A cubical block (of sie 2m) of mass 20 kg slides on inclined plane lubricated with the oil of viscosity `eta=10^(-1)` poise with constant velocity of 10 m/s find out the thickness of layer of liquid (`g=10m//s^(2)`)

Answer

4

Explanation

` F = F ' eta A (dv )/ (dz ) = m g sin theta , (dv )/(dz ) = (v)/(h) `
` 20 xx 10 xx sin 30 ^@ = eta xx 4 xx (10)/(h) `
` h = (40 xx 10 ^(-2))/(100) - [eta = 10 ^(-1) ` poise ` = 10 ^(-2) N - sec - m ^(-2) ] `
` = 4 xx 10 ^(-3 ) m = 4mm `