Consider a perfectly conducting uniform disc of mass `m` and radius `'a'` hinged in a vertical plane from its centre and free to rotate with respect to hinge. A resistance `R` is connected between centre of the disc annd periphery by using two sliding contacts `C_(1)` & `C_(2)`. A long non -conducting massless string is wrapped around the disk, whose another end is attached with a block of mass `m`. There exists a uniform horizontal magnetic field `B`, whose arrangement is shown in figure.

Given system is released from rest at `t=0`. Assume friction between string and disc is sufficient so that there is no slipping between them let at any instant `t`, velocity of block is `v`, angular velocity of the disc is `omega` and current in resitance is `i`
Find the terminal speed of the block of mass `m` (Terminal speed `=` constant speed attained by the block after very long time)

Options

A. `(mgR)/(B^(2)a^(2))`
B. `(2mgR)/(B^(2)a^(2))`
C. `(3mgR)/(B^(2)a^(2))`
D. `(4mgR)/(B^(2)a^(2))`

Answer

D

Explanation

Writing energy equation
`mgv=mv(dv)/(dt)+I(omega d omega)/(dt)+i^(2)R`………..i
`v=aomega`
`(dv)/(dt)=a(domega)/(dt)`
`i=(Bomegaa^(2))/(2R)=(Bva)/(2R)`………ii
From equation i and ii
`(3m)/2 (dv)/(dt)=mg-(B^(2)a^(2))/(6mR) v`
`v=(4mgR)/(B^(2)a^(2)) (1-e^(-((B^(2)a^(2))/(6mr))t))`
`(dv)/(dt)=(2g)/3 e^(-(B^(2)a^(2))/(6mR)t)`