AIIMS BSc Nursing Physics Most Expected Questions

Preparing Physics for the AIIMS BSc Nursing entrance exam requires a strong grasp of concepts as well as regular practice of exam-oriented questions. Solving AIIMS BSc Nursing Physics most expected questions helps candidates identify important topics, understand question patterns, and improve their problem-solving speed. 

This collection of carefully selected questions covers key concepts from Electrostatics, including Coulomb’s Law, Gauss’s Law, electric flux, electric field properties, and dimensional analysis, helping aspirants strengthen their preparation and boost their confidence for the exam.

Problem 1: Force between Two Equal Charges

Question: The force between two equal charges is F. If the amount of charges and separation between them becomes double, then find a new force.

Solution:

1. Initial State:

• Charges: q, q; Separation: r; Force: F

• According to Coulomb's Law, F = k * (q * q) / r² = kq²/r².

2. Modified State:

• New charges: 2q, 2q; New separation: 2r

• New force F' = k * (2q * 2q) / (2r)² = k * (4q²) / (4r²) = k * q² / r².

3. Result:

• Since F' = kq²/r² and F = kq²/r², therefore, F' = F.

• The new force remains the same as the initial force.

Problem 2: Force after 25% Charge Transfer

Question: The force between a given system of charges is F. If 25% charge of one of them is transferred to the other, then find the new force.

 Solution:

1. Initial State:

• Charges: q₁ = q, q₂ = q; Initial force: F = k * (q * q) / r² = kq²/r².

2. Charge Transfer:

• 25% transfer means 1/4 of the charge.

• One charge becomes: q₁' = q - (q/4) = 3q/4.

• The other charge becomes: q₂' = q + (q/4) = 5q/4.

3. New Force Calculation:

• New force F' = k * (q₁' * q₂') / r²

• F' = k * (3q/4 * 5q/4) / r² = k * (15q²/16) / r².

• F' = (15/16) * (kq²/r²).

4. Result:

• Since F = kq²/r², the new force F' = (15/16) F.

Problem 3: Find Dimension of Coulombian Constant (k)

Question: Find the dimension of Coulombian Constant (k).

 Solution:

1. Coulomb's Law:

• The Coulombian force F = k * q² / r².

2. Rearranging for k:

• k = F * r² / q².

3. Substituting Dimensions:

• Dimension of Force (F) = [M¹L¹T⁻²]

• Dimension of Distance (r²) = [L²]

• Dimension of Charge (q²) = [A²T²] (since Charge q = [AT])

4. Dimension of k:

• k = [M¹L¹T⁻²] * [L²] / [A²T²] = [M¹L³T⁻²A⁻²T⁻²]

• k = [M¹L³T⁻⁴A⁻²].

Problem 4: Find Dimension of Permittivity of Free Space (ε₀)

Question: Find the dimension of ε₀.

 Solution:

1. Relationship between k and ε₀:

• k = 1 / (4πε₀). Therefore, ε₀ = 1 / (4πk).

2. Using Coulomb's Law and Rearranging for ε₀:

• F = (1 / 4πε₀) * q² / r²

• ε₀ = q² / (4πF r²).

3. Substituting Dimensions:

• Dimension of Charge (q²) = [A²T²]

• Dimension of Force (F) = [M¹L¹T⁻²]

• Dimension of Distance (r²) = [L²]

• (4π is dimensionless).

4. Dimension of ε₀:

• ε₀ = [A²T²] / ([M¹L¹T⁻²] * [L²]) = [A²T²] / [M¹L³T⁻²]

• ε₀ = [M⁻¹L⁻³T⁴A²].

5. Observtion:

• The dimension of ε₀ is the inverse of the dimension of k.

Problem 5: Total Electric Flux from a Unit Positive Charge

Question: Total electric flux coming out of a unit positive charge in air is:

 Solution:

1. Gauss's Law:

• Total electric flux (Φ) = Q_enclosed / ε₀.

2. Given Information:

• Q_enclosed = 1 unit positive charge (i.e., +1 Coulomb).

• Medium is air (implies ε₀).

3. Calculation:

• Φ = 1 / ε₀. This can be written as ε₀⁻¹.

Problem 6: Total Flux for a System of Charges

Question: Find total flux for the given system of charges.

(Assumed charges inside: +2C, -4C, +5C. Charges outside are not relevant).

 Solution:

1. Gauss's Law:

• Total electric flux (Φ) = Q_inside / ε₀.

• Only the net charge enclosed within the Gaussian surface contributes to the flux.

2. Calculating Q_inside:

• Q_inside = (+2 C) + (-4 C) + (+5 C) = +3 C.

3. Result:

• Φ = 3 / ε₀.

Problem 7: Find Dimension of √(μ₀ε₀)

Question: Find the dimension of √(μ₀ε₀).

 Solution:

1. Relationship with Speed of Light (c):

• The speed of light (c) in a vacuum is given by the formula: c = 1 / √(μ₀ε₀).

2. Rearranging:

• Therefore, √(μ₀ε₀) = 1 / c.

3. Dimension of Speed (c):

• Dimension of Speed = [LT⁻¹].

4. Dimension of √(μ₀ε₀):

• Dimension of 1/c = 1 / [LT⁻¹] = [L⁻¹T¹].

• (Memory Tip: Remember the relation between the speed of light (c) and the constants μ₀ and ε₀, as this is a commonly asked question.)

Problem 8: Total Flux for 8 Dipoles Inside a Cube

Question: Eight dipoles of charge 'e' are placed inside a cube. Find the total flux associated with it.

 Solution:

1. Definition of a Dipole:

• A dipole consists of two charges of equal magnitude but opposite polarity (e.g., +q and -q).

2. Net Charge of a Single Dipole:

• The net charge of one dipole is q + (-q) = 0.

3. Net Charge of Eight Dipoles:

• Total enclosed charge (Q_in) = 8 * (net charge of one dipole) = 8 * 0 = 0.

4. Gauss's Law:

• Total flux (Φ) = Q_in / ε₀.

5. Result:

• Since Q_in = 0, the total flux Φ = 0 / ε₀ = 0.

Problem 9: Total Flux for an Infinite Wire Passing Through a Cube

Question: A wire of infinite length with charge density λ Coulomb per meter is passing from a cube of side length L, parallel to one of its sides. Find the total flux.

 Solution:

1. Gauss's Law:

• Total flux (Φ) = Q_in / ε₀.

• We need to find the charge enclosed by the cube.

2. Identifying Enclosed Charge (Q_in):

• Only the portion of the wire inside the cube contributes.

• The length of the wire inside the cube is equal to the side length (L).

• Charge density (λ) means λ Coulombs per meter.

3. Calculating Q_in:

• Q_in = charge density * enclosed length = λL.

4. Result:

• Φ = λL / ε₀.

Problem 10: Gauss Law Validity Condition

Question: Gauss's Law is valid only if force due to a point charge varies with:

 Solution:

1. Force due to a Point Charge:

• The Coulombian force (F) between two point charges is inversely proportional to the square of the distance 'r': F = k * q₁q₂ / r².

2. Inverse Square Law:

• This shows that the force is inversely proportional to r² (F ∝ 1/r²).

• This can be expressed as F ∝ r⁻².

3. Relation to Gauss's Law:

• Gauss's Law fundamentally relies on the inverse square nature of the electrostatic force. If the force varied differently, Gauss's Law would not hold in its current form.

4. Result:

• Gauss's Law is valid if the force varies with r⁻².

Problem 11: Electric Field Due to Conducting Sheet

Question: Electric field due to a thin conducting sheet is:

 Solution:

Understanding the electric field generated by charged sheets:

• Thin Conducting Sheet:

• • For a thin conducting sheet, the electric field at a point near it is E = σ / 2ε₀. This is applicable when considering the field from a single, infinitely large, uniformly charged thin sheet.

• Thick Conducting Sheet:

• • For a thick conducting sheet (where charge distributes on both surfaces), the total electric field at a point outside is E = σ / ε₀. Each surface contributes σ / 2ε₀.

Result:

Based on the specific distinction, for a thin sheet, the answer is σ / 2ε₀.

Problem 12: Direction of Electric Field Lines for Negative Charge

Question: The direction of the electric lines of force for a negative charge will be:

 Solution:

1. Nature of Electric Field Lines:

• Electric field lines originate from positive charges and terminate on negative charges.

2. Direction for a Negative Charge:

• For a negative point charge, the electric field lines point towards the charge.

• As they point directly towards the center from all directions, this direction is described as radially inwards.

3. Result:

• The direction is radially inwards.

Problem 13: Choose the Wrong Option (Properties of Electric Field Lines)

Question: Choose the wrong option regarding electric field lines:

 Options:

1. Electric field lines go from high potential to low potential.

2. Electric field lines can intersect.

3. Electric field lines never form a closed loop.

4. Electric field is directly proportional to electric field lines.

 Solution:

Let's analyze each option:

• 1. Electric field lines go from high potential to low potential.

• Correct Statement: Field lines start at positive charges (high potential) and end at negative charges (low potential).

• 2. Electric field lines can intersect.

• Incorrect Statement: Electric field lines can never intersect each other. If they did, it would mean the electric field has two different directions at that point, which is physically impossible. This is the wrong option.

• 3. Electric field lines never form a closed loop.

• Correct Statement: Electrostatic field lines are conservative, implying they do not form closed loops.

• 4. Electric field is directly proportional to electric field lines.

• Correct Statement: The density of electric field lines (number of lines per unit area) is directly proportional to the magnitude of the electric field. Closer lines indicate a stronger field.

Result:

The wrong option is that electric field lines can intersect. Therefore, Option 2 is the correct answer to the question.