question_answer

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

A)  \[0.25\text{ }cm\]     

B)  \[2\text{ }cm\]

C)  \[4\text{ }cm\]

D)  \[2.5\text{ }cm\]                         

Correct Answer: D

Solution :

The various forces acting on the particle, are its weight mg acting vertically downwards,  formal 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,  \[v=0.5m/s,\,\,\,g=10m/{{s}^{2}}\] \[\therefore \] \[h=\frac{{{(0.5)}^{2}}}{10}=0.025m\] Since, \[100\,cm=1m\] \[\therefore \] \[h=2.5\,\,cm\]