question_answer

A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 m/s. What is the height of the plane of circle from vertex of the funnel?

A)  0.25 cm                               

B)  2 cm

C)  4 cm                                     

D)  2.5 cm

Correct Answer: D

Solution :

                 The various forces acting on the particle, are its weight mg acting vertically downwards, normal reaction N. Equating the vertical forces, we have \[N\,\sin \theta =mg\]                       ... (i) Also, centripetal force, \[\frac{m{{v}^{2}}}{R}=N\,\cos \theta \]                                     ...(ii) From Eqs. (i) and (ii), we get \[\tan \theta =\frac{Rg}{{{v}^{2}}}\]                                         ...(iii) Also, from triangle OAB, \[\tan \theta =\frac{R}{h}\]                                    ...(iv) Equating Eqs. (iii) and (iv), we get                 \[h=\frac{{{v}^{2}}}{g}\] Given,   \[v=0.5\text{ }m/s,\text{ }g=10\text{ }m/{{s}^{2}}\] \[\therefore \]  \[h=\frac{{{(0.5)}^{2}}}{10}=0.025\,m\] Since,     \[100\text{ }cm=1\text{ }m\] \[\therefore \] \[h=2.5\,cm\]