Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Water flows at a speed of 6 cm s^-1 through a tube of radius 1 cm. Coefficient of viscosity of water at room temperature is 0.01 poise. Calculate the Reynolds number. Is it a steady flow?

60, yes

60, no

120, yes

120, no

SOLUTION

Solution : C

You know,
Reynolds number, $N = \frac{ρ v D}{η}$
Where, $ρ$ = density of fluid

v = velocity of flow of the fluid

D = diameter of the tube
$η$ = coefficient of viscosity of the fluid
$\Rightarrow N = \frac{1000 \times \frac{6}{100} \times \frac{2}{100}}{0.01}$
{all values in SI}
N = 120 {Dimensionless}, yes the flow is steady

Question 2

A spherical ballof radius $3.0 \times 10^{- 4} m$ and density $10^{4}$ $k g / m^{3}$ falls freely under gravity through a distance h it before entering a tank of water (coefficient of viscosity = $η$ = $9.8 \times 10^{- 6} N s m^{- 2}$ and density = $10^{3} k g / m^{3}$ . If after entering the water, the velocity of the ball does not change, find~h.(Take g =9.8 $m s^{- 2}$ )

10.24 m

1.65 km

32.36 m

50.12 m

SOLUTION

Solution : B

Terminal velocity,

$u = \sqrt{2 g h} = \frac{2 r^{2} (ρ - σ) g}{9 η} \Rightarrow h = \frac{2 g r^{4} (ρ - r)^{2}}{81 η^{2}} = 1.65 k m$

Question 3

A long cylindrical rod of radius $R_{1}$ is placed co-axially inside a cylindrical tube of radius $R_{2} (> R_{1})$ filled withviscous liquid and pulled along the axis with a constant velocity. In steady state the velocity gradient $(\frac{d v}{d r})$ depends on the distance from the common axis 'r' as proportional to

$r^{\circ}$

$r^{- 1}$

$r^{- 2}$

$r^{- 3}$

SOLUTION

Solution : B

$f = η A \frac{d v}{d x} = c o n s t η 2 π r l \frac{d v}{d x} = c o n s t \frac{d v}{d x} α \frac{1}{r}$

Question 4

Which of the following processes will be least affected by the viscosity of water ?

Water flowing through a pipe

Air bubble rising up through water

A wide, shallow sheet of water flowing on a flat surface

Water flowing out through a hole in the side of a tank

SOLUTION

Solution : D

When water flows out of a hole in a tank, it is in contact with the edges of the hole over a very small area, and therefore viscosity effects are small

Question 5

An ice berg of density 900 Kg/ $m^{3}$ is floating in water of density 1000 Kg/ $m^{3}$ . The percentage of volume of ice-cube outside the water is

20%

35%

10%

25%

SOLUTION

Solution : C

Let the total volume of ice-berg is V and its density is $ρ$ . If this ice-berg floats in water with volume $V_{i n}$ inside

it then $V_{i n} σ g = V ρ g \Rightarrow V_{i n} = ⟮ \frac{ρ}{σ} ⟯ V$ [ $σ$ = density of water]

or $V_{o u t} = V - V_{i n} = ⟮ \frac{σ - ρ}{σ} ⟯ V$

$\Rightarrow \frac{V_{o u t}}{V} = ⟮ \frac{σ - ρ}{σ} ⟯ = \frac{1000 - 900}{1000} = \frac{1}{10}$

$∴ V_{o u t} = 10 %$ of V

Question 6

A body of density $d_{1}$ is counterpoised by Mg of weights of density $d_{2}$ in air of density d. Then the true mass of the body is

$M ⟮ 1 - \frac{d}{d_{2}} ⟯$

$M ⟮ 1 - \frac{d}{d_{1}} ⟯$

$\frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})}$

SOLUTION

Solution : D

Let $M_{O}$ =mass of body in vacuum.

Apparent weight of the body in air = Apparent weight of standard weights in air

Actual weight of the body - upthrust due to displaced air

= Actual weight of the weights- upthrust due to displaced air

$\Rightarrow M_{0} g - ⟮ \frac{M_{0}}{d_{1}} ⟯ d g = M g - ⟮ \frac{M}{d_{2}} ⟯ d g \Rightarrow M_{0} = \frac{M [1 - \frac{d}{d_{2}}]}{[1 - \frac{d}{d_{1}}]}$

Question 7

Density of ice is $ρ$ and that of water is $σ$ . What will be the decrease in volume when a mass M of ice melts

$\frac{M}{σ - ρ}$

$\frac{σ - ρ}{M}$

$M [\frac{1}{ρ} - \frac{1}{σ}]$

$\frac{1}{M} [\frac{1}{ρ} - \frac{1}{σ}]$

SOLUTION

Solution : C

Volume of ice = $\frac{M}{ρ}$ , volume of water = $\frac{M}{σ}$

$∴$ Change in volume = $\frac{M}{ρ} - \frac{M}{σ} = M ⟮ \frac{1}{ρ} - \frac{1}{σ} ⟯$

Question 8

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / $c m^{3}$ . If the mass of the other is 48 g, its density in g / $c m^{3}$ is

4/3

3/2

SOLUTION

Solution : C

Apparent weight = V $(ρ - σ) g$ = $\frac{m}{ρ} (ρ - σ) g$

where m = mass of the body,

$ρ$ = density of the body

$σ$ = density of water

If two bodies are in equilibrium then their apparent weight must be equal.

$∴ \frac{m_{1}}{ρ_{1}} (ρ_{1} - σ)$ = $\frac{m_{2}}{ρ_{2}} (ρ_{2} - σ)$

$\Rightarrow \frac{36}{9} (9 - 1)$ = $\frac{48}{ρ_{2}} (ρ_{2} - 1)$

By solving we get $ρ_{2}$ = 3

Question 9

A large tank filled with water to a height 'h' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to h/2 and from h/2 to zero is

$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

$\sqrt{2} - 1$

$\frac{1}{\sqrt{2} - 1}$

SOLUTION

Solution : C

Time taken for the level to fall from H to H'      $t = \frac{A}{A_{0}} \sqrt{\frac{2}{g}} [\sqrt{H} - \sqrt{H^{'}}]$

According to problem - the time taken for the level to fall from h to $\frac{h}{2}$      $t_{1} = \frac{A}{A_{0}} \sqrt{\frac{2}{g}} [\sqrt{h} - \sqrt{\frac{h}{2}}]$

and similarly time taken for the level to fall from $\frac{h}{2}$ to zero      $t_{2} = \frac{A}{A_{0}} \sqrt{\frac{2}{g}} [\sqrt{\frac{h}{2}} - 0] ∴ \frac{t_{1}}{t_{2}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - 0} = \sqrt{2} - 1$

Question 10

Air is streaming past a horizontal air plane wing such that its speed is 120m/s over the upper surface and 90 m/s at the lower surface. If the density of air is 1.3 kg per $m e t r e^{3}$ and the wing is 10 m long and has an average width of 2 m, then the difference of the pressure on the two sides of the wing is

4095.0 Pascal

409.50 Pascal

40.950 Pascal

4.0950 Pascal

SOLUTION

Solution : A

From the Bernoulli's theorem

$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} \times 1.3 \times [(120)^{2} - (90)^{2}]$

$= 4095 N / m^{2}$ or Pascal

Question 11

A manometer connected to a closed tap reads 3.5 $\times$ $10^{5}$ N/ $m^{2}$ . When the valve is opened, the reading of manometer falls to 3.0 $\times$ $10^{5}$ N/ $m^{2}$ , then velocity of flow of water is

100 m/s

10 m/s

1 m/s

10 .10m/s

SOLUTION

Solution : B

Bernoulli's theorem for unit mass of liquid

$\frac{P}{ρ} + \frac{1}{2} v^{2}$ = constant

As the liquid starts flowing, it pressure energy decreases $\frac{1}{2} v^{2} = \frac{P_{1} - p_{2}}{ρ} \Rightarrow \frac{1}{2} v^{2} = \frac{3.5 \times 10^{5} - 3 \times 10^{5}}{10^{3}}$
$\Rightarrow v^{2} = \frac{2 \times 0.5 \times 10^{5}}{10^{3}} \Rightarrow v^{2} = 100 \Rightarrow v = 10 m / s$

Question 12

In the following fig. is shown the flow of liquid through a horizontal pipe. Three tubes A, B and C are connected to the pipe. The radii of the tubes A, B and C at the junction are respectively 2 cm, 1 cm and 2 cm. It can be said that the

Height of the liquid in the tube A is maximum

Height of the liquid in the tubes A and B is the same

Height of the liquid in all the three tubes is the same

Height of the liquid in the tubes A and C is the same

SOLUTION

Solution : D

As cross-section areas of both the tubes A and C are same and tube is horizontal. Hence according to equation of continuity $v_{A} = v_{C}$ and therefore according to Bernoulli's theorem $P_{A} = P_{C}$ i.e. height of liquid is same in both the tubes A and C.

Question 13

Water is moving with a speed of 5.18 ms $^{-} 1$ through a pipe with a cross-sectional area of 4.20 $c m^{2}$ . The water gradually descends 9.66 m as the pipe increase in area to 7.60 $c m^{2}$ . The speed of flow at the lower level is

3.0 $m s^{- 1}$

5.7 $m s^{- 1}$

3.82 $m s^{- 1}$

2.86 $m s^{- 1}$

SOLUTION

Solution : D

$a_{1} v_{1} = a_{2} v_{2}$

$\Rightarrow 4.20 \times 5.18 = 7.60 \times v_{2} \Rightarrow v_{2} = 2.86 m / s$

Question 14

For which of the two pairs, the angle of contact is same

Water and glass; glass and mercury

Pure water and glass; glass and alcohol

Silver and water; mercury and glass

Silver and chromium; water and chromium

SOLUTION

Solution : B

Both liquids water and alcohol have same nature (i.e. wet the solid). Hence angle of contact for both is acute.

Question 15

A soap bubble assumes a spherical surface. Which of the following statement is wrong

The soap film consists of two surface layers of molecules back to back

The bubble encloses air inside it

The pressure of air inside the bubble is less than the atmospheric pressure; that is why the atmospheric pressure has compressed it equally from all sides to give it a spherical shape

Because of the elastic property of the film, it will tend

SOLUTION

Solution : C

The pressure of air inside the bubble is less than the atmospheric pressure; that is why the atmospheric pressure has compressed it equally from all sides to give it a spherical shape

Question 16

The excess of pressure inside a soap bubble than that of the outer pressure is

$\frac{2 T}{r}$

$\frac{4 T}{r}$

$\frac{T}{2 r}$

$\frac{T}{r}$

SOLUTION

Solution : B

The pressure inside the bubble is more than outside. Inside pressure -outside pressure =excess pressure. In the case of bubble excess pressure = 4S/R. where s is the surface tension and r is the radius.

$△ P = \frac{4 T}{r}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Question 16

SOLUTION

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