question_answer

A heavy small-sized sphere is suspended by a string of length \[l.\]The sphere .rotates uniformly in a horizontal circle with the string making an angle \[\theta \] with the vertical. Then, the time-period of this conical pendulum is

A) \[t=2\pi \sqrt{\frac{l}{g}}\]                         

B) \[t=2\pi \sqrt{\frac{l\sin \theta }{g}}\]

C) \[t=2\pi \sqrt{\frac{l\cos \theta }{g}}\]  

D) \[t=2\pi \sqrt{\frac{l}{g\cos \theta }}\]

Correct Answer: C

Solution :

Radius of circular path in the horizontal plane\[r=l\sin \theta \] Resolving T along the vertical and horizontal directions, we get \[T=\cos \theta =Mg\]    ?(i) \[T\sin \theta =Mr{{\omega }^{2}}=M(l\sin \theta ){{\omega }^{2}}\]or\[T=Ml{{\omega }^{2}}\](ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{1}{\cos \theta }=\frac{l{{\omega }^{2}}}{g}\]or\[{{\omega }^{2}}=\frac{g}{l\cos \theta }\] \[\therefore \]Time period \[t=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l\cos \theta }{g}}\]