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

A, B, C, D

Explanation

`implies T = mg`
`T + N mu = mg sin theta`
`N = m g cos theta`
`implies Mg + mg cos theta mu = m g sin theta`
`implies (m)/(M) = (1)/(sin theta - mu cos theta) = 5`
(B) `implies M = m (sin theta + mu cos theta)`
`implies M = m (sin theta + mu cos theta)`
`implies (M)/(m) = (3)/(5) + ((1)/(2)) (4)/(5) = 1`
(C ) `T = m g sin theta = Mg implies m//M = (1)/(sin theta) = (5)/(3)` `implies` vertical component of acceleration of m
(D ) `= a sin theta` And vertical acc of M = a
`implies` ratio `= sin theta = (3)/(5)`