Q. When a capillary tube of diameter 0.8 mm is dipped in a liquid having density 800 kg $m^{- 3},$ then the height of liquid in the capillary tube rises to 4 cm. The surface tension of liquid is $\left( g = 9.8\ m/s^{2} \right)\$
A) $4.3 \times 10^{- 2}\ {Nm}^{- 1}$
B) $5.6 \times 10^{- 2}\ {Nm}^{- 1}$
C) $6.3 \times 10^{- 2}\ {Nm}^{- 1}$
D) $7.3 \times 10^{- 2}\ {Nm}^{- 1}$
Explanation
(c)
Given that,
Density of liquid $(D) = 800\ {kgm}^{- 3}$
Height of liquid $(h) = 4\ cm = 0.04\ m$
Acceleration due to gravity $(g) = 9.8\ ms^{- 2}$
Diameter of tube $(d) = 0.8\ mm$
Radius of tube $(r) = 0.4\ mm = 4 \times 10^{- 4}\ m$
Surface tension $(T) = ?$
By using
$T = \frac{rh\ Dg}{2}$
$\ \ \ \ = \frac{\left( 4 \times 10^{- 4} \right) \times (0.04) \times 800 \times 9.8}{2}$
$\ \ \ \ = 4 \times 10^{- 4} \times 0.04 \times 400 \times 9.8$
$\ \ \ \ = 4 \times 4 \times 4 \times 98 \times 10^{- 5}$
Hence, $T = 6.272 \times 10^{- 2} \approx 6.3 \times 10^{- 2}\ {Nm}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.