As shown in the figure a block of mass 1 kg connected with a pulley of mass 1 kg and radius 10 cm is acted upon by a variable force of F = kt, as shown. Angular acceleration of pulley when block leaves the contact with ground is:

Options

A. `50sqrt(3) "rad"//s^(2)`
B. `20/sqrt(3) "rad"//s^(2)`
C. `200sqrt(3) "rad"//s^(2)`
D. `5sqrt(3) "rad"//s^(2)`

Answer

C

Explanation

Resolving F
`F cos 30^(@) - T = ma`….. (i)
`F sin 30^(@) + N = mg`……. (ii)
For the time it leaves the contact
`F sin 30^(@) = mg`
F = 2mg
T.R. `= Ialpha`
`2mg cos 30 - T = mr alpha`
`T = sqrt(3) mg - mralpha`
`T.R. = I.alpha`.
`{sqrt(3) xx 1 xx 10 - 1 xx 0.1 alpha} xx 0.1 = (1 xx 0.01)/2 xx alpha`
`alpha = 200sqrt(3) "rad"//s^(2)`