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

Question 1. Choose the correct answer from the following: (1 X 5 = 10 Marks)

(a) The Earth frame is

(i) inertial

(ii) non-inertial

(iii) both inertial and non-inertial

(iv) None of the above

Ans: (ii) non-inertial

The Earth rotates around its own axis and revolves around the Sun. Due to these rotational motions, it possesses centripetal acceleration and gives rise to Coriolis forces, making it a non-inertial frame of reference.

(b) For a non-conservative system, force is

(i) negative gradient of potential function

(ii) positive gradient of potential function

(iii) both negative and positive gradients of potential function

(iv) None of the above

Ans: (iv) None of the above

The force is expressed as the negative gradient of a scalar potential function F⃗ = −∇U strictly for conservative systems. For non-conservative systems, forces like friction are path-dependent and cannot be derived from a scalar potential.

(c) If ω = 1 rad/sec, then the relation between moment of inertia and kinetic energy is

(i) I = 3E

(ii) I = 2E

(iii) I = 3E/2

(iv) None of the above

Ans: (ii) I = 2E

The rotational kinetic energy (E) is given by E = ½Iω². Substituting ω = 1 rad/sec gives E = ½I, which simplifies to I = 2E.

(d) Torsional couple per unit angular twist is

(i) C = πηr³/(2l)

(ii) C = πηr⁴/(3l)

(iii) C = πηr⁴/(2l)

(iv) None of the above

Ans: (iii) C = πηr⁴/(2l)

This is the standard formula for the restoring torsional couple per unit angle of twist for a solid cylinder or wire of length l, radius r, and modulus of rigidity η.

(e) Poisson's ratio cannot have the value

(i) 0.7

(ii) 0.2

(iii) 0.5

(iv) None of the above

Ans: (i) 0.7

Theoretically, the value of Poisson's ratio (σ) for a stable, isotropic material must lie in the range −1 ≤ σ ≤ 0.5. Therefore, it cannot have a value of 0.7.

Question 2. Answer the following questions: (2 X 6 = 12 Marks)

(a) Distinguish inertial and non-inertial frames of reference.

Ans:

(b) Define stable equilibrium and unstable equilibrium.

Ans: Stable Equilibrium: A system is in stable equilibrium if, when displaced slightly from its equilibrium position, a net restoring force arises that pushes the system back toward the equilibrium position. The potential energy of the system is at a local minimum.

Unstable Equilibrium: A system is in unstable equilibrium if a slight displacement produces a net force that pushes the system further away from the equilibrium position. The potential energy of the system is at a local maximum.

(c) Write the physical significance of moment of inertia.

Ans: The moment of inertia plays the same role in rotational motion as mass does in translational motion. Its physical significance is that it measures the rotational inertia of a rigid body—meaning it represents the resistance of the body to any change in its state of rotational motion. A larger moment of inertia means it is harder to start or stop the body's rotation.

(d) Define power dissipation and quality factor.

Ans: Power Dissipation: In a damped harmonic oscillator, power dissipation is the rate at which the mechanical energy of the system is lost, usually as heat, due to resistive or damping forces. It is given by P = −bv², where b is the damping coefficient and v is the velocity.

Quality Factor (Q-factor): It is a dimensionless parameter that describes how under-damped an oscillator or resonator is. It characterizes the sharpness of resonance. Mathematically,

Q = 2π × (Energy Stored / Energy Dissipated per cycle)

(e) Define Coriolis force and mention one application of it.

Ans: Definition: The Coriolis force is a fictitious (pseudo) force that appears to act on a mass when its motion is observed from a rotating frame of reference. For a particle of mass m moving with velocity v⃗ in a frame rotating with angular velocity ω⃗,

F⃗_c = −2m(ω⃗ × v⃗)

Application/Effect: It is responsible for the rotation of the plane of oscillation of a Foucault pendulum, which serves as a direct experimental proof of Earth's rotation. It also explains the deflection of trade winds and the rotational direction of cyclones.

(f) Write the postulates of special theory of relativity.

Ans: Principle of Relativity: The laws of physics are identical in all inertial frames of reference. There is no preferred or "absolute" inertial frame.

Constancy of Speed of Light: The speed of light in a vacuum (c) is the same for all observers in inertial frames, regardless of the motion of the light source or the observer.

Question 3.

(a) State and prove work-energy theorem. (1 + 3 = 4 Marks)

Ans: Principle: The Work-Energy Theorem states that the net work done by all the forces acting on a particle is equal to the change in the kinetic energy of the particle.

Derivation: Consider a particle of mass m moving with an initial velocity u. A net force F acts on it, changing its velocity to v.

The small amount of work done dW for a small displacement dx is:

dW = F dx

From Newton's Second Law, F = m(dv/dt).

Therefore, dW = m(dv/dt)dx

Since dx/dt = v,

dW = mv dv

Integrating both sides from initial velocity u to final velocity v:

∫dW = ∫_u^v mv dv

W = m[v²/2]_u^v

W = ½mv² − ½mu²

Result:

W = K_f − K_i = ΔK

Thus, the net work done equals the change in kinetic energy.

(b) Write the difference between conservative and non-conservative forces. Discuss work done by a non-conservative force. (2 + 3 = 5 Marks)

Ans:

Work done by a non-conservative force: When a non-conservative force such as kinetic friction or air resistance acts on a system, it removes mechanical energy from the system, usually transforming it into thermal energy or sound.

The work-energy theorem for a system experiencing both conservative (W_c) and non-conservative (W_nc) forces is:

W_net = W_c + W_nc = ΔK

Since the work done by conservative forces is the negative change in potential energy,

W_c = −ΔU

Therefore,

−ΔU + W_nc = ΔK

W_nc = ΔK + ΔU = ΔE_mech

Thus, the work done by a non-conservative force equals the change in the total mechanical energy of the system.

Or Calculate the moment of inertia of a rectangular lamina about an axis passing through its centre perpendicularly. (5 Marks)

Ans: Principle: The moment of inertia of a 2D planar object (lamina) about an axis perpendicular to its plane can be determined using the Perpendicular Axis Theorem: I_z = I_x + I_y, where x and y are mutually perpendicular axes in the plane of the lamina passing through the same origin.

Derivation: Let the rectangular lamina have mass M, length a along the x-axis, and width b along the y-axis. Let the origin be at its centre.

The area of the lamina is A = ab, and its uniform mass surface density is:

σ = M/(ab)

Consider an infinitesimal area element dx dy at coordinates (x, y). The mass of this element is:

dm = σ dx dy

Moment of inertia about the x-axis:

I_x = ∫y² dm

I_x = σ(∫_−a/2^a/2 dx) (∫_−b/2^b/2 y²dy)

I_x = σa(b³/12)

I_x = Mb²/12

Moment of inertia about the y-axis:

I_y = Ma²/12

By the perpendicular axis theorem:

I_z = I_x + I_y

I_z = Mb²/12 + Ma²/12

I_z = M(a² + b²)/12

(c) Calculate the extension of a steel wire of length 4 m and diameter 2 mm when loaded with a weight of 8 kg. Young's modulus of steel is 2 × 10¹² dyne/cm². (4 Marks)

Ans: Given:

Length of wire, L = 4 m = 400 cm

Diameter of wire, D = 2 mm = 0.2 cm

Radius of wire, r = D/2 = 0.1 cm

Mass of the load, m = 8 kg = 8000 g

Acceleration due to gravity, g = 980 cm/s²

Young's modulus, Y = 2 × 10¹² dyne/cm²

Formula:

Y = Stress/Strain = (F/A)/(ΔL/L)

Y = FL/(πr²ΔL)

Therefore,

ΔL = mgL/(πr²Y)

Substitution:

ΔL = (8000 × 980 × 400) / (π × (0.1)² × 2 × 10¹²)

ΔL = (3.136 × 10⁹) / (π × 0.01 × 2 × 10¹²)

ΔL = (3.136 × 10⁹) / (6.28318 × 10¹⁰)

ΔL ≈ 0.0499 cm

Final Answer: The extension of the steel wire is 0.0499 cm, or roughly 0.5 mm.

Or Establish a relation among Young's modulus (Y), bulk modulus (K) and Poisson ratio (σ). (4 Marks )

Ans: Principle: When a uniform, isotropic cube is subjected to a uniform hydrostatic stress on all its faces, the total volumetric strain is the linear combination of the strains along the three mutually perpendicular axes, factoring in both longitudinal extensions and lateral contractions described by Poisson's ratio.

Consider a cube of unit volume. Apply a uniform outward normal stress P on all six faces along the x, y, and z directions.

The stress P acting in the x-direction produces a longitudinal extension strain in the x-direction equal to P/Y.

The stress P in the y-direction produces a lateral contraction strain in the x-direction equal to −σP/Y.

The stress P in the z-direction also produces a lateral contraction strain in the x-direction equal to −σP/Y.

Therefore, the net linear strain along the x-direction is:

e_x = P/Y − σP/Y − σP/Y

e_x = (P/Y)(1 − 2σ)

By symmetry:

e_y = e_z = (P/Y)(1 − 2σ)

Total volumetric strain:

ΔV/V = e_x + e_y + e_z

ΔV/V = 3(P/Y)(1 − 2σ)

By definition of bulk modulus:

K = P/(ΔV/V)

Therefore,

K = Y/[3(1 − 2σ)]

Result:

Y = 3K(1 − 2σ)

Question 4.

(a) Explain damped oscillation with its general solution and find the frequency of damped oscillation. (6 Marks)

Ans: Explanation of Damped Oscillation: A damped oscillation is a harmonic motion in which the amplitude continuously decreases over time due to dissipative forces such as friction or air drag.

The restoring force is proportional to displacement (−kx), and the damping force is proportional to velocity (−bv).

From Newton's second law:

m(d²x/dt²) = −kx − b(dx/dt)

m(d²x/dt²) + b(dx/dt) + kx = 0

Dividing by m:

d²x/dt² + 2γ(dx/dt) + ω₀²x = 0

where 2γ = b/m is the damping factor and ω₀ = √(k/m) is the natural angular frequency.

For an under-damped system (ω₀ > γ), the roots of the auxiliary equation are complex, giving the general solution:

x(t) = A₀e^−γt cos(ω_dt + φ)

Here, A₀e^−γt represents the exponentially decaying amplitude, and φ is the initial phase.

Frequency of Damped Oscillation:

ω_d = √(ω₀² − γ²)

Therefore,

ω_d = √[k/m − (b/2m)²]

The ordinary frequency is:

f_d = (1/2π)√[k/m − b²/(4m²)]

Or Deduce the differential equation of simple harmonic motion and solve it. (6 Marks)

Ans: Differential Equation of SHM: Simple Harmonic Motion occurs when the restoring force acting on a particle is directly proportional to its displacement from the mean position and is directed toward the mean position.

F = −kx

Using Newton's Second Law:

m(d²x/dt²) = −kx

d²x/dt² + (k/m)x = 0

Let ω² = k/m. Then:

d²x/dt² + ω²x = 0

The auxiliary equation is:

r² + ω² = 0

Therefore, r = ±iω

The general solution is:

x(t) = C₁e^iωt + C₂e^−iωt

Using Euler's formula: e^±iθ = cos θ ± i sin θ

x(t) = A cos(ωt) + B sin(ωt)

In amplitude-phase form:

x(t) = X_m sin(ωt + φ)

Alternatively, x(t) = X_m cos(ωt + φ) may also be used depending on the choice of phase constant.

(b) What are uniformly rotating and non-uniformly rotating frames of reference? Deduce the following operator: d/dt = d′/dt + ωx
(2 + 4 = 6 Marks)

Ans: Uniformly Rotating Frame: A frame of reference that rotates with a constant angular velocity ω⃗ = constant. There is no angular acceleration.

Non-Uniformly Rotating Frame: A frame of reference whose angular velocity changes with time: dω⃗/dt ≠ 0. It possesses angular acceleration.

Deduction of the Operator:

Let S be a fixed frame with orthogonal unit vectors î, ĵ, k̂. Let S′ be a frame rotating with angular velocity ω⃗, sharing the same origin, with unit vectors î′, ĵ′, k̂′.

An arbitrary vector A⃗ can be expressed in the rotating frame as:

A⃗ = A_x′î′ + A_y′ĵ′ + A_z′k̂′

Taking the time derivative with respect to the fixed frame:

dA⃗/dt = d′A⃗/dt + ω⃗ × A⃗

This follows because the unit vectors of the rotating frame satisfy:

dî′/dt = ω⃗ × î′

dĵ′/dt = ω⃗ × ĵ′

dk̂′/dt = ω⃗ × k̂′

Therefore,

dA⃗/dt = d′A⃗/dt + ω⃗ × A⃗

Since this applies to any arbitrary vector A⃗, the operator relation is:

d/dt = d′/dt + ω⃗ ×

(c) Discuss the difference between streamline and turbulent flow with proper diagram. (4 Marks)

Ans:

Diagram:

Question 5.

(a) "A moving clock runs slower." Justify the statement with the help of special theory of relativity. (4 Marks)

Ans: This statement refers to Time Dilation, a direct consequence of the postulates of the Special Theory of Relativity.

Consider a "light clock" consisting of two parallel mirrors separated by a distance L. A photon bounces back and forth between them.

In the rest frame of the clock, the time interval for one round trip, the proper time Δt₀, is:

Δt₀ = 2L/c

Now consider an observer in a different inertial frame observing the clock moving horizontally at a constant velocity v. To this observer, the photon travels along a diagonal path because the mirrors are moving horizontally while the photon is in transit.

Since the speed of light c is the same for all inertial observers and the diagonal path is longer than the vertical path, the observed time interval Δt is longer.

Mathematically:

Δt = Δt₀ / √(1 − v²/c²) = γΔt₀

Because γ > 1 for any velocity v > 0, Δt > Δt₀.

Thus, a moving clock is observed to tick less frequently. Hence, "a moving clock runs slower."

(b) Deduce the relativistic transformation of velocities. (5 Marks)

Ans: Principle: The relativistic velocity transformations are derived by taking the differentials of the Lorentz coordinate transformation equations and dividing the spatial differentials by the time differential.

Let a frame S′ move with a constant velocity v in the positive x-direction relative to frame S.

The Lorentz transformations are:

x′ = γ(x − vt)

y′ = y

z′ = z

t′ = γ(t − vx/c²)

Taking differentials:

dx′ = γ(dx − v dt)

dy′ = dy

dz′ = dz

dt′ = γ(dt − v dx/c²)

For the x-component:

u′_x = dx′/dt′

u′_x = (u_x − v) / (1 − u_xv/c²)

For the y-component:

u′_y = u_y / [γ(1 − u_xv/c²)]

Similarly,

u′_z = u_z / [γ(1 − u_xv/c²)]

Result:

u′_x = (u_x − v) / (1 − u_xv/c²)

u′_y = u_y / [γ(1 − u_xv/c²)]

u′_z = u_z / [γ(1 − u_xv/c²)]

Or Find the expression for relativistic kinetic energy of a particle.

Ans: Principle: Kinetic energy is the work done on a particle to accelerate it from rest to a velocity v. In relativity, the force is the rate of change of relativistic momentum.

Relativistic momentum is:

p = mv = m₀v/√(1 − v²/c²)

Kinetic energy is:

K = ∫F dx

Since F = dp/dt and dx = v dt,

K = ∫v dp

Using integration by parts:

∫v dp = vp − ∫p dv

After substitution and integration:

K = m₀c²/√(1 − v²/c²) − m₀c²

Therefore,

K = (γ − 1)m₀c²

where γ = 1/√(1 − v²/c²) .

(c) Obtain a relation between the mass and the velocity of a particle moving with relativistic velocity. (5 Marks)

Ans: Principle: The relation can be obtained by applying the law of conservation of momentum to an inelastic collision between two identical particles, viewed from two different inertial reference frames.

Consider two inertial frames S and S′. Frame S′ moves with velocity V in the +x direction.

In S′, two identical particles, each with rest mass m₀, move toward each other with equal speeds u′ and −u′.

They collide entirely inelastically and form a stationary object of mass M₀ in S′.

Choose V = u′. From frame S, particle B has zero velocity and mass m₀, while particle A has velocity:

v = (V + V)/(1 + V²/c²)

Conservation of momentum gives:

mv = MV

Hence,

M = mv/V

Applying conservation of mass-energy:

m + m₀ = M

Therefore,

m = m₀(1 + V²/c²)/(1 − V²/c²)

Also,

√(1 − v²/c²) = (1 − V²/c²)/(1 + V²/c²)

Therefore, the mass-velocity relation becomes:

m = m₀/√(1 − v²/c²)

Thus, as v approaches c, the relativistic mass m increases toward infinity.

Question 6.

(a) The displacement equation of simple harmonic motion is y = 10 sin(10t − π/6) m. Then, the maximum velocity is

(i) 10 m/sec

(ii) 100 m/sec

(iii) 10³ m/sec

(iv) None of the above

Ans: (ii) 100 m/sec

Velocity is the time derivative of displacement:

v = dy/dt = 10 × 10 cos(10t − π/6)

v = 100 cos(10t − π/6)

The maximum value of the cosine function is 1. Therefore,

v_max = 100 m/sec

(b) Which of the following is variant in Galilean transformation?

(i) Velocity

(ii) Length

(iii) Acceleration

(iv) None of the above

Ans: (i) Velocity

In a Galilean transformation between a stationary frame and one moving with velocity V:

v⃗′ = v⃗ − V⃗

Therefore velocity changes. However, space intervals (length) and acceleration remain invariant:

a⃗′ = a⃗

Question 7.

(a) What is elasticity? What is the cause of it?

Ans: Elasticity: Elasticity is the physical property of a material by virtue of which it resists deformation, changes in size or shape, when under the influence of an external force and successfully regains its original dimensions immediately after the deforming force is removed.

Cause: The fundamental cause of elasticity is the interatomic or intermolecular forces within the solid. In an undeformed state, atoms remain at an equilibrium distance where net forces are zero. When a deforming force is applied, the atoms are either pulled apart or pushed together. This creates an internal electrostatic restoring force that attempts to pull or push the atoms back to their stable equilibrium positions.

(b) Explain briefly the law of conservation of angular momentum.

Ans: The law 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.

Torque is the rate of change of angular momentum:

τ⃗_ext = dL⃗/dt

If τ⃗_ext = 0, then:

dL⃗/dt = 0

Therefore,

L⃗ = constant

Since L⃗ = Iω⃗, if a system's moment of inertia changes internally without external torque, its angular velocity must change so that angular momentum remains constant.

Question 8.

(a) Derive Poiseuille's equation for the flow of liquid through a capillary tube.

Ans: Principle: For a steady, streamlined flow of an incompressible viscous liquid through a horizontal capillary tube, the forward force driving the fluid due to a pressure difference is exactly balanced by the backward viscous drag force between adjacent fluid layers.

Let a liquid of viscosity η flow through a horizontal capillary tube of length l and uniform radius R. Let P be the pressure difference across the two ends of the tube.

Consider a coaxial liquid cylinder of radius r and length l.

The driving force due to pressure difference is:

F_P = Pπr²

According to Newton's law of viscosity, the opposing viscous force is:

F_v = −η(2πrl)(dv/dr)

For steady flow:

Pπr² = −η(2πrl)(dv/dr)

Therefore,

dv = −(P/2ηl)r dr

Using the boundary condition r = R, v = 0:

v = P(R² − r²)/(4ηl)

The volume of liquid flowing per second through a cylindrical shell is:

dQ = v(2πr dr)

dQ = [P/(4ηl)](R² − r²)(2πr dr)

Integrating from r = 0 to r = R:

Q = ∫₀ᴿ [πP/(2ηl)](R²r − r³)dr

Q = πPR⁴/(8ηl)

Result — Poiseuille's equation:

Q = πPR⁴/(8ηl)

Or Explain relativistic Doppler effect.

Ans: The Doppler effect is the apparent change in the frequency of a wave caused by the relative motion between the source and observer.

For light in a vacuum, there is no medium. The relativistic Doppler effect therefore depends on the relative velocity between source and observer and incorporates Special Relativity, particularly time dilation.

When a light source moves relative to an observer, wave crests become compressed when the source approaches and spread out when the source recedes. Time dilation also affects the frequency emitted by the moving source.

For a source emitting light of proper frequency f₀ and moving directly toward the observer with velocity v:

f = f₀√[(1 + v/c)/(1 − v/c)]

This produces a blueshift. If the source is receding, the signs are reversed and a redshift occurs.

For the transverse Doppler effect:

f = f₀/γ = f₀√(1 − v²/c²)

(b) Find the moment of inertia of a solid cylinder about an axis passing through its centre parallel to it.

Ans: The moment of inertia of a continuous volume can be found by dividing the body into infinitesimally thin coaxial cylindrical shells.

Let the solid cylinder have mass M, radius R and length L.

Its uniform volume density is:

ρ = M/(πR²L)

Consider an elementary coaxial cylindrical shell of radius r, thickness dr and length L.

dV = 2πrL dr

Therefore,

dm = ρ(2πrL dr)

Its moment of inertia is:

dI = dm r² = 2πρLr³ dr

Integrating from r = 0 to r = R:

I = ∫₀ᴿ 2πρLr³ dr

I = 2πρL(R⁴/4)

Since M = ρπR²L,

I = ½MR²

Result: The moment of inertia of the solid cylinder about its central longitudinal axis is I = ½MR².

(c) What is time dilation? Explain briefly.

Ans: Definition: Time dilation is a relativistic phenomenon where a time interval measured by an observer moving relative to a clock is longer than the time interval measured by an observer at rest with respect to the clock.

According to Einstein's Special Theory of Relativity, the speed of light is constant for all observers. Therefore, time is not absolute.

If an event takes a proper time Δt₀ in its rest frame, an observer moving with relative velocity v measures a dilated time interval Δt:

Δt = Δt₀/√(1 − v²/c²)

Since the denominator is less than 1 for v > 0, Δt > Δt₀.

Thus, moving clocks appear to tick slower relative to a stationary observer. This effect also influences the observed decay rates of high-speed subatomic particles such as muons.