1.

In the above question what is the angular velocity of cylinder after the masses fall down from height 'h'?

Answer»

`sqrt((8mgh//M+4m)/(R ))`
`sqrt((8mgh//M+m)/(R))`
`sqrt((4mgh//M+m)/(R))`
`sqrt((2mgh//M+2m)/(R))`

Solution :From the equation (II) in the above question, we get
`a=(4T)/(M)=(4Mmg)/(M(M+4m))`
`=(4mg)/(M+4m)`
Now `omega=(V)/(R)" but "v^(2)=2ah`
`therefore omega=((2ah)^(1/2))/(R)` PUTTING for .a.
or `omega=(((2xx4mgh)/(M+4m))^(1//2))/(R)`
`=(((8mgh)/(M+4m))^(1//2))/(R)`


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