As per the shown figure the central solid cylinder starts with initial angular velocity `omega_(0)` Find the time after which the angular velocity becomes half.

Answer

`t=(m(R_(2)-R_(1))//ln2)/(4pi eta lnaR_(1))`

Explanation

`F=etaA(dv)/(dz),` where `(dv)/(dz)=(omegaR_(1)-0)/(R_(2)-R_(1))`
`F=eta(2piR_(1)lomegaR_(1))/(R_(1)-R_(1))`

and `tau=FR_(1)=(2pietaR_(1)^(3)omegal)/(R_(2)-R_(1))`
`I prop(2pietaR_(1)^(3)omegal)/(R_(2)-R_(1))`
`rArr (MR_(1)^(2))/(2)(-(domega)/(dt))=(2pietaR_(1)^(3)omegal)/(R_(2)-R_(1))`
or `-underset(omega_(0))overset(omega_(0)//2)int(domega)/(omega)=(4pietaR_(1)l)/(m(R_(2)-R_(1))) underset(0)overset(t)intdt`
`rArr t=(m(R_(2)-R_(1))ln2)/(4pietalR_(1))`

`h=(40xx10^(-2))/(100)-[eta=10^(-1)`poise `=10^(-2)N-sec-m^(2)]`
`=4xx10^(-3)m=4mm`