Question

Find the depression of the meniscus in the capillary tube of diameter $0.4mm$, dipped in a beaker containing mercury. (Density of mercury $=13.6\times {10}^{3}kg/m^3$) ; surface tension of the mercury is $0.49N/m$ ; angle of contact is ${135}^o$ ).

• A. $0.025cm$
• B. $0.021cm$
• C. $0.020cm$
• D. $0.027cm$

Answer 1

The correct option is A $0.025cm$

$\text{We know that}$

$h=\left|\frac{2T\cos \theta }{\rho gr}\right|$

$H=\frac{2\times 0.49\times 1/\sqrt{2}}{13.6\times {10}^3\times 10\times 0.2\times {10}^{-3}}=0.025cm$

$\text{Answerl:- (A)}0.025cm$