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

A capillary tube made of glass is dipped into mercury. Then

1. Mercury rises in the capillary tube

2. Mercury rises and flows out of the capillary tube

3. Mercury descends in the capillary tube

4. Mercury neither rises nor descends in the capillary tube

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