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 center is vertically below 0. The speed of the pendulum, in its circular path, will be:   (Take \[g=10m{{s}^{-2}}\])

A) 0.4 m/s

B) 4 m/s

C) 0.2 m/s

D) 2 m/s

Correct Answer: D

Solution :

[d] \[\operatorname{Given},\theta =45{}^\circ ,r=0.4m,g=10m/{{s}^{2}}\] \[T\sin \theta =\frac{m{{v}^{2}}}{r}\]          ...... (i) \[Tcos\theta =mg\]         ...... (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, \[v=\sqrt{rg}=\sqrt{0.4\times 10}=2\,m/s\]