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.  $\frac{{\mathrm{r}}_1 + {\mathrm{r}}_2}{2}$

2.  $\frac{{\mathrm{r}}_1{\mathrm{r}}_2}{{\mathrm{r}}_1 - {\mathrm{r}}_2}$

3.  $\sqrt{{\mathrm{r}}_1{\mathrm{r}}_1}$

4.  $\sqrt{{\mathrm{r}}_1^2 + {\mathrm{r}}_2^2}$

The energy needed in breaking of a drop of liquid of radius R into n drops of radius r is given by (T is surface tension and P is atmospheric pressure) :

1.  $\left(4{\mathrm{\pi r}}^2\mathrm{n}-4{\mathrm{\pi R}}^2\right)\mathrm{T}$

2.  $\left(4/3{\mathrm{\pi r}}^2\mathrm{n}-4/3{\mathrm{\pi R}}^3\right)\mathrm{T}$

3.  $\left(4{\mathrm{\pi R}}^2\mathrm{n}-4{\mathrm{\pi r}}^2\right)\mathrm{T}$

4.  $\left(8{\mathrm{\pi r}}^2\mathrm{n}-8{\mathrm{\pi R}}^2\right)\mathrm{T}$

A tank is filled with water up to a height H. A hole is made at a depth h below the free surface of the water in the tank. The water coming out of the orifice strikes at a distance S from the walls of the tank. Then S is given by :

$\left(1\right) \sqrt{2g\left(H-h\right)}$
$\left(2\right) \sqrt{g\left(H-h\right)}$
$\left(3\right) \sqrt{2gh}$
$\left(4\right) \sqrt{4h\left(H-h\right)}$

An incompressible liquid travels as shown in the figure. The speed of the fluid in the lower branch will :

1. 1 m/s

2. 1.5 m/s

3. 2.25 m/s

4. 3 m/s

The weight of an aeroplane flying in the air is balanced by 

(1) Vertical component of the thrust created by air currents striking the lower surface of the wings

(2) Force due to reaction of gases ejected by the revolving propeller

(3) Upthrust of the air which will be equal to the weight of the air having the same volume as the plane

(4) Force due to the pressure difference between the upper and lower surfaces of the wings created by different air speeds on the surfaces

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 - 

$1. 4{\mathrm{\pi R}}^3\mathrm{T} \left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
$2. 2{\mathrm{\pi R}}^3\mathrm{T} \left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
$3. \frac{4}{3}{\mathrm{\pi r}}^2\mathrm{T} \left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
$4. 2\mathrm{\pi RT} \left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$

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

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

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 $\mathrm{\pi }$