question_answer

                A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 Kg   It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just ager the bullet embeds into it will be:                     JEE Main Online Paper (Held On 09 April 2013)

A)                 6.25 rad/sec                      

B)                 0.625 rad/sec                    

C)                 3.35 rad/sec                      

D)                        0.335 rad/sec                

Correct Answer: C

Solution :

                                From conservation of angular momentum                 \[mv\,({{r}_{axis}})=\frac{1}{2}l{{\omega }^{2}}\]                 \[10\times {{10}^{-3}}\times 500\times 0.5=\frac{1}{2}\times \left( \frac{M{{l}^{2}}}{12}+M{{\left( \frac{l}{2} \right)}^{2}} \right)\times {{\omega }^{2}}\]                 (from parallel axes theorem)                 \[2.5\times 2=\frac{M{{l}^{2}}}{4}\,\left[ \frac{1}{3}+1 \right]{{\omega }^{2}}\]                 \[5\times 4=12\times {{(1)}^{2}}\times \frac{4}{3}\times {{\omega }^{2}}\]                 \[\frac{5\times 3}{12}={{\omega }^{2}}\]                 \[\frac{15}{12}={{\omega }^{2}}\]                 \[1.25={{\omega }^{2}}\]                 \[\omega =1.12\,\,\text{rad/s}\]