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3-2.Motion in Plane
hard
A conical pendulum of length $1\,m$ makes an angle $\theta \, = 45^o$ 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 centre is vertically below $O$. The speed of the pendulum, in its circular path, will be ..... $m/s$ (Take $g\, = 10\, ms^{-2}$)
A$0.4$
B$4$
C$0.2$
D$2$
(JEE MAIN-2017)
Solution
$\begin{array}{l} Given,\theta = {45^ \circ },r = 0.4m,g = 10m/{s^2}\\ T\,\sin \,\theta = \frac{{m{v^2}}}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\ T\,\cos \,\theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,….\left( {ii} \right)\\ From\,equation\,\left( i \right) and \left( {ii} \right)\\ We\,have,\,\tan \theta = \frac{{{v^2}}}{{rg}}\\ {V^2} = rg\,\,\,\,\,\,\,\,\,\,\,\,\theta = {45^ \circ }\\ Hence,\,speed\,of\,the\,pendulum\,in\,its\,circular\\ path,\\ V = \sqrt {rg} = \sqrt {0.4 \times 10} = 2m/s \end{array}$
Standard 11
Physics
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