Three capillaries of internal radii 2r, 3r and 4r, all of the same length, are joined end to end. A liquid passes through the combination and the pressure difference across this combination is 20.2 cm of mercury. The pressure differene across the capillary of internal radius 2r is

Three capillaries of internal radii 2r, 3r and 4r, all of the same len

1.	Three capillaries of internal radii 2r, 3r and 4r, all of the same length, are joined end to end. A liquid passes through the combination and the pressure difference across this combination is 20.2 cm of mercury. The pressure differene across the capillary of internal radius 2r is
Answer» 2cm of Hg 4cm of Hg 8cm of Hg 16cm of Hg Solution :As capillaries of same length are joined end to end, the volume rate V across each of the capillaries is same. Here, `P_(1).r_(1)^(4) = P_(2). r_(2)^(4) = P_(3).r_(3)^(4)` or `P_(1) (2r)^(4) = P_(2)(3r)^(4) = P_(3)(4r)^(4) rArr 16P_(1) = 81P_(2)= 256P_(3)` `P_(1) = (810/(16) P_(2)= 16P_(3) rArr P_(2) = (16)/(81) P_(1) and P_(3) = (P_(1))/(16)` It is given that: `(P_(1) + P_(2) + P_(3)) = 20.2 CM` of Hg `rArr P_(1) (1 + (16)/(81) + (1)/(16))= 20.2` `or P_(1) (((81 xx 16) + (16 xx 16) + 81)/(81 xx 16))= 20.2` `P_(1)= 20.2 [(1296)/(1296 + 256 + 81)]= 16 cm "of" Hg` `rArr` (d) is CORRECT