Try each question on your own first. Use the Show Answer button only after you have selected your option, then read the explanation and revise the related concept from your chapter notes.
Explanation: Electric current is defined as rate of flow of charge (I = Q/t), and its SI unit is Ampere (A), equivalent to Coulomb per second.
2. Electric current is a scalar quantity because:
A. It has magnitude only
B. It does not obey laws of vector addition
C. It flows in all directions
D. Its unit is Ampere
Answer: B. It does not obey laws of vector addition
Explanation: Although electric current has direction, it adds up algebraically like scalars rather than obeying vector addition rules, so it is a scalar quantity.
3. What is current density (J)?
A. Scalar quantity (J = I × A)
B. Vector quantity (J = I / A)
C. Scalar quantity (J = I / A)
D. Dimensionless quantity
Answer: B. Vector quantity (J = I / A)
Explanation: Current density is defined as electric current per unit cross-sectional area (J = I/A) and is a vector quantity directed along current flow.
4. According to Ohm's Law, the relation between potential difference (V) and current (I) is:
A. V ∝ I
B. V ∝ 1/I
C. V ∝ I²
D. V is independent of I
Answer: A. V ∝ I
Explanation: Ohm's Law states that current flowing through a conductor is directly proportional to potential difference across its ends (V = IR), provided physical conditions remain constant.
5. The microscopic form of Ohm’s Law is given by:
A. J = σ E
B. V = IR
C. I = n e A v_d
D. E = ρ J²
Answer: A. J = σ E
Explanation: The vector form/microscopic form of Ohm's Law relates current density (J), conductivity (σ), and electric field (E) as J = σ E.
6. What is the relation between electric current (I) and drift velocity (v_d)?
A. I = n e A v_d
B. I = n e / (A v_d)
C. I = v_d / (n e A)
D. I = n A v_d
Answer: A. I = n e A v_d
Explanation: Electric current flowing in a conductor is related to drift velocity by I = n e A v_d, where n is electron density and A is area of cross section.
7. When temperature of a metallic conductor increases, its resistance:
A. Decreases
B. Increases
C. Remains constant
D. Becomes zero
Answer: B. Increases
Explanation: Increasing temperature increases thermal vibrations of metal ions, decreasing relaxation time (τ) between electron collisions, thereby increasing resistance.
8. When temperature of a semiconductor increases, its resistance:
A. Increases
B. Decreases
C. Remains same
D. First increases then decreases
Answer: B. Decreases
Explanation: In semiconductors, raising temperature breaks covalent bonds, significantly increasing free charge carrier density (n), which lowers resistance.
9. What is the SI unit of Specific Resistance (Resistivity, ρ)?
A. Ohm (Ω)
B. Ohm-meter (Ω m)
C. Ohm / meter
D. Mho
Answer: B. Ohm-meter (Ω m)
Explanation: Resistance R = ρ L / A, which gives ρ = R A / L. The unit is (Ω × m²) / m = Ohm-meter (Ω m).
10. Electrical conductance is defined as the reciprocal of resistance. Its SI unit is:
A. Siemens (S) or Ohm^-1
B. Ohm-meter
C. Henry
D. Tesla
Answer: A. Siemens (S) or Ohm^-1
Explanation: Electrical conductance G = 1/R, so its SI unit is Siemens (S) or mho (Ω^-1).
11. If a wire of resistance R is stretched to double its original length keeping mass constant, its new resistance will be:
A. 2 R
B. 4 R
C. R / 2
D. R / 4
Answer: B. 4 R
Explanation: When length doubles (L' = 2L), area becomes half (A' = A/2) to keep volume constant. Thus R' = ρ (2L) / (A/2) = 4 (ρ L / A) = 4R.
12. What is the order of drift velocity of free electrons in a conductor?
A. 10^8 m/s
B. 10^-4 m/s
C. 10^3 m/s
D. 10^-10 m/s
Answer: B. 10^-4 m/s
Explanation: The average drift velocity of electrons under an applied electric field in standard conductors is very small, typically around 10^-4 m/s (or a few mm/s).
13. Kirchhoff’s First Law (Junction Rule), ∑ I = 0, is based on the conservation of:
A. Energy
B. Charge
C. Momentum
D. Mass
Answer: B. Charge
Explanation: Kirchhoff's Junction Rule states total current entering a junction equals total current leaving, which directly reflects conservation of electric charge.
14. Kirchhoff’s Second Law (Loop Rule), ∑ ΔV = 0, is based on the conservation of:
A. Charge
B. Energy
C. Linear momentum
D. Angular momentum
Answer: B. Energy
Explanation: The loop rule states sum of potential changes around any closed loop is zero, which is a statement of conservation of energy for electric circuits.
15. Wheatstone bridge is balanced when:
A. P / Q = R / S
B. P × Q = R × S
C. P + Q = R + S
D. P - Q = R - S
Answer: A. P / Q = R / S
Explanation: When no current flows through the galvanometer, the bridge is balanced and ratio of adjacent resistances satisfies P/Q = R/S.
16. Wheatstone bridge is most sensitive when all four resistances are:
A. Of equal magnitude order
B. Very high
C. Very low
D. Zero
Answer: A. Of equal magnitude order
Explanation: A Wheatstone bridge offers maximum sensitivity when all four arm resistances are nearly equal in order of magnitude.
17. Meter Bridge works on the principle of:
A. Potentiometer
B. Wheatstone Bridge
C. Faraday's Law
D. Joule's Heating Effect
Answer: B. Wheatstone Bridge
Explanation: A meter bridge is a practical setup operating on the balanced Wheatstone bridge principle to measure an unknown resistance.
18. The terminal potential difference (V) of a cell of EMF 'E' and internal resistance 'r' during discharging is:
A. V = E + Ir
B. V = E - Ir
C. V = E / Ir
D. V = E
Answer: B. V = E - Ir
Explanation: When a cell delivers current (discharging), terminal potential difference drops below EMF due to internal resistance loss: V = E - Ir.
19. When a cell is being charged by an external source, its terminal potential difference (V) is:
A. V = E - Ir
B. V = E + Ir
C. V = 0
D. V = E
Answer: B. V = E + Ir
Explanation: During charging, current enters positive terminal of the cell, making terminal voltage higher than its EMF: V = E + Ir.
20. In parallel combination of two cells of EMFs E₁, E₂ and internal resistances r₁, r₂, equivalent EMF is:
A. E_eq = (E₁r₂ + E₂r₁) / (r₁ + r₂)
B. E_eq = E₁ + E₂
C. E_eq = (E₁r₁ + E₂r₂) / (r₁ + r₂)
D. E_eq = E₁ - E₂
Answer: A. E_eq = (E₁r₂ + E₂r₁) / (r₁ + r₂)
Explanation: For two cells connected in parallel aiding each other, equivalent EMF is given by E_eq = (E₁/r₁ + E₂/r₂) / (1/r₁ + 1/r₂) = (E₁r₂ + E₂r₁) / (r₁ + r₂).
21. Potentiometer is preferred over a voltmeter for measuring EMF because:
A. It draws no current from the cell at null point
B. It is very small and portable
C. It measures internal resistance directly
D. It works on AC only
Answer: A. It draws no current from the cell at null point
Explanation: Potentiometer acts as an ideal voltmeter with infinite resistance at null balance point, measuring true open-circuit EMF without drawing current.
22. Sensitivity of a potentiometer can be increased by:
A. Decreasing length of potentiometer wire
B. Increasing length of potentiometer wire
C. Increasing current in primary circuit
D. Using thicker wire
Answer: B. Increasing length of potentiometer wire
Explanation: Increasing wire length decreases potential gradient (K = V/L), making the potentiometer more sensitive to small voltage differences.
23. According to Joule's Law of heating, heat produced in a resistor of resistance R carrying current I for time t is:
A. H = I R t
B. H = I² R t
C. H = I R² t
D. H = I² R / t
Answer: B. H = I² R t
Explanation: Joule's heating effect states heat generated H = I² R t (in Joules), directly proportional to square of current, resistance, and time.
24. What is the electric power consumed by a device of resistance R connected across potential difference V?
A. P = V I = I² R = V² / R
B. P = V / I
C. P = V² R
D. P = I R²
Answer: A. P = V I = I² R = V² / R
Explanation: Electric power is work done per unit time: P = V × I. Substituting V = IR gives P = I²R = V²/R.
25. Two electric bulbs rated (60 W, 220 V) and (100 W, 220 V) are connected in series across 220 V supply. Which bulb will glow brighter?
A. 100 W bulb
B. 60 W bulb
C. Both glow equally
D. Neither will glow
Answer: B. 60 W bulb
Explanation: R = V²/P, so 60 W bulb has higher resistance. In series, current I is same, power consumed P_consumed = I²R, so higher resistance bulb (60 W) glows brighter.
This page contains a chapter-wise objective practice set. Try the questions without checking the answer first, then use the explanation to understand the point you missed. The current page contains 25 questions, so you can practise in smaller sets when you are revising.
Quick revision tip
After the practice set, write down the question numbers you missed and revise only those concepts once more. This makes MCQ practice more useful than repeatedly attempting the same questions without reviewing mistakes.
