question_answer

A conical pendulum of length 1 m makes an angle \[\theta =45{}^\circ \] w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its centre is vertically below O. The speed of the pendulum, in its circular path, will be: (Take\[g=10\,m/{{s}^{-2}}\])

A) 0.4 m/s 

B)        4 m/s

C) 0.2 m/s 

D)         2 m/s

Correct Answer: D

Solution :

[d] Given, \[\theta =45{}^\circ ,\]\[r=0.4\,m,\]\[g=10m/{{s}^{2}}\] \[T\sin \theta =\frac{m{{v}^{2}}}{r}\]                          ?(i) \[T\cos \theta =mg\]                                ?(ii) From equation (i) & (ii) we have, \[\tan \theta =\frac{{{v}^{2}}}{rg}\] \[{{v}^{2}}=rg\] \[\because \] \[\theta =45{}^\circ \] Hence, speed of the pendulum in its circular path, \[v=\sqrt{rg}=\sqrt{0.4\times 10}=2\,\,m/s\]