The radius of a soap bubble is increased from R to 2 R. Work done in this process (T = surface tension) is: 

A cubical vessel of height 1 m is full of water. Find the work done in pumping out whole water.

1. 49 J

2. 490 J

3. 4900 J

4. 49000 J

The flow rate from a tap of diameter 1.25 cm is 3 lit/min. The coefficient of viscosity of water is ${10}^{-3}$ Pas. The nature of flow is :

1. Turbulent

2. Laminar

3. Neither laminar nor turbulent

4. data inadequate

Water flowing from a hose pipe fills a 15-liter container in one minute. The speed of water from the free opening of radius 1 cm is (in ms${}^{-1}$) :

1. 2.5

2. $\frac{\mathrm{\pi }}{2.5}$

3. $\frac{2.5}{\mathrm{\pi }}$

4. 5 $\mathrm{\pi }$

When a cylindrical tube is dipped vertically into a liquid the angle of contact is 140^o. When the tube is dipped with an inclination of 40^o, the angle of contact is- 

1.  $100°<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$  

2.  $140°<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

3.  $180°<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

4.    $60°$

Two liquid drops have their diameters as 1 mm and 2 mm. The ratio of excess pressure in them is :

1.  1:2

2.  2:1

3.  4:1

4.  1:4

If a soap bubble of radius 3 cm coalesce with another soap bubble of radius 4 cm under isothermal conditions, the radius of the resultant bubble formed is in cm-

1. 7

2. 1

3. 5

4. 12

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1.  5:3:7

2.  7:3:5

3.  21:35:15

4.  1:1:1

Several spherical drops of a liquid each of radius r coalesce to form a single drop of radius R. If T is the surface tension, then the energy liberated will be - 

$1.<mspace\; width="1em"></mspace>4{\mathrm{\pi R}}^3\mathrm{T}<mspace\; width="1em"></mspace>\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
$2.<mspace\; width="1em"></mspace>2{\mathrm{\pi R}}^3\mathrm{T}<mspace\; width="1em"></mspace>\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
$3.<mspace\; width="1em"></mspace>\frac{4}{3}{\mathrm{\pi r}}^2\mathrm{T}<mspace\; width="1em"></mspace>\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
$4.<mspace\; width="1em"></mspace>2\mathrm{\pi RT}<mspace\; width="1em"></mspace>\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$

The rate of flowing of water from the orifice in a wall of a tank will be more if the orifice is:

1.  Near the bottom

2.  Near the upper end

3.  Exactly in the middle

4.  Does not depend upon the position of the orifice