Given, Let, height rise in tube A,HA=10cm Height fall in tube B,HB=2cm Density of fluid in tube A,B ρA=1g∕cm3,ρB=10g∕cm3. Contact angle in A and B be θA and θB i.e. θA=0∘ and θB=135∘, and surface tensions in tube A and B are SA,SB As, we know that, H( capillary rise )=
2Scosθ
ρgr
Here, ρ is density, r is radius of capillary tube. ∴HA=
2SAcosθA
ρAgr
⇒10=
2SAcos0∘
1gr
⇒10gr=2SA and HB=
2SBcosθB
ρBgr
⇒2SB=
HBρBgr
cosθB
=
2×10gr
cos135∘
=−20√2gr On dividing Eq. (ii) by Eq. (i), we get