Turpentine oil is flowing thorough a tube of length l and radius r. the pressure difference between the two ends of the tube is P. the viscocity of oil is given by n=P(r2-v2)/4vl,where vis the velocity of oil at a difference x from the axis of the tube. find the dimensions of eeta

Dear Student

The dimension of p is [ML-1 T-2]  The dimension of r is [L]  The dimension of v is [LT-1]  The dimension of I is [L]  Therefore,  the dimension of η will be  η= [P][r2-x2]4vl  =[ML-1 T-2] L2 [LT-1]  [L]  Soη=  [ML-1 T-1] 

  • 11
The dimension of p is [ML^-1 T^-2]

The dimension of r is [L]

The dimension of v is [LT^-1]

The dimension of I is [L]


the dimension of eeta will be 


= [ML^-1 T^-2] [L^2]/[LT^-1] [L]

So answer will be [ML^-1 T^-1]
  • 6
What are you looking for?