Question 1

A physical quantity X is defined as ε0L(ΔV/Δt), where ε0 denotes the permittivity of free space, L is a length, ΔV is a potential difference, and Δt is a time interval. The dimensional formula of X matches that of which of the following?

Accepted Answer

electric current. eps0 times a length has the dimensions of capacitance (C = eps0*Area/d ~ eps0*length). Then X = C*(dV/dt) = I, the displacement current, so X has the dimensions of electric current.

Question 2

A set of n identical resistors is arranged once in a series combination and once in a parallel combination. What is the ratio of the larger equivalent resistance to the smaller equivalent resistance?

Accepted Answer

n². Series equivalent = nR (larger); parallel equivalent = R/n (smaller). Ratio of larger to smaller = nR/(R/n) = n^2.

Question 3

A resistance coil is connected to a battery. In which arrangement will the maximum heat be produced?

Accepted Answer

When the coil is split into four equal sections and all four sections are connected in parallel across the battery. Connecting the coil in parallel allows each section to receive the full voltage of the battery, which maximizes the current through each section. Since heat produced in a resistor is proportional to the square of the current (I²R), having more sections in parallel increases the total current and thus the total heat generated.

Question 4

A bulb is rated at 220 V and 100 W. If the applied voltage falls by 2.5% from its rated value, by what percentage of the rated power will the power output decrease?

Accepted Answer

5%. For a bulb of fixed resistance, P = V^2/R so percentage change in power = 2 * percentage change in voltage = 2 * 2.5% = 5%.

Question 5

An ammeter has a coil resistance of R. To extend its measuring range by a factor of n, the shunt resistance needed should be:

Accepted Answer

R/(n-1). To extend an ammeter's range by factor n, the shunt resistance must be S = R/(n-1), where R is the coil resistance.

Question 6

How does the drift velocity Vd depend on the strength of the applied electric field?

Accepted Answer

Vd is directly proportional to E. The drift velocity (Vd) of charge carriers in a conductor increases linearly with the strength of the applied electric field (E), meaning that as the electric field strength increases, the drift velocity also increases proportionally.

Question 7

The product of the conductivity and resistivity of a metallic conductor is determined by which factor?

Accepted Answer

None of these. Conductivity is the reciprocal of resistivity, so their product is exactly 1 regardless of temperature, area, or pressure. Hence it depends on none of these factors.

Question 8

An iron wire has resistance R. If both its length and its radius are made twice their original values, what happens?

Accepted Answer

The resistance becomes half, while the resistivity stays the same.. Doubling L and r: R' = rho*(2L)/(pi*(2r)^2) = rho*(2L)/(4*pi*r^2) = R/2. Resistance halves; resistivity is intrinsic and unchanged.

Question 9

A potentiometer has a wire 100 cm long, and its standard cell has an e.m.f. of E volt. It is used to determine the e.m.f. of a battery whose internal resistance is 0.5 Ω. If the null point is found at 30 cm from the positive end, the battery’s e.m.f. is

Accepted Answer

30E/100. Potential gradient = E/100 per cm; at null point (no current through the cell) internal resistance has no effect. EMF = (E/100)*30 = 30E/100.

Question 10

In a potentiometer wire carrying current i, if the current through the conductor is altered by 1%, by what percentage does the power consumed change?

Accepted Answer

2%. The power consumed in a conductor is proportional to the square of the current (P ∝ i²). Therefore, a 1% change in current results in a change of approximately 2% in power, since the relationship is quadratic.

Question 11

To obtain the greatest current through an external resistance of 3 Ω, a battery is formed by connecting n parallel branches, each branch containing m cells in series. If the battery uses 24 cells in all and each cell has an internal resistance of 0.5 Ω, which arrangement should be chosen?

Accepted Answer

m = 12, n = 2. Max current needs external R = internal resistance m*r/n: 3 = 0.5m/n -> m = 6n. With m*n = 24, n = 2 and m = 12.

Question 12

In a neon discharge tube, 2.9 × 10¹⁸ Ne+ ions cross a given cross-section each second toward the right, and at the same time 1.2 × 10¹⁸ electrons cross the same section each second toward the left. If the electronic charge is 1.6 × 10⁻¹⁹ C, what is the resulting electric current?

Accepted Answer

0.66 A toward the right. The resulting electric current is calculated by considering the contributions from both the Ne+ ions and the electrons. The current due to the Ne+ ions is positive as they move to the right, while the current due to the electrons moving to the left is negative. The net current is the sum of these two contributions, resulting in a total of 0.66 A toward the right.

Question 13

A battery in a car has an emf of 12 V and an internal resistance of 5 × 10⁻² Ω. When it supplies a current of 60 A, what is the terminal potential difference across the battery?

Accepted Answer

9 volt. The terminal potential difference is calculated by subtracting the voltage drop across the internal resistance from the emf. With an internal resistance of 0.05 Ω and a current of 60 A, the voltage drop is 3 V (0.05 Ω × 60 A), leading to a terminal potential difference of 12 V - 3 V = 9 V.

Question 14

Two cells with identical emf are connected to an external resistor R. Their internal resistances are R1 and R2, with R1 > R2. If the potential difference across the cell having internal resistance R2 becomes zero, then which relation for R is correct?

Accepted Answer

R = R2 - R1. With both cells in series, current I = 2E/(R+R1+R2). The terminal PD across the cell of internal resistance R2 is E - I*R2 = 0, giving I = E/R2. Equating: R + R1 + R2 = 2*R2, so R = R2 - R1.

Question 15

A galvanometer has a resistance of 50 Ω and a scale of 25 divisions. If a current of 4 × 10⁻⁴ A produces a deflection of one division, what resistance should be added in series to use it as a voltmeter of 25 V range?

Accepted Answer

2450 Ω in series. Ig = 25 * 4e-4 = 0.01 A. Series resistance R = V/Ig - G = 25/0.01 - 50 = 2500 - 50 = 2450 ohm.

Question 16

A galvanometer having resistance G is connected in parallel with a shunt of resistance S ohm. If the total current in the circuit is to remain the same, the resistance that must be connected in series with the galvanometer is

Accepted Answer

G² / (S + G). To keep the total current unchanged, the series resistance P plus the parallel combination must equal the original G: P + GS/(G+S) = G, giving P = G - GS/(G+S) = G^2/(S+G).

Question 17

A battery is joined across two points A and B on the rim of a uniform circular conducting ring of radius r and resistance R. Suppose one of the two arcs AB makes an angle θ at the centre. The magnetic field at the centre produced by the current in the ring is

Accepted Answer

zero for every value of θ. The two arcs are in parallel; each arc's current is inversely proportional to its length (resistance) while its field contribution is proportional to the angle it subtends. The products are equal and oppositely directed, so the net field at the centre is zero for every value of theta.

Question 18

A galvanometer shows a deflection of 50 divisions without shunt. When a 12 Ω shunt is connected across it, the deflection reduces to 20 divisions. What is the resistance of the galvanometer?

Accepted Answer

18 Ω. Deflection is proportional to galvanometer current. With shunt, Ig = (20/50)I = 0.4I and shunt current Is = 0.6I. Equating voltages: 0.4I*G = 0.6I*12, so G = (0.6*12)/0.4 = 18 ohm.

Question 19

A dc motor has an armature resistance of 20 a9. When connected to a 220 V dc source, it takes a current of 1.5 A. What is the back emf developed in the motor?

Accepted Answer

190 V. The back emf can be calculated using the formula: Back emf = Supply voltage - (Armature current x Armature resistance). Substituting the values, we get Back emf = 220 V - (1.5 A x 20 Ω) = 220 V - 30 V = 190 V, confirming that option D is correct.

Question 20

A 10 H ideal inductor is joined in series with a 5 Ω resistor and a 5 V battery. What is the current in the circuit 2 s after the circuit is closed?

Accepted Answer

(1 - e⁻¹) A. I = (V/R)(1 - e^(-Rt/L)) = (5/5)(1 - e^(-5*2/10)) = (1 - e^(-1)) A.

Question 21

An experiment uses two identical capacitors, a resistor R, and a 6 V source in an RC circuit. When the capacitors are connected in parallel, the time taken for the voltage across the fully charged combination to fall to one-half of its initial value is 10 s. If the same two capacitors are

Accepted Answer

2.5 s. Parallel: C_p = 2C, t = R*2C*ln2 = 10 s -> RC*ln2 = 5 s. Series: C_s = C/2, t = R*(C/2)*ln2 = 2.5 s.

Question 22

An AC generator rated at 220 V has an internal resistance of 10 Ω and is connected to an external resistance of 100 Ω. What power is dissipated in the external circuit?

Accepted Answer

400 W. Total resistance = 10 + 100 = 110 ohm, so I = 220/110 = 2 A. Power in external resistor = I^2 * 100 = 4 * 100 = 400 W.

Question 23

When setting up a circuit to verify Ohm’s law, which arrangement of the meters is correct?

Accepted Answer

Connect the voltmeter in parallel and the ammeter in series.. The correct arrangement is to connect the voltmeter in parallel to measure the voltage across the component, while the ammeter must be in series to measure the current flowing through the circuit, ensuring accurate readings for verifying Ohm's law.

Question 24

Two potentiometers, A and B, are each 100 cm long. Potentiometer A has 4 wires, while potentiometer B has 10 wires. They are used to compare the e.m.f.s of two cells. Which potentiometer will produce the greater balancing length?

Accepted Answer

Potentiometer B. Potentiometer B, having more wires, allows for a finer division of the potential along its length, which can lead to a greater balancing length when comparing e.m.f.s. This increased number of wires enhances the sensitivity of the measurement, making it more effective in achieving a balance.

Question 25

A 220 V d.c. source is joined to a storage battery of emf 200 V through a 1 Ω resistor. The battery is also connected to an external resistor R at its terminals. What is the least value of R for which current flows into the battery and it gets charged?

Accepted Answer

11 Ω. For current to flow into the battery, the voltage across the external resistor R must be greater than the battery's emf. The effective voltage from the source is 220 V, and the voltage drop across the 1 Ω resistor must be considered. The minimum resistance R can be calculated using Ohm's law, leading to the conclusion that R must be at least 11 Ω to ensure that the current flows into the battery.

Question 26

The current in a conductor varies with time according to i = α₀t + βt², where α₀ = 20 A s⁻¹ and β = 8 A s⁻². Determine the total charge that passes through a cross-section of the wire in 15 s.

Accepted Answer

11250 C. Q = integral_0^15 (20t + 8t^2) dt = 10t^2 + (8/3)t^3 at t=15 = 2250 + 9000 = 11250 C.

Question 27

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is

Accepted Answer

6%. Since R = V/I, the fractional error is dR/R = dV/V + dI/I = 3% + 3% = 6%.

Question 28

A bulb rated at 220 V, 1000 W is connected to a 110 V supply. The electrical power it will draw is

Accepted Answer

250 W. R = 220^2/1000 = 48.4 ohm. At 110 V, P = 110^2/48.4 = 250 W.

Question 29

Two resistors are connected in series and their equivalent resistance is S. When the same two resistors are connected in parallel, their equivalent resistance becomes P. If S = nP, what is the smallest possible value of n?

Accepted Answer

4. S/P = (R1+R2)^2/(R1*R2). By AM-GM this is minimum when R1 = R2, giving S/P = 4. So the smallest value of n is 4.

Question 30

Two wires made of the same substance are joined in parallel in a circuit. If the ratio of their lengths is 4/3 and the ratio of their radii is 2/3, what is the ratio of the currents flowing through the two wires?

Accepted Answer

1/3. With equal voltage, I = V/R and R = rho L/(pi r^2), so I is proportional to r^2/L. I1/I2 = (r1/r2)^2 x (L2/L1) = (2/3)^2 x (3/4) = (4/9)(3/4) = 1/3.

Question 31

In a meter bridge setup, the balance point is found 20 cm from one end of the wire when resistance X is compared with resistance Y. Given that X < Y, what will be the balance point measured from the same end if the resistance 4X is now balanced against Y?

Accepted Answer

50 cm. When resistance X is less than Y, the balance point is closer to X. If we now use 4X against Y, the increased resistance of 4X will shift the balance point further away from X, resulting in a new balance point at 50 cm from the same end.

Question 32

Thermistors are generally fabricated from which type of material?

Accepted Answer

Metal oxides that have a large temperature coefficient of resistivity. Thermistors are made from metal oxides (semiconducting) that have a large temperature coefficient of resistivity, giving a strong, usually negative, change of resistance with temperature.

Question 33

A heating element is divided into two identical halves, and then just one of those halves is used in the heater. The amount of heat produced will become

Accepted Answer

twice as much. Using one half of the heating element means that the total surface area available for heat production is reduced, but since the two halves are identical, the remaining half can still produce heat at the same rate, resulting in twice the heat output compared to using just a quarter of the original element.

Question 34

Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are Z1 and Z2 respectively the charge which flows through the sil

Accepted Answer

q/(1 + Z2/Z1). The correct option is derived from the principle that the mass of metal deposited is proportional to the charge passed and the electrochemical equivalent. Since equal amounts of metals are deposited, the relationship between the charges in the two voltameters can be expressed as the ratio of their electrochemical equivalents, leading to the formula q/(1 + Z2/Z1) for the charge through the silver voltameter.

Question 35

In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. The internal resistance of the cell is

Accepted Answer

2Ω. The internal resistance of the cell can be determined using the change in balancing length when the cell is shunted. The initial length of 240 cm corresponds to the full voltage of the cell, while the reduced length of 120 cm indicates that the voltage across the potentiometer has halved due to the shunt, which implies that the internal resistance of the cell is equal to the shunt resistance of 2Ω.

Question 36

The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp when not in use?

Accepted Answer

40Ω. The resistance of the lamp when not in use is calculated using the formula for power, P = V²/R. Rearranging gives R = V²/P. Plugging in the values of 200 V and 100 W results in a resistance of 40Ω.

Question 37

An energy source will supply a constant current into the load if its internal resistance is

Accepted Answer

very large as compared to the load resistance. A constant current independent of the load is delivered by a current source, which has internal resistance very large compared with the load. (Zero internal resistance gives a constant-voltage source instead.)

Question 38

The Kirchhoff's first law (Σi = 0) and second law (ΣiR = ΣE), where the symbols have their usual meanings, are respectively based on

Accepted Answer

conservation of charge, conservation of energy. Kirchhoff's first law is based on the conservation of charge, stating that the total current entering a junction must equal the total current leaving it. The second law is based on the conservation of energy, indicating that the sum of the potential differences in a closed loop must equal the sum of the electromotive forces.

Question 39

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. then for the two wires to have the same resistance, the ratio lB/lA is

Accepted Answer

2. The resistance of a wire is directly proportional to its length and resistivity, and inversely proportional to its cross-sectional area. Since material 'B' has twice the resistivity of 'A' and its diameter is twice that of 'A', its cross-sectional area is four times larger. To maintain equal resistance, the length of wire 'B' must be twice that of wire 'A', resulting in a ratio of lB/lA = 2.

Question 40

A bulb filament has a resistance of 100 Ω at 100°C. If the temperature coefficient of resistance is 0.005 per °C, at what temperature will its resistance rise to 200 Ω?

Accepted Answer

400°C. 200/100 = (1+0.005*T2)/(1+0.005*100) = (1+0.005*T2)/1.5, so 1+0.005*T2 = 3, giving T2 = 400 C.

Question 41

In a Wheatstone bridge, the three arms contain resistances P, Q, and R, while the fourth arm consists of two resistors S1 and S2 connected in parallel. For the bridge to be in balance, which relation must hold?

Accepted Answer

P/Q = R(S1 + S2)/(S1S2). The correct option establishes the condition for balance in a Wheatstone bridge by relating the ratios of the resistances. When the bridge is balanced, the voltage across the parallel combination of S1 and S2 must equal the voltage across R, leading to the derived relationship that incorporates the product of the parallel resistances.

Question 42

An electric lamp is marked 220 V, 100 W. If it is connected to a 110 V supply, the power it will consume is

Accepted Answer

75 watt. Lamp resistance R = V^2/P = 220^2/100 = 484 ohm. At 110 V, P = 110^2/484 = 25 W.

Question 43

If a current I enters at point A, what is the magnitude of the electric field at a point located a distance r from A?

Accepted Answer

ρI/(2πr²). Current entering at a point on the surface of a conducting medium spreads radially through a hemisphere of area 2*pi*r^2, giving current density J = I/(2*pi*r^2). The field is E = rho*J = rho*I/(2*pi*r^2).

Question 44

Two resistors have equal resistance at 0°C, and their temperature coefficients of resistance are α1 and α2. The temperature coefficients of resistance for their combination in series and in parallel are approximately

Accepted Answer

(α1 + α2)/2, (α1 + α2)/2. The temperature coefficient of resistance for resistors in series and parallel can be averaged when the resistors have equal resistance, leading to the same value for both configurations. Thus, the coefficients for both combinations are approximately (α1 + α2)/2.

Question 45

A wire is elongated by 0.1%. What happens to its electrical resistance?

Accepted Answer

It increases by 0.2%. When a wire is elongated, its length increases while its cross-sectional area decreases. Since resistance is directly proportional to length and inversely proportional to area, the overall effect leads to an increase in resistance, which in this case is calculated to be 0.2%.

Question 46

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω, a balance is found when the cell is connected across 40 cm of the w

Accepted Answer

1.5 Ω. The internal resistance of the cell can be determined using the ratio of the lengths of the potentiometer wire at balance points before and after adding the shunt resistance. The change in balance length indicates how the effective voltage across the cell changes due to the internal resistance, leading to the calculation that results in an internal resistance of 1.5 Ω.

Question 47

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1kΩ. How much was the resistance on the left slot before interchanging the resistances?

Accepted Answer

550 Ω. The shift in the balance point indicates a change in the ratio of the resistances. Since the total resistance in series is 1kΩ, and the balance point shifts left by 10 cm, it suggests that the resistance on the left was lower than the right. The only option that satisfies the condition of the series combination and the shift is 550 Ω.

Question 48

A copper wire is elongated so that its length increases by 0.5%. If its volume is assumed to stay constant, what is the percentage increase in its electrical resistance?

Accepted Answer

1.0%. When the length of a conductor increases while its volume remains constant, its cross-sectional area decreases. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, a 0.5% increase in length leads to a 1.0% increase in resistance due to the combined effects of lengthening and area reduction.

Question 49

A copper wire has a cross-sectional area of 5 mm² and carries a current of 1.5 A. If the electron number density in copper is 9 × 10²⁸ m⁻³, the drift speed of the electrons is v. Taking the electron charge as 1.6 × 10⁻¹⁹ C, the value of v in mm/s is approximately:

Accepted Answer

0.02. The drift speed of electrons can be calculated using the formula v = I / (n * A * e), where I is the current, n is the electron number density, A is the cross-sectional area, and e is the charge of an electron. Substituting the given values results in a drift speed of approximately 0.02 mm/s, confirming that option A is correct.

Question 50

An ammeter has a full-scale reading of 1 A and an internal resistance of 0.81 Ω. What shunt resistance must be connected to extend its measuring range to 10 A?

Accepted Answer

0.09 Ω. To extend the ammeter's range to 10 A, a shunt resistor is needed to bypass most of the current. The correct shunt resistance can be calculated using the formula for shunt resistors, which shows that a lower resistance allows more current to pass through the shunt while keeping the ammeter within its safe operating range.

JEE Main Physics: Current Electricity questions with solutions

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