The glycerin of density 1.25 × 10 3 k g m - 3 is flowing through a conical tube with end radii 0 . 1 m and 0 . 04 m respectively. The pressure difference across the ends is 10 N m - 2 . The rate of flow of glycerine through the tube is

Step-by-step solution

V = a 1 a 2 2 ( p 1 - p 2 ) ρ ( a 1 2 - a 2 2 ) = π r 1 2 × π r 2 2 2 ( p 1 - p 2 ) ρ π r 1 2 2 - π r 2 2 2 = π r 1 2 r 2 2 2 ( p 1 - p 2 ) ρ ( r 1 4 - r 2 4 ) = 22 7 × 0.1 2 × 0.04 2 2 × 10 1.25 × 10 3 [ 0.1 4 - 0.04 4 ] = 6.4 × 10 - 4 m 3 s - 1