A plastic circular disc of radius 10cm is placed on a thin oil film of thickness 2mm, spread over a flat horizontal surface. The torque (N-m) required to spin the disc about its central vertical axis with a constant angular velocity 8 rad/sec is (coefficient of viscosity of oil is 1.0 kg/m-s) (take p = 3.14)

  $\tau ={\displaystyle \int F.r={\displaystyle \underset{0}{\overset{R}{\int }}\eta \left(2\pi rdr\right)\left(\frac{rw}{t}\right).r}}$  

  $=\frac{2\pi \eta W{R}^4}{4t}=\frac{2\times 3.14\times 1\times 8\times 10^{-4}}{4\times 2\times 10^{-3}}$          f

  $=0.625N-m?0.63N-m$            

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