| 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 |
| \[{{K}_{f}}=\frac{1}{2}\,mv{{'}^{2}}\,=\frac{1}{2}\,m{{(2{{v}_{0}})}^{2}}\] |
| \[=4\frac{1}{2}mv_{0}^{2}=2mv_{0}^{2}\] |
You need to login to perform this action.
You will be redirected in
3 sec
