JEE Physics Current Electricity MCQs Set 07

Answer: (b) 24 C

Explanation:
1. q = ∫ I dt from 2 to 3 = [t² + t³] evaluated between 2 and 3.
2. At t = 3: 9 + 27 = 36. At t = 2: 4 + 8 = 12.
3. q = 36 − 12 = 24 C.

Teacher's Note:
1. Charge is the time integral of current.
2. Do not substitute t = 3 − 2 = 1 directly.

Answer: (b) \( 1.6\times10^{-13}\ \text{A} \)

Explanation:
1. I = (number per second) × e = 10⁶ × 1.6 × 10⁻¹⁹ A.
2. I = 1.6 × 10⁻¹³ A.
3. The area is not needed for the current itself.

Teacher's Note:
1. Current is charge crossing per second.
2. The cross-section matters only for current density.

Answer: (c) \( 1.56\times10^{-5}\ \text{m/s} \)

Explanation:
1. n = 8.4 × 10²² cm⁻³ = 8.4 × 10²⁸ m⁻³ and A = 1 mm² = 10⁻⁶ m².
2. v = I/(neA) = 0.21/(8.4 × 10²⁸ × 1.6 × 10⁻¹⁹ × 10⁻⁶).
3. v = 0.21/(1.344 × 10⁴) = 1.56 × 10⁻⁵ m/s.

Teacher's Note:
1. Convert cm³ to m³ (factor 10⁶) and mm² to m² (factor 10⁻⁶).
2. Drift speeds in metals are extremely small.

Answer: (b) \( 2.50\times10^{-3}\ \text{m/s} \)

Explanation:
1. v = I/(neA) = 40/(10²⁹ × 1.6 × 10⁻¹⁹ × 10⁻⁶).
2. Denominator = 1.6 × 10⁴, so v = 2.5 × 10⁻³ m/s.

Teacher's Note:
1. Use I = neAv_d.
2. Check the powers of ten carefully.

Answer: (c) the average velocity of a free electron over a long period is zero

Explanation:
1. Free electrons are always in random thermal motion, so they move and their speed is not zero.
2. The motion is random, so the average velocity over a long period is zero (no net drift).
3. The mean of v² is not zero.

Teacher's Note:
1. Distinguish speed (scalar, not zero) from velocity (vector, average zero).
2. Drift appears only when an electric field is applied.

Answer: (d) current only

Explanation:
1. In steady flow the same charge passes every cross-section per second, so I is constant.
2. Since I = neAv_d, the drift speed varies inversely with area, and E = J/σ also varies with area.

Teacher's Note:
1. Charge cannot pile up anywhere in steady flow.
2. Current density and drift velocity depend on the local area.

Answer: (b) 8 A

Explanation:
1. Electrons per metre of wire = nA = 2 × 10²¹/0.01 = 2 × 10²³ m⁻¹.
2. I = (nA) e v = 2 × 10²³ × 1.6 × 10⁻¹⁹ × 0.25 × 10⁻³.
3. I = 8 A.

Teacher's Note:
1. nA is the number of free electrons per unit length of the wire.
2. Convert 0.25 mm/s to 0.25 × 10⁻³ m/s.

Answer: (b) 0.66 A towards the right

Explanation:
1. Positive ions moving right give conventional current to the right: 2.9 × 10¹⁸ × 1.6 × 10⁻¹⁹ = 0.464 A.
2. Electrons moving left also give conventional current to the right: 1.2 × 10¹⁸ × 1.6 × 10⁻¹⁹ = 0.192 A.
3. Total current = 0.464 + 0.192 ≈ 0.66 A towards the right.

Teacher's Note:
1. Electrons moving left are equivalent to positive charge moving right.
2. The two contributions add.

Answer: (c) 4 : 1

Explanation:
1. The same current flows through both, and I = neAv_d with equal A and e.
2. So v_d ∝ 1/n.
3. v₁ : v₂ = 4 : 1.

Teacher's Note:
1. In series the current is the same everywhere.
2. The wire with fewer carriers needs faster drift.

Answer: (d) \( 4v \)

Explanation:
1. Area ∝ d², so halving the diameter makes the area A/4.
2. I = neAv_d is constant, so v_d ∝ 1/A.
3. The new drift velocity is 4v.

Teacher's Note:
1. Area depends on the square of the diameter.
2. n and e are unchanged for the same material.

Answer: (c) minimum for the wire of cross-section 2A

Explanation:
1. R = ρl/A.
2. R₁ = ρl/A, R₂ = ρ(2l)/(A/2) = 4ρl/A, R₃ = ρ(l/2)/(2A) = ρl/4A.
3. R₃ is the smallest, and it belongs to the wire of area 2A.

Teacher's Note:
1. Compare all three in terms of ρl/A.
2. Short and thick means low resistance.

Answer: (b) \( R/4 \)

Explanation:
1. The two points divide the circle into two semicircles, each of resistance R/2.
2. They are in parallel: (R/2)(R/2)/(R/2 + R/2) = R/4.

Teacher's Note:
1. Each half of the loop is a separate path between the two points.
2. Equal resistors in parallel give half the value of one.

Answer: (a) the resistance is halved and the specific resistance is unchanged

Explanation:
1. R' = ρ(2l)/(π(2r)²) = ρ(2l)/(4πr²) = R/2.
2. Specific resistance is a property of the material and does not change.

Teacher's Note:
1. Resistivity depends on material and temperature, not on dimensions.
2. Doubling radius multiplies the area by 4.

Answer: (d) sixteen times

Explanation:
1. Volume stays constant: A' = A/4 so l' = 4l.
2. R' = ρl'/A' = ρ(4l)/(A/4) = 16 ρl/A.
3. The resistance becomes 16 times.

Teacher's Note:
1. When a wire is drawn, volume (lA) is constant, so R ∝ l² ∝ 1/A².
2. Halving the diameter gives (4)² = 16.

Answer: (d) remains constant

Explanation:
1. Conductivity σ = 1/ρ.
2. So ρσ = 1 at every temperature.

Teacher's Note:
1. Conductivity is defined as the reciprocal of resistivity.
2. The individual values change but the product does not.

Answer: (c) 44%

Explanation:
1. For constant volume R ∝ l², so R'/R = (1.2)² = 1.44.
2. The resistance rises by 44%.

Teacher's Note:
1. Stretching reduces the area in proportion, so the effect on R is squared.
2. Do not simply double 20%.

Answer: (a) 300%

Explanation:
1. Length doubles, so R' = 2² R = 4R.
2. Increase = 3R, i.e. 300%.

Teacher's Note:
1. For a wire stretched at constant volume, R ∝ l².
2. Percentage change = (new − old)/old.

Answer: (c) increases and decreases respectively

Explanation:
1. In a conductor, more frequent collisions increase resistivity.
2. In a semiconductor, more charge carriers are freed, so resistivity decreases.

Teacher's Note:
1. Metals have a positive temperature coefficient; semiconductors a negative one.
2. Carrier density matters most in semiconductors.

Answer: (a) increase in temperature

Explanation:
1. Specific resistance (resistivity) is a property of the material and depends on temperature, not on dimensions.
2. For a metal it rises with temperature.

Teacher's Note:
1. Dimensions affect resistance, not resistivity.
2. ρ = ρ₀[1 + α(T − T₀)] for metals.

Answer: (b) 1127 K

Explanation:
1. Take R_t = R₀(1 + αt) with t in °C. At 300 K (27 °C): R₀ = 1/(1 + 0.00125 × 27) = 0.9674 Ω.
2. For R = 2 Ω: 1 + αt = 2/0.9674 = 2.0675, so t = 0.0675/0.00125 = 854 °C = 1127 K.

Teacher's Note:
1. The coefficient α is referred to the resistance at 0 °C.
2. Convert to kelvin at the end by adding 273.

Answer: (d) \( 3\ \Omega \)

Explanation:
1. After the break only one wire remains, with resistance 2 Ω.
2. 1/R_parallel = 1/2 + 1/R_broken = 5/6, so 1/R_broken = 5/6 − 1/2 = 1/3.
3. The broken wire had resistance 3 Ω.

Teacher's Note:
1. The remaining wire alone gives 2 Ω.
2. Use the parallel formula to find the other.

Answer: (c) 500 °C

Explanation:
1. R = R₀(1 + αt) = 3R₀.
2. 1 + 0.004t = 3, so t = 2/0.004 = 500 °C.

Teacher's Note:
1. Set the ratio R/R₀ = 3.
2. α is per degree Celsius.

Answer: (a) decreases in copper but increases in silicon

Explanation:
1. Copper is a metal: resistivity falls as temperature falls.
2. Silicon is a semiconductor: fewer carriers at low temperature, so resistivity rises.

Teacher's Note:
1. Metals and semiconductors behave oppositely with temperature.
2. At very low temperature a semiconductor becomes nearly an insulator.

Answer: (d) \( \beta/\alpha \)

Explanation:
1. Change in R₁ is R₁α ΔT and in R₂ is −R₂β ΔT.
2. For no net change R₁α = R₂β.
3. R₁/R₂ = β/α.

Teacher's Note:
1. The increase of one must be cancelled by the decrease of the other.
2. Series resistances add.

Answer: (c) \( R/9 \)

Explanation:
1. Each piece has resistance R/3.
2. Three equal resistors in parallel: (R/3)/3 = R/9.

Teacher's Note:
1. Cutting divides the resistance; paralleling divides it again.
2. The combined effect is a factor of 9.