Question
A vertical capillary tube with inside diameter a $$0.5\,mm$$ is submerged into water so that the length of its part protruding over the surface of water is equal to $$2.5\,mm.$$ Find the radius of curvature of the meniscus.
A.
$$0.3\,mm$$
B.
$$0.6\,mm$$
C.
$$0.9\,mm$$
D.
$$1.2\,mm$$
Answer :
$$0.6\,mm$$
Solution :
Let the liquid rise to a height $$h.$$
$$\eqalign{ & T = \frac{{h\rho rg}}{2} \cr & \therefore hr = \frac{{2T}}{{\rho g}} \cr} $$
If the tube is of height $${h_1} < h$$
$$\eqalign{ & {h_1}{r_1} = hr = \frac{{2T}}{{\rho g}} \cr & \therefore {r_1} = \frac{{2T}}{{{h_1}\rho g}} = \frac{{2 \times 0.073}}{{\frac{{25}}{{1000}} \times 1000 \times 9.8}}\left\{ {25\,mm = \frac{{25}}{{1000}}m} \right\} \cr & = \frac{{2 \times 0.073}}{{9.8 \times 25}} = 0.0006\,m = 0.6\,mm \cr} $$
Let the liquid rise to a height $$h.$$
$$\eqalign{ & T = \frac{{h\rho rg}}{2} \cr & \therefore hr = \frac{{2T}}{{\rho g}} \cr} $$
If the tube is of height $${h_1} < h$$
$$\eqalign{ & {h_1}{r_1} = hr = \frac{{2T}}{{\rho g}} \cr & \therefore {r_1} = \frac{{2T}}{{{h_1}\rho g}} = \frac{{2 \times 0.073}}{{\frac{{25}}{{1000}} \times 1000 \times 9.8}}\left\{ {25\,mm = \frac{{25}}{{1000}}m} \right\} \cr & = \frac{{2 \times 0.073}}{{9.8 \times 25}} = 0.0006\,m = 0.6\,mm \cr} $$
