A conical pendulum of length 1 m makes an angle \[\theta ={{45}^{\text{o}}}\] 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 O. The speed of the pendulum, in its circular path, will be - (Take \[g=10\,m{{s}^{-1}}\]) [JEE Online 09-04-2017] |
A) 0.2 m/s
B) 0.4 m/s
C) 2 m/s
D) 4 m/s
Correct Answer: C
Solution :
[c] |
\[T\sin \theta \,=\frac{m{{v}^{2}}}{r}\] |
\[T\cos \theta \,=mg\] |
\[\tan \theta \,=\frac{{{v}^{2}}}{rg}\] |
\[\theta ={{45}^{\text{o}}}\] |
\[{{v}^{2}}=rg\] |
\[\upsilon =\sqrt{rg}\,=\sqrt{0.4\,\times 10}\] |
\[=2\,m/s\] |
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