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Concept: Excess pressure inside a soap bubble is given by 4S / R due to two surfaces (inner and outer) under surface tension.

Let:

P₀ = Atmospheric pressure

P₁ = Pressure inside the larger bubble (radius 2R)

P₂ = Pressure inside the smaller bubble (radius R)

Pressure inside the larger bubble:

P₁ = P₀ + 4S / (2R) = P₀ + 2S / R

Pressure difference between smaller and larger bubble:

P₂ - P₁ = 4S / R

Substituting from (i) and (ii):

P₂ - P₀ = (P₂ - P₁) + (P₁ - P₀)

= (4S / R) + (2S / R) = 6S / R

Answer: Option 3) 6S / R

CONCEPT:

Surface tension is the property of a liquid that makes its surface behave like a stretched elastic membrane. For a soap film, there are two free surfaces. Therefore, the work done in increasing the surface area is:

W = 2TΔA

where:

• W = Work done
• T = Surface tension
• ΔA = Increase in area of one surface

EXPLANATION:

Given:

• Initial dimensions = 10 cm × 6 cm
• Final dimensions = 10 cm × 11 cm
• Work done, W = 3 × 10⁻⁴ J

Step 1: Calculate the initial area

A₁ = 10 × 6 = 60 cm²

Step 2: Calculate the final area

A₂ = 10 × 11 = 110 cm²

Step 3: Increase in area

ΔA = A₂ − A₁

= 110 − 60 = 50 cm²

Convert into SI unit:

50 cm² = 50 × 10⁻⁴ m² = 5 × 10⁻³ m²

Step 4: Use the formula for a soap film

W = 2TΔA

Rearranging,

T = W / 2ΔA

Substitute the values:

T = (3 × 10⁻⁴) / (2 × 5 × 10⁻³) = (3 × 10⁻⁴) / (10 × 10⁻³)

= (3 × 10⁻⁴) / (10⁻²) = 3 × 10⁻² N/m

Final Answer: The surface tension of the film is:  3 × 10−2 N/m

Correct Option: (2)

CONCEPT:

Isothermal Process

An isothermal process is a thermodynamic process in which the temperature of a system remains constant.

In such a process, the heat given to the system is used entirely to perform work, as the internal energy of an ideal gas depends only on temperature.

EXPLANATION:

Let's examine the given options:

Option 1: Net increase in internal energy

In an isothermal process, the internal energy of the system remains constant because the temperature does not change. Hence, this option is incorrect.

Option 2: Net work done by the system

In an isothermal process, the heat given to the system is entirely converted into work done by the system. Hence, this option is correct.

Option 3: Net decrease in internal energy

Similar to option 1, the internal energy does not change in an isothermal process. Hence, this option is incorrect.

Option 4: Net change in volume

While the volume may change in an isothermal process, this option does not directly relate to the heat given to the system. Hence, this option is incorrect.

Therefore, the correct answer is option 2: Net work done by the system.

CONCEPT:

Carnot Engine

A Carnot engine is an idealized thermodynamic engine that operates on the Carnot cycle. It is the most efficient possible engine, as it works on reversible processes.

The efficiency of a Carnot engine depends on the temperatures of the heat source and the heat sink.

EXPLANATION:

Given:

Heat absorbed from the reservoir (Q_H) = 3000 kcal

Temperature of the heat source (T_H) = 627°C = 900 K

Temperature of the heat sink (T_C) = 27°C = 300 K

The efficiency of a Carnot engine is given by:

η = 1 - (T_C/T_H)

Substitute the values to find the efficiency:

η = 1 - (300/900)

η = 1 - 1/3

η = 2/3

Work done by the engine (W) is given by:

W = η × Q_H

Q_H in joules = 3000 kcal × 4184 J/kcal = 12552 × 10³ J

W = (2/3) × 12552 × 10³ J

W = 8370.66 × 10³ J ≈ 8.4 × 10⁶ J

Therefore, the correct answer is option 2: 8.4 × 10⁶ J.

CONCEPT:

Translational Kinetic Energy and Pressure-Volume Relation

Translational kinetic energy is the energy possessed by an object due to its motion from one place to another.

In thermodynamics, the relationship between the pressure (P), volume (V), and the translational kinetic energy (E) of an ideal gas can be derived from the ideal gas law.

EXPLANATION:

Let's examine the given options:

Option 1: PV = E

This is incorrect. The pressure-volume product is not equal to the translational kinetic energy.

Option 2: PV = 323232

E

This is incorrect. While 323232

E is related to the total kinetic energy per mole of gas, it is not the correct relation for PV.

Option 3: PV = 3E

This is incorrect. The factor of 3 is not applicable in this context.

Option 4: PV = 232323

E

This is correct. The pressure-volume product (PV) of an ideal gas is directly proportional to the translational kinetic energy (E) with the relation PV = 232323

E.

Therefore, the correct answer is option 4: PV = 232323

E.

CONCEPT:

Properties of Sound Waves

Sound waves are mechanical waves that require a medium to travel through. They can propagate through solids, liquids, and gases.

When a sound wave travels from one medium to another, several properties of the wave can change, such as its speed, wavelength, and amplitude.

The frequency of a sound wave, however, remains unchanged when it moves from one medium to another.

EXPLANATION:

Let's examine the given options:

Option 1: Frequency

The frequency of a sound wave is determined by the source of the sound and does not change when the wave moves from one medium to another.

Option 2: Amplitude

The amplitude of a sound wave can change as it moves between different media due to the varying densities and elastic properties of the media.

Option 3: Wavelength

The wavelength of a sound wave changes when it enters a different medium because the speed of sound is different in different media.

Option 4: Speed

The speed of a sound wave varies depending on the medium through which it is traveling. For example, sound travels faster in solids than in liquids and gases.

Therefore, the correct answer is option 1: Frequency.

Concept:

Wave Equation:

The general form of a wave equation is y = A sin(kx - ωt), where:

A = Amplitude of the wave

k = Wave number (k = 2π / λ)

ω = Angular frequency (ω = 2πf)

t = Time

x = Position

The wave velocity v can be calculated using the relation:

v = ω / k

Calculation:

Given the wave equation: y = 0.5 sin(2π / λ (400t - x)) m, we can identify the following:

ω = 2π × 400 = 800π rad/s

k = 2π / λ

The velocity of the wave is given by:

v = ω / k = (800π) / (2π / λ) = 400λ

Since the equation is in the standard wave form, we can conclude that the velocity of the wave is 400 m/s.

∴ The velocity of the wave is 400 m/s, which corresponds to Option 3.

Concept:

Energy Stored in a Capacitor:

The energy U stored in a capacitor is given by the formula:

U = 1/2 × Q × V, where:

Q = Charge on the capacitor (Coulombs)

V = Voltage across the capacitor (Volts)

The graph given in the question represents the relationship between charge Q and voltage V on the capacitor. The area of the triangle OAB represents the energy stored in the capacitor, as the energy stored is the product of charge and voltage divided by 2.

Calculation:

The area of triangle OAB is given by:

Area = 1/2 × base × height

Here, base = Q (charge) and height = V (voltage). Thus, the area is equivalent to the energy stored in the capacitor.

∴ The area of triangle OAB represents the energy stored in the capacitor, which corresponds to Option 4.

Concept:

Capacitance of a Parallel Plate Capacitor:

• The capacitance C of a parallel plate capacitor with a dielectric material between the plates is given by the formula:
• C = (K × ε₀ × A) / d, where:
• K = Dielectric constant of the material (dimensionless)
• ε₀ = Permittivity of free space (8.85 × 10⁻¹² F/m)
• A = Area of the plates (m²)
• d = Distance between the plates (m)
• When a dielectric material (oil) is inserted between the plates, the capacitance increases by a factor of K (dielectric constant).
• If the oil is removed (K = 1 for air or vacuum), the capacitance will be reduced by a factor of K.

Calculation:

Initially, the capacitance is C = K × C₀, where C₀ is the capacitance without the oil.

Given that K = 2 with the oil, the capacitance becomes C = 2 × C₀.

When the oil is removed (K = 1), the capacitance reduces to:

C' = C / K = 2C / 2 = C / 2

∴ The capacitance of the capacitor becomes C / 2, which corresponds to Option 4.

Concept:

Effect of Inserting a Dielectric Slab:

• When a dielectric slab is inserted between the plates of a charged capacitor, the following effects are observed:
• The capacitance increases due to the dielectric constant of the material.
• If the capacitor is isolated (not connected to a battery), the charge on the plates remains constant, as no external source is available to change the charge.
• The potential difference between the plates decreases, as the dielectric reduces the effective electric field between the plates.
• The stored energy in the capacitor decreases because energy is proportional to the square of the voltage, and the voltage decreases.
• The electric field inside the capacitor also decreases, since the dielectric reduces the field between the plates.

Explanation:

Since the capacitor is isolated, the charge remains constant. The insertion of the dielectric only affects the electric field, voltage, and energy stored in the capacitor. The charge on the capacitor does not change because there is no external circuit to either supply or remove charge.

∴ The charge on the capacitor remains unchanged, which corresponds to Option 1.