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"):}`

Explanation

Gain in Kinetic Energy = Work done by external forces
implies gain in KE = Work done by gravity + work done by electrostatic force
`implies(1)/(2)m(v_(2)^(2)-v_(1)^(2))=(mg)(2l)-(qE)(2l)" "...(1)`
For the ball at the lowermost position Q :
`qE+T-mg=(mv_(2)^(2))/(l)" "...(2)`
Putting T = Tension = 15 mg,
`qE+15mg-mg=(mv_(2)^(2))/(l)" "...(3)`
From (1) and (3) :
`implies(1)/(2)m(v_(2)^(2)-v_(1)^(2))=2mgl-2qEl`
`implies mv_(2)^(2)-mv_(1)^(2)=4mgl-4qEl`
`implies (mv_(2)^(2))/(l)=(mv_(1)^(2))/(l)+4mg-4qE`
Hence, `15mg+qE-mg=(mv_(1)^(2))/(l)+4m-4qE`
`v_(1)=sqrt((l)/(m)(10mg+5qE))`.