Two parallel glass plates are dipped partly in the liquid of density 'd' keeping them vertical. If the distance between the plates is 'x', surface tension for liquids is T and the angle of contact is $\mathrm{\theta }$, then rise of liquid between the plates due to capillary will be 

(1) $\frac{\mathrm{T} \mathrm{cos\theta }}{\mathrm{xd}}$

(2) $\frac{2\mathrm{T} \mathrm{cos\theta }}{\mathrm{xdg}}$

(3) $\frac{2\mathrm{T} }{\mathrm{xdgcos\theta }}$

(4) $\frac{\mathrm{T} \mathrm{cos\theta }}{\mathrm{xdg}}$

Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75\times {10}^{-4}$ $\mathrm{N}$ force due to the weight of the liquid. If the surface tension of water is $6\times {10}^{-2}$ ${\mathrm{Nm}}^{-1}$, the inner circumference of the capillary must be:

1. $1.25\times {10}^{-2}$ $\mathrm{m}$ 

2. $0.50\times {10}^{-2}$ $\mathrm{m}$

3. $6.5\times {10}^{-2}$ $\mathrm{m}$

4. $12.5\times {10}^{-2}$ $\mathrm{m}$

Two capillary tubes P and Q are dipped in water. The height of water level in capillary P is 2/3 to the height in Q capillary. The ratio of their diameters is

(1) 2 : 3                           

(2) 3 : 2

(3) 3 : 4                           

(4) 4 : 3

A vessel, whose bottom has round holes with a diameter of 0.1mm, is filled with water. The maximum height to which the water can be filled without leakage is
(S.T. of water =75 dyne/cm, g =1000 $\mathrm{cm}/{\mathrm{s}}^2$)

(1) 100 cm

(2) 75 cm

(3) 50 cm

(4) 30 cm

By inserting a capillary tube up to a depth l in water, the water rises to a height of h ( h<l). If the lower end of the capillary is closed inside the water and the capillary is taken out and closed-end opened, to what height the water will remain in the tube?

(1) Zero

(2) l+h 

(3) 2h

(4) h

A capillary tube when immersed vertically in liquid records a rise of 3 cm. If the tube is immersed in the liquid at an angle of $60°$ with the vertical. The length of the liquid column along the tube is 

(1) 9cm

(2) 6cm

(3) 3cm

(4) 2cm

The action of a nib split at the top is explained by

1. Gravity flow

2. Diffusion of fluid

3. Capillary action

4. Osmosis of liquid

The correct relation is

(1) $\mathrm{r}=\frac{2\mathrm{Tcos\theta }}{\mathrm{hdg}}$

(2) $\mathrm{r}=\frac{\mathrm{hdg}}{2\mathrm{Tcos\theta }}$

(3) $\mathrm{r}=\frac{2\mathrm{Tdgh}}{\mathrm{cos\theta }}$

(4) $\mathrm{r}=\frac{\mathrm{Tcos\theta }}{2\mathrm{hdg}}$

During the capillary rise of a liquid in a capillary tube, the surface of contact that remains constant is of

(1) Glass and liquid

(2) Air and glass

(3) Air and liquid

(4) All of these

A shell having a hole of radius r is dipped in water. It holds the water up to a depth of h then the value of r is

(1) $\mathrm{r}=\frac{2\mathrm{T}}{\mathrm{hdg}}$

(2) $\mathrm{r}=\frac{\mathrm{T}}{\mathrm{hdg}}$

(3) $\mathrm{r}=\frac{\mathrm{Tg}}{\mathrm{hd}}$

(4) None of these