Answer:
40.96%
Solution:
As per the question:
If the original radius be 'r' and the initial pressure difference be 'P':
After the drop, radius:
r' = 0.8r
After rise in pressure:
P' = 0.1 P
Now,
Rate of flow is given by:
R ∝ [tex]r^{4}P[/tex]
Thus
[tex]\frac{R'}{R} = (\frac{0.8r}{r})^{4}\frac{0.1P}{P}[/tex]
R' = 40.96%
