Question

Figure shows a rod of length $L$ pivoted near an end and which is made to rotate in a horizontal plane with a constant angular speed. A ball of mass $m$ is suspended by a string also of length $L$ from the other end of the rod. If $\theta$ is the angle made by string with the vertical, then which is wrong?

• A. $T\sin \theta =m{\omega }^{2}L(1+\sin \theta )$
• B. $T\cos \theta =mg$
• C. $\tan \theta =\frac{{\omega }^{2}L(1+\sin \theta )}{g}$
• D. None of the above

Answer 1

The correct option is D None of the above

For motion of ball


$T\cos \theta =mg\dots \dots \left(1\right)$
$T\sin \theta =m{\omega }^2(L+L\sin \theta )\dots \dots \left(2\right)$
From equation (1) and (2)
$\therefore \tan \theta =\frac{{\omega }^2(L+L\sin \theta )}{g}=\frac{{\omega }^{2}L(1+\sin \theta )}{g}$