IIT JAM Physics MCQ Test 5 - Physics MCQ

divide the semicircular arc into small elements dq, each subtending a small angle dθ at the center.
dq = λrdθ (since dq = λ⋅arc length,arc length = rdθ).
the electric field due to the charge element dqdqdq at the center is given by coulomb's law:

1. Components of dE:
   Since the arc is symmetric about the vertical axis:

   Vertical component:

   

2. Total Electric Field:
   Integrate dE_y​ over the semicircular arc (θ ranges from −π/2 to π/2):

   

   Since cos⁡θ is an even function, the integral simplifies:

   

3. Evaluate the Integral:

   

   Substituting this:

   

4. The electric field is directed vertically downward due to the symmetry of the setup.
   Final Answer: 

The interaction energy between two dipoles p₁ and p₂ separated by a distance r is given by:

where is the unit vector along the line joining the two dipoles, and r is the distance between them.

• The two dipoles p_1​ and p_2​ are aligned along the +z - direction.
• Before moving:
  r = (2,0,0)
  Distance r = 2 units
• After moving:
  r = (0,4, 0)
  Distance r = 4 units

Since the dipoles are aligned in the +z - direction:

p₁⋅p₂ = p₁p₂

p1 ⋅  = 0 (no component of p₁ along ).

The interaction energy simplifies to:

1. Before Moving:

   • r = 2r
   • 

2. After Moving:

   • r = 4
   • 

3. Ratio:

   

Final Answer: B: 1 : 8

The spherical shell has a uniform charge density σ. When a small spherical piece (radius much smaller than the radius of the shell) is removed, the electric field at the midpoint of the aperture can be calculated using the principle of superposition:

1. The electric field due to the complete spherical shell (before the piece is removed).
2. The electric field due to the "removed" spherical piece (considered as a sphere with charge −σ).

• Inside a spherical shell, the electric field is zero due to symmetry.
• Therefore, the field at the midpoint of the aperture (before removing the small piece) is zero.

When the small spherical piece is removed:

• The removed piece is considered to have a uniform charge density σ\sigmaσ, and its charge generates an electric field.
• At the midpoint of the aperture, the field due to this removed spherical piece can be calculated using the expression for the electric field near the surface of a uniformly charged shell:
  E  =  σ / 2ε₀

The direction of the electric field is away from the removed piece because the removed piece effectively has a charge opposite to the rest of the shell.

Since the field inside the shell due to the remaining portion of the shell is zero, the net field at the midpoint of the aperture is solely due to the removed piece:

E = σ / 2ε₀

Gauss's law states:

Φ_net = q_enclosed / ε₀

where:

• Φ_net​ is the total electric flux,
• q_enclosed​ is the total charge enclosed within the Gaussian surface (the cubic box here),
• ε₀​ is the permittivity of free space.

In this problem, the cubic box is placed in a uniform electric field. There is no mention of any charge inside the box, so:

q_enclosed = 0

For a uniform electric field, the flux entering the box from one side is equal and opposite to the flux leaving the box from the opposite side. This means the total flux through the surface of the box is:

Φ_net = 0

This conclusion aligns with Gauss's Law, as there is no charge enclosed within the box.

The force on q is due to the two charges Q. Since q is equidistant from both charges, the magnitudes of the forces due to Q on q are equal, and they are directed oppositely. These forces cancel each other out, so the net force on q is zero.

Each charge Q experiences a repulsive force due to the other charge Q, and an attractive force due to the charge q. For equilibrium:

F_repulsion = F_attraction

1. Repulsive Force Between Q and Q:

   

2. Attractive Force Between Q and q:

   F_attraction = kQ_q / d²

Equating the repulsive and attractive forces:

Simplify:

Q / 4 = q

Thus:

q = −Q / 4

The negative sign indicates that q must have an opposite charge to Q for the forces to balance.

The correct option is D: Q₁ = Q/2, Q₂ = Q/2

Q₁ + Q₂ = Q (i) and F = k * Q₁ * Q₂ / r² (ii)

From (i) and (ii):
F = Q₁ * Q₂ = QF = k * Q₁ * (Q - Q₁) / r²

For F to be maximum:
dF/dQ₁ = 0 ⇒ Q₁ = Q₂ = Q/2

Laplace's equation is given by:

∇²ϕ = 0

where:

The potential is:

ϕ = q(x² + y² + z²)^−1/2

Let:

r² = x² + y² + z² so that ϕ = qr⁻¹

Using the formula for the Laplacian in spherical coordinates (since ϕ depends only on r):

First, calculate ∂ϕ / ∂r:

Now calculate r²∂ϕ / ∂r:

Take the derivative with respect to r:

Thus:

∇²ϕ = 0

Since ∇²ϕ = 0, the given potential satisfies Laplace's equation.
Final Answer: A:Yes

The capacitor is divided into three regions, each filled with dielectric materials having constants k_1​, k₂, and k₃. The goal is to replace these dielectrics with a single dielectric constant k that gives the same capacitance.

The capacitance for each region is given by:

where i represents each region, A is the area, and d_i is the thickness of the dielectric in the region.

Regions with k₁​ and k₂​ are in parallel. The equivalent capacitance of C₁​ and C₂ is in series with the capacitance C₃​.

The combined capacitance is:

Simplifying:

The total equivalent capacitance is:

Substituting values:

Simplify:

For a single dielectric material, the capacitance is:

Substituting for C in terms of k:

Final Answer: 

The capacitance of a parallel plate capacitor filled with a uniform dielectric is given by:

where:

• ε₀ is the permittivity of free space,
• K is the dielectric constant,
• A is the area of the plates,
• d is the separation between the plates.

The figure shows the two dielectric slabs arranged in series within the capacitor. This means:

• Each dielectric slab contributes to the overall capacitance as part of a series combination.
• The total separation d is divided equally, so:
  • Thickness of each slab is d₁ = d₂ = d/2

1. For the slab with dielectric constant K₁:

   

2. For the slab with dielectric constant K2K_2K2​:

   

For capacitors in series, the equivalent capacitance C_eq​ is given by:

Substitute C₁​ and C_2​:

Simplify:

Combine terms:

Thus:

The original capacitance C (without dielectrics) is:

Substitute this into C_eq​:

The magnetic field is given as , pointing into the page (negative z-direction).

The electron is moving initially with velocity along the x-axis.

The magnetic force acting on a charged particle is:

F = q(v × B)

For an electron (q = −e):

F = −e(v × B)

Substituting  and :

Using the cross-product rule:

The force is along the negative y-direction, causing the electron to move in a circular trajectory.

In a uniform magnetic field, the charged particle undergoes circular motion due to the Lorentz force, with the centripetal force provided by the magnetic force:

Equating the forces:

Solve for the radius r:

The electron moves in a circular trajectory in the x-y plane.

• The electron enters the magnetic field at y = 0.
• Due to the negative charge and the magnetic field direction, the electron curves downward (y < 0).
• The speed of the electron remains constant (v = u) because the magnetic force does no work (it is always perpendicular to the velocity).

The electron exits the region of the magnetic field with:

• Speed: v = u

Position: y < 0.
Final Answer: C: v = u; y < 0

For a circular loop of radius r carrying a current I, the magnetic field at the center of the loop is given by:

This applies to a full circular loop. If only a fraction of the loop is present, the contribution to the magnetic field will be proportional to the fraction of the loop.

The conductor consists of:

1. A full circular loop of radius r, carrying current I, contributing to the magnetic field.
2. Straight portions of the wire, which extend radially outward from the center of the loop.

The straight portions of the conductor do not contribute to the magnetic field at the center of the loop. This is because the magnetic field due to a straight conductor is zero along the line of the conductor at the point of interest.

Thus, the magnetic field at the center of the loop is entirely due to the circular portion.

Using the right-hand rule:

• Curl your fingers in the direction of the current in the loop.
• The thumb points into the page at the center of the loop.

Thus, the magnetic field is directed into the page.

For a bent conductor that forms a loop, the effective magnetic field accounts for the current distribution. Using the given formula:

Here:

The factor (1 − 1/π) represents the fractional contribution of the current configuration to the magnetic field at the center.
Final Answer: 

1. The total current III in the circuit is given by Ohm's Law:

   I = V / R

2. The coil is divided into two semicircles carrying currents I₁ and I₂​, respectively, due to their symmetrical nature.

   • For the upper semicircle:

     

   • For the lower semicircle:

     

 Since B₁ and B₂​ are equal in magnitude but opposite in direction:

B = B₁ − B₂ = 0.
Final Answer: 0

1. Relation between B and H: The magnetic field B is related to the magnetizing field H by:

   B = μ₀H

   where μ₀ is the permeability of free space.

2. Dimensions of B: The magnetic field B has dimensions:

   [B] = Force per unit charge per unit velocity = 

3. Dimensions of μ₀​: The permeability of free space μ₀​ has dimensions:

   

   From the electromagnetic relation:

   

   Substituting the dimensional analysis:

   [μ₀] = M¹L¹Q⁻²

Dimensions of H: Using B = μ₀H, rearranging gives:

[H] = [B] / [μ₀]

Substituting the dimensions:

Final Answer: L⁻¹T⁻¹Q¹
Correct Option: A.

A magnetic needle is a dipole, consisting of a north and a south pole separated by a small distance. In a non-uniform magnetic field, the forces on the two poles differ in magnitude because the field strength varies across the dipole.

This difference in force creates a net force on the magnetic needle, causing it to move toward the region of higher magnetic field strength.

In addition to the net force, a torque is exerted on the dipole if the magnetic needle is not aligned with the direction of the magnetic field. This torque tries to align the dipole moment of the magnetic needle with the magnetic field.

In a non-uniform magnetic field, the magnetic needle:

1. Experiences a force due to the variation in field strength.
2. Experiences a torque if it is not aligned with the magnetic field direction.

The work done by the electric field is equal to the change in kinetic energy of the particle. Using the work-energy theorem:

Rearranging:

Solving for E:

The instantaneous rate of work done by the electric field at P is:

Substituting:

Using the value of E:

 At Q, the velocity v⃗ is perpendicular to E⃗(θ = 90^∘):

The work done by the magnetic field is always zero because the force due to the magnetic field is perpendicular to the velocity of the particle:

W_B = 0.

Given Data

• Resistance (R) = 120 Ω
• Angular resonance frequency (ω_0​) = 4 × 10⁵ rad/s,
• Voltage across R = 60 V
• Voltage across L = 40 V

At resonance:

1. The inductive reactance X_L = ω₀L,
2. The capacitive reactance X_C = 1 / ω₀C,
3. X_L = X_C and the impedance is purely resistive Z = R.

The voltage across the inductance is related to the inductive reactance:

V_L = I X_L

At resonance:

Substitute into V_L = Iω₀L:

40 = 0.5⋅4 × 10⁵⋅L

Thus, Statement A is correct.

At resonance:

Substitute ω₀ = 4×  10⁵ rad/s and L = 0.2 mH = 0.2×10⁻³ H:

Simplify:

Thus, Statement B is incorrect because it refers to m F instead of μF.

At a frequency where the current lags the voltage by 45^∘, the condition is:

X_L − X_C = R

The angular frequency is:

Substitute

Simplify:

Thus, Statement C is approximately correct for 8 × 10⁵ rad/s.

The electric field along the z-axis due to a uniformly charged ring of radius R and total charge Q at a point (0,0,z) is:

This field is directed towards the xy-plane (negative z-direction) if z > 0.

The negatively charged particle P (charge −q) experiences a force:

The force is directed along the z-axis and depends on the position z.

1. For Small Displacements (z₀≪R): When z₀≪R, the electric field can be approximated as:

   E_z ∝ z

   which results in a restoring force proportional to z. This implies approximately simple harmonic motion (SHM) for small initial displacements (z₀≪R).

   Thus, Statement C is correct.

2. For Arbitrary Values of z₀​: The electric field is not linear for z₀∼R or z₀ > R. However, the force remains symmetric about the xy-plane, and the particle oscillates periodically about z = 0. The motion remains periodic but is not SHM.

   Thus, Statement A is correct.

• Statement B (Simple Harmonic for All z₀​):
  This is incorrect because the force is not proportional to z for large displacements (z₀∼R).

• Statement D (P crosses 0 and moves indefinitely):
  This is incorrect because the particle is bound to the electric field of the ring and cannot escape.

Final Answer: A,C

Given Data

• Resistance (R) = 300 Ω
• Inductance (L) = 60 mH = 60 × 10⁻³ H
• Capacitance (C) = 0.50 μF = 0.50 × 10⁻⁶ F
• Angular frequency (ω) = 10000 rad/s1
• Voltage (V) = 50 V

The inductive reactance is given by:

X_L = ωL

Substitute ω = 10000 rad/s and L= 60 × 10⁻³ H:

X_L = 10000 × 60 × 10⁻³ = 600 Ω

Thus, Statement A is correct.

The capacitive reactance is given by:

X_C = 1/ωC

Substitute ω = 10000 rad/s and C = 0.50 × 10⁻⁶ F:

​=200Ω.

Thus, Statement B is correct.

The impedance of a series RLC circuit is given by:

Substitute R = 300 Ω, X_L= 600 Ω, and X_C = 200 Ω:

Thus, Statement C is correct.

The phase angle is given by:

Substitute X_L = 600 Ω, X_C = 200 Ω, and R=300 Ω:

The angle is approximately:

ϕ = 53^∘

Thus, Statement D is correct.

From the wave equation:

E = 4sin⁡(2π⋅10⁷t−0.8x) a_y,

• The amplitude of the electric field (E₀) is 4 V/m
• The angular frequency (ω) is 2π⋅10⁷ rad/s
• The wavenumber (β) is 0.8 rad/m

For a nonmagnetic medium (μ_r = 1):

The intrinsic impedance of free space is:

η₀ = 120π Ω ≈ 377 Ω.

In the medium, with ε_r = 14.59

Thus, Statement A is correct.

The time-average power density of an electromagnetic wave is given by:

Substitute E₀ = 4 V / m and η = 98.7 Ω:

= 81μW/cm².

Thus, the time-average power density is 81 μW/cm², and Statement B is correct.

For a plane with area A = 100 cm² =0.01 m², the total power crossing is:

P_total = P_avg⋅A

Substitute P_avg = 81 μW/cm² = 8100 μW/m² and A = 100 cm² = 0.01 m²:

P_total = 81 × 100 = 724.4 μW

Thus, Statement C is correct.

If the area or medium differs, the power might vary, but based on the given details, Statement D is not consistent with the results.

Figure is a diagram of the step with q₁ = 2 μC, q₂ = −3 μC, and q₃ = −5 μC. The force on q₂​ is the vector sum of two contributions, the attractive force due to q_1​ (toward q₁​) and the repulsive force due to q_3​ (also toward q₁​). The sum of these two forces, taken algebraically since they are along the same line, is:

F = F₁ + F₂

Substituting values (k = 9 × 10⁹):

Simplify:

F = −9 × 10⁹[37.5×10⁻¹²]

F =  −0.55 N

The negative sign indicates the direction is to the left.
Final Answer: 0.55 N or 0.55 N to the left

All the capacitors are in parallel and P.D. across each is 200 V. 
∴ O = CV = 25 × 10^-6 × 200 = 5 × 10^-3 C

The initial energy of the system is given by:

When the dielectric of constant 3 is inserted into capacitor A, the capacitance becomes 3C. The energy stored in A is:

Since the switch S is opened, the total charge is conserved, and the voltage across capacitor B is determined by the charge redistribution. The voltage across B becomes:

The energy stored in B is:

The total energy after the dielectric is introduced is:

Simplify:

The ratio of final energy to initial energy is:

The surface charge density σ\sigmaσ is related to the normal component of the electric field E_n​ at the surface of the conductor by the equation:

σ = ε₀E_n

where:

• ε₀ = 8.85 × 10⁻¹² F/m is the permittivity of free space,
• E_n = ∣E∣ is the magnitude of the electric field vector.

The electric field is given as:

The magnitude E_n​ is:

E_n ≈ 0.412 V/m.

Using σ = ε₀E_n

σ = (8.85 × 10⁻¹²)⋅(0.412).

σ ≈ 3.64 × 10⁻¹² C/m² = 3.64 pC/m².

Magnetic Field Due to a Long Straight Wire

The magnetic field B at a distance x from a long straight wire carrying current III is given by:

Induced e.m.f. in the Moving Rod

The rod is moving in this magnetic field with velocity vvv, and the induced e.m.f. is calculated as:

dE = vBdx.

Substituting B into the expression:

To find the total induced e.m.f., integrate between the limits x₁ = 0.01 m and x₂ = 0.2 m:

Given:

• μ₀ = 4π × 10⁻⁷ H/m\
• I = 5 A
• v = 10 m/s
• x₁ = 0.01 m
• x₂ = 0.2 m

Substitute into the equation:

E = 10⁻⁵ . ln⁡(20).

Using ln⁡(20) ≈ 2.9957:

E = 10⁻⁵⋅2.9957 = 3 × 10⁻⁵ V

Convert to microvolts:

E = 30 μV.

In an LCR circuit, the resonance condition is defined by:

where:

• L is the inductance,
• C is the capacitance,
• ω is the angular resonance frequency.

Initially, the circuit is in resonance with the original capacitance C and inductance L:

When the capacitance is reduced to one-fourth of its original value (C′ = C/4​), the new resonance condition is:

To maintain resonance (ω′ = ω):

Square both sides:

Thus:

L′ = 4L

The inductance must be increased by a factor of 4 to maintain resonance.

Frequency of A.C. n = 50 Hz
time period T  = 1/n - 1/50s
Time in which current changes from peak = T/4 = 1/50 x 1/4 = 1/200s
value (i_i) to zero (i_i) 
ΔT = 1/200S.
i.e. 
then Δi- change in current 
= i_i — i_i = 0 — 1 = - 1 A .
∴ mean induced e.m.f.

e = -MΔI / ΔT

= -1.5 (-1/1/200)
= 300 V.

The magnetic field inside a solenoid is given by:

B_s = μ₀nI

where:

• μ₀ = 4π × 10⁻⁷ H/m is the permeability of free space,
• n = 2 × 10⁴ turns/m is the number of turns per meter,
• I is the current through the solenoid.

The flux linked with the 100-turn coil is:

Φ_c = N_c⋅B_s⋅A_c,

where:

• N_c = 100 is the number of turns in the coil,
• A_c = πr² is the area of the coil,
• B_s = μ₀nI is the magnetic field inside the solenoid.

Substitute A_c = π(0.01)² = π × 10⁻⁴ m²:

Φ_c = 100⋅(μ₀nI)⋅(π × 10⁻⁴).

Substitute μ₀ = 4π×10⁻⁷, n = 2 × 10⁴:

Φ_c = 100⋅(4π × 10⁻⁷)⋅(2 × 10⁴)⋅(π × 10⁻⁴)I

Simplify:

Φ^c = 8π² × 10⁻⁵I

The induced emf is given by Faraday's law:

From the flux expression:

Φ_c = 8π² × 10⁻⁵I

Given the current changes from +2 A to −2 A in Δt = 0.05 s:

ΔI = −2−(+2) = −4 A

The rate of change of current is:

Substitute dI / dt into the emf equation:

e = 8π² × 10⁻⁵⋅(−80)

Simplify:

e = −640π² × 10⁻⁵ V

The current in the coil is given by:

I = e/R

where R = 10π² Ω.

Substitute:

Simplify:

I = −64 × 10⁻⁵ = −0.0064 A

The total charge q is:

q = I⋅Δt

Substitute  I= 0.0064 A and Δt = 0.05 s:

q = 0.0064 x 0.05 = 32 × 10⁻⁶ C

Convert to microcoulombs:

q = 32 μC

The sawtooth waveform varies linearly with time. The equation for voltage in the time period T is:

The average value over one period is given by:

Substitute V(t):

Split the integral:

Solve each term:

Substitute:

Simplify:

The RMS value over one period is given by:

Substitute V(t):

Expand the expression for V²(t)

so,

The RMS value becomes:

Now,  Split the integral

Solve each term:

Substitute these:

Simplify each term:

Combine terms:

Ratio of V_av to V_rms

From the previous calculations:

• V_av = V₀/2V
• V_rms = V₀/√3

The ratio is:

Numerically:

The reactance of the inductor (X_L​) and capacitor (X_C​) are given by:

Substitute the given values:

• f = 750 Hz
• L = 0.1803 H
• C = 10 μF = 10 × 10⁻⁶ F

X_L = 2π⋅750⋅0.1803 = 849.2 Ω.

X_C = 1 / 2π⋅750⋅10 × 10⁻⁶ = 21.2 Ω.

The net reactance (X) is:

X = X_L − X_C = 849.2 − 21.2 = 828.0 Ω.

The total impedance (Z) of the circuit is:

where R=100 Ω

Substitute:

The power dissipated in the resistor is:

where 

Given V_peak = 20 V, calculate V_rms​:

Substitute into the power formula:

Simplify:

The energy required to heat the resistor is:

E = mcΔθ

where:

• mc = 2 J/°C (thermal capacity),
• Δθ = 10^∘ C

Substitute:

E = 2⋅10 = 20 J

The time required is:

t = E / P

Substitute:

• The charge density is proportional to r, so ρ = kr where k is a constant.
• Consider a Gaussian surface inside the cylinder at radius r (r < R).
• The charge enclosed by the Gaussian surface, Q_enc = ∫ρ dV = ∫kr dV.
• Volume of a thin cylindrical shell is dV = 2πrLdr. Integrate to find Q_enc.
• Q_enc = 2πkL ∫r² dr = (2/3)πkr³L.
• Electric field, E, inside a cylinder: E(2πrL) = Q_enc/ε₀.
• Simplifying, E ∝ r².

The correct answer is: B: E ∝ r²