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CONCEPT:

Torque on an Electric Dipole in a Uniform Electric Field

• Electric Dipole Moment (P): It is the product of the magnitude of one of the charges and the distance between them. It is given by:

P = q × d

where 'q' is the charge and 'd' is the dipole length.

• Torque (τ): When an electric dipole is placed in a uniform electric field (E) at an angle (θ), it experiences a torque given by:

τ = P × E × sin(θ)

EXPLANATION:

• Given values:
• Charge (q) = 5 μC = 5 × 10^-6 C
• Length of dipole (d) = 0.2 m
• Electric field (E) = 20 V/m
• Angle (θ) = 30°
• Step 1: Calculate the electric dipole moment (P):
• P = q × d
• P = (5 × 10^-6 C) × (0.2 m)
• P = 1 × 10^-6 C.m
• Step 2: Calculate the torque (τ):
• τ = P × E × sin(θ)
• τ = (1 × 10^-6) × 20 × sin(30°)
• τ = 20 × 10^-6 × 0.5 (Since sin 30° = 0.5)
• τ = 10 × 10^-6 N.m
• τ = 1 × 10^-5 N.m

Therefore, the torque is 1 × 10^-5 N.m.

CONCEPT:

Polarization and Malus Law

• When unpolarized light of intensity I₀ passes through a polarizer, the intensity of the transmitted light (I₁) becomes half of the incident intensity:

I₁ = I₀ / 2

• Malus Law: It states that when completely plane-polarized light is incident on an analyzer, the intensity I of the light transmitted through the analyzer is proportional to the square of the cosine of the angle θ between the transmission axes of the polarizer and the analyzer:

I₂ = I₁ cos²θ

EXPLANATION:

• Given data:
• Initial intensity of unpolarized light, I₀ = 40 W/m²
• Angle between the two polaroids, θ = 30°
• Step 1: Calculate the intensity of light after passing through the first polaroid (I₁):

I₁ = I₀ / 2

I₁ = 40 / 2 = 20 W/m²

• Step 2: Calculate the intensity of light after passing through the second polaroid (I₂) using Malus Law:

I₂ = I₁ cos²θ

I₂ = 20 × cos²(30°)

• Since cos(30°) = √3 / 2, we have cos²(30°) = 3 / 4:

I₂ = 20 × (3 / 4)

I₂ = 5 × 3 = 15 W/m²

Therefore, the intensity after passing through the second polaroid is 15 W/m².

CONCEPT:

Average Power in an AC Circuit

• The average power (P_avg) dissipated in an R-L-C circuit depends on the RMS voltage, RMS current, and the power factor.
• The formula for average power is:

P_avg = V_rmsI_rms cosφ

• The power factor (cosφ) is the ratio of resistance (R) to impedance (Z):

cosφ = R / Z ⇒ Z = R / cosφ

• Since I_rms = V_rms / Z, we can substitute Z to get:

I_rms = V_rms / (R / cosφ) = (V_rms cosφ) / R

• Substituting I_rms back into the power formula:

P_avg = V_rms × [(V_rms cosφ) / R] × cosφ = (V_rms² / R) cos²φ

EXPLANATION:

• Given values:
• Voltage (V) = 220 V
• Resistance (R) = 10 Ω
• Power factor (cosφ) = 0.5
• Using the derived formula for average power:
• P_avg = (V² / R) × cos²φ
• Placing the values into the equation:
• P_avg = (220² / 10) × (0.5)²
• P_avg = (48400 / 10) × 0.25
• P_avg = 4840 × 0.25
• P_avg = 1210 W

Therefore, the average power is 1210 W.

CONCEPT:

Magnetic Energy Density (u)

• Magnetic energy density represents the magnetic energy stored per unit volume in a given region.
• For a solenoid or any magnetic field B in a vacuum or air, the energy density is given by the formula:

u = B² / (2μ₀)

• Where:
• u is the energy density (J/m³).
• B is the magnetic field strength (Tesla, T).
• μ₀ is the permeability of free space (μ₀ ≈ 4π × 10^-7 T·m/A).

EXPLANATION:

• From the given problem:
• The energy density u = 10⁵ J/m³.
• The permeability of free space μ₀ ≈ 4π × 10^-7 T·m/A.
• We need to find the magnetic field B. Rearranging the energy density formula:
• B² = 2 × μ₀ × u
• B = √(2 × μ₀ × u)
• Substituting the values:
• B = √(2 × 4π × 10^-7 × 10⁵)
• B = √(8π × 10^-2)
• B = √(0.08 × 3.14159)
• B = √(0.251327)
• B ≈ 0.5013 T

Therefore, the value of the magnetic field (B) in the solenoid is approximately 0.5013 T.

CONCEPT:

Refractive Index of a Thin Prism

• For a thin prism, where the angle of the prism is small, the relationship between the refractive index (n), the angle of the prism (A), and the angle of minimum deviation (δ_m) is given by the formula:

δ_m = (n - 1)A

• In this equation:
• n represents the refractive index of the prism material.
• A represents the angle of the prism.
• δ_m represents the angle of minimum deviation.

EXPLANATION:

• Given values:
• Refractive Index (n) = 1.6
• Applying the thin prism formula:
• δ_m = (n - 1)A
• δ_m = (1.6 - 1)A
• δ_m = 0.6A
• The question asks for the ratio of the angle of the prism (A) to the angle of minimum deviation (δ_m):
• Ratio = A / δ_m
• A / δ_m = 1 / 0.6
• A / δ_m = 10 / 6
• A / δ_m = 5 / 3
• The ratio can be written as 5:3.

Therefore, the ratio of the angle of prism to the angle of minimum deviation is 5:3.

CONCEPT:

Acceleration in Simple Harmonic Motion (SHM)

• In Simple Harmonic Motion, the magnitude of acceleration (a) of a particle at a given displacement (x) from its mean position is directly proportional to the displacement and is governed by the angular frequency (ω).
• The mathematical relation is:

a = ω²x

• Where:
• a = acceleration (m/s²)
• ω = angular frequency (rad/s)
• x = displacement from the mean position (m)

EXPLANATION:

• Given data:
• Angular frequency (ω) = 0.5 rad/sec
• Amplitude (A) = 5 cm
• Displacement (x) = 4 cm
• To find the acceleration in standard units (m/s²), we must first convert the displacement from centimeters (cm) to meters (m):
• x = 4 cm = 4 / 100 m = 0.04 m
• Using the acceleration formula:
• a = ω²x
• a = (0.5)² × 0.04
• a = 0.25 × 0.04
• a = 0.01 m/s²

Therefore, the acceleration at the position x = 4 cm is 0.01 m/s².

STRICT FORMAT. DO NOT ADD EXTRA TEXT.

Question: A lens power 5 D forms virtual image having magnification 2.5 find position of object.

Options: A. 12 cm B. 30 cm C. 8 cm D. 16 cm

CONCEPT:

• Power of a Lens (P): The power of a lens is defined as the reciprocal of its focal length (f). When focal length is measured in centimeters, the relation is:

f = 100 / P

• Magnification (m): For a lens, the linear magnification is the ratio of the image distance (v) to the object distance (u). It can also be expressed using focal length (f) and object distance (u):

m = f / (f + u)

• Sign Convention:
• For a virtual image formed by a lens, the magnification is positive.
• A convex lens has a positive power and a positive focal length.

EXPLANATION:

• Given data:
• Power of the lens, P = 5 D
• Magnification, m = 2.5 (positive for a virtual image)
• Step 1: Calculate the focal length (f) of the lens.

f = 100 / P

f = 100 / 5 = 20 cm

• Step 2: Apply the magnification formula.

m = f / (f + u)

2.5 = 20 / (20 + u)

• Step 3: Solve the equation for the object distance (u).

20 + u = 20 / 2.5

20 + u = 8

u = 8 - 20

u = -12 cm

• The negative sign indicates that the object is placed in front of the lens at a distance of 12 cm.

Therefore, the position of the object is 12 cm.

CONCEPT:

Einstein's Photoelectric Equation

• According to Einstein's photoelectric equation, the energy of an incident photon (E) is used in two parts: overcoming the work function (Φ) of the metal and providing maximum kinetic energy (K_max) to the emitted photoelectron.

E = Φ + K_max

• The energy of a photon (E) in electron-volts (eV) can be calculated using the wavelength (λ) in nanometers (nm) as:

E (eV) = 1240 / λ (nm)

• The maximum kinetic energy (K_max) is numerically equal to the stopping potential (V_s) when expressed in electron-volts:

K_max = e × V_s

EXPLANATION:

• Given values:
• Wavelength of incident light (λ) = 400 nm
• Stopping Potential (V_s) magnitude = 2.5 V
• Maximum Kinetic Energy (K_max) = 2.5 eV
• Step 1: Calculate the energy of the incident photon (E):
• E = 1240 / λ
• E = 1240 / 400
• E = 3.1 eV
• Step 2: Use the photoelectric equation to find the work function (Φ):
• E = Φ + K_max
• 3.1 eV = Φ + 2.5 eV
• Φ = 3.1 eV - 2.5 eV
• Φ = 0.6 eV

Therefore, the work function of the material is 0.6 eV.

CONCEPT:

Capacitors in Series

• In a series connection, the charge (Q) on each capacitor is the same.
• The equivalent capacitance (C_eq) for a series combination is calculated using the formula:

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

• The total potential difference (V) across the combination is the sum of the potential differences across individual capacitors:

V = V₁ + V₂ + V₃

• The potential difference across an individual capacitor is given by:

V_i = Q / C_i

EXPLANATION:

• Given parameters:
• Capacitance C₁ = 1 μF
• Capacitance C₂ = 2 μF
• Capacitance C₃ = 3 μF
• Total Voltage V = 10 V
• Step 1: Calculate the equivalent capacitance (C_eq):
• 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
• 1/C_eq = 1/1 + 1/2 + 1/3
• 1/C_eq = (6 + 3 + 2) / 6
• 1/C_eq = 11/6 μF^-1
• C_eq = 6/11 μF
• Step 2: Calculate the total charge (Q) in the circuit:
• Q = C_eq × V
• Q = (6/11 μF) × 10 V
• Q = 60/11 μC
• Step 3: Calculate the potential difference across C₂ (V₂):
• V₂ = Q / C₂
• V₂ = (60/11 μC) / 2 μF
• V₂ = 30/11 V
• V₂ ≈ 2.73 V

Therefore, the potential difference across C₂ is 2.73 V.

CONCEPT:

Newton's Second Law of Motion

• According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration:

F = m × a

• When two blocks are placed in contact on a frictionless horizontal surface and a force is applied, they move together with a common acceleration.
• The contact force (or the force between the two bodies) can be calculated by analyzing the individual blocks using their respective free-body diagrams.

EXPLANATION:

• Given:
• Mass of the first block (m₁) = 3 kg
• Mass of the second block (m₂) = 1 kg
• Applied force (F) = 5 N (from the left side on the 3 kg block)
• Step 1: Calculate the common acceleration of the system.
• Total mass (M) = m₁ + m₂ = 3 kg + 1 kg = 4 kg
• Acceleration (a) = Total Force / Total Mass
• a = 5 N / 4 kg = 1.25 m/s²
• Step 2: Find the force between the two blocks.
• The force between the two blocks (let it be 'f') is the contact force.
• Consider the 1 kg block. The only horizontal force acting on it is the contact force 'f' exerted by the 3 kg block.
• Using F = ma for the 1 kg block:
• f = m₂ × a
• f = 1 kg × 1.25 m/s²
• f = 1.25 N
• Alternatively, for the 3 kg block:
• F - f = m₁ × a
• 5 N - f = 3 kg × 1.25 m/s²
• 5 N - f = 3.75 N
• f = 5 N - 3.75 N = 1.25 N

Therefore, the force between the two bodies is 1.25 N.