Properties of Matter — Page 13 | JEE Physics

Q121

Under isothermal condition, the pressure of a gas is given by

\mathrm{P}=a \mathrm{~V}^{-3}

, where

a

is a constant and

\mathrm{V}

is the volume of the gas. The bulk modulus at constant temperature is equal to

A

\dfrac{P}{2}

B 2 P

C 3 P

D P

Correct Answer

Option C

Solution

The bulk modulus (

B

) of a substance is defined as the ratio of the infinitesimal pressure increase (

\Delta P

) to the relative decrease in volume (

\frac{-\Delta V}{V}

) at constant temperature:

B = -V \frac{\Delta P}{\Delta V}

To find the bulk modulus for the given pressure-volume relationship, we first need to find the differential change in pressure with respect to volume:

P = aV^{-3}

Differentiate

P

with respect to

V

:

\frac{dP}{dV} = -3aV^{-4}

Now we can use the definition of the bulk modulus:

B = -V \frac{\Delta P}{\Delta V} = -V \frac{dP}{dV}

Plug in the value for

\frac{dP}{dV}

:

B = -V(-3aV^{-4})

Simplify the expression:

B = 3aV^{-3}

Notice that

3aV^{-3}

is equal to

3P

, since

P = aV^{-3}

:

B = 3P

Therefore, the bulk modulus at constant temperature is equal to 3P.

Q122

Correct Bernoulli's equation is (symbols have their usual meaning) :

A

P+\dfrac{1}{2} \rho g h+\dfrac{1}{2} \rho v^2=

constant

B

P+m g h+\dfrac{1}{2} m v^2=

constant

C

P+\rho g h+\rho v^2=

constant

D

P+\rho g h+\dfrac{1}{2} \rho v^2=

constant

Correct Answer

Option D

Solution

Bernoulli's equation relates the pressure, velocity, and height in a flowing fluid and is derived from the principle of conservation of energy.

The correct form of Bernoulli's equation is:

P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}

where: P is the pressure within the fluid. $\rho$ is the density of the fluid. g is the acceleration due to gravity. h is the height above a reference point. v is the velocity of the fluid.

Looking at the options given: Option A:

P + \frac{1}{2} \rho g h + \frac{1}{2} \rho v^2 = \text{constant}

Option B:

P + m g h + \frac{1}{2} m v^2 = \text{constant}

Option C:

P + \rho g h + \rho v^2 = \text{constant}

Option D:

P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}

Option D correctly represents Bernoulli's equation in its proper form. Therefore, the correct answer is: Option D

Q123

A body cools from

80^{\circ} \mathrm{C}

to

60^{\circ} \mathrm{C}

in 5 minutes. The temperature of the surrounding is

20^{\circ} \mathrm{C}

. The time it takes to cool from

60^{\circ} \mathrm{C}

to

40^{\circ} \mathrm{C}

is :

A 500 s

B

\dfrac{25}{3} \mathrm{~S}

C 450 s

D 420 s

Correct Answer

Option A

Solution

We are given the rate of cooling is proportional to the temperature difference between the body and the surroundings.

Mathematically, it can be expressed as:

\frac{\Delta T_{body}}{\Delta t} \propto (T_{body} - T_{surroundings})

Here,

\Delta T_{body}

is the change in temperature of the body,

\Delta t

is the time taken for the change in temperature,

T_{body}

is the temperature of the body,

T_{surroundings}

is the temperature of the surroundings. Now, introducing a proportionality constant

c

, we can write:

\frac{\Delta _{body}}{\Delta t} = c (T_{body} - T_{surroundings})

For the first cooling interval (from

80^{\circ} \mathrm{C}

to

60^{\circ} \mathrm{C}

in 5 minutes):

\frac{20}{5} = c (70 - 20)

For the second cooling interval (from

60^{\circ} \mathrm{C}

to

40^{\circ} \mathrm{C}

in

x

minutes):

\frac{20}{x} = c (50 - 20)

Now, we have two equations: 1)

\frac{20}{5} = c (50)

2)

\frac{20}{x} = c (30)

Solve equation (1) for

c

:

c = \frac{20}{5 \cdot 50} = \frac{1}{25}

Substitute the value of

c

into equation (2):

\frac{20}{x} = \frac{1}{25}(30)

Solve for

x

:

x = \frac{20 \cdot 25}{30} = \frac{500}{30} = \frac{25}{3}

So, the time it takes for the body to cool from

60^{\circ} \mathrm{C}

to

40^{\circ} \mathrm{C}

is

\frac{25}{3}

minutes, which is equal to 500 seconds.

Q124

Eight equal drops of water are falling through air with a steady speed of

10 \mathrm{~cm} / \mathrm{s}

. If the drops coalesce, the new velocity is:-

A

40 \mathrm{~cm} / \mathrm{s}

B

16 \mathrm{~cm} / \mathrm{s}

C

10 \mathrm{~cm} / \mathrm{s}

D

5 \mathrm{~cm} / \mathrm{s}

Correct Answer

Option A

Solution

In this problem, we need to consider the terminal velocity of the droplets, which is reached when the gravitational force is balanced by the drag force acting on the droplet.

Terminal velocity is related to the square of the droplet's radius.

The relationship between the terminal velocity (v) and the radius (r) of the droplet is given by:

v \propto r^2

Initially, there are 8 equal drops of water, each with radius r and velocity 10 cm/s.

When these droplets coalesce, they form a single droplet with a larger radius R.

The volume of the new droplet should be equal to the total volume of the 8 smaller droplets.

Using the volume formula for spheres, we can write the relationship between the radii as:

8 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3

Solving for R, we get:

R = 2r

Now, we can use the relationship between the terminal velocities and radii of the droplets:

\frac{v_1}{v_2} = \left(\frac{r}{R}\right)^2

Given the initial terminal velocity of 10 cm/s for the smaller droplets ( $v_1$ ) and the relationship between r and R:

\frac{10}{v_2} = \left(\frac{1}{2}\right)^2

Solving for the new terminal velocity ( $v_2$ ):

v_2 = 40 \mathrm{~cm} / \mathrm{s}

The new terminal velocity after the droplets coalesce is 40 cm/s.

Q125

Young's moduli of the material of wires A and B are in the ratio of

1: 4

, while its area of cross sections are in the ratio of

1: 3

. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires

\mathrm{A}

and

\mathrm{B}

will be in the ratio of [Assume length of wires A and B are same]

A 1 : 12

B 1 : 36

C 12 : 1

D 36 : 1

Correct Answer

Option C

Solution

Given the formula for elongation in a material due to a force:

\Delta L = \frac{FL}{AY}

where: F is the force applied, L is the original length, A is the cross-sectional area of the material, and Y is Young's modulus of the material.

The ratio of the elongations in the two wires A and B is given by:

\frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2} \times \frac{A_2}{A_1} \times \frac{Y_2}{Y_1}

Since the same force is applied on both wires (i.e., ( $F_1/F_2$ = 1)), the areas are in the ratio 1:3 (i.e., ( $A_2/A_1$ = 3)), and the Young's moduli are in the ratio 1:4 (i.e., ( $Y_2/Y_1$ = 4)), substituting these values into the equation gives:

\frac{\Delta L_1}{\Delta L_2} = 1 \times 3 \times 4 = 12

So, the ratio of the elongations is 12:1, which indicates that wire A will elongate 12 times more than wire B when the same force is applied.

Q126

Given below are two statements: Statement I : Pressure in a reservoir of water is same at all points at the same level of water. Statement II : The pressure applied to enclosed water is transmitted in all directions equally. In the light of the above statements, choose the correct answer from the options given below:

A Both Statement I and Statement II are false

B Statement I is false but Statement II is true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

Statement I: Pressure in a reservoir of water is same at all points at the same level of water.

This statement is true.

According to the principle of fluid statics, in a body of static fluid, the pressure is the same at all points at the same horizontal level.

This is because the pressure at any point in a static fluid is determined by the weight of the fluid above it.

Therefore, at any given level in the reservoir, the pressure is the same because the weight of the water above each point is the same.

Statement II: The pressure applied to enclosed water is transmitted in all directions equally.

This statement is also true.

It is a direct statement of Pascal's law, which states that any change in pressure applied at any point in a fluid in a closed system is transmitted undiminished to all points in the fluid and acts in all directions.

Therefore, both Statement I and Statement II are true.

Q127

A hydraulic automobile lift is designed to lift vehicles of mass

5000 \mathrm{~kg}

. The area of cross section of the cylinder carrying the load is

250 \mathrm{~cm}^{2}

. The maximum pressure the smaller piston would have to bear is

\left[\right.

Assume

\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]

A

20 \times 10^{+6} \mathrm{~Pa}

B

200 \times 10^{+6} \mathrm{~Pa}

C

2 \times 10^{+5} \mathrm{~Pa}

D

2 \times 10^{+6} \mathrm{~Pa}

Correct Answer

Option D

Solution

A hydraulic lift works based on Pascal's principle, which states that the pressure applied at one point in an incompressible fluid is transmitted equally in all directions.

The force exerted by the car on the hydraulic fluid is equal to the weight of the car, which is

F = mg

, where

m

is the mass of the car and

g

is the acceleration due to gravity. Substituting the given values, we get:

F = 5000 \, \text{kg} \times 10 \, \text{m/s}^2 = 50000 \, \text{N}

The pressure exerted by the car on the hydraulic fluid is equal to the force divided by the area over which the force is distributed, which is

P = \frac{F}{A}

, where

A

is the cross-sectional area of the cylinder carrying the load.

However, the given area is in cm², so we need to convert it to m².

We know that 1 m² = 10,000 cm², so:

A = 250 \, \text{cm}^2 \times \frac{1 \, \text{m}^2}{10000 \, \text{cm}^2} = 0.025 \, \text{m}^2

Substituting the values of force and area into the formula for pressure, we get:

P = \frac{50000 \, \text{N}}{0.025 \, \text{m}^2} = 2 \times 10^6 \, \text{Pa}

Q128

An air bubble of volume

1 \mathrm{~cm}^{3}

rises from the bottom of a lake

40 \mathrm{~m}

deep to the surface at a temperature of

12^{\circ} \mathrm{C}

. The atmospheric pressure is

1 \times 10^{5} \mathrm{~Pa}

the density of water is

1000 \mathrm{~kg} / \mathrm{m}^{3}

and

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}

. There is no difference of the temperature of water at the depth of

40 \mathrm{~m}

and on the surface. The volume of air bubble when it reaches the surface will be:

A

4 \mathrm{~cm}^{3}

B

3 \mathrm{~cm}^{3}

C

2 \mathrm{~cm}^{3}

D

5 \mathrm{~cm}^{3}

Correct Answer

Option D

Solution

The volume of the air bubble changes due to the change in pressure as it rises from the bottom of the lake to the surface.

We can use Boyle's Law to calculate the change in volume, which states that the product of pressure and volume is constant for a given mass of confined gas held at a constant temperature: $P_1V_1 = P_2V_2$ where $P_1$ and $V_1$ are the pressure and volume at the bottom of the lake and $P_2$ and $V_2$ are the pressure and volume at the surface of the lake.

At the bottom of the lake, the pressure is the atmospheric pressure plus the pressure due to the water column above the bubble: $P_1 = P_{\text{atm}} + \rho gh$ where $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $h$ is the height of the water column.

Substituting the given values, we get: $P_1 = 1 \times 10^{5} \text{ Pa} + 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 40 \text{ m} = 5 \times 10^{5} \text{ Pa}$ At the surface of the lake, the pressure is the atmospheric pressure: $P_2 = P_{\text{atm}} = 1 \times 10^{5} \text{ Pa}$ The initial volume of the bubble is: $V_1 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3$ Substituting these values into Boyle's Law and solving for $V_2$ , we get: $V_2 = \dfrac{P_1V_1}{P_2} = \dfrac{5 \times 10^{5} \text{ Pa} \times 1 \times 10^{-6} \text{ m}^3}{1 \times 10^{5} \text{ Pa}} = 5 \times 10^{-6} \text{ m}^3 = 5 \text{ cm}^3$ So, the volume of the air bubble when it reaches the surface is 5 cm³.

Q129

An aluminium rod with Young's modulus

Y=7.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}

undergoes elastic strain of

0.04 \%

. The energy per unit volume stored in the rod in SI unit is:

A 5600

B 2800

C 11200

D 8400

Correct Answer

Option A

Solution

The strain energy stored per unit volume in a material under stress can be calculated using the following formula: $U = \dfrac{1}{2} \sigma \epsilon$ where $\sigma$ is the stress and $\epsilon$ is the strain.

For an elastic material, stress is proportional to strain (Hooke's law), and the constant of proportionality is the Young's modulus (Y).

So we can write: $\sigma = Y \epsilon$ Substituting this into the energy density equation we get: $U = \dfrac{1}{2} Y \epsilon^2$ The strain given in the problem is 0.04%, which needs to be converted to a decimal for use in this formula.

Therefore, $\epsilon = 0.04/100 = 0.0004$ .

Substituting the values into the equation gives: $U = \dfrac{1}{2} \times 7.0 \times 10^{10} N/m^2 \times (0.0004)^2 = 5600 ~J/m^3$

Q130

A small ball of mass

\mathrm{M}

and density

\rho

is dropped in a viscous liquid of density

\rho_{0}

. After some time, the ball falls with a constant velocity. What is the viscous force on the ball ?

A

\mathrm{F}=\mathrm{Mg}\left(1-\dfrac{\rho_{\mathrm{O}}}{\rho}\right)

B

\mathrm{F}=\mathrm{Mg}\left(1+\dfrac{\rho}{P_{o}}\right)

C

\mathrm{F}=\mathrm{Mg}\left(1+\dfrac{\rho_{\mathrm{o}}}{\rho}\right)

D

F=M g\left(1 \pm \rho \rho_{0}\right)

Correct Answer

Option A

Solution

When the ball is falling with a constant velocity, it means the net force acting on the ball is zero.

This is because it's in a state of dynamic equilibrium - the downward force equals the upward force.

The downward force is the gravitational force (weight of the ball), which is given by $F_g = Mg$ .

The upward force is the sum of buoyant force and the viscous drag.

The buoyant force is the weight of the fluid displaced by the ball, which is given by $F_b = Vg\rho_0 = Mg\rho_0/\rho$ where $V = M/\rho$ is the volume of the ball.

The viscous force, $F_v$ , is the force that we need to find.

Since the net force is zero, we have: $F_g = F_b + F_v$ or $Mg = Mg\rho_0/\rho + F_v$ which simplifies to $F_v = Mg - Mg\rho_0/\rho = Mg(1 - \rho_0/\rho)$