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

By inserting a capillary tube upto a depth l in water, the water rises to a height h ( h<l). If the lower end of the capillary is closed inside 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

If the surface tension of water is \(0.06~\text{N/m}^2,\) then the capillary rise in a tube of diameter \(1~\text{mm}\) is:
\((\theta = 0^{\circ})\)

1. \(1.22~\text {m}\)
2. \(2.44~\text {cm}\)
3. \(3.12~\text {cm}\)
4. \(3.86~\text {cm}\) 

If the capillary experiment is performed in a vacuum, then for a liquid the capillary will:

If liquid level falls in a capillary then radius of capillary will

1. Increase                     

2. Decrease

3. Unchanged                 

4. None of these

Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are 0.8 and 0.6 and surface tensions are 60 and 50 dyne/cm respectively Ratio of heights of liquids in the two tubes $\frac{{\mathrm{h}}_1}{{\mathrm{h}}_2}$ is

1. $\frac{10}{9}$                     

2. $\frac{3}{10}$

3. $\frac{10}{3}$                   

4. $\frac{9}{10}$

Water rises in a vertical capillary tube upto a height of 2.0 cm . If the tube is inclined at an angle of $60°$ with the vertical, then upto what length the water will rise in the tube ?

1. 2.0 cm                             

2. 4.0 cm
3. $\frac{4}{\sqrt{3}}\mathrm{cm}$                             

4.$2\sqrt{2}\mathrm{cm}$

A capillary tube is immersed vertically in the water and the height of the water column is x. When this arrangement is taken into a mine of depth d, the height of the water column is y. If R is the radius of the earth, the ratio x/y is :

1.  $\left(1<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\frac{\mathrm{d}}{\mathrm{R}}\right)$

2.  $\left(1<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\frac{2\mathrm{d}}{\mathrm{R}}\right)$

3.  $\left(\frac{\mathrm{R}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\mathrm{d}}{\mathrm{R}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{d}}\right)$

4.  $\left(\frac{\mathrm{R}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{d}}{\mathrm{R}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\mathrm{d}}\right)$

When an air bubble of radius r rises from the bottom to the surface of the lake, its radius becomes $\raisebox{1ex}{$5r$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$. Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :

1.  11.2 m

2.  8.7 m

3.  9.5 m

4.  10.5 m

A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a?

1. $\frac{2\mathrm{ma}}{\mathrm{g}<mspace\; width="1em"></mspace>+\hspace{0.17em}\mathrm{a}}$

2. $\frac{2\mathrm{ma}}{\mathrm{g}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\mathrm{a}}$

3. $\frac{\mathrm{ma}}{\mathrm{g}<mspace\; width="1em"></mspace>+\hspace{0.17em}\mathrm{a}}$

4. $\frac{\mathrm{ma}}{\mathrm{g}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\mathrm{a}}$