25 questions in 25 minutes on the NTA computer-based test interface, all from Current Electricity. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Current Electricity, of which 8 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
Revising first? Read the Current Electricity chapter notes, then come back and take this test to check it stuck.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Current Electricity mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 8 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Current Electricity?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Physics
- Q1To get the minimum possible resistance from a set of resistors, they should be connected:
- all in series
- in a mixed arrangement
- all in parallel
- two in series, rest in parallel
Answer: (C) all in parallel
Parallel combination always yields an equivalent resistance smaller than the smallest individual resistor, giving the minimum possible value.
- Q2The specific resistance of a conductor increases with:
- decrease in length
- increase in temperature
- decrease in cross-sectional area
- increase in cross-sectional area
Answer: (B) increase in temperature
Specific resistance (resistivity) is independent of the dimensions (length, area) of the sample. For a metallic conductor it rises with temperature because more frequent lattice collisions reduce the relaxation time τ (ρ = m/ne²τ).
- Q3A cell is balanced against 110 cm and 100 cm of a potentiometer wire respectively, without and with being short-circuited through a 10 Ω resistance. Its internal resistance is:
- 1.0 Ω
- zero
- 2.0 Ω
- 0.5 Ω
Answer: (A) 1.0 Ω
The open-circuit balance (110 cm) reads the EMF; the balance with the 10 Ω across the cell (100 cm) reads the terminal voltage. r = R((l₁ – l₂)/(l₂)) = 10×(110 – 100)/(100) = 10×0.1 = 1.0 Ω.
- Q4Across a metallic conductor of non-uniform cross-section a constant potential difference is applied. The quantity that remains constant along the conductor is the:
- current
- current density
- drift velocity
- electric field
Answer: (A) current
By the junction rule (charge conservation) the same current flows through every cross-section in a steady state. Because the area varies, the current density J = I/A, the drift velocity v_d = J/ne and the field all change; only the current I stays constant.
- Q5From four identical resistors, the maximum equivalent resistance obtainable is 16 times the minimum equivalent resistance. This factor of 16 arises because:
- Rₛₑᵣᵢₑₛ/Rₚₐᵣₐₗₗₑₗ = n³, with n = 4
- Rₛₑᵣᵢₑₛ/Rₚₐᵣₐₗₗₑₗ = n², with n = 4
- Rₛₑᵣᵢₑₛ/Rₚₐᵣₐₗₗₑₗ = n, with n = 4
- Rₛₑᵣᵢₑₛ/Rₚₐᵣₐₗₗₑₗ = 2n, with n = 4
Answer: (B) Rₛₑᵣᵢₑₛ/Rₚₐᵣₐₗₗₑₗ = n², with n = 4
For n equal resistors R: max = all in series = nR; min = all in parallel = R/n. Ratio = nR/(R/n) = n². For n = 4, n² = 16.
- Q6n equal resistors are first connected in series and then connected in parallel. The ratio of the maximum (series) to the minimum (parallel) equivalent resistance is:
- n
- (1)/(n²)
- (1)/(n)
- n²
Answer: (D) n²
For n equal resistors R: series gives nR (maximum) and parallel gives R/n (minimum). The ratio is (nR)/(R/n) = n².
- Q7A ’60 W, 220 V’ bulb and a ‘100 W, 220 V’ bulb are connected in series across a 220 V supply. The total power consumed by the two bulbs together is approximately:
- 37.5 W
- 160 W
- 75 W
- 80 W
Answer: (A) 37.5 W
R₆₀ = 220²/60 ≈ 806.7 Ω, R₁₀₀ = 220²/100 = 484 Ω. In series Rₜₒₜₐₗ ≈ 1290.7 Ω, I = 220/1290.7 ≈ 0.1705 A, total P = I×220 ≈ 37.5 W. (In series the combined power is even less than the smaller bulb’s rating.)
- Q8If the voltage across a bulb rated ‘220 V, 100 W’ drops by 2.5% of its rated value, the percentage by which the power decreases is (assume constant resistance):
- 2.5%
- 10%
- 20%
- 5%
Answer: (D) 5%
Since P = V²/R with R constant, (Δ P)/(P) = 2(Δ V)/(V). For a 2.5% drop in V, the power decreases by 2×2.5% = 5%.
- Q9A wire of resistance 12 Ω is bent to form a complete circle. The resistance between two diametrically opposite points is:
- 3 Ω
- 12 Ω
- 24 Ω
- 6 Ω
Answer: (A) 3 Ω
Each half of the ring is 6 Ω. The two halves are in parallel between diametrically opposite points: (6×6)/(6+6) = 3 Ω.
- Q10A cell has an EMF of 1.5 V. When connected across an external resistance of 2 Ω, the terminal potential difference falls to 1.0 V. The internal resistance of the cell is:
- 1.5 Ω
- 0.5 Ω
- 1.0 Ω
- 2 Ω
Answer: (C) 1.0 Ω
r = ((ε – V)/(V))R = ((1.5 – 1.0)/(1.0))×2 = 0.5×2 = 1.0 Ω.
- Q11The velocity of charge carriers of current (about 1 A) in a metal under normal conditions is of the order of:
- a fraction of mm/s
- several thousand m/s
- the velocity of light
- a few hundred m/s
Answer: (A) a fraction of mm/s
Free electrons in a metal have very high random thermal speeds, but under an applied field their net drift speed is extremely small. From I = neAv_d with n≈10²⁹ m⁻³, v_d for a 1 A current works out to about 10⁻⁴ m/s, i.e. a fraction of a millimetre per second.
- Q12In a potentiometer, a standard cell of EMF 1.08 V gives a balance length of 50 cm. An unknown cell balances at 75 cm on the same wire. The EMF of the unknown cell is:
- 1.62 V
- 1.08 V
- 0.72 V
- 2.16 V
Answer: (A) 1.62 V
EMF is proportional to balancing length: ε = 1.08 × (75/50) = 1.08 × 1.5 = 1.62 V.
- Q13The terminal voltage of a cell equals its EMF when:
- internal resistance is very large
- current drawn is maximum
- no current is drawn (open circuit)
- the cell is short-circuited
Answer: (C) no current is drawn (open circuit)
V = ε − Ir. When I = 0 (open circuit), the Ir drop vanishes and V = ε. This is the basis of measuring EMF with a potentiometer.
- Q14Five resistors, each 10 Ω, are connected as the four arms of a Wheatstone bridge with the fifth resistor as the galvanometer (bridge) arm. The equivalent resistance between the two opposite (input) corners is:
- 25 Ω
- 15 Ω
- 5 Ω
- 10 Ω
Answer: (D) 10 Ω
With all four arms equal (10 Ω), the bridge is balanced, so no current flows through the central 10 Ω – it can be removed. The two arms on each side form 10 + 10 = 20 Ω, and the two 20 Ω branches are in parallel: (20×20)/(20+20) = 10 Ω.
- Q15In a meter bridge, with a known resistance of 6 Ω in the right gap, the balance point is found at 40 cm from the left end (where the unknown X is). The value of X is:
- 12 Ω
- 6 Ω
- 9 Ω
- 4 Ω
Answer: (D) 4 Ω
X/R = l/(100−l) with l = 40 cm: X = 6×40/(100−40) = 6×40/60 = 4 Ω.
- Q16In a metallic conductor, if the relaxation time τ is halved (other factors unchanged), the resistivity ρ:
- is doubled
- remains the same
- becomes one-fourth
- is halved
Answer: (A) is doubled
ρ = m/(ne²τ), so ρ ∝ 1/τ. Halving τ doubles ρ. (Heating a metal reduces τ via more frequent collisions, which is exactly why metallic resistivity rises with temperature.)
- Q17An electric kettle is rated ‘1000 W, 250 V’. The current it draws at rated operation, and hence the minimum fuse rating suitable for it, is:
- 0.25 A
- 2.5 A
- 4 A
- 8 A
Answer: (C) 4 A
I = P/V = 1000/250 = 4 A. The fuse must carry at least this, so a 4 A (or slightly higher) fuse is required.
- Q18A potentiometer wire is 10 m long and has a steady potential difference of 2 V maintained across it. The balancing length for a cell of EMF 1 V is:
- 2 m
- 5 m
- 10 m
- 0.5 m
Answer: (B) 5 m
Potential gradient k = 2 V / 10 m = 0.2 V/m. Balancing length = EMF/k = 1/0.2 = 5 m.
- Q19In a potentiometer experiment, the null point shifts towards the higher-potential end of the wire when a small resistance is introduced in series in the primary (driver) circuit. This happens because adding resistance in the primary:
- reverses the current
- has no effect on the gradient
- decreases the potential gradient of the wire
- increases the potential gradient of the wire
Answer: (C) decreases the potential gradient of the wire
Adding series resistance in the primary lowers the current through the potentiometer wire, so the potential gradient (V per unit length) decreases. A smaller gradient means a longer balancing length for the same test EMF – the null point moves further along the wire.
- Q20The resistance of a metallic wire increases from 100 Ω to 115 Ω when its temperature rises by 30°C. The temperature coefficient of resistance α is:
- 0.05 °C⁻¹
- 0.005 °C⁻¹
- 0.0015 °C⁻¹
- 0.15 °C⁻¹
Answer: (B) 0.005 °C⁻¹
Rₜ = R₀(1 + αΔT) → 115 = 100(1 + α×30) → 1.15 = 1 + 30α → 30α = 0.15 → α = 0.005 °C⁻¹.
- Q21An electric kettle has two heating coils. With one coil the water boils in 10 min; with the other it boils in 40 min. If both coils are connected in parallel, the same quantity of water will boil in:
- 15 min
- 25 min
- 4 min
- 8 min
Answer: (D) 8 min
For a fixed heat at fixed voltage, time ∝ R (since H = V²t/R). So R₂ = 4R₁ (40 min vs 10 min). In parallel Rₚ = (R₁·4R₁)/(R₁+4R₁) = (4R₁)/(5). Time ∝ R, so t = (4)/(5)×10 = 8 min.
- Q22Kirchhoff’s junction rule (KCL) is a direct consequence of conservation of:
- mass
- charge
- energy
- momentum
Answer: (B) charge
The junction rule states the algebraic sum of currents at a node is zero, meaning charge neither accumulates nor disappears – it follows from conservation of charge.
- Q23A battery of EMF 2 V and internal resistance 0.1 Ω is being charged by an external source that pushes 5 A through it. The terminal voltage across the battery during charging is:
- 3.0 V
- 2.5 V
- 1.5 V
- 2.0 V
Answer: (B) 2.5 V
During charging the current is forced into the cell, so V = ε + Ir = 2 + (5)(0.1) = 2.5 V. (During discharge it would be ε − Ir = 1.5 V – note the sign flip.)
- Q24A potentiometer is preferred over a voltmeter for measuring the EMF of a cell because the potentiometer:
- has very low resistance
- gives a larger reading
- draws no current from the cell at the balance point
- is cheaper
Answer: (C) draws no current from the cell at the balance point
At the null point the potentiometer draws zero current from the test cell, so there is no Ir drop and it measures the true EMF. A voltmeter always draws some current and reads terminal voltage.
- Q25With rising temperature, the resistivity of a semiconductor:
- decreases
- first increases then decreases
- increases
- remains constant
Answer: (A) decreases
In semiconductors, heating greatly increases the number of charge carriers (n), which dominates over reduced relaxation time, so resistivity decreases (negative temperature coefficient).