A viscous clutch as shown in figure tranmits torque. Radius of each clutch plate is R and separation between the plates is a and is completely filled with liquid of coefficient of viscous `mu`. If `omega_(i)` and `omega_(0)` are angular velocities of plates connected to input and output respectively.
If efficiency of transmission is ratio of output power to input power, then efficiency is given by

Options

A. `1-(omega_(0))/(omega_(i))`
B. `(omega_(0))/(omega_(i))`
C. `(omega_(0)^(2))/(omega_(i)^(2))`
D. `(omega_(0)^(2))/(omega_(i)^(2))`

Answer

B

Explanation

Let's consider a circular element of radius r and width dr, co-axial with the system.
Velocity difference across the element
`dv=r(omega_(i)-omega_(0))`
Hence `dF_(v)=muxx2pirdrxx(r(omega_(i)-omega_(0)))/(a)`
These force are tangential and hence
`dtau_(v)=dF_(v)xxr=(2pimu(omega_(i)-omega_(0)))/(a)r^(3)dr`
Hence `tau=int_(0)^(R)dtau_(v)=(pimu(omega_(i)-omega_(0))R^(4))/(2a)`
Power (P) `=tauomega`
Hence efficiency `=(tauomega_(0))/(tauomega_(i))=(omega_(o))/(omega_(i))`