A drum of mass m_(1) and radius r_(1) rotates freely with initially angular velocity omega_(0). A second drum with mass m_(2) and radius r_(2)(r_(2)gtr_(1)) is mounted on same axle and is at rest although it is free to rotate. A thin layer of sand with mass m is distributed on iner surface of smaller drum. At t=0 small perforations in the inner drum are opened. The sand starts to fly out at a consant rate lamdakg//sec and sticks to the outer drum. Ignore the transit time of the sand. Choose the corect alternatives.

Class 12 Physics in Class 12 3 years ago

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Angular speed of outer DRUM at time is `(lamda t omega_(0))/(m_(2)+lamdat)((r_(1))/(r_(2)))^(2)`
Difference in final angular speeds of two DRUMS is 0
Difference in final angular speeds of two drums is `((m(r_(2)^(2)-r_(1)^(2))+m_(2)r_(2)^(2))/((m+m_(2))r_(2)^(2)))omega_(0)`
Angular speed of inner drum remains constant

Solution :
As the sand leaves the inner drum through the OPEN holes, it does not exert any force of the drum, angular MOMENTUM remains conserved.
At time `t`,
`m_(1)r_(1)^(2)omega_(0)=(m_(1)-lamdat)r_(1)^(2)omega_(0)+(m_(2)+lamdat)r_(2)^(2) omega_(2)`
`omega_(2)-(lamdat r_(1)^(2)omega_(0))`
`((m_(2)+lamdat)r_(2)^(2))`
The speed of inner drum does not change.