As shown in the figure, two bobs of masses m & 2m are tied by a massless string which is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob of mass 2m starts falling vertically. When it has covered distance of h, the angular speed of the wheel will be:
A
`r sqrt((3)/(2 gh ))`
B
`1/r sqrt((7gh)/(3))`
C
`1/r sqrt((4gh)/(7))`
D
`r sqrt((3)/(7 gh ))`
Text Solution
Verified by Experts
The correct Answer is:
C
Loss in P.E. = Gain in K.E. `2mgh -mgh =1/2 mv ^(2) =1/2 mv ^(2) + 1/2 I omega ^(2) implies mgh =3/2m omega ^(2) r ^(2) + 1/2. (mr^(2))/(2) . Omega ^(2)` `impliesmgh =7/4 momega ^(2) r ^(2) , " " omega = sqrt((4 gh)/(7r ^(2)))=1/r sqrt((4gh)/(7))`
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VMC MODULES ENGLISH-JEE MAIN REVISON TEST-23-PHYSICS (SECTION 2)