A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by liquid at the other end is

1.  $\frac{{\mathrm{M\omega }}^2\mathrm{L}}{2}$

2.  ${\mathrm{ML\omega }}^2$

3.  $\frac{{\mathrm{M\omega }}^2\mathrm{L}}{4}$

4.  $\frac{M{L}^2{\omega }^2}{2}$

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

When a body of mass 'm', density ${\mathrm{d}}_{\mathrm{B}}$ is suspended from a wire, its elongation is 'e' when the body is in air. If the body is completely immersed in non-viscous liquid of density ${\mathrm{d}}_l$ ,then its elongation is

1. $\mathrm{e}\left(1-\frac{{\mathrm{d}}_{\mathcal{l}}}{{\mathrm{d}}_{\mathrm{B}}}\right)$

2. $\mathrm{e}\left(1-\frac{{\mathrm{d}}_{\mathrm{B}}}{{\mathrm{d}}_{\mathcal{l}}}\right)$

3. $\mathrm{e}\left(\frac{{\mathrm{d}}_{\mathrm{B}}}{{\mathrm{d}}_{\mathcal{l}}}-1\right)$

4. $\mathrm{e}\left(\frac{{\mathrm{d}}_{\mathrm{B}}}{{\mathrm{d}}_{\mathcal{l}}}\right)$

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

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 $\mathrm{\pi }$

Work of 6.0 x 10${}^{-4}$ Joule is required to be done in increasing the size of a soap film from 10cm x 6cm to 10cm x 11cm. The surface tension of the film is :

1. 5 x 10${}^{-2}$ N/m

2. 6 x 10${}^{-2}$ N/m

3. 1.5 x 10${}^{-2}$ N/m

4. 1.2 x 10${}^{-1}$ N/m

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

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 - 

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

Bernoulli's theorem is based on :

1.  conservation of energy

2.  conservation of mass

3.  conservation of momentum

4.  conservation of angular momentum

Two spherical soap bubbles of radii r₁ and r₂ in vaccum collapse under isothermal condition. The resulting bubble has radius equal to :

1.  $\frac{{\mathrm{r}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{r}}_2}{2}<mspace\; width="1em"></mspace>$

2.  $\frac{{\mathrm{r}}_1{\mathrm{r}}_2}{{\mathrm{r}}_1<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>{\mathrm{r}}_2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

3.  $\sqrt{{\mathrm{r}}_1{\mathrm{r}}_1}<mspace\; width="1em"></mspace>$

4.  $\sqrt{{\mathrm{r}}_1^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{r}}_2^2}$