In all the four situations depicted in Column-I, a ball of mass m is connected to a string. In each case, find the tension in the string and match the appropriate entries in Column-II.
`{:((A) (##VMC_PHY_XI_WOR_BOK_01_C04_E03_050_Q01##) "Conical pendulum", (P)T= mg cos theta) ,((B) (##VMC_PHY_XI_WOR_BOK_01_C04_E03_050_Q02##)"Pendulum is swinging. Angular position is the extreme position". "T is tension in extreme position", (Q) T cos theta = mg) ,((C) (##VMC_PHY_XI_WOR_BOK_01_C04_E03_050_Q03##) "The car is moving with constant acceleration." "The ball is at rest with respect to car",(R) "Speed of ball with respect to ground is constant"),((D) (##VMC_PHY_XI_WOR_BOK_01_C04_E03_050_Q04##) "The car is moving with constant velocity"." The ball is at rest with respect to car", (S) "Velocity of ball with respect to ground is changing continuously"):}`

Answer

5

Explanation

When the string is cut, the weight of the rod constitutes torque about the hinge, so
`tau_(A)=mgl/2`……..i
According to Newton's second law
`tau_(A)=Ialpha`……….ii
Where `alpha` is the angular acceleration of the rod about the end `A`. From ean i and ii we get
`Ialpha=mgl/2`
or `alpha=(mgl/2)/I`
Here `I=ml^(2)//3`, therefore
`:. alpha=(mgl//2)/(ml^(2)//3)=3/2 g/l`
Acceleration of the `CM` of the rod is
`a_(CM)=alphar = 3/2 g/l xxl/2=(3g)/4`
Again by Newtons' second law
`mg-R_(A)=ma_(CM)`
or `mg-R_(A)=mxx(3g)/4`
`:. R_(A)=(Mg)/4=(2xx10)/4=5N`