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Physics NEET MCQ
Quiz 1
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Q.1
Three resistances P, Q, R each of 2 Ω and an unknown resistances S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 Ω is connected in parallel to S the bridges gets balanced. What is the value of S?
0%
3 ohm
0%
6 ohm
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1 ohm
0%
2 ohm
Explanation
For a balanced Wheatstone bridge with P = Q = 2Ω (equal), the balance condition P/Q = R/S_eff requires S_eff = R = 2Ω (since P/Q = 1). With 6Ω connected in parallel to S: S_eff = 6S/(S+6) = 2 6S = 2S + 12 → 4S = 12 → S = 3Ω.
Q.2
The n rows each containing m cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of 3 Ω. If the total number of cells used is 24 and internal resistance of each cell is 0.5 Ω, then
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m = 8, n = 3
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m = 6, n = 4
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m = 12, n = 2
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m = 2, n = 12
Explanation
For n rows of m cells each (total cells nm = 24, each with internal resistance r = 0.5 Ω) delivering maximum current to an external resistance R = 3 Ω, the condition for maximum current is that the battery combination's total internal resistance (mr/n) equals the external resistance: mr/n = R → m(0.5)/n = 3 → m/n = 6 → m = 6n Combined with nm = 24: n(6n) = 24 → n² = 4 → n = 2, m = 12.
Q.3
When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite point will be
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R/4
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4R
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R/8
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R/2
Explanation
Bending the wire into a circle splits it into two semicircular arcs between the diametrically opposite points, each carrying half the total length and so each having resistance R/2. These two R/2 arcs form two parallel paths between the two points: R_eq = (R/2 × R/2)/(R/2 + R/2) = (R²/4)/R = R/4.
Q.4
Two identical galvanometers are converted into an ammeter and a milliammeter. The shunt, which has more resistance due to the current passing through the coil will be
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less
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equal
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more
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zero
Explanation
Shunt resistance follows S = IgG/(I − Ig), where Ig is the galvanometer's own full-scale current and I is the full-scale current of the converted instrument. A milliammeter has a much smaller full-scale current I than an ammeter, making the denominator (I − Ig) smaller — so the milliammeter's shunt resistance works out larger (more) than the ammeter's shunt.
Q.5
In the circuit shown, current flowing through 25 V cell is
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8 A
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10 A
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14.2 A
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12 A
Explanation
This is a multi-branch network of cells (each with its own internal/series resistance) all connected between the same pair of nodes. The standard method is Millman's theorem: treat each branch as an EMF source Eᵢ in series with resistance Rᵢ, and find the common node voltage V = (ΣEᵢ/Rᵢ) / (Σ1/Rᵢ). Once V is known, the current through any one branch (here, the 25 V cell) follows from (E − V)/R for that branch. Carrying this through for the circuit shown gives a current of 12 A through the 25 V cell.
Q.6
Two resistance filaments of same length are connected first in series and then in parallel. Find the ratio of power dissipated in both cases assuming that equal current flows in the main circuit.
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1 : 4
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4 :1
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1 : 2
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1 :2
Explanation
With the SAME main-circuit current I flowing in both cases: in series, total resistance is 2R, so power = I²(2R). In parallel, the equivalent resistance is R/2, so power = I²(R/2). Ratio (series/parallel) = I²(2R) / [I²(R/2)] = 2R / (R/2) = 4, i.e. 4:1.
Q.7
Kirchhoff’s first law of electric circuits is based on the law of conservation of
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only on mass
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only on charge
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on charge as well as on energy
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on charge as well as on mass
Explanation
Kirchhoff's current law (junction rule) — the sum of currents into a junction equals the sum out — is simply a statement that charge cannot accumulate at a point; it's a direct consequence of conservation of charge alone.
Q.8
An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. A resistance is now joined parallel with the voltmeter. Then,
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A will increase, V will decrease
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Both A and V will decrease
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A will decrease, V will increase
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Both A and V will increase
Explanation
Adding a resistor in parallel with the voltmeter reduces the combined resistance of that branch, lowering the total resistance of the series circuit — which increases the total current drawn from the cell, so the ammeter reading A increases. At the same time, because that branch's resistance has decreased, the voltage drop across it decreases, so the voltmeter reading V decreases.
Q.9
Find the equivalent resistance between point a and b
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10 ohm
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7.5 ohm
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9.5 ohm
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8 ohm
Explanation
This is a series-parallel resistor network reducible step by step from the outer branches inward (combining series resistors by adding them, and parallel resistors using 1/R_eq = 1/R₁+1/R₂+...), which works out to an equivalent resistance of 7.5Ω between the marked terminals.
Q.10
Kirchhoff’s I and II laws are based on conservation of
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energy and charge
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mass and charge
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charge and mass
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charge and energy
Explanation
Kirchhoff's first (current/junction) law reflects conservation of charge, and Kirchhoff's second (voltage/loop) law reflects conservation of energy (the total energy gained and lost by a charge going around any closed loop must balance) — together, the two laws rest on charge and energy conservation.
Q.11
A constant voltage is applied between the two ends of a uniform metallic wire and some heat is developed in it . The heat developed is doubled if
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both length and radius of wire are halved
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both length and radius of wire are doubled
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radius of wire is doubled
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length of wire is doubled
Explanation
Heat developed at constant voltage: H = V²t/R, so H is inversely proportional to resistance R. Resistance R = ρL/A. If both length and radius are doubled: new length = 2L, new area = 4A (area scales with radius squared), giving new R = ρ(2L)/(4A) = R/2 — resistance is halved. Since H ∝ 1/R, halving R doubles H.
Q.12
Assertion reason type question. (a) If both assertion and reason are true and reason is correct explanation of assertion. (b) If both assertion and reason are true and reason is not correct explanation of assertion. (c) Assertion is true, but reason is false. (d) both assertion and reason are false. Assertion: Wire carrying current is not charged. Reason: It is because , at any instant number of electrons leaving wire is sometimes equal to the number of electrons flowing in from the battery.
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(a)
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(b)
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(c)
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(d)
Explanation
A steady current-carrying wire genuinely stays electrically neutral overall, so the Assertion is true. But the Reason's wording — that the numbers of electrons entering and leaving are "sometimes" equal — isn't quite right: for the wire to remain consistently uncharged during steady current flow, this balance must hold continuously, at every instant, not merely "sometimes".
Q.13
Each of the resistance in the network shown in the figure is equal to R. The resistance between terminal A and B is
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R
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5R
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3R
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6R
Explanation
By the symmetry of this network (an outer triangle of three R resistors plus an inner Y of three more R resistors meeting at a central point, with A and B both near that central junction), the classic reduction for this well-known symmetric configuration gives an equivalent resistance of exactly R between the two terminals.
Q.14
The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then
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the resistance will be doubled and the specific resistance will be halved
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the resistance will be halved and the specific resistance will remains unchanged
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the resistance will be halved and the specific resistance will be doubled.
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the resistance and the specific resistance, will both remain unchanged.
Explanation
R = ρL/A. Doubling both length and radius gives new length 2L and new area 4A (area scales as radius squared): R' = ρ(2L)/(4A) = R/2 — resistance is halved. Resistivity (specific resistance) ρ is an intrinsic material property and doesn't depend on the wire's dimensions at all, so it stays unchanged.
Q.15
If the length of the potentiometer wire is increased, then the accuracy in determination of null point
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will increase
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will decrease
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will remain unaffected
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cannot be decided , until EMF of auxiliary battery is known
Explanation
A longer potentiometer wire spreads the same total potential difference over a greater length, giving a smaller potential drop per unit length (finer resolution). This makes it possible to locate the exact balance (null) point more precisely, improving accuracy.
Q.16
Assertion reason type question. (a) If both assertion and reason are true and reason is correct explanation of assertion. (b) If both assertion and reason are true and reason is not correct explanation of assertion. (c) Assertion is true, but reason is false. (d) both assertion and reason are false. Assertion: Drift velocity of electrons decreases on increasing temperature of the conductor. Reason: It is because, on increasing temperature of a conductor , the value of resistivity of its material increases.
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(a)
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(b)
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(c)
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(d)
Explanation
Drift velocity vd = eEτ/m, where τ is the average time between electron collisions (the relaxation time). As temperature rises, lattice vibrations increase, causing more frequent electron collisions — shortening τ, and hence reducing vd. This shorter relaxation time is also exactly why resistivity (ρ = m/(ne²τ)) rises with temperature — both effects trace back to the same underlying cause, so the Reason correctly explains the Assertion.
Q.17
Assertion reason type question. (a) If both assertion and reason are true and reason is correct explanation of assertion. (b) If both assertion and reason are true and reason is not correct explanation of assertion. (c) Assertion is true, but reason is false. (d) both assertion and reason are false. Assertion: A potentiometer measures the potential difference more accurately then a voltmeter. Reason: Because it has a wire of high resistance and draws a heavy current from external circuit.
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(a)
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(b)
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(c)
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(d)
Explanation
A potentiometer genuinely measures potential difference more accurately than a voltmeter, because at the balance (null) point it draws NO current at all from the circuit being measured — unlike a voltmeter, which always draws some current and slightly disturbs the very voltage it's trying to measure. So the Reason here is false: it's the ABSENCE of current draw at balance that makes a potentiometer accurate, not "drawing a heavy current", which is the opposite of what actually happens.
Q.18
Assertion reason type question. (a) If both assertion and reason are true and reason is correct explanation of assertion. (b) If both assertion and reason are true and reason is not correct explanation of assertion. (c) Assertion is true, but reason is false. (d) both assertion and reason are false. Assertion: Resistors are connected in series combination in order to increase the resistance of the circuit. Reason: In series combination potential difference across any resistor is proportional to its resistance.
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(a)
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(b)
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(c)
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(d)
Explanation
Connecting resistors in series does increase total resistance (resistances simply add: R_total = R₁+R₂+...), so the Assertion is true. The Reason — that in series the potential difference across each resistor is proportional to its resistance (since the same current flows through all of them, V=IR) — is also true, but it describes how voltage is DISTRIBUTED among the resistors, not why the TOTAL resistance increases when they're combined in series (which follows simply from resistances adding directly). So the Reason doesn't explain the Assertion.
Q.19
The current voltage graph for a given metallic wire at two different temperatures T1 and T2 are shown in the fig. which one of the following option is true
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T1 = T2
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T1 < T2
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T1 > T2
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none of the above
Explanation
The slope of an I-V graph is 1/R — a steeper line means lower resistance. T1's line is steeper than T2's, so T1 corresponds to the lower resistance. Since a metallic conductor's resistance increases with temperature, the lower-resistance curve (T1) must correspond to the lower temperature: T1 < T2.
Q.20
The temperature coefficient of resistance of wire is 0.00125 per ° C. At 300K, its resistance is 1 ohm. The resistance of wire would be 2 ohms at
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1154K
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1127K
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1167K
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1176K
Explanation
Using R_T = R₀(1 + αT), with T measured in °C from a 0°C reference: At 300 K = 27°C: 1 = R₀(1 + 0.00125×27) = R₀ × 1.03375 → R₀ ≈ 0.9674 Ω Find T where R = 2Ω: 2 = 0.9674(1 + 0.00125T) 2/0.9674 ≈ 2.0675 = 1 + 0.00125T T ≈ 1.0675/0.00125 ≈ 854°C Converting to Kelvin: 854 + 273 ≈ 1127 K.
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