🔹 Topic: Ohm’s Law| MCQs: 50 🔹
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1. The SI unit of electrical conductivity is:
Correct Answer: B) Ω⁻¹m⁻¹
Electrical conductivity (σ) is the reciprocal of resistivity (ρ), and since resistivity has units of Ωm, conductivity has units of Ω⁻¹m⁻¹.
2. A wire of resistance 4Ω is stretched to double its length. The new resistance is:
Correct Answer: C) 16Ω
When a wire is stretched to double its length, its cross-sectional area becomes half (volume remains constant).
Resistance R = ρL/A. New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R = 16Ω.
3. Ohm's law is obeyed by:
Correct Answer: D) Only when physical conditions remain unchanged
Ohm's law (V = IR) is obeyed only when the physical conditions (like temperature) remain constant. Many materials, including semiconductors and electrolytes, don't obey Ohm's law under varying conditions.
4. The resistance of a conductor depends on:
Correct Answer: D) All of the above
Resistance R = ρL/A, where ρ is resistivity (depends on material), L is length, and A is cross-sectional area. Thus, resistance depends on all these factors.
5. The resistivity of a wire depends on:
Correct Answer: C) Material
Resistivity (ρ) is an intrinsic property of the material and does not depend on the dimensions (length or cross-section) of the conductor.
6. The temperature coefficient of resistance is positive for:
Correct Answer: A) Metals
Metals have positive temperature coefficient of resistance (resistance increases with temperature), while semiconductors and insulators have negative temperature coefficient.
7. If the length of a wire is halved and its diameter is doubled, the new resistance will be:
Correct Answer: C) 1/8th of original
Resistance R = ρL/A = ρL/(πr²). New length L' = L/2, new radius r' = 2r (since diameter is doubled).
New resistance R' = ρ(L/2)/(π(2r)²) = (ρL/2)/(4πr²) = (1/8)(ρL/πr²) = R/8.
8. The resistance of a conductor increases with temperature because:
Correct Answer: B) Relaxation time decreases
With increase in temperature, the thermal vibrations of lattice ions increase, which decreases the relaxation time (average time between collisions) of electrons, thereby increasing resistance.
9. A 10Ω resistor carries a current of 2A. The power dissipated is:
Correct Answer: C) 40W
Power dissipated P = I²R = (2A)² × 10Ω = 4 × 10 = 40W.
10. The resistivity of a superconductor is:
Correct Answer: C) Zero
Superconductors have exactly zero resistivity below their critical temperature, allowing current to flow without any energy loss.
11. The resistance of a wire is R. If it is melted and recast into a wire of half its original length, the new resistance will be:
Correct Answer: B) R/4
When melted and recast, volume remains same. New length L' = L/2 ⇒ new area A' = 2A (since V = AL = A'L').
Resistance R' = ρ(L/2)/(2A) = (1/4)ρL/A = R/4.
12. The current-voltage graph for a non-ohmic conductor is:
Correct Answer: C) Curve passing through origin
For non-ohmic conductors, the I-V graph is not linear but still passes through the origin (zero current at zero voltage).
13. The equivalent resistance between points A and B in the given circuit (two 2Ω resistors in series connected in parallel with another two 2Ω resistors in series) is:
Correct Answer: B) 2Ω
Each series branch has 2Ω + 2Ω = 4Ω resistance. Two 4Ω resistors in parallel give equivalent resistance of (4×4)/(4+4) = 2Ω.
14. The drift velocity of electrons in a conductor is of the order of:
Correct Answer: B) 10⁻⁶ m/s
Drift velocity is typically very small, of the order of mm/s or 10⁻⁶ m/s, even though the electric signal propagates at nearly the speed of light.
15. The resistance of a human body is typically about:
Correct Answer: C) 10kΩ
The resistance of dry human skin is typically about 10,000Ω to 100,000Ω, but can drop significantly when wet.
16. A potential difference V is applied across a conductor of length L and cross-section A. If the potential difference is doubled, the drift velocity becomes:
Correct Answer: C) Double
Drift velocity \( v_d = \frac{eEτ}{m} = \frac{eVτ}{mL} \), where E is electric field (V/L). Thus, v_d is directly proportional to V.
17. The resistivity of a semiconductor at 0K is:
Correct Answer: B) Infinity
At 0K, semiconductors behave as perfect insulators because no electrons have enough energy to jump to the conduction band, making resistivity infinite.
18. The current in a wire varies with time as I = 2t + 3t². The charge flowing through a cross-section in 2 seconds is:
Correct Answer: D) 16 C
Charge \( Q = \int I \, dt = \int_0^2 (2t + 3t²) dt = [t² + t³]_0^2 = (4 + 8) - 0 = 12 \) C. (Note: There seems to be a discrepancy in the options vs calculation)
19. The resistance of a bulb filament is 100Ω at room temperature. When it glows, its resistance becomes 500Ω. The temperature coefficient of resistance of the filament material is 4.5 × 10⁻³/°C. The temperature of the glowing filament is about:
Correct Answer: A) 900°C
Using \( R = R_0(1 + αΔT) \), \( 500 = 100(1 + 4.5×10⁻³×ΔT) \) ⇒ \( ΔT = (5-1)/(4.5×10⁻³) ≈ 889°C \). Assuming room temperature as 27°C, final temperature ≈ 916°C ≈ 900°C.
20. A wire carries a current of 1A. The number of electrons passing through a cross-section per second is:
Correct Answer: A) 6.25 × 10¹⁸
Current I = ne/t ⇒ n = It/e = (1A × 1s)/(1.6×10⁻¹⁹C) = 6.25 × 10¹⁸ electrons.
21. The resistance of a wire is 10Ω. It is stretched so that its length increases by 25%. Its resistance will now be:
Correct Answer: C) 15.6Ω
New length L' = 1.25L ⇒ new area A' = A/1.25 (volume constant). New resistance R' = ρ(1.25L)/(A/1.25) = ρL/A × (1.25)² = 10 × 1.5625 = 15.625Ω ≈ 15.6Ω.
22. The equivalent resistance between points A and B in an infinite ladder network of 1Ω resistors is:
Correct Answer: A) (1+√5)/2 Ω
For infinite ladder, let equivalent resistance be R. The circuit remains identical if one section is removed, so R = 1 + (1×R)/(1+R). Solving gives R² - R - 1 = 0 ⇒ R = (1+√5)/2 Ω.
23. A 100W, 200V bulb is connected to a 100V supply. The power consumed will be:
Correct Answer: C) 25W
Resistance R = V²/P = (200)²/100 = 400Ω. At 100V, power P' = V'²/R = (100)²/400 = 25W.
24. The resistivity of a material is 2×10⁻⁸ Ωm. Its conductivity is:
Correct Answer: A) 5×10⁷ Ω⁻¹m⁻¹
Conductivity σ = 1/ρ = 1/(2×10⁻⁸) = 0.5×10⁸ = 5×10⁷ Ω⁻¹m⁻¹.
25. The resistance of a conductor is 5Ω at 50°C and 6Ω at 100°C. Its resistance at 0°C is:
Correct Answer: D) 4Ω
Using \( R = R_0(1 + αT) \), we get two equations:
5 = R₀(1 + 50α) and 6 = R₀(1 + 100α).
Solving gives α = 0.01/°C and R₀ = 4Ω.
26. A current of 1A flows through a copper wire. How many electrons pass through a cross-section in 1.6 seconds?
Correct Answer: A) 10¹⁹
Charge Q = It = 1A × 1.6s = 1.6C. Number of electrons n = Q/e = 1.6C/(1.6×10⁻¹⁹C) = 10¹⁹.
27. The resistance of a wire is R. It is bent into a circle. The resistance between two diametrically opposite points is:
Correct Answer: C) R/4
When bent into a circle, the two halves (each of length L/2) are in parallel. Resistance of each half is R/2. Parallel combination gives (R/2 × R/2)/(R/2 + R/2) = R/4.
28. The current density in a wire is 10⁶ A/m² and the drift velocity is 10⁻⁴ m/s. The number of free electrons per unit volume is:
Correct Answer: C) 6.25 × 10²⁸ m⁻³
Current density J = nevd ⇒ n = J/(evd) = 10⁶/(1.6×10⁻¹⁹ × 10⁻⁴) = 6.25 × 10²⁸ m⁻³.
29. The resistance of a wire is R. If its length is doubled by stretching, its new resistance will be:
Correct Answer: B) 4R
When length doubles, area becomes half (volume constant). New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R.
30. The resistance of a wire is R. It is cut into n equal parts. All parts are connected in parallel. The equivalent resistance is:
Correct Answer: B) R/n²
Each part has resistance R/n. Parallel combination of n resistors each of R/n gives equivalent resistance (R/n)/n = R/n².
31. The current in a wire is given by I = (3t² + 2t + 1)A. The charge flowing through a cross-section between t = 1s and t = 2s is:
Correct Answer: B) 11 C
Charge \( Q = \int_{1}^{2} (3t² + 2t + 1) dt = [t³ + t² + t]_1^2 = (8 + 4 + 2) - (1 + 1 + 1) = 14 - 3 = 11 \) C.
32. The resistance of a wire is R. If its radius is halved, its new resistance will be:
Correct Answer: B) 4R
Resistance R = ρL/A = ρL/(πr²). If radius is halved (r' = r/2), new resistance R' = ρL/(π(r/2)²) = 4ρL/(πr²) = 4R.
33. The resistivity of a material depends on:
Correct Answer: C) Temperature
Resistivity is an intrinsic property that depends on the material and temperature, but not on dimensions (length or cross-section).
34. A wire of resistance 12Ω is bent into an equilateral triangle. The equivalent resistance between any two vertices is:
Correct Answer: C) 8Ω
Each side has resistance 4Ω (total 12Ω for 3 sides). Between two vertices, two sides are in series (4+4=8Ω) and this combination is in parallel with the third side (4Ω). Equivalent resistance = (8×4)/(8+4) = 32/12 ≈ 2.67Ω. (Note: There seems to be inconsistency in the question/answer)
35. The resistance of a wire is R. If its length is doubled and diameter is halved, its new resistance will be:
Correct Answer: D) 8R
R = ρL/A = ρL/(πr²). New length L' = 2L, new radius r' = r/2 ⇒ new area A' = π(r/2)² = πr²/4. New resistance R' = ρ(2L)/(πr²/4) = 8ρL/(πr²) = 8R.
36. The temperature coefficient of resistance of a wire is 0.00125/°C. At 300K, its resistance is 1Ω. The resistance at 500K will be:
Correct Answer: A) 1.25Ω
ΔT = 500K - 300K = 200K = 200°C. R = R₀(1 + αΔT) = 1Ω(1 + 0.00125×200) = 1(1 + 0.25) = 1.25Ω.
37. The current in a conductor varies with time as I = 2√t. The charge flowing through a cross-section between t = 0 to t = 4s is:
Correct Answer: C) 10.67 C
Charge \( Q = \int_{0}^{4} 2\sqrt{t} \, dt = 2 \times \left[ \frac{2}{3} t^{3/2} \right]_0^4 = \frac{4}{3} (8 - 0) = 32/3 ≈ 10.67 \) C.
38. The resistance of a wire is R. If its material volume remains the same but length is increased by 10%, the new resistance will be:
Correct Answer: B) 1.21R
New length L' = 1.1L ⇒ new area A' = A/1.1 (volume constant). New resistance R' = ρ(1.1L)/(A/1.1) = ρL/A × (1.1)² = R × 1.21 = 1.21R.
39. The equivalent resistance between points A and B in the given circuit (three 3Ω resistors forming a delta connection) is:
Correct Answer: B) 2Ω
For a delta connection with three equal resistors R, the equivalent resistance between any two points is (2/3)R. Here, (2/3)×3Ω = 2Ω.
40. A wire has resistance R. It is cut into five equal parts which are then connected in parallel. The equivalent resistance is:
Correct Answer: B) R/25
Each part has resistance R/5. Parallel combination of five R/5 resistors gives equivalent resistance (R/5)/5 = R/25.
41. The resistance of a wire is R. If its length is increased by 50% and diameter is doubled, its new resistance will be:
Correct Answer: A) 3R/8
New length L' = 1.5L, new diameter D' = 2D ⇒ new radius r' = 2r ⇒ new area A' = π(2r)² = 4πr². New resistance R' = ρ(1.5L)/(4πr²) = (3/8)ρL/(πr²) = 3R/8.
42. The current density in a wire is related to electric field E and conductivity σ as:
Correct Answer: A) J = σE
This is the microscopic form of Ohm's law, where current density J = σE, with σ being conductivity and E electric field.
43. The resistance of a wire is R. If its temperature is increased, its new resistance will be:
Correct Answer: C) Increase for metals, decrease for semiconductors
Metals have positive temperature coefficient (resistance increases with temperature), while semiconductors have negative temperature coefficient.
44. The equivalent resistance between points A and B in the given circuit (two 4Ω resistors in parallel connected in series with a 3Ω resistor) is:
Correct Answer: A) 5Ω
Parallel combination of two 4Ω resistors gives 2Ω. This is in series with 3Ω, giving total resistance 2 + 3 = 5Ω.
45. A wire carries a current of 2A. The charge flowing through a cross-section in 30 minutes is:
Correct Answer: C) 3600 C
Charge Q = It = 2A × (30 × 60)s = 2 × 1800 = 3600 C.
46. The resistance of a wire is R. If its length is tripled and cross-sectional area is doubled, its new resistance will be:
Correct Answer: B) 3R/2
New length L' = 3L, new area A' = 2A. New resistance R' = ρ(3L)/(2A) = (3/2)ρL/A = 3R/2.
47. The resistivity of copper is 1.7 × 10⁻⁸ Ωm. The resistance of a copper wire of length 1m and cross-sectional area 1mm² is:
Correct Answer: A) 1.7 × 10⁻² Ω
R = ρL/A = (1.7×10⁻⁸ × 1)/(1×10⁻⁶) = 1.7 × 10⁻² Ω (Note: 1mm² = 10⁻⁶ m²).
48. The resistance of a wire is R. If its length is increased by 20% and radius is decreased by 10%, its new resistance will be:
Correct Answer: C) 1.48R
New length L' = 1.2L, new radius r' = 0.9r ⇒ new area A' = π(0.9r)² = 0.81πr². New resistance R' = ρ(1.2L)/(0.81πr²) = (1.2/0.81)ρL/(πr²) ≈ 1.48R.
49. The equivalent resistance between points A and B in the given circuit (three 6Ω resistors forming a star connection) is:
Correct Answer: A) 2Ω
For a star connection with three equal resistors R, the equivalent resistance between any two points is (2/3)R. Here, (2/3)×6Ω = 4Ω. (Note: There seems to be inconsistency in the question/answer)
50. A current of 2A flows through a resistor of 10Ω for 5 minutes. The heat produced is:
Correct Answer: D) 12000 J
Heat H = I²Rt = (2)² × 10 × (5 × 60) = 4 × 10 × 300 = 12000 J.
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