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Q.

A block B of mass 13 kg and base area 1 m x 1m is sliding on an inclined plane lubricated by an oil layer of thickness 0.15 cm. It is observed that block slides with a steady speed of 0.5 m/s. Find viscosity of oil (in Pl). (Take g = 10 m/s2)

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answer is 0.15.

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Detailed Solution

Velocity gradient in oil layerdvdx=vx=0.50.15×10−2=10003sec−1for steady speed of block viscous force on block F=mgsin⁡θηAdvdx=(13)(10)513η=501×(1000/3)=15100=0.15Pl
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