Question

Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.5cm in the same capillary tube. If the density of mercury is 13.6 gm/cc and its angle of contact is 135^o and density of water is 1 gm/cc and its angle of contact is $\left( \cos 135 ^ { \circ } = 0.7 \right)$

• A. 1 : 14
• B. 5 : 34
• C. 1 : 5
• D. 5 : 27

Answer 1

Answer: (B)

5 : 34

Answer 2

$\therefore \frac { h _ { W } } { h _ { H g } } = \frac { T _ { W } } { T _ { H g } } \frac { \cos \theta _ { W } } { \cos \theta _ { H g } } \frac { d _ { H g } } { d _ { W } }$ [as r and g are constants]

$\Rightarrow \frac { 10 } { 3.5 } = \frac { T _ { W } } { T _ { H g } } \cdot \frac { \cos 0 ^ { \circ } } { \cos 135 } \frac { 13.6 } { 1 }$