Answer:
The forces acting on the
particle are: weight (mg) of particle acting vertically downwards. Reaction (R)
of funnel is acting perpendicular to wall of the funnel.
As is clear
from Fig. 5(HT).7,
\[R\,\cos
\theta =mg\] ?..(i)
\[R\,\sin
\text{ }\!\!\theta\!\!\text{ }=\frac{m{{\upsilon }^{2}}}{r}\] ?..
(ii)
Dividing
(ii) by (i), we get
\[\tan
\text{ }\!\!\theta\!\!\text{ =}\frac{{{\upsilon }^{2}}}{rg}\]
\[\upsilon
=\sqrt{r\,g\,\tan \text{ }\!\!\theta\!\!\text{ }}\]
From
Fig. 5(HT).7, \[\tan \text{ }\!\!\theta\!\!\text{ =}\frac{h}{r}\]
\[\therefore
\] \[\upsilon =\sqrt{rg\times \frac{h}{r}}=\sqrt{gh}=\sqrt{9.8\times 9.8\times
{{10}^{-2}}}=0.98\,m\,{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec