{2}$$
B.
$$\frac{{M{L^2}\omega }}{2}$$
C.
$$ML{\omega ^2}$$
D.
$$\frac{{M{L^2}{\omega ^2}}}{2}$$
Answer :
$$\frac{{ML{\omega ^2}}}{2}$$
Solution :
Let the length of a small element of tube be $$dx.$$ Mass of this element $$dm = \frac{M}{L}dx$$
where, $$M$$ is mass of filled liquid and $$L$$ is length of tube, Force on this element
$$\eqalign{ & dF = dm \times x{\omega ^2} \cr & \int_0^F {dF = \frac{M}{L}{\omega ^2}} \int_0^L x dx \cr & {\text{or}}\,\,F = \frac{M}{L}{\omega ^2}\left[ {\frac{{{L^2}}}{2}} \right] = \frac{{ML{\omega ^2}}}{2} \cr & {\text{or}}\,\,F = \frac{1}{2}ML{\omega ^2} \cr} $$
Let the length of a small element of tube be $$dx.$$ Mass of this element $$dm = \frac{M}{L}dx$$
where, $$M$$ is mass of filled liquid and $$L$$ is length of tube, Force on this element
$$\eqalign{ & dF = dm \times x{\omega ^2} \cr & \int_0^F {dF = \frac{M}{L}{\omega ^2}} \int_0^L x dx \cr & {\text{or}}\,\,F = \frac{M}{L}{\omega ^2}\left[ {\frac{{{L^2}}}{2}} \right] = \frac{{ML{\omega ^2}}}{2} \cr & {\text{or}}\,\,F = \frac{1}{2}ML{\omega ^2} \cr} $$
