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3-2.Motion in Plane
medium
A hemispherical bowl of radius $R$ is rotated about its axis of symmetry which is kept vertical with angular velocity $\omega $ . A small block is kept in the bowl. It remains stationary relative to the bowl surface at a position where the radius makes an angle $\theta $ with the vertical. The friction is absent. The value of $\theta $ is
A${\cos ^{ - 1}}\,\left( {\frac{g}{{R{\omega ^2}}}} \right)$
B${\sin ^{ - 1}}\,\left( {\frac{g}{{R{\omega ^2}}}} \right)$
C${\tan ^{ - 1}}\,\left( {\frac{g}{{R{\omega ^2}}}} \right)$
Dnone of these
Solution

For horizontal forces.
$\mathrm{N} \sin \theta=\mathrm{mx} \omega^{2}$ ……$(i)$
For vertical forces.
$\mathrm{N} \cos \theta=\mathrm{mg}$ ………$(ii)$
$\tan \theta=\frac{x \omega^{2}}{g}$
$\text { But } \quad x=R \sin \theta$
$\therefore $ $\frac{\sin \theta}{\cos \theta}=\frac{R \sin \theta \cdot \omega^{2}}{g}$
${\rm{ or }}\cos \theta = g/{{\mathop{\rm Rg}\nolimits} ^2}$
${\rm{ or\,\, }}\theta = {\cos ^{ – 1}}\left( {\frac{g}{{{\mathop{\rm R}\nolimits} {\omega ^2}}}} \right)$
Standard 11
Physics
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