2 glass capillary tubes of 1mm radius are put in mercury and water.
compare liquid levels of the 2 tubes sp gr=13.6 surface tension of water=7310^-3 n/m surface tension of mercury=47310^-3 n/m and angle of contact 139 deg

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Solution

$h = \frac{2 s \cos θ}{r ρ g} a n g l e o f c o n t a c t f o r w a t e r i s n o t g i v e n l e t i t b e i d e a l c a s e 0 d e g r e e . d e n s i t y 100 k g / m^{3} h_{w} = \frac{2 (73 * 10^{- 3}) \cos 0}{(10^{- 3}) (1000) 10} = 146 * 10^{- 4} m w a t e r w i l l r i s e i n c a p i l l a r y h_{h g} = \frac{2 (473 * 10^{- 3}) \cos 139}{(10^{- 3}) (13.6 * 1000) 10} = - 52.5 * 10^{- 4} m m a e r c u r y w i l l f a l l s d o w n i n c a p i l l a r y t h a t s s h o w b y - v e s i g n . R e g a r d s$

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