Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

1.

Water rises in a capillary upto a extension height such that upward force of surface tension balances the force of 75 x 10^{-4} N due to weight of water. If surface tension of water is 6x10^{-2} N/m.The internal circumference of the capillary must be

12.5$\times {10}^{-2}$m

6.5$\times {10}^{-2}$m

0.50$\times {10}^{-2}$m

1.25$\times {10}^{-2}$m

A.

12.5$\times {10}^{-2}$m

F=TL

$\mathrm{L}=\frac{\mathrm{F}}{\mathrm{T}}=\frac{75\times {10}^{-4}}{6\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=12.5\times {10}^{-2}\mathrm{m}$

2.

Bernoulli's equation is a consequence of conservation of

energy

linear momentum

angular momentum

mass

A.

energy

The Bernoulli's equation can be considered to be a statement of the conservation of energy principle appropriate for flowing fluids. It is one of the most important/useful in fluid mechanics. It puts into a relation pressure and velocity in an inviscid ( negligible viscosity ) incompressible flow.

The general energy equation is simplified to

p_{1} + $\frac{1}{2}\mathrm{p}{\mathrm{v}}_{1}^{2}$ + ρgh_{1} = p_{2} + $\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}$ + ρgh_{2}

3.

Scent sprayer is based on

Charle's law

Archimedes principle

Boyle's law

Bernoulli's theorem

D.

Bernoulli's theorem

A scent sprayer is an example of fall in pressure due to increase in velocity. Also as per Bernoulli's theorem that datum head, pressure head and velocity head of a flowing liquid is constant. Therefore scent sprayer is based on Bernoulli's theorem.

4.

Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after change is

1 : 2

^{1/3}2

^{1/3}: 12 : 1

1 : 2

B.

2^{1/3} : 1

Radius of one drop of mercury is R

∴ The volume of one drop = $\frac{4}{3}\mathrm{\pi}{\mathrm{R}}^{3}$

∴ Total volume of the two drops,

V = 2 × $\frac{4}{3}\mathrm{\pi}{\mathrm{R}}^{3}$

V = $\frac{8}{3}{\mathrm{\pi R}}^{3}$

Let the radius of the large drop formed be R'

The total volume of the large is also V

$\frac{4}{3}\mathrm{\pi}\mathrm{R}{\text{'}}^{2}=\frac{8}{3}\mathrm{\pi}{\mathrm{R}}^{3}$

R'^{3} = 2R^{3}

⇒ R' = 2^{1/3} R

Now the surface area of the resultant drop is

S_{1}_{ }= 2 × 4$\mathrm{\pi}$R^{2}

S_{1}_{ }= 8$\mathrm{\pi}$R^{2 }

and the surface area of the resultant drop is

S_{2} = 4$\mathrm{\pi}$R'^{2 }

S_{2} = 4$\mathrm{\pi}$ 2^{2/3 }R^{2}

Let T be the surface tension of the mercury. Therefore the surface energy of the two drops before coalescing is

U_{1} = S_{1} T

U_{1}_{ }= 8$\mathrm{\pi}$R^{2}T

and the surface energy after coalescing,

U_{2} = S_{2}T

U_{2}_{ }= 2^{2/3} × 4$\mathrm{\pi}$R^{2}T

∴ $\frac{{\mathrm{U}}_{1}}{{\mathrm{U}}_{2}}=\frac{8\mathrm{\pi}{\mathrm{R}}^{2}\mathrm{T}}{{2}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\times 4\mathrm{\pi}{\mathrm{R}}^{2}\mathrm{T}}$

= $\frac{2}{{2}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}$

$\frac{{\mathrm{U}}_{1}}{{\mathrm{U}}_{2}}$ = 2^{1/3}

5.

For liquid to rise in a capillary tube, the angle of contact should be

acute

obtuse

right

none of these

A.

acute

When impure water or kerosene is taken in a glass vessel, it is found that the surface near the walls is curved concave upwards.

Consider a molecule of water M on the free surface close to the wall of the glass vessel. The magnitude of adhesive force A is greater than the magnitude of cohesive force C and resultant of the two molecular forces of attraction R is or outside the liquid. Hence the molecule A is attracted towards the walls of glass vessel. The free surface of water adjust itself at right angle to the resultant R. Therefore molecules like M creep upwards and the angle of contact is acute.

6.

Which of the following physical wuantity do not have same dimensions?

pressure and stress

tension and surface tension

strain and angle

energy and work

B.

tension and surface tension

Pressure and stress both have the dimensions Force/area. Strain and angle are both dimensionless. Energy and work have the dimension force × distance.

Tension and surface tension refer to two different physical quantites and their dimensions are different. Tension is a force and surface tension is force per unit length.

7.

**Assertion: ** Thin films such a soap bubble or a thin layer of oil on water show beautiful colours when illuminated by white light.

**Reason: ** It happens due to the interference of light reflected from upper surface of the thin film.

If both the assertion and reason are true and reason is a correct explanation of assertion.

If both assertion and reason are true but assertion is not a correct explanation of the assertion.

If both the assertion is true but the reason is false.

If both assertion and reason are false.

C.

If both the assertion is true but the reason is false.

The beautiful colours are seen on account of interference of light reflected from the upper and the lower surfaces of the thin film. As conditions for constructive and destructive interference depend upon the wavelength of light, therefore coloured interference fringes are observed.

8.

The property utilized in the manufacture of lead shots is

specific weight of liquid lead

compressibility of liquid lead

specific gravity of liquid lead

surface tension of liquid lead

D.

surface tension of liquid lead

The property utilized in the manufacture of lead shots is surface tension of liquid lead. In this process, molten lead is made to pass through a sieve from a high tower and allowed to fall in water. The molten lead particles, while descending, assume a spherical shape and solidify in this form, before falling into water.

9.

The soap bubbles have radii in the ratio 2: 1 ratio of excess pressure in side there is

1:4

4:1

2:1

1:2

D.

1:2

$\mathrm{P}\propto \frac{1}{\mathrm{r}}\phantom{\rule{0ex}{0ex}}\mathrm{hence},\frac{{\mathrm{P}}_{1}}{{\mathrm{P}}_{2}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{r}}_{1}$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{r}}_{2}$}\right.}}=\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}=\frac{1}{2}...\mathrm{since}{\mathrm{r}}_{1}:{\mathrm{r}}_{2}=2:1\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{1}:{\mathrm{P}}_{2}=1:2$

10.

A capillary tube of radius r can support a liquid of weight 6.28 x 10^{-4} N. If the surface tension of the liquid is 5 x 10^{-2}N/m. The radius of capillary must be

$2.0\times {10}^{-3}\mathrm{m}$

$2.0\times {10}^{-4}$m

1.5$\times {10}^{-3}\mathrm{m}$

$12.5\times {10}^{-4}\mathrm{m}$

A.

$2.0\times {10}^{-3}\mathrm{m}$

Given:- weight =6.24$\times $10^{-4}N,

W=T.$2\mathrm{\pi r}$

r=$\frac{\mathrm{W}}{2\mathrm{\pi T}}$=$\frac{6.28\times {10}^{-4}}{2\times 3.14\times 5\times {10}^{-2}}$

r = 2$\times {10}^{-3}$m