A mass m moves in a circle on a smooth horizontal plane with velocity \[{{v}_{0}}\] at a radius \[{{R}_{0}}\]. The mass is attached to a string which passes through a smooth hole in the plane as shown. [NEET 2015 ] |
The tension in the string is increased gradually and finally m moves in a circle of radius \[\frac{{{R}_{0}}}{2}\]. The final value of the kinetic energy is |
A) \[mv_{0}^{2}\]
B) \[\frac{1}{4}mv_{0}^{2}\]
C) \[2\,mv_{0}^{2}\]
D) \[\frac{1}{2}mv_{0}^{2}\]
Correct Answer: C
Solution :
Conserving angular momentum |
\[{{L}_{i}}={{L}_{t}}\] |
\[\Rightarrow \] \[m{{v}_{0}}{{R}_{0}}=mv'\left( \frac{{{R}_{0}}}{2} \right)\] |
\[\Rightarrow \] \[v'=2{{v}_{0}}\] |
So, final kinetic energy of the particle is |
\[{{K}_{f}}=\frac{1}{2}\,mv{{'}^{2}}\,=\frac{1}{2}\,m{{(2{{v}_{0}})}^{2}}\] |
\[=4\frac{1}{2}mv_{0}^{2}=2mv_{0}^{2}\] |
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