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, \[v=\sqrt{rg}=\sqrt{0.4\times 10}=2\,m/s\]You need to login to perform this action.
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