TS EAMCET 5-Aug-2021 Shift 2 Question Paper

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
‌

SB

SA

‌‌=‌

−20√2gr

10gr

=−2√2
⇒‌‌|‌

SB

SA

|=2√2