Test Compilation PHY 102 13

Study Summary of Physics Questions

This document contains a series of multiple-choice physics questions covering various topics in electricity, magnetism, and basic electronics. Below is a structured summary of each question, including the topic, given information, relevant formulas, calculations or explanations, and the most appropriate answer.

Question 2 (Page 1): Capacitors in Series

• Topic: Capacitance, Series Connection
• Given: Two capacitors, C1 = 2 Farads, C2 = 4 Farads, connected in series.
• Formula: For capacitors in series, the equivalent capacitance (Ceq) is given by: 1/Ceq = 1/C1 + 1/C2
• Calculation: 1/Ceq = 1/2 + 1/4 = 2/4 + 1/4 = 3/4 Ceq = 4/3 F ≈ 1.33 F
• Answer: d. 1.33 Farads

Question 3 (Page 1 & Page 29): Gauss's Law Application

• Topic: Electrostatics, Gauss's Law
• Question: Gauss's law can be applied to the following except...
• Explanation: Gauss's law is a fundamental principle in electrostatics.
  • a. It helps to determine the location of excess charge on a conductor (charges reside on the surface).
  • b. Coulomb's law can be derived from Gauss's law.
  • c. It is used to calculate the field just outside any charged conductor.
  Since options a, b, and c are all valid applications or derivations related to Gauss's Law, the question implies that all listed items are within its scope.
• Answer: d. no correct option (meaning all other options are valid applications/derivations)

Question 4 (Page 1): Direction of Electric Field

• Topic: Electric Fields, Field Lines
• Question: The direction of field at any point can be determined by drawing a?
• Explanation: Electric field lines are drawn such that the tangent to a field line at any point gives the direction of the electric field at that point.
• Answer: b. tangent to the line of force passing through that point

Question 7 (Page 2): Transformer Current

• Topic: Transformers, AC Circuits
• Given: Primary turns (Np) = 200, Secondary turns (Ns) = 600, Primary current (Ip) = 12.0 A.
• Formula: For an ideal transformer, the ratio of turns is inversely proportional to the ratio of currents: Np/Ns = Is/Ip
• Calculation: 200/600 = Is/12 1/3 = Is/12 Is = 12/3 = 4 A
• Answer: b. 4.0 A

Question 8 (Page 2 & Page 20): Magnetic Force on Conductor

• Topic: Magnetism, Force on Current-Carrying Wire
• Given: Current (I) = 3 A, Total length = 1 m, Length perpendicular to field (L) = Total length / 4 = 0.25 m, Magnetic field (B) = 0.5 T, Angle (θ) = 90° (perpendicular).
• Formula: The magnetic force (F) on a current-carrying wire is: F = I * L * B * sin(θ)
• Calculation: F = 3 A * 0.25 m * 0.5 T * sin(90°) F = 3 * 0.25 * 0.5 * 1 = 0.375 N
• Answer: c. 0.375N

Question 13 (Page 3): RLC Circuit RMS Current

• Topic: AC Circuits, RLC Series Circuit
• Given: Resistance (R) = 100 Ω, Inductance (L) = 1 H, Capacitance (C) = 10 μF = 10 × 10-6 F, Source voltage V = 200sin(ωt) (so peak voltage Vp = 200 V), angular frequency (ω) = 400 rad/s.
• Formulas:
  • Inductive Reactance (XL) = ωL
  • Capacitive Reactance (XC) = 1/(ωC)
  • Impedance (Z) = √(R2 + (XL - XC)2)
  • RMS voltage (Vrms) = Vp/√2
  • RMS current (Irms) = Vrms/Z
• Calculation: 1. XL = 400 rad/s * 1 H = 400 Ω 2. XC = 1 / (400 rad/s * 10 × 10-6 F) = 1 / (4 × 10-3) = 250 Ω 3. Z = √(1002 + (400 - 250)2) = √(10000 + 1502) = √(10000 + 22500) = √32500 ≈ 180.28 Ω 4. Vrms = 200 V / √2 ≈ 141.42 V 5. Irms = 141.42 V / 180.28 Ω ≈ 0.7845 A
• Answer: b. 0.78A

Question 14 (Page 3 & Page 4): Magnetic Induction in a Wire

• Topic: Magnetism, Magnetic Field from a Current
• Given: Magnetic induction (B) = 20 μT = 20 × 10-6 T, Radius (r) = 120 mm = 0.12 m, Permeability of free space (μ₀) = 4π × 10-7 Tm/A.
• Formula: For a long straight current-carrying wire, the magnetic field is: B = (μ₀ * I) / (2π * r)
• Calculation: I = (B * 2π * r) / μ₀ I = (20 × 10-6 T * 2π * 0.12 m) / (4π × 10-7 Tm/A) I = (20 × 10-6 * 0.12) / (2 × 10-7) // (2π/4π = 1/2) I = 10 * 0.12 * 10(-6 - (-7)) = 10 * 0.12 * 101 = 12 A
• Answer: c. 12A

Question 15 (Page 4 & Page 31 & Page 32): Semiconductor Current Flow / Doping

• Topic: Semiconductors, Doping
• Question (Page 4, incomplete): In a semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called...
• Explanation: This describes intrinsic semiconductors where electron-hole pairs are generated by thermal energy breaking covalent bonds.
• Question (Page 31 & 32): Which of the following is true? i. Doping of semiconductors improves its conductivity. ii.doping a semiconductor renders it totally immune to ambient temperature. Doping of semiconductor alters the composition of carriers in it.
• Explanation:
  • i. Doping indeed improves conductivity by increasing the number of free charge carriers (electrons or holes). (True)
  • ii. Doping does not make a semiconductor totally immune to temperature; temperature still affects carrier concentration and mobility. (False)
  • iii. Doping introduces impurities that act as donors or acceptors, thereby changing the concentration of electrons and holes, thus altering the composition of carriers. (True)
• Answer (Page 31 & 32): c. i, iii.

Question 10 (Page 5): Charged Particle in Magnetic Field (Speed and Period)

• Topic: Magnetism, Motion of Charged Particles
• Given: Mass (m) = 10 kg, Charge (q) = 5 C, Magnetic field (B) = 5 T, Radius of path (r) = 80 m.
• Formulas:
  • Magnetic force = Centripetal force: qvB = mv2/r => v = (qBr)/m
  • Period (T) = 2πr/v OR T = 2πm/(qB)
• Calculation: 1. Speed (v) = (5 C * 5 T * 80 m) / 10 kg = (25 * 80) / 10 = 200 m/s 2. Period (T) = (2 * π * 10 kg) / (5 C * 5 T) = 20π / 25 = 4π / 5 ≈ 2.51 s
• Answer: d. 200 m/s, 2.5 s

(Page 6 & Page 8): Circuit Element Storing Electromagnetic Energy

• Topic: Circuit Components, Energy Storage
• Question: Which of following circuit element stores energy in electromagnetic field?
• Explanation: Inductors store energy in the magnetic field generated by the current flowing through them (electromagnetic energy). Capacitors store energy in an electric field (electrostatic energy). Resistors dissipate energy as heat.
• Answer: a. Inductor

Question 7 (Page 7): Coulomb's Law, Force and Distance

• Topic: Electrostatics, Coulomb's Law
• Given: Initial separation (r1) = 6 cm, New separation (r2) = 3 cm.
• Formula: Coulomb's Law: F = k * |q1q2| / r2. The force is inversely proportional to the square of the distance (F ∝ 1/r2).
• Calculation: F2/F1 = (r1/r2)2 = (6 cm / 3 cm)2 = (2)2 = 4 The force changes by a factor of 4.
• Answer: c. 4

Question 7 (Page 8): Capacitors in Parallel, Stored Energy

• Topic: Capacitance, Parallel Connection, Energy Storage
• Given: Two capacitors, C1 = 15 F, C2 = 5 F, connected in parallel across a 20 V battery.
• Formulas:
  • For capacitors in parallel, Ceq = C1 + C2
  • Energy stored (U) = 0.5 * Ceq * V2
• Calculation: 1. Ceq = 15 F + 5 F = 20 F 2. U = 0.5 * 20 F * (20 V)2 = 10 * 400 = 4000 J
• Answer: a. 4000J

Question 9 (Page 9): Parallel Plate Capacitor with Dielectric

• Topic: Capacitance, Dielectrics
• Given: Air capacitor (C0, Q0) connected to a constant voltage source (V). Dielectric with constant K=2 inserted.
• Explanation: 1. Since the capacitor remains connected to a constant voltage source, the voltage (V) across it remains unchanged. 2. When a dielectric is inserted, the capacitance increases by a factor of K: C = K * C0 = 2 * C0 3. The charge stored (Q) is given by Q = CV. Since C doubles and V remains constant: Q = C * V = (2 * C0) * V = 2 * (C0 * V) = 2 * Q0
• Answer: d. C=2C₀ and Q=2Q₀ (assuming C and Q in the option refer to C0 and Q0)

Question 10 (Page 9 & Page 10 & Page 16): Galvanometer Equilibrium Condition

• Topic: Magnetism, Galvanometer Principle
• Given: Galvanometer with spring constant (k), magnetic field (B), deflection angle (θ, represented by 'Ö'), N turns, area (A), current (I).
• Explanation: In a galvanometer, the magnetic torque (τmag) on the coil is proportional to the current and given by NIAB (for a radial magnetic field, sin(angle) is effectively 1). The restoring torque (τspring) from the spring is proportional to the deflection angle (kθ). At equilibrium, these torques balance: τmag = τspring NIAB = kθ
• Answer: b. KÖ = BIAN (interpreting KÖ as kθ and BIAN as NIAB)

Question 11 (Page 9 & Page 10): Field without Magnetic Poles

• Topic: Magnetism, Magnetic Field Lines
• Question: Field which does not have magnetic poles is...
• Explanation: Magnetic monopoles do not exist in classical electromagnetism; therefore, magnetic field lines always form closed loops, never starting or ending at a point. Examples include the circular field lines around a current-carrying wire or inside a solenoid.
• Answer: d. circular

(Page 11 & Page 42 & Page 43): Copper Conductor with Least Resistance

• Topic: Electrical Resistance
• Question: Which of the following copper conductor that has the least resistance is:
• Formula: Resistance R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. For metals, resistance also generally increases with temperature.
• Explanation: To minimize resistance:
  • Reduce length (L)
  • Increase cross-sectional area (A) (make it thick)
  • Reduce temperature (make it cool)
  Thus, a thick, short, and cool conductor will have the least resistance.
• Answer: b. thick, short and cool

Question 15 (Page 12): Capacitor Charging (Max Current and Stored Charge)

• Topic: RC Circuits, Capacitor Charging
• Given: Capacitance (C) = 4.0 μF = 4.0 × 10-6 F, Resistance (R) = 2.5 MΩ = 2.5 × 106 Ω. (Assumed charging voltage V for calculation).
• Formulas:
  • Maximum charging current (Imax) = V/R (at t=0)
  • Maximum stored charge (Qmax) = C * V (when fully charged)
• Calculation: Let's work backward from option 'a' (4 μA, 40 μC). If Imax = 4 μA = 4 × 10-6 A, then V = Imax * R = (4 × 10-6 A) * (2.5 × 106 Ω) = 10 V. With V = 10 V, Qmax = C * V = (4.0 × 10-6 F) * 10 V = 40 × 10-6 C = 40 μC. This consistency confirms option 'a'.
• Answer: a. 4uA, 40 μC μC

Question 2 (Page 13 & Page 14): Electric Field and Potential

• Topic: Electrostatics, Electric Potential
• Question: Electric field is steepest in the direction in which the potential...
• Explanation: The electric field (E) is the negative gradient of the electric potential (V), E = -∇V. This means the electric field points in the direction of the steepest decrease in electric potential.
• Answer: d. Decrease

Question 1 (Page 14): Positive Charge in Electrostatic Field (Kinetic Energy)

• Topic: Electrostatics, Energy Conservation
• Question: When a positive charge is moved in an electrostatic field from a point at high potential to a low potential, its kinetic energy...
• Explanation: For a positive charge, moving from high potential to low potential means its electric potential energy decreases (ΔPE < 0). By the principle of conservation of energy (assuming no other forces or energy losses), this decrease in potential energy is converted into kinetic energy, so its kinetic energy increases.
• Answer: c. Increases

Question 2 (Page 15): Galvanometer to Ammeter Conversion

• Topic: DC Circuits, Galvanometer, Ammeter Shunt
• Given: Galvanometer resistance (Rg) = 50 Ω, Full scale galvanometer current (Ig) = 2 mA = 2 × 10-3 A. Desired ammeter range (I) = 10 A.
• Formula: Shunt resistance (Rs) for an ammeter conversion is: Rs = (Ig * Rg) / (I - Ig)
• Calculation: Rs = (2 × 10-3 A * 50 Ω) / (10 A - 2 × 10-3 A) Since I >> Ig, (I - Ig) ≈ I. Rs ≈ (2 × 10-3 * 50) / 10 = (100 × 10-3) / 10 = 10 × 10-3 = 0.01 Ω
• Answer: c. 0.01 Ù (Ω)

Question 14 (Page 17): Electric Flux

• Topic: Electrostatics, Electric Flux
• Given: Area is circular, radius (r) = 0.025 m (assuming "0.025m²" is a typo and should be radius 0.025m), Electric field (E) = 100 V/m, Angle with surface = 30°.
• Formula: Electric Flux (Φ) = E * A * cos(θ), where A is the area and θ is the angle between the electric field vector and the normal to the surface.
• Calculation: 1. Area (A) = πr2 = π * (0.025 m)2 ≈ π * 0.000625 m2 ≈ 0.001963 m2. 2. If the angle with the surface is 30°, then the angle with the normal (θ) is 90° - 30° = 60°. 3. Φ = 100 V/m * 0.001963 m2 * cos(60°) = 100 * 0.001963 * 0.5 ≈ 0.09815 Nm2/C.
• Answer: a. 0.098Nm²/C

Question 14 (Page 18): AC Circuit RMS Current

• Topic: AC Circuits, Ohm's Law
• Given: Peak voltage (Vp) = 80 V, Resistance (R) = 60 Ω.
• Formulas:
  • RMS voltage (Vrms) = Vp/√2
  • RMS current (Irms) = Vrms/R
• Calculation: 1. Vrms = 80 V / √2 ≈ 56.57 V 2. Irms = 56.57 V / 60 Ω ≈ 0.9428 A
• Answer: b. 0.94 A

Question 8 (Page 19): Electric Flux (Gauss's Law)

• Topic: Electrostatics, Gauss's Law, Electric Flux
• Given: Charge (Q) = 50 μC = 50 × 10-6 C, Sphere radius (r) = 4 m (radius is irrelevant for net flux through a closed surface enclosing the charge).
• Formula: Gauss's Law states that the total electric flux (Φ) through any closed surface is proportional to the enclosed electric charge (Qenc): Φ = Qenc / ε₀ where ε₀ (permittivity of free space) ≈ 8.854 × 10-12 C2/(N·m2).
• Calculation: Φ = (50 × 10-6 C) / (8.854 × 10-12 C2/(N·m2)) Φ ≈ 5.647 × 106 Nm2/C
• Answer: a. 5.56 x 106 Nm²/C (closest value)

Question 9 (Page 20, incomplete): Series R-L-C Circuit (Instantaneous Current)

• Topic: AC Circuits, RLC Series Circuit, Instantaneous Current
• Given: R = 100 Ω, L = 0.5 H, C = 10 μF (assuming "10iC" is a typo for 10 μF = 10 × 10-6 F), f = 50 Hz, v = 120 sin(ωt) (so Vp = 120 V), t = 0.10 s.
• Formulas:
  • ω = 2πf
  • XL = ωL
  • XC = 1/(ωC)
  • Z = √(R2 + (XL - XC)2)
  • Peak current (Ip) = Vp/Z
  • Phase angle (φ) = arctan((XL - XC)/R)
  • Instantaneous current i(t) = Ip sin(ωt - φ)
• Calculation (steps shown, actual answer depends on options which are not visible): 1. ω = 2π * 50 = 100π rad/s ≈ 314.16 rad/s 2. XL = 100π * 0.5 = 50π Ω ≈ 157.08 Ω 3. XC = 1 / (100π * 10 × 10-6) = 1000/π Ω ≈ 318.31 Ω 4. Z = √(1002 + (157.08 - 318.31)2) = √(10000 + (-161.23)2) ≈ √36000 ≈ 189.74 Ω 5. φ = arctan((-161.23)/100) ≈ -58.18° (current leads voltage because XC > XL) 6. Ip = 120 V / 189.74 Ω ≈ 0.632 A 7. i(t) = 0.632 sin(100π * 0.10 - (-58.18° * π/180°)) = 0.632 sin(10π + 1.015 rad) = 0.632 sin(0 + 1.015 rad) ≈ 0.632 * 0.849 ≈ 0.537 A (Instantaneous current at t=0.10s would be calculated from this, paying attention to angle units).

Question 5 (Page 22): Electric Field Direction (Positive Point Charge)

• Topic: Electrostatics, Electric Field Lines
• Question: The diagram shows a positive point charge Q. Which of the following describes the magnitude and direction of the electric field at points r and s respectively?
• Explanation: Electric field lines originate from positive charges and point radially outward. The magnitude of the electric field decreases with distance from the charge (E ∝ 1/r2). Thus, at points r and s, the field will point away from Q, and if s is further than r, the magnitude at s will be different (smaller) than at r.
• Answer: c. different and Away from Q

Question 13 (Page 23): ElectronVolt Definition

• Topic: Energy, ElectronVolt
• Question: One electronVolt is...
• Explanation: An electronVolt (eV) is defined as the amount of kinetic energy gained by a single electron when it is accelerated through an electric potential difference of one volt.
• Answer: d. the kinetic energy of an electron which has been accelerated through 1V

(Page 24): Infinite Series of Charges (Potential and Electric Field)

• Topic: Electrostatics, Electric Potential, Electric Field, Infinite Series
• Given: Infinite number of charges, each equal to q, placed along the x-axis at x = 1, 2, 4, 8, ... Point of interest is x = 0.
• Formulas:
  • Electric Potential (V) at x=0: V = Σ (kq / xi)
  • Electric Field (E) at x=0: E = Σ (kq / xi2) (E points towards negative x-axis)
• Calculation: 1. Potential (V): V = kq * (1/1 + 1/2 + 1/4 + 1/8 + ...) This is an infinite geometric series with first term a = 1 and common ratio r = 1/2. The sum is a / (1 - r) = 1 / (1 - 1/2) = 2. V = 2kq = 2q / (4πε₀) 2. Electric Field (E): E = kq * (1/12 + 1/22 + 1/42 + 1/82 + ...) This is an infinite geometric series with a = 1 and r = 1/4. The sum is a / (1 - r) = 1 / (1 - 1/4) = 4/3. E = (4/3)kq = (4/3)q / (4πε₀) = 4q / (12πε₀)
• Answer: a. V = 2q/(4πε₀) and E = 4q/(12πε₀)

Question 6 (Page 25): Force between Two Protons

• Topic: Electrostatics, Coulomb's Law
• Given: Two protons, each with charge e = 1.6 × 10-19 C. Distance (r) = 1 μm = 1 × 10-6 m. Coulomb's constant (k) = 9 × 109 Nm2/C2.
• Formula: Coulomb's Law: F = k * q1q2 / r2
• Calculation: F = (9 × 109 Nm2/C2) * (1.6 × 10-19 C)2 / (1 × 10-6 m)2 F = (9 × 109) * (2.56 × 10-38) / (1 × 10-12) F = 9 * 2.56 * 10(9 - 38 + 12) = 23.04 * 10-17 = 2.304 × 10-16 N Since protons are like charges, the force is repulsive.
• Answer: d. 2.3*10-16N, repulsion

Question 9 (Page 26): Magnetic Field Pattern of a Coil

• Topic: Magnetism, Magnetic Fields
• Question: The magnetic field pattern produced by a coil carrying a direct current is similar to the magnetic field pattern of...
• Explanation: A current-carrying coil (especially a solenoid) produces a magnetic field pattern that is very similar to that of a permanent bar magnet. Both have distinct North and South poles and field lines that emerge from one pole and enter the other, forming closed loops outside the magnet/coil and running straight through its interior.
• Answer: c. a permanent bar magnet

Question 8 (Page 27): Potential Difference and Distance

• Topic: Electrostatics, Electric Potential
• Given: Charge (Q) = 5 μC = 5 × 10-6 C. Initial potential difference (V1) = 10 V at distance (r1) = 120 m. New potential difference (V2) = 30 V.
• Formula: Electric potential due to a point charge: V = kQ/r. Therefore, r = kQ/V.
• Calculation: 1. From the first set of values, we find kQ: 10 V = k * (5 × 10-6 C) / 120 m => kQ = 10 * 120 = 1200 V·m 2. Now, use this kQ to find the new distance r2 for V2 = 30 V: r2 = kQ / V2 = 1200 V·m / 30 V = 40 m
• Answer: b. 40m

Question 9 (Page 27): Electron in Electric Field (Time to Travel)

• Topic: Kinematics, Electrodynamics
• Given: Electric field (E) = 1.2 × 105 N/C, Distance (d) = 20 mm = 0.02 m. Electron charge (e) = 1.6 × 10-19 C, Electron mass (me) = 9.1 × 10-31 kg. Starts from rest.
• Formulas:
  • Force (F) = qE
  • Acceleration (a) = F/m
  • Kinematic equation: d = v0t + 0.5at2 (since v0=0, d = 0.5at2)
• Calculation: 1. Force (F) = (1.6 × 10-19 C) * (1.2 × 105 N/C) = 1.92 × 10-14 N 2. Acceleration (a) = (1.92 × 10-14 N) / (9.1 × 10-31 kg) ≈ 2.1099 × 1016 m/s2 3. Time (t) = √(2d/a) = √(2 * 0.02 m / (2.1099 × 1016 m/s2)) = √(0.04 / (2.1099 × 1016)) t = √(1.8968 × 10-18) ≈ 1.377 × 10-9 s The calculated value is approximately 1.38 ns. The option A.3.25x10-6 s (3.25 μs) is significantly different. There might be an error in the question's values or the provided options.
• Answer: Calculated: 1.38 × 10-9 s

(Page 28): Resonant Frequency of LC Circuit

• Topic: AC Circuits, Resonance
• Given: Initial resonant frequency (fres1) = 1000 KHz. Inductance (L) and Capacitance (C) are both "reduced by 5".
• Formula: Resonant frequency (fres) = 1 / (2π√(LC))
• Explanation: If "reduced by 5" means L' = L/5 and C' = C/5 (reduced by a factor of 5), then: fres2 = 1 / (2π√((L/5)(C/5))) = 1 / (2π√(LC/25)) = 1 / (2π * (1/5)√(LC)) = 5 * (1 / (2π√(LC))) = 5 * fres1 fres2 = 5 * 1000 KHz = 5000 KHz None of the given options (1000, 2000, 100, 200 KHz) match 5000 KHz. If "reduced by 5" implies L and C are *increased* by a factor of 5 (a common misunderstanding), then fres2 = fres1/5 = 1000/5 = 200 KHz, which is option d. Given the mismatch, this scenario suggests an ambiguity in the question's phrasing.
• Answer: Based on "reduced by a factor of 5", the calculated answer is 5000 KHz. If interpreting "reduced by 5" as "L and C are *increased* by a factor of 5", then d. 200.0KHz.

Question 6 (Page 28): Potential Difference in Parallel Plate Capacitor

• Topic: Capacitance, Parallel Plate Capacitor
• Question: The potential difference between the two plates of a parallel plate capacitor is...
• Formulas: Capacitance (C) = Q/V and C = ε₀A/d (for parallel plate capacitor).
• Explanation: From C = Q/V, we get V = Q/C. Substituting the formula for C: V = Q / (ε₀A/d) = Qd / (ε₀A) (Interpreting option 'a' as representing Qd/(ε₀A)).
• Answer: a. Qd/ε₀A

Question 14 (Page 30): Electric Field at Center of Equilateral Triangle

• Topic: Electrostatics, Electric Field, Vector Addition
• Given: Charges +1mC at A, -1mC at B, and -1mC at C of an equilateral triangle.
• Explanation: * The electric field (EA) due to +1mC at A points away from A (towards the midpoint of BC). * The electric field (EB) due to -1mC at B points towards B. * The electric field (EC) due to -1mC at C points towards C. * If the triangle is oriented such that A is at the top, B to the bottom-left, and C to the bottom-right: * EA points downwards. * EB points upwards and to the left. * EC points upwards and to the right. * The vector sum of EB and EC will have their horizontal components cancel out, and their vertical components will add up to point upwards. * However, if we consider EB and EC individually, they point towards B and C respectively. * A more intuitive way: Consider the field due to -1mC at B and -1mC at C. Their resultant field at the center points in the direction *away* from A (towards the midpoint of BC). EA also points away from A. Thus, the total electric field at the center points towards the midpoint of side BC.
• Answer: b. towards side BC

Question 14 (Page 32): Wire Resistance and Stretching

• Topic: Electrical Resistance, Resistivity, Stretching a Wire
• Given: Initial resistance (R1) = 2 Ω. Wire is stretched uniformly to twice its original length (L2 = 2L1).
• Formulas:
  • Resistance R = ρL/A (where ρ is resistivity, L is length, A is cross-sectional area).
  • Volume (V) = A * L (remains constant during stretching).
• Calculation: 1. Since volume is constant, A1L1 = A2L2. If L2 = 2L1, then A2 = A1/2. 2. New resistance R2 = ρL2/A2 = ρ(2L1) / (A1/2) = 4 * (ρL1/A1) = 4 * R1. 3. R2 = 4 * 2 Ω = 8 Ω.
• Answer: d. 8? (Ω)

Question 11 (Page 33): Power Line and Light Bulbs

• Topic: DC/AC Circuits, Power, Fuses
• Given: Power line voltage (V) = 120 V, Fuse maximum current (Ifuse) = 15 A. Each light bulb is rated 120 V, 500 W.
• Formulas:
  • Power (P) = V * I
  • Total current = n * Ibulb (where n is the number of bulbs)
• Calculation: 1. Current drawn by one bulb (Ibulb) = P / V = 500 W / 120 V ≈ 4.167 A. 2. Maximum number of bulbs (n) = Ifuse / Ibulb = 15 A / 4.167 A ≈ 3.6. Since you can only connect whole bulbs, the maximum number is 3.
• Answer: c. 3

Question 12 (Page 33): R-L AC Circuit (Inductance)

• Topic: AC Circuits, R-L Series Circuit
• Given: Apparent power (S) = 30 VA. Source voltage (V) = 36 V (RMS). Frequency (f) = 50 Hz. Resistance (R) = 40 Ω.
• Formulas:
  • S = V * Irms
  • Impedance (Z) = V / Irms
  • Z = √(R2 + XL2)
  • XL = ωL = 2πfL
• Calculation: 1. RMS current (Irms) = S / V = 30 VA / 36 V = 5/6 A ≈ 0.8333 A. 2. Impedance (Z) = V / Irms = 36 V / (5/6 A) = 36 * 6 / 5 = 216 / 5 = 43.2 Ω. 3. Inductive Reactance (XL) = √(Z2 - R2) = √(43.22 - 402) = √(1866.24 - 1600) = √266.24 ≈ 16.317 Ω. 4. Inductance (L) = XL / (2πf) = 16.317 Ω / (2π * 50 Hz) = 16.317 / (100π) ≈ 0.0519 H. The calculated value is approximately 0.052 H, which does not match option a. 1.05H. There may be a typo in the question's values or options.
• Answer: Calculated: 0.052 H

Question 8 (Page 35 & Page 40): Magnetic Force on a Charge

• Topic: Magnetism, Force on Moving Charge
• Question: Which of the following is/are correct about a charge q, moving with velocity, v, in a uniform magnetic field? I. The magnitude and direction of the force, FB depends on v and q. II. The magnitude and direction of the force, FB independent of v and q. III. The magnitude and direction of the force, FB is zero always.
• Formula: Magnetic force FB = q(v × B) = qvBsinθ
• Explanation: * I. The formula shows FB is directly proportional to q and v, and its direction is determined by the cross product of v and B, which depends on the direction of v and the sign of q. (True) * II. This is false as FB clearly depends on q and v. (False) * III. This is false; FB is zero only if q=0, v=0, B=0, or v is parallel/anti-parallel to B (sinθ=0). (False)
• Answer: a. I only

(Page 36 & Page 37): Reduce Circuit Resistance

• Topic: Electrical Resistance, Parallel Resistors
• Given: Original circuit resistance = 200 ohms. Desired new resistance = 120 ohms.
• Formula: For two resistors in parallel: 1/Rtotal = 1/R1 + 1/R2
• Explanation: To reduce the total resistance of a circuit, a resistor must be added in parallel. 1/120 = 1/200 + 1/Radd 1/Radd = 1/120 - 1/200 = (5 - 3) / 600 = 2 / 600 = 1/300 Radd = 300 ohms
• Answer: d. 300 ohm resistor in parallel

Question 5 (Page 36): Potentiometer Function

• Topic: DC Circuits, Potentiometer
• Question: A potentiometer is a device used to...
• Explanation: A potentiometer is primarily used for precise measurement and comparison of electromotive forces (EMFs) or potential differences, without drawing any current from the source under test at balance.
• Answer: a. compare two voltages

Question 3 (Page 37): Demagnetize Steel Needle

• Topic: Magnetism, Demagnetization
• Question: Which of the following is the best way to demagnetise a magnetised steel needle?
• Explanation: To demagnetize a material, its magnetic domains must be randomized. This can be achieved by: * Heating it above its Curie temperature and allowing it to cool without an external magnetic field. * Placing it in a strong alternating magnetic field and slowly withdrawing it, causing the domains to reorient rapidly and eventually randomize as the field weakens.
• Answer: d. slowly pull it out of a solenoid carrying alternating current

Question 1 (Page 38): Doping of Semiconductor (Not True)

• Topic: Semiconductors, Doping
• Question: Which of the following is not true about doping of semiconductor?
• Explanation: * a. Doping with Group III elements (e.g., Boron) leads to p-type semiconductors (creates holes). (True) * b. Doping with Group V elements (e.g., Phosphorus) leads to n-type semiconductors (creates excess electrons). (True) * c. Doping with Group III elements leads to excess holes in the valency band (they are acceptor impurities). (True) * d. Doping with Group V elements leads to excess holes in the conduction band. (False, Group V elements lead to excess *electrons* in the conduction band, not holes).
• Answer: d. doping with group V elements leads to excess holes in the conduction band.

Question 2 (Page 39): Electric Field of Uniformly Charged Hollow Metallic Sphere

• Topic: Electrostatics, Electric Field, Conductors
• Question: The electric field associated with a uniformly charged hollow metallic sphere is the greatest at:
• Explanation: For a uniformly charged hollow metallic sphere: * The electric field inside the conductor is zero. * The electric field is maximum at the outer surface of the sphere. * Outside the sphere, the electric field decreases with the square of the distance from the center.
• Answer: b. the sphere's outer surface.

Question 3 (Page 39): RC Circuit (Voltage across Resistor)

• Topic: RC Circuits, Capacitor Charging, Time Constant
• Given: Capacitance (C) = 10 μF = 10 × 10-6 F, Battery voltage (Vbattery) = 24 V, Resistance (R) = 400 kΩ = 400 × 103 Ω. Time (t) = 1 s.
• Formulas:
  • Time constant (τ) = RC
  • Voltage across resistor (VR) during charging: VR(t) = Vbattery * e(-t/τ)
• Calculation: 1. τ = RC = (400 × 103 Ω) * (10 × 10-6 F) = 4 s. 2. VR(1s) = 24 V * e(-1s / 4s) = 24 * e(-0.25). 3. e(-0.25) ≈ 0.7788 4. VR(1s) = 24 * 0.7788 ≈ 18.69 V.
• Answer: d. 18.7V

Question 5 (Page 41): Straight Conductor in Magnetic Field (Current)

• Topic: Electromagnetic Induction, Motional EMF, Ohm's Law
• Given: Length of conductor (L) = 25 cm = 0.25 m, Velocity (v) = 30 m/s, Magnetic field (B) = 0.4 T. Resistance (R) = 0.8 Ω. (Assuming B, L, v are mutually perpendicular).
• Formulas:
  • Motional EMF (ε) = B L v
  • Current (I) = ε / R
• Calculation: 1. ε = 0.4 T * 0.25 m * 30 m/s = 3 V. 2. I = 3 V / 0.8 Ω = 3.75 A.
• Answer: d. 3.75A

Question 6 (Page 42): Valence Electrons Absorb Energy

• Topic: Semiconductors, Energy Bands
• Question: When valence electrons in an atom absorb energy, they jump to...
• Explanation: In solid-state physics, valence electrons are in the valence band. When they absorb sufficient energy (e.g., thermal or photon energy), they can overcome the energy gap and jump to the conduction band, where they become free to conduct electricity.
• Answer: c. conduction band

Question 8 (Page 43 & Page 44): Force when Current Parallel to Magnetic Field

• Topic: Magnetism, Force on Current-Carrying Wire
• Question: If the flow of electric current is parallel to the magnetic field, the force will be.......
• Formula: The magnetic force (F) on a current-carrying wire is: F = I * L * B * sin(θ) where θ is the angle between the current direction (L) and the magnetic field (B).
• Explanation: If the current is parallel to the magnetic field, θ = 0°. Since sin(0°) = 0, the magnetic force on the wire will be zero.
• Answer: a. zero

Question (Page 45): Conducting Bar Sliding, Magnetic Flux Change

• Topic: Electromagnetic Induction, Magnetic Flux
• Question: How is the magnetic flux through the conducting loop changing, if at all, as the bar slides to the right?
• Explanation: The magnetic flux (Φ) through a loop is given by Φ = B * A * cos(θ). Here, the magnetic field (B) is uniform and perpendicular to the area (θ=0, cos(0)=1), so Φ = B * A. As the conducting bar slides to the right, the area (A) enclosed by the conducting loop increases. Since B is constant and A is increasing, the magnetic flux (Φ) through the loop increases.
• Answer: d. yes, the magnetic flux increases

Question 10 (Page 46): Capacitors in Parallel (Effective Plate Area)

• Topic: Capacitance, Parallel Connection
• Question: When capacitors are connected in parallel, what happens to the effective plate area?
• Explanation: When capacitors are connected in parallel, their individual plate areas effectively add up. The equivalent capacitance (Ceq) for parallel capacitors is Ceq = C1 + C2 + ... . Since C = ε₀A/d, this addition implies an increase in the total effective plate area (A) for a given plate separation (d).
• Answer: a. Increases

Question 11 (Page 47): Transformer Core Losses

• Topic: Transformers, Energy Losses
• Question: Transformer core losses include:
• Explanation: Transformer losses are generally categorized into core losses and copper losses. * Core losses (or iron losses) occur in the transformer core and consist of: * Hysteresis loss: Energy lost due to the repeated magnetization and demagnetization of the core material as the alternating magnetic field changes. * Eddy current loss: Energy lost due to circulating currents induced in the core material itself by the changing magnetic flux. * Copper losses (or I²R losses) occur in the windings due to the resistance of the coil wires.
• Answer: b. Hysteresis loss and Eddy current loss

Question 12 (Page 48): R-L Circuit (Power Factor)

• Topic: AC Circuits, R-L Series Circuit, Power Factor
• Given: Inductance (L) = 0.5 H, Resistance (R) = 250 Ω, Source voltage (V) = 100 V (RMS), Frequency (f) = 50 Hz.
• Formulas:
  • Angular frequency (ω) = 2πf
  • Inductive Reactance (XL) = ωL
  • Impedance (Z) = √(R2 + XL2)
  • Power factor (cos φ) = R/Z
• Calculation: 1. ω = 2π * 50 Hz = 100π rad/s ≈ 314.16 rad/s. 2. XL = 100π rad/s * 0.5 H = 50π Ω ≈ 157.08 Ω. 3. Z = √(2502 + (50π)2) = √(62500 + 24674) = √87174 ≈ 295.25 Ω. 4. cos φ = 250 Ω / 295.25 Ω ≈ 0.8467.
• Answer: b. 0.85

Question 13 (Page 49): Electric Flux (Rectangular Surface)

• Topic: Electrostatics, Electric Flux
• Given: Electric field (E) = 4.50 kN/C = 4.50 × 103 N/C. Rectangular surface area (A) = 12 cm × 14 cm = 0.12 m × 0.14 m = 0.0168 m2. Angle (θ) = 60° with the normal to the area.
• Formula: Electric Flux (Φ) = E * A * cos(θ)
• Calculation: Φ = (4.50 × 103 N/C) * (0.0168 m2) * cos(60°) Φ = 4500 * 0.0168 * 0.5 = 75.6 * 0.5 = 37.8 Nm2/C The calculated value is 37.8 Nm2/C. Options b and d are 378 Nm2/C, implying a potential typo in the area (e.g., if the area was 0.168 m2 or E was 45 kN/C). Assuming a typo in the question or options for the final numerical match.
• Answer: Calculated: 37.8 Nm2/C. Matching option (assuming error in problem statement): b. 378Nm²/C

Question 14 (Page 49): Force between Similar Charges

• Topic: Electrostatics, Coulomb's Law
• Question: In case the charges are similar the force is of?
• Explanation: According to Coulomb's Law, like charges (both positive or both negative) repel each other, while opposite charges attract.
• Answer: c. Repulsion

Question 15 (Page 50): Value of Two Equal Charges

• Topic: Electrostatics, Coulomb's Law
• Given: Two equal charges (q1 = q2 = q). Repulsive force (F) = 0.2 N. Distance (r) = 50 cm = 0.5 m. In vacuum (k = 9 × 109 Nm2/C2).
• Formula: F = k * q2 / r2
• Calculation: q2 = (F * r2) / k = (0.2 N * (0.5 m)2) / (9 × 109 Nm2/C2) q2 = (0.2 * 0.25) / (9 × 109) = 0.05 / (9 × 109) = (5/9) × 10-11 = 5.55... × 10-12 C2 q = √(5.55... × 10-12) ≈ 2.357 × 10-6 C The calculated value is approximately 2.36 μC. None of the options match this result directly. If the force was 0.1 N, then q would be approximately 1.67 × 10-6 C, which is close to option a. 1.7 X 10-6 C. There appears to be a numerical discrepancy in the problem or options.
• Answer: Calculated: 2.36 × 10-6 C