Question

A liquid flowing with speed v through a horizontal pipe of cross sectional area A enters into another pipe of double the area of cross section. Now, the speed of the liquid is

• A. v
• B. 2 v
• C. v/2
• D. v/4

Answer 1

The correct option is C v/2

initially area$A;velocity=v;Discharge=Q=A.v$
$v_2=2v$
discharge will remain same.
so $v_2=\frac{Q}{2A}=\frac{v.A}{2A}=\frac{v}{2}$