Dibrugarh University B.Sc. Physics Minor 2024 Question Paper with Answers 1st Semester

Q1. Choose the correct answer from the following. (5 × 2 = 10 Marks)

1. The vector sum of mass moments of a system of particles about the centre of mass is

(i) zero
(ii) positive
(iii) negative
(iv) Cannot be found

Ans: (i) zero

Explanation: The position vector of the centre of mass R⃗_CM is defined as R⃗_CM = (1/M) ∑ m_ir⃗_i. The mass moment of a particle i about the centre of mass is m_i(r⃗_i − R⃗_CM). Summing this over all particles gives ∑m_ir⃗_i − ∑m_iR⃗_CM = MR⃗_CM − MR⃗_CM = 0.

2. If the mechanical energy of a particle is conserved, then the particle is acted upon by a

(i) non-conservative force
(ii) conservative force
(iii) drag force
(iv) None of the above

Ans: (ii) conservative force

Explanation: Mechanical energy, which is the sum of kinetic and potential energy, remains constant when the forces doing work on the system are conservative. Non-conservative forces such as drag or friction dissipate mechanical energy.

3. The radius of gyration of a solid sphere about the diameter is

(i) 2r/5
(ii) 5r/2
(iii) r√(2/5)
(iv) r√(5/2)

Ans: (iii) r√(2/5)

Explanation: The moment of inertia of a solid sphere about its diameter is I = (2/5)Mr². The radius of gyration k is defined by I = Mk². Therefore, Mk² = (2/5)Mr², giving k = r√(2/5).

4. The ratio of Young's modulus and bulk modulus of a substance is

(i) 3(1 − σ)
(ii) 3(1 + σ)
(iii) 3(1 − 2σ)
(iv) 3(1 + 2σ)

Ans: (iii) 3(1 − 2σ)

Explanation: The theoretical relationship between Young's modulus Y, bulk modulus K, and Poisson's ratio σ is Y = 3K(1 − 2σ). Dividing both sides by K gives Y/K = 3(1 − 2σ).

5. The differential equation of a damped harmonic oscillator is given by d²x/dt² + 0.4 dx/dt + 36x = 0. Its time period is nearly

(i) π/4
(ii) π/3
(iii) π/6
(iv) π/2

Ans: (ii) π/3

Explanation: Given: d²x/dt² + 0.4 dx/dt + 36x = 0.

The standard equation is d²x/dt² + 2b dx/dt + ω₀²x = 0. Comparing coefficients: 2b = 0.4, so b = 0.2, and ω₀² = 36, giving ω₀ = 6.

The angular frequency of the damped oscillator is ω′ = √(ω₀² − b²).

Therefore, ω′ = √(36 − 0.2²) = √35.96 ≈ 6.

Hence, T = 2π/ω′ ≈ 2π/6 = π/3.

Q2. Answer the following questions.

1. Differentiate between inertial and non-inertial frames of reference.

Ans:

2. State Newton's law of viscosity. What is the SI unit of viscosity?

Ans: Newton's Law of Viscosity: The viscous shear stress τ acting between two adjacent layers of a fluid is directly proportional to the velocity gradient dv/dy perpendicular to the direction of flow.

Mathematically, τ = −η(dv/dy), where η is the coefficient of viscosity. The negative sign indicates that the viscous force opposes relative motion.

SI Unit: The SI unit of the coefficient of viscosity η is Pa·s, equivalently N·s/m² or kg/(m·s).

3. Briefly explain the principle of conservation of angular momentum.

Ans: The principle of conservation of angular momentum states that if the net external torque acting on a system is zero, the total angular momentum of the system remains constant in both magnitude and direction.

The time rate of change of angular momentum L⃗ is equal to the net external torque: dL⃗/dt = τ⃗_ext.

If τ⃗_ext = 0, then dL⃗/dt = 0, which implies that L⃗ is constant.

This principle explains why an ice skater spins faster when pulling their arms inward: decreasing the moment of inertia increases angular velocity so that L = Iω remains constant.

4. Show that the quality factor of a damped harmonic oscillator is inversely proportional to the damping coefficient.

Ans: The equation of motion for a damped harmonic oscillator is m(d²x/dt²) + c(dx/dt) + kx = 0, where c is the damping coefficient.

This can be written as d²x/dt² + 2b(dx/dt) + ω₀²x = 0, where b = c/(2m) and ω₀ = √(k/m).

The quality factor Q for a lightly damped oscillator is Q = ω₀/(2b).

Substituting b = c/(2m), we obtain Q = ω₀/[2(c/2m)] = mω₀/c.

Since m and ω₀ are constants for a given system, Q ∝ 1/c.

Therefore, the quality factor is inversely proportional to the damping coefficient.

5. Briefly explain the relativity of simultaneity of events.

Ans: The relativity of simultaneity is a core concept in Einstein's Special Theory of Relativity. It states that two spatially separated events that appear simultaneous to an observer in one inertial frame may not appear simultaneous to an observer in another inertial frame moving relative to the first.

This occurs because the speed of light is constant for all observers. If lightning strikes the front and back of a moving train simultaneously according to an observer standing on the ground midway between the strikes, an observer riding in the middle of the train will perceive the strike at the front first.

Hence, absolute simultaneity does not exist; time is relative to the observer's frame of reference.

6. What is resonance in a forced oscillator? What is the necessary condition for its occurrence?

Ans: Resonance in a forced oscillator is the phenomenon that occurs when a system is subjected to an external periodic driving force whose frequency closely matches the natural frequency of the system. At resonance, the system absorbs maximum energy from the driving source, resulting in a very large amplitude of oscillation.

For amplitude resonance, the condition is ω = √(ω₀² − 2b²), where ω₀ is the natural angular frequency and b is the damping parameter.

For a lightly damped system, b is very small and therefore ω ≈ ω₀.

3(a). What is centre of mass? Obtain the position vector of the centre of mass of a system of n particles.

Ans: Definition: The centre of mass of a system of particles is a unique hypothetical point where the entire mass of the system can be assumed to be concentrated for describing its translational motion as a whole.

Consider a system consisting of n particles with masses m₁, m₂, m₃, …, mₙ. Let their respective position vectors from an origin O be r⃗₁, r⃗₂, r⃗₃, …, r⃗ₙ.

The total mass of the system is M = m₁ + m₂ + … + mₙ = ∑_i=1ⁿm_i.

Let the position vector of the centre of mass be R⃗_CM. By the principle of moments,

MR⃗_CM = m₁r⃗₁ + m₂r⃗₂ + … + mₙr⃗ₙ

In summation notation,

MR⃗_CM = ∑_i=1ⁿm_ir⃗_i

Therefore,

R⃗_CM = [∑_i=1ⁿm_ir⃗_i] / [∑_i=1ⁿm_i]

3(b). What are conservative forces? Show that the force field F⃗ = (y² − x²)î + 2xyĵ is conservative.

Ans: A force is conservative if the total work done by the force on a particle moving between two points is independent of the path taken between those points.

Equivalently, the work done by a conservative force along any closed path is zero: ∮F⃗ · dr⃗ = 0. A force field is conservative if its curl is zero: ∇ × F⃗ = 0.

Given, F⃗ = (y² − x²)î + 2xyĵ + 0k̂.

Therefore, F_x = y² − x², F_y = 2xy, and F_z = 0.

The curl is ∇ × F⃗.

For the k̂ component, ∂F_y/∂x − ∂F_x/∂y = 2y − 2y = 0.

The other two components are also zero because F_z = 0 and there is no z-dependence.

Hence, ∇ × F⃗ = 0.

Therefore, the given force field is conservative.

4(a)(i). What do you mean by modulus of rigidity?

Ans: The modulus of rigidity, also known as the shear modulus and denoted by η, is defined as the ratio of tangential or shear stress to the corresponding shear strain within the elastic limit.

η = Shear Stress / Shear Strain

4(a)(ii). Obtain an expression for the work done per unit volume in shearing strain.

Ans: Consider a solid metallic cube of side L. Its volume is V = L³ and the area of each face is A = L².

Let the lower face be fixed rigidly and a tangential force F be applied to the upper face. The upper face is displaced laterally through a distance x.

For small deformation, tan θ ≈ θ = x/L.

The shear stress is τ = F/A, while the shear strain is θ = x/L.

Therefore, the modulus of rigidity is η = τ/θ = (F/A)/(x/L).

Hence, F = (ηA/L)x.

The work done is dW = F dx.

Therefore, W = ∫₀ˣ (ηA/L)x′ dx′ = ηAx²/(2L).

Since Fx = ηAx²/L, we obtain W = Fx/2.

Dividing by the volume V = AL,

W/V = (1/2)(F/A)(x/L)

Therefore, the work done per unit volume is

u = 1/2 × Shear Stress × Shear Strain

4(b). Obtain an expression for the kinetic energy of a body rolling on a plane surface. Calculate the acceleration of a solid sphere about its diameter rolling down a plane inclined at angle θ with the horizontal surface.

Ans: For a rigid body rolling without slipping, the motion consists of translational motion of its centre of mass and rotational motion about its centre of mass.

The translational kinetic energy is K_trans = 1/2 Mv².

The rotational kinetic energy is K_rot = 1/2 Iω².

Hence, K = 1/2 Mv² + 1/2 Iω².

If I = Mk² and the body rolls without slipping, v = ωR.

Therefore, K = 1/2 Mv²[1 + k²/R²] .

For a solid sphere rolling down an inclined plane, the equation of translational motion is Mg sin θ − f = Ma.

The rotational equation is fR = Iα.

For rolling without slipping, a = αR, hence α = a/R.

Therefore, f = Ia/R².

Substituting into the translational equation,

Mg sin θ − Ia/R² = Ma

Thus, a = g sin θ / [1 + I/(MR²)] .

For a solid sphere, I = 2MR²/5.

Therefore, a = g sin θ/[1 + 2/5] = 5g sin θ/7 .

Hence, the acceleration is a = (5/7)g sin θ.

5(a). Explain the mass-energy equivalence and obtain Einstein's mass-energy relation.

Ans: Einstein's mass-energy equivalence principle states that mass and energy are two forms of the same physical entity and can be converted into each other. Even an object at rest possesses rest energy.

The rest energy is E₀ = m₀c².

According to special relativity, m = m₀/√(1 − v²/c²).

Squaring and rearranging, m²c² − m²v² = m₀²c².

Differentiating, c²dm − m v dv − v²dm = 0 .

Hence, c²dm = m v dv + v²dm .

From the work-energy theorem, dE_k = mv dv + v²dm.

Therefore, dE_k = c²dm.

Integrating from rest mass m₀ to relativistic mass m,

E_k = ∫_m₀^mc²dm = c²(m − m₀) .

The total energy is E = m₀c² + E_k.

Therefore, E = m₀c² + mc² − m₀c².

Hence, Einstein's mass-energy relation is E = mc².

5(b). What is a damped harmonic oscillator? Obtain an expression for the general solution of a damped harmonic oscillator.

Ans: A damped harmonic oscillator is a system that oscillates under a restoring force and a resistive force that opposes motion. The damping force continuously dissipates mechanical energy, causing the amplitude to decrease with time.

The restoring force is F_r = −kx, while the damping force is F_d = −γ(dx/dt).

From Newton's second law, m(d²x/dt²) = −kx − γ(dx/dt) .

Therefore, d²x/dt² + (γ/m)(dx/dt) + (k/m)x = 0 .

Let γ/m = 2b and k/m = ω₀².

The equation becomes d²x/dt² + 2b(dx/dt) + ω₀²x = 0 .

Assume a trial solution x = Ae^αt.

Substitution gives the characteristic equation α² + 2bα + ω₀² = 0 .

Therefore, α = −b ± √(b² − ω₀²) .

Hence, the general solution is

x(t) = C₁e^{[−b + √(b² − ω₀²)]t} + C₂e^{[−b − √(b² − ω₀²)]t} .

The nature of the motion depends on whether b² − ω₀² is positive, zero, or negative, corresponding to overdamped, critically damped, or underdamped motion respectively.

5(c). What are the Galilean transformation equations in space and time? Show that the law of conservation of energy is invariant to Galilean transformation.

Ans: Let two inertial frames S and S′ be moving with relative velocity V along the positive x-axis.

The Galilean transformation equations are:

x′ = x − Vt
y′ = y
z′ = z
t′ = t

Consider an elastic collision between masses m₁ and m₂. Conservation of momentum in S gives

m₁u⃗₁ + m₂u⃗₂ = m₁v⃗₁ + m₂v⃗₂ .

Conservation of kinetic energy gives

1/2 m₁u₁² + 1/2 m₂u₂² = 1/2 m₁v₁² + 1/2 m₂v₂² .

Under Galilean transformation, u⃗ = u⃗′ + V⃗.

Substituting this relation into the energy equation and expanding the squares, the terms containing V² cancel from both sides.

The terms containing V⃗ also cancel because linear momentum is conserved: m₁u⃗₁′ + m₂u⃗₂′ = m₁v⃗₁′ + m₂v⃗₂′ .

Therefore,

1/2m₁u₁′² + 1/2m₂u₂′² = 1/2m₁v₁′² + 1/2m₂v₂′² .

Hence, conservation of energy is invariant under Galilean transformation.

5(d). Show that σ = (3K − 2η)/(6K + 2η), where σ is Poisson's ratio, K is bulk modulus and η is modulus of rigidity.

Ans: The standard relations are Y = 3K(1 − 2σ) and Y = 2η(1 + σ).

Equating them, 3K(1 − 2σ) = 2η(1 + σ) .

Expanding, 3K − 6Kσ = 2η + 2ησ .

Rearranging, 3K − 2η = 6Kσ + 2ησ .

Therefore, 3K − 2η = σ(6K + 2η) .

Hence, σ = (3K − 2η)/(6K + 2η) .

5(e). Show that the torsional rigidity of a hollow cylinder is greater than that of a solid cylinder if both cylinders have the same mass, density and length.

Ans: Consider a solid cylinder of radius R and length L. Its torsional rigidity is C_solid = πηR⁴/(2L).

Its volume is V_solid = πR²L.

For a hollow cylinder with inner radius R₁ and outer radius R₂,

C_hollow = πη(R₂⁴ − R₁⁴)/(2L)

and V_hollow = π(R₂² − R₁²)L .

Since mass, density and length are the same, their volumes are equal:

R² = R₂² − R₁² .

Therefore,

C_hollow/C_solid = (R₂⁴ − R₁⁴)/R⁴ .

Using the difference of squares, R₂⁴ − R₁⁴ = (R₂² − R₁²)(R₂² + R₁²) .

Hence, C_hollow/C_solid = (R₂² + R₁²)/R² .

Since R₂² = R² + R₁²,

C_hollow/C_solid = 1 + 2R₁²/R² .

Since R₁²/R² > 0, C_hollow/C_solid > 1.

Therefore, C_hollow > C_solid. Thus, the hollow cylinder has greater torsional rigidity.

6(a). State the postulates of Einstein's special theory of relativity.

Ans: Einstein's Special Theory of Relativity is based on two fundamental postulates:

1. Principle of Relativity: The laws of physics are the same in all inertial frames of reference. There is no preferred or absolute state of rest.

2. Constancy of the Speed of Light: The speed of light in vacuum has the same constant value c ≈ 3 × 10⁸ m/s in all inertial frames of reference, independent of the relative motion of the source or observer.

6(b). Write down Hooke's law of elasticity. Give the dimension and SI unit of modulus of elasticity.

Ans: Hooke's law states that, for small deformations within the elastic limit, stress is directly proportional to strain.

Stress = E × Strain, where E is the modulus of elasticity.

Since strain is dimensionless, the dimension of the modulus of elasticity is the same as that of stress:

[E] = [Force]/[Area] = [MLT⁻²]/[L²] = [ML⁻¹T⁻²] .

The SI unit is N/m², or Pascal (Pa).

6(c). Write Poiseuille's equation. Draw and explain the profile or velocity distribution curve of the advancing liquid through a tube.

Ans: Poiseuille's equation gives the volume of liquid flowing per second through a capillary tube under streamline flow:

V = πPr⁴/(8ηl)

where P is the pressure difference between the two ends, r is the inner radius of the tube, l is its length, and η is the coefficient of viscosity.

The velocity distribution of the liquid is given by

v = P(r² − x²)/(4ηl) .

At the wall of the tube, x = r, so v = 0. At the central axis, x = 0, so the velocity is maximum.

Therefore, the velocity distribution curve is parabolic. The velocity increases gradually from zero at the walls to a maximum value at the central axis.

7(a). Obtain the relativistic formula for the addition of velocities. Show that a photon moving with the velocity of light has an absolute constant velocity.

Ans: Consider two inertial frames S and S′, where S′ moves with velocity v relative to S.

The Lorentz transformations for differentials are

dx = γ(dx′ + vdt′)

dt = γ(dt′ + vdx′/c²)

where γ = 1/√(1 − v²/c²).

The velocity in S is u_x = dx/dt.

Therefore,

u_x = (dx′ + vdt′) / (dt′ + vdx′/c²) .

Dividing by dt′,

u_x = (u′_x + v) / (1 + vu′_x/c²) .

This is the relativistic velocity addition formula.

For a photon, u′_x = c.

Therefore,

u_x = (c + v)/(1 + v/c) = c .

Thus, the velocity of light remains c in every inertial frame, irrespective of the relative velocity of the observers.

7(b). What is simple harmonic motion? Derive the general differential equation of motion of a simple harmonic oscillator and obtain its solutions.

Ans: Simple Harmonic Motion (SHM) is a periodic oscillatory motion in which the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction.

Therefore, F = −kx.

From Newton's second law, F = m(d²x/dt²).

Hence, m(d²x/dt²) = −kx .

Therefore, d²x/dt² + (k/m)x = 0 .

Putting ω² = k/m, the differential equation becomes

d²x/dt² + ω²x = 0 .

Since v = dx/dt, we have v(dv/dx) = −ω²x.

Integrating, v² = ω²(A² − x²) .

Therefore, dx/√(A² − x²) = ωdt .

On integration, sin⁻¹(x/A) = ωt + φ .

Hence, the general solution is

x(t) = A sin(ωt + φ) .

An equivalent form is x(t) = A cos(ωt + δ).

7(c). Prove that 9/Y = 3/η + 1/K, where Y is Young's modulus, K is bulk modulus and η is modulus of rigidity.

Ans: The standard relations between elastic constants are

Y = 3K(1 − 2σ)

and

Y = 2η(1 + σ) .

From the first equation, Y/(3K) = 1 − 2σ .

Hence, 2σ = 1 − Y/(3K) .

From the second equation, σ = Y/(2η) − 1 .

Substituting, 2[Y/(2η) − 1] = 1 − Y/(3K) .

Therefore, Y/η − 2 = 1 − Y/(3K) .

Rearranging, Y/η + Y/(3K) = 3 .

Dividing by Y, 1/η + 1/(3K) = 3/Y .

Multiplying by 3, 3/η + 1/K = 9/Y .

Hence proved: 9/Y = 3/η + 1/K.