Solution: Using Coulomb's Law: F=k(q1q2)/r². When both charges are doubled, the force becomes 4F. To maintain the same force, the distance must increase by a factor of 2. Therefore, r' = 2d.

Quick Tip: To keep the force constant when charges are doubled, the distance between them should be increased by a factor of 2.

Solution: In a series combination, the same charge Q flows through both capacitors. The potential difference divides inversely proportional to capacitance: V₁/V₂ = C₂/C₁ = 3/2. For capacitors in series, since Q is the same, U₁/U₂ = C₂/C₁ = 3/2.

Quick Tip: For capacitors in series, the potential across each capacitor is inversely proportional to the capacitance, and the energy stored is proportional to C × V².

Solution:The electric field is zero outside the parallel plates (at P₁ and P₃) because the fields cancel. Inside the plates (at P₂), the electric field is uniform, and q experiences a force F₂ ≠ 0.

Quick Tip: When two parallel plates have equal and opposite charge densities, the electric field inside the plates is uniform, but outside the plates, the field cancels out.

Solution:When the spheres are brought into contact, charge redistributes in proportion to their capacitances. Capacitance of a sphere is proportional to its radius: Q₁/Q₂ = R₁/R₂

Quick Tip: When metallic spheres are brought into contact, the charge distribution results in a ratio of charges proportional to the radii of the spheres.

Solution:

Total resistance = 300Ω. Current = 20V/300Ω = 1/15 A. Voltage across the 200Ω resistor = (1/15 A) * 200Ω = 10 V. The voltmeter reading is equal to the voltage drop across the 200Ω resistor. The voltage across the 200 Ω resistance is V = IR = ¹⁄₁₅ × 200 = ²⁰⁄₃V.

Quick Tip: When resistors are connected in series, the total resistance is the sum of the individual resistances, and the same current flows through both resistors.

Solution: Equivalent emf = (2*2 + 1*1)/(1+2) = 5/3 V. Equivalent resistance = (1*2)/(1+2) = 2/3 Ω. Total resistance = 2/3Ω + 4/3Ω = 2Ω. Current = (5/3 V) / (2Ω) = 5/6 A.

Quick Tip: When cells are connected in parallel, the total emf and internal resistance are calculated using reciprocal sums.

Solution:Stretching the wire reduces its radius to half, increasing its length to four times. R' = ρ(4L)/[π(r/2)²] = 16R. Resistivity does not change (ρ' = ρ). Power rating at constant voltage: P' = V²/R' = (V²/R)/16 = P/16.

Quick Tip: Stretching a wire increases its resistance by the square of the ratio of the new length to the original length, while resistivity remains unchanged.

Solution: Diamagnetic materials have the lowest susceptibility, followed by paramagnetic, and ferromagnetic with the highest.
Quick Tip: Magnetic susceptibility increases from diamagnetic to paramagnetic to ferromagnetic materials.

Solution:The force per unit length between two parallel current-carrying conductors is given by: F/L = (μ₀I₁I₂)/(2πd). For a length L, the force is: F = (μ₀I₁I₂L)/(2πd). The force is directly proportional to I₁, I₂, and L.

Quick Tip: The force between two parallel current-carrying conductors depends on the product of the currents, the length of the conductors, and the distance between them.

Solution: Current splits inversely proportional to resistances: I_ABC = I, I_ADC = 2I. Magnetic field at the center for a semicircular wire is (μ₀I)/(4r). Fields add vectorially; since currents are in opposite directions, they add up: B_net = (μ₀I)/(4r) + (2μ₀I)/(4r) = (3μ₀I)/(4r). Direction is out of the plane using the right-hand rule. 

Quick Tip: The direction of the magnetic field is determined by the right-hand rule, with the net magnetic field pointing into the plane of the loop.

Solution:Torque (τ) = NIABsinθ, where N=1, I=10A, A=(0.01m)²=10⁻⁴m², B=0.2T, and θ=90° (since the field is parallel to the plane). τ = 1 * 10 * 10⁻⁴ * 0.2 * sin90° = 2 × 10⁻⁴ Nm.

Quick Tip: If the magnetic field is parallel to the plane of a current-carrying loop, the torque on the loop is zero.

Solution: In a purely capacitive circuit, the current leads the voltage by 90° (π/2).
Quick Tip: In a purely capacitive circuit, the current leads the voltage by 90°.

Solution: Mutual inductance (M) = ΔΦ/ΔI = 15Wb / 10A = 1.5 H.

Quick Tip: Mutual inductance is the ratio of the change in magnetic flux to the change in current in a pair of coupled coils.

Solution: According to Lenz's law, the induced current will oppose the change in magnetic flux. In figure (a), as the wire turns into a circular loop, the area of the loop increases, increasing the magnetic flux into the plane. To oppose this increase, the induced current will be clockwise. In figure (b), the wire is turning into a straight line, decreasing the area and the magnetic flux out of the plane, so the induced current will be anticlockwise to oppose this change.

Quick Tip: Lenz's law states that the direction of induced current will always oppose the change in magnetic flux.

Solution: Impedance depends on both reactance and resistance. Capacitive reactance decreases with frequency; inductive reactance increases with frequency; resistance is frequency independent.

Quick Tip: Remember that resonance occurs when inductive and capacitive reactances cancel out, and the opposition to AC is determined by the reactance and resistance in the circuit.

Solution: The energy density of an electromagnetic wave is equal for the electric and magnetic fields because E = cB.
Quick Tip: In electromagnetic waves, the energy is equally divided between the electric and magnetic fields, and their energy densities are in a 1:1 ratio.

Solution: The electromagnetic spectrum is ordered by wavelength from radio to gamma rays.

Quick Tip: Remember: As wavelength decreases, the energy of the electromagnetic wave increases. X-rays have shorter wavelengths and higher energy compared to visible light and microwaves.

Remember the acronym "Raging Martians Invaded Venus Using X-ray Guns" to recall the order of the electromagnetic spectrum.

Solution:Microwaves are produced by magnetrons. Infrared waves arise from vibrations of atoms/molecules. X-rays are generated by bombarding heavy nuclei with electrons. Radio waves are created by LC oscillators.

Quick Tip: Different electromagnetic waves have distinct production mechanisms: microwaves by a magnetron, X-rays by bombarding heavy metal targets, and radio waves by oscillators.

Solution: Using the refraction formula for curved surfaces: (μ₂/v) - (μ₁/u) = (μ₂ - μ₁)/R. Substituting the values, v = -16 cm, indicating the image is 16 cm to the left of P in air.

Quick Tip: To solve refraction problems at curved surfaces, use the formula for spherical refraction, and remember that a negative value of 'v' indicates that the image is on the same side as the object.

Solution: Lens power (P) = (μ - 1)[(1/R₁) + (1/R₂)]. For fixed radii, power is directly proportional to (μ - 1).

Quick Tip: In lens formulas, the power depends on the refractive index μ and the shape of the lens. For a given shape (fixed radii of curvature), the refractive index primarily determines the lens's power.

Solution: The lens equation (1/v) - (1/u) = 1/f produces a hyperbolic relation between v and u, with asymptotes along the axes.

Quick Tip: For a convex lens, the image distance *v* increases as the object distance *u* decreases, and the relationship follows a hyperbolic-like curve. The focal point *f* is the limit.

Solution: Linear width of central maximum (Δx) = 2λL/a, where λ is wavelength, L is distance to screen, and a is slit width. Solving for λ: λ = (Δx * a)/(2L) = (5 × 10⁻³ m * 0.1 × 10⁻³ m) / (2 * 0.5 m) = 5 × 10⁻⁷ m.

Quick Tip: In single-slit diffraction, the width of the central maximum depends on the wavelength, slit width, and distance to the screen. Use w = ^2λD⁄_a for calculations.

Solution:Photoelectric equation: K_max = hf - hν₀. Since f = 2ν₀, K_max = 2hν₀ - hν₀ = hν₀.

Quick Tip: In the photoelectric effect, the maximum kinetic energy of photoelectrons depends on the difference between the frequency of the incident light and the threshold frequency.

Solution: Photoelectric current depends on light intensity, which follows the inverse square law (I ∝ 1/d²). However, at saturation, all emitted electrons are collected, making the current independent of distance (as long as intensity is above the threshold).

Quick Tip: The photoelectric current decreases as the distance from the light source increases, following the inverse square law of intensity. As the light weakens, the emission stops.

Solution: de Broglie wavelength (λ) = h/(√2m_pV), where m_p is proton mass and V is potential difference. Doubling V reduces λ by a factor of √2.

Quick Tip: For charged particles like protons, the de Broglie wavelength is inversely proportional to the square root of the accelerating potential, so increasing the potential decreases the wavelength.

Solution: In the Bohr model, total energy (E) = -K (negative because it's a bound state). Potential energy (U) = 2E = -2K. Kinetic energy (K) = -E = K.

Quick Tip: For an electron in a hydrogen atom, the potential energy is -2 times the kinetic energy, and the total energy is equal to the negative of the kinetic energy.

Solution: Nuclear density is constant and independent of mass number because volume is proportional to A^1/3 and mass is proportional to A.

Quick Tip: The density of nuclei is nearly constant and does not depend on the mass number A, leading to the ratio of 1:1 for different nuclei.

Solution: Shortest wavelength corresponds to the transition to the lowest energy level. The order of series in decreasing wavelength is Pfund (n=5), Brackett (n=4), Balmer (n=2), Lyman (n=1).

Quick Tip: For hydrogen spectral series, Lyman emits the shortest wavelengths (UV region), followed by Balmer (visible), Brackett, and Pfund (infrared region).

Solution:For n-type semiconductors, dopants from group 15 (Arsenic and Phosphorus) are used as they have one more valence electron than silicon.

Quick Tip: To form n-type semiconductors, silicon is doped with Group V elements (like arsenic and phosphorus), which donate extra electrons, making the material negatively charged.

Solution:
- Graph (A) represents a forward-biased p-n junction, showing an exponential increase in current with forward voltage.
- Graph (C) represents a Zener diode, showing sharp breakdown in reverse bias.
- Graph (D) represents a photodiode, where current increases after a threshold voltage under illumination.
- Graph (B) represents a solar cell with open circuit voltage (Voc) and short circuit current (Isc).

Quick Tip: For understanding different diode characteristics, remember:
- Forward-biased p-n junctions have exponential curves.
- Zener diodes exhibit breakdown in reverse bias.
- Photodiodes show an increase in current when exposed to light.
- Solar cells have open circuit voltage and short circuit current characteristics.

Solution: Force (F) = IL x B. For straight segments, F = ILB sinθ = ILB (θ = 90°). For the semicircular segment, the force is equivalent to that on a straight wire of length 2R. Total force = 2(ILB) + 2(ILB) = 4BIR, vertically downward.

Quick Tip: The net force on a wire bent in a semicircular shape in a uniform magnetic field is proportional to the radius R and the current I. The straight sections cancel each other out, and the force is directed downward.

Solution: The formula for the angle of minimum deviation (δ_min) of a prism is given by: μ = ^{sin(A+δ_min⁄₂)}⁄_sin(^A⁄2)
where A is the angle of the prism and μ is the refractive index of the material. For an equilateral prism, A = 60°. Given μ = √2, solving the equation results in the angle of minimum deviation being δ_min = 60°.

Quick Tip: For equilateral prisms, use the formula involving the refractive index μ and the prism angle A to calculate the minimum deviation angle.

Solution:    

Electric charge is quantized in integral multiples of the elementary charge (e), carried by electrons.

Quick Tip: Electric charge is quantized, meaning it exists in integer multiples of the elementary charge carried by electrons.

Solution: With dielectric, effective separation (d_eff) = d - t + t/K, where d is initial separation, t is dielectric thickness, and K is dielectric constant. d_eff = 7.2 mm, d = 4 mm, t = 4 mm. Solving for K gives K = 5.
Quick Tip: The dielectric constant *K* reduces the effective separation between capacitor plates, affecting capacitance.

Solution: The condition for the ball to be suspended is that the electric force must balance the gravitational force. By solving using F_e = F_g, we get the charge as 1 × 10^-5 C.

Quick Tip: The electric force acting on a charged object in an electric field is F_e = Q · E, where Q is the charge and E is the electric field.

Solution: Drift velocity (v_d) = (eEτ)/m, where τ is relaxation time. τ decreases with increasing temperature due to increased scattering, thus decreasing v_d. Thermal velocity increases with temperature.

Quick Tip: In metals, increasing temperature increases the resistance and reduces the drift velocity of electrons, while thermal agitation increases.

Solution: Simplify the circuit step-by-step, combining parallel resistors first, then adding series resistances. The equivalent resistance between A and B is 18Ω.

Quick Tip: For complex resistor combinations, simplify the circuit step-by-step by reducing series and parallel connections.

Solution: Initially, I = 1.1V/(0.5Ω + 0.5Ω) = 1.1A. After adding the second cell, total emf = 2.2V. For the current to remain 1.1A, total resistance must be 2Ω. Therefore, 0.5Ω + 0.5Ω + r = 2Ω, where r is the internal resistance of the second cell; r = 1Ω.

Quick Tip: For cells in series, the total internal resistance and emf should be considered when analyzing the current through a circuit.

Solution: To balance the Wheatstone bridge, the ratio of the resistances must be equal. By solving for the required shunt resistance, we get 7 Ω.

Quick Tip: For a balanced Wheatstone bridge, the ratio of resistances in both branches must be equal.

Solution: Magnetic moment (M) = m * l, where m is pole strength and l is length. When bent into a semicircle, effective length becomes 2r/π (where r is the radius). Therefore M' = m * (2r/π) = (2M)/π.

Quick Tip: The magnetic moment of a bar magnet is given by the product of its pole strength and the effective length between the poles. Bending the magnet changes the effective length.

Solution: Ferromagnetic materials in transformers must have high permeability to increase magnetic flux and low hysteresis loss to minimize energy loss during magnetization cycles.

Quick Tip: Ferromagnetic cores in transformers should exhibit low energy loss and high magnetic permeability to improve efficiency.

Solution: The electric field induced in a conducting loop is proportional to the rate of change of the magnetic flux through the loop. In this case, the electric field intensity E = rx.

Quick Tip: Faraday's law of electromagnetic induction states that the induced electric field is proportional to the rate of change of magnetic flux.

Solution: The voltage induced in the secondary is given by V = M^dI⁄_dt, where M is the mutual inductance and ^dI⁄_dt is the rate of change of current. Using the given values and solving, the crest voltage induced in the secondary is 150 V.

Quick Tip: The induced voltage in the secondary of a transformer depends on the mutual inductance and the rate of change of current in the primary.

Solution: The induced emf in the coil is given by Faraday's law: E = -N^dΦ⁄_dt
where N = 100 is the number of turns in the coil and Φ = B⋅A is the magnetic flux. The magnetic field B inside the solenoid is given by: B = μ₀nI
where n = 2 × 10⁴ turns/m and I is the current in the solenoid. The change in magnetic flux is due to the reduction in current, and the total charge Q flowing through the coil is related to the induced emf and resistance by: Q = ^EΔt⁄_R
Substituting the given values and solving, the total charge flowing through the coil is 32π μC.

Quick Tip: Use Faraday's law to calculate the induced emf in a coil, and relate the emf to the total charge using Q = ^EΔt⁄_R.

Solution: When half of a convex lens is blocked, the entire image can still be formed, but the intensity of the image is reduced. The image will appear less bright because fewer rays reach the image.

Quick Tip: Blocking part of a lens reduces the number of rays forming the image, lowering the intensity but still producing a complete image.

Solution: The fringe separation Δy in a double-slit experiment is given by: Δy = ^λD⁄_d
where λ = 500 nm = 500 × 10^-9 m, D = 2 m, and d = 0.1 mm = 1 × 10^-4 m. Substituting these values, we get Δy = 1.5 cm.

Quick Tip: The fringe separation in a double-slit experiment increases with wavelength and screen distance and decreases with slit separation.

Solution: The tube length of an astronomical telescope is the sum of the focal lengths of the objective and the eyepiece: L = f_objective + f_eyepiece = 10 m + 0.1 m = 10.1 m = 1010 cm.
The magnification is given by: M = ^f_objective⁄_feyepiece = ¹⁰⁄_0.1 = 100.

Quick Tip: For astronomical telescopes, magnification is given by the ratio of the focal lengths of the objective and eyepiece lenses.

Solution: In Bohr's model:
- (A) is incorrect because the radius is proportional to n², not n.
- (B) is correct because the speed of the electron is inversely proportional to ¹⁄_n.
- (C) is correct because the total energy is proportional to ¹⁄_n².
- (D) is correct because the radius of the orbit is proportional to n².

Quick Tip: In Bohr's model, the radius of the electron's orbit is proportional to n², the speed of the electron is inversely proportional to n, and the total energy is inversely proportional to ¹⁄_n².

Solution: In a full-wave rectifier, the frequency of the output is double the input frequency because both halves of the AC waveform are utilized. So, if the input frequency is 50 Hz, the output frequency will be 100 Hz.

Quick Tip: A full-wave rectifier doubles the input frequency because it rectifies both halves of the AC waveform.

Solution: In a non-uniform field, there's a net force due to varying field strength, but no torque if the dipole is aligned with the field (τ = pEsinθ, θ = 0).

Quick Tip: For an electric dipole in a non-uniform electric field, a net force arises if the field is non-uniform, but the torque is zero if the dipole moment is aligned with the field.