question_answer

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad \[{{\text{s}}^{\text{-1}}}\text{,}\]the magnitude of its angular momentum about a point on the ground right under the centre of the circle is : [JEE Main Online Paper (Held On 11 April 2015)]  

A) \[\text{8}\text{.64}\,\text{kg}\,{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}}\]

B) \[\text{11}\text{.52}\,\text{kg}\,{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}}\]

C) \[\text{14}\text{.4}\,\text{kg}\,{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}}\]

D) \[\text{20}\text{.16}\,\text{kg}\,{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}}\]

Correct Answer: C

Solution :

\[\overline{R}=0.8\hat{k}+0.6\hat{i}m\] \[\overline{V}=-7.2\hat{j}m/s\] \[\overline{L}=m\overline{R}\times \overline{V}\]          \[\overline{L}=2\left( 5.76\hat{i}-4.32\hat{k} \right)\] \[\therefore \,\,\,\,\left| \overline{L} \right|=14.4kg\,{{m}^{2}}{{s}^{-1}}\]