Laws Of Motion Question 347
Question: 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}} $ )
Options:
A) 0.4 m/s
B) 4 m/s
C) 0.2 m/s
D) 2 m/s
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ Given,\theta =45{}^\circ ,r=0.4m,g=10m/s^{2} $
$ T\sin \theta =\frac{mv^{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}=2m/s $
BETA
