Let h cm be the head of pressure at B over C. Then pressure difference between A and B,

`p_(1) = (5-h)" cm of "H_(2)O`

For tube AB : `L_(1) = 12 cm`,

`r_(1) = 2mm -=0.2 cm`,

`p_(1) = (5-h)" cm of water"`

`= (5-h) xx 1xx 980 dyn e//cm^(2)`

Rate of flow of water through AB is

`Q_(1) =(pi p_(1)r_(1)^(4))/(8 eta l_(1)) = (pi(5-h) xx 1 xx 980 xx (0.2)^(4))/(8 eta xx 12) cm^(3) s^(-1)`

For tube BC: `l_(2) = 3 cm` ,

`r_(2) 1 mm = 0.1 cm`,

`p_(2)" h cm of water "= h xx 1 xx 980 dyn e//cm^(2)`

Rate of flow of water through BC is

`Q_(2) =(pi p_(2)r_(2)^(4))/(8 eta l_(2)) = (pi xx h xx 1 xx 980 xx (0.1)^(4))/(8 eta xx 3) cm^(3) s^(-1)`

As flow through the two tubes in series is steady, so `Q_(1) = Q_(2)`

or `(pi(5-h) xx 1 xx 980 xx (0.2)^(4))/(8 eta xx 12) = (pi h xx 1 xx 980 xx (0.1)^(4))/(8 eta xx 3)`

`3(5-h) xx 16 = 12 h`

On solving, `h = 4" cm of water column"`.