Two tubes P and Q of lengths 100 cm and 30 cm have radii 0.1 mm and 0.2 mm, respectively. A liquid passing through the two tubes is entering P at a pressure of 60 cm of mercury-and leaving Q at a pressure of 55 cm of mercury. What is the pressure at the junction of Pand Q?

Explanation

Given,
Tube P, Tube Q
Length `_(l) =100cm" "l_(Q)=30cm`
Radius, `r_(p)=0.1 mm =0.01cm, r_(Q)=0.2 mm=0.02cm`
Pressure at which liquid enters `P, P_(A)=60" cm of Hg"`
Pressure at which liquid leaves `Q, P_(Q)="55 cm of Hg"`
Let `P_(j)` be the pressure at the junction of P and Q Rate of flow of liquid through tube P `Q_(p)=pi ((P_(P)-P_(j))r_(P)^(4))/(8 eta l_(P))`
`=(pi (60-P_(j)) (0.01)^(4))/(8 eta xx 100) cm^(3) s^(-1)`
Rate of flow of liquid through tube Q, `Q_(Q)=(pi (P_(j) -P_(Q))r_(Q)^(4))/(8 eta l_(Q))`
`=(pi (P_(j)-55) (0.02)^(4))/(8 eta xx 30) cm^(3) s^(-1)`
Since `Q_(p)=Q_(Q)`
`therfore (pi (60-P_(j)) (0.01)^(4))/( 8 eta xx 100) =(pi (P_(j) -55) (0.02)^(4))/(8 eta xx 30)`
`rArr 3(60-P_(j)) xx (0.01)^(4)=10(P_(j) -55) xx (0.02)^(4)`
`rArr 3(60-P_(j)) =160 (P_(j)-55)`
`rArr 180-3P_(j)=160P_(j)-8800`
`rArr 180+8800=163 P_(j)`
`rArr P_(j)=(8980)/(163) ="55.09 cm of Hg"`