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

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

If the capillary experiment is performed in a vacuum, then for a liquid there capillary rise 

(1) It will rise

(2) Will remain the same

(3) It will fall

(4) Rise to the top