Question

A capillary tube of diameter 0.4 mm is dipped in a beaker containing mercury of density 13.6x10${}^3$ kgm${}^{-3}$ and surface tension 0.49Nm${}^{-1}$.The angle of contact of mercury w.r.t. glass is 130${}^o$. [Cos 130${}^o=$ -0.64280]. The depression of the meniscus in the capillary tube is (g $=$ 9.8 ms${}^{-2}$).

• A. 0.24cm
• B. 24 cm
• C. 2.4 cm
• D. 0.024 cm

Answer 1

The correct option is C 2.4 cm

The depression of meniscus in the capillary tube is
= $\frac{2Tcos\theta }{\rho gr}$
= $\frac{2\times 0.49\times 0.64280}{13.6\times {10}^3\times \frac{0.4}{2}\times {10}^{-3}\times 9.8}$
= 0.024m
= 2.4 cm