Q: For a metallic wire, the ratio $ \frac{V}{i} $ (where, $ \mathbf{V}= $ applied potential difference and $ \mathbf{i}= $ current flowing

increases as the temperature rises. For a metallic wire, $V/i$ is the resistance. In metals, increasing temperature increases lattice vibrations, which makes electron flow harder, so resistance rises. Therefore $V/i$ increases as temperature rises.

Question 1

For a metallic wire, the ratio $ \frac{V}{i} $ (where, $ \mathbf{V}= $ applied potential difference and $ \mathbf{i}= $ current flowing

Accepted Answer

increases as the temperature rises. For a metallic wire, $V/i$ is the resistance. In metals, increasing temperature increases lattice vibrations, which makes electron flow harder, so resistance rises. Therefore $V/i$ increases as temperature rises.

Question 2

In a metallic conductor, electric current is thought to be due to the movement of

Accepted Answer

electrons. Metallic conduction occurs because some electrons are delocalized and can move through the lattice when a potential difference is applied. Ions and protons remain bound in the metal structure, and amperes are units of current, not particles.

Question 3

Assertion If the length of the conductor is doubled, the drift velocity will become half of the original value (keeping potential difference unchanged). Reason At constant potential difference, drift velocity is inversely proportional to the length of the conductor.

Accepted Answer

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion. For a conductor, drift velocity is proportional to the electric field, and the electric field is V/L when the potential difference is fixed. So if length doubles while voltage stays the same, the field halves and the drift velocity also halves. Thus both statements are true, and the reason correctly explains the assertion.

Question 4

The resistance of an iron wire is $ 10 \Omega $ and its temperature coefficient of resistance is $ 5 \times 10^{-3} /^{o} C . $ A current of $ 30 \mathrm{mA} $ is flowing in it at $ 20^{\circ} \mathrm{C} $. Keeping potential difference across its ends constant, if its temperature i

Accepted Answer

35. At constant potential difference, current is inversely proportional to resistance. The resistance increases with temperature by the factor $1+\alpha\Delta T$, so the current must decrease accordingly; this gives 35 mA.

Question 5

Wire of a certain material is stretched slowly be dep10. Its new resistance and specific resistance becomes respectively.

Accepted Answer

1.21 times, same. When a wire is stretched slowly, its volume is taken as constant, so its length increases and cross-sectional area decreases. Resistance depends on geometry, so it increases, but specific resistance (resistivity) is a material property and remains unchanged.

Question 6

Which of the following statements are wrong w.r.t the specific resistance of wire? This question has multiple correct options

Accepted Answer

It varies with its length. Specific resistance (resistivity) is an intrinsic property of a material, so it does not depend on the wire’s length, cross-section, or mass. It depends on the nature of the material (and temperature), so the statement about length is wrong.

Question 7

Two wires each of radius of cross section $ r $ but of different materials are connected together end to end (in series). If the densities of charge carriers in the two wires are in the ration $ 1: 4, $ the drift velocity of electrons in the two wires will be in the ratio:

Accepted Answer

4: 1. Since the wires are in series, the same current flows through both. With equal cross-sectional area, drift velocity satisfies $v_d = \frac{I}{nAe}$, so it is inversely proportional to carrier density. Therefore, if the densities are in the ratio 1:4, the drift velocities are in the ratio 4:1.

Question 8

In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire $ x $ cm long. If the length of the potentiometer wire is increased without changing the cell, the balancing length will (Driving source is not changed)

Accepted Answer

not change. In a potentiometer, the balancing length depends on the potential gradient of the wire and the cell’s emf. Since the driving source is unchanged, the potential gradient remains the same, so the null point stays at the same length.

Question 9

When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two diametrically opposite points will be

Accepted Answer

R/4. When the wire is bent into a circle, it forms two equal resistances (R/2 each) in parallel between the diametrically opposite points. The equivalent resistance of two parallel resistances R/2 is given by 1/Req = 1/(R/2) + 1/(R/2), which simplifies to Req = R/4.

Question 10

Resistances, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to

Accepted Answer

nr. In parallel, the equivalent resistance is given by R = r/n. When connected in series, the total resistance is nr, as the resistances simply add up. Thus, the correct answer is nr.

Question 11

Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is 10Ω. The total resistance of the combination will be

Accepted Answer

20Ω. The resistance of a wire is inversely proportional to its cross-sectional area. Since the cross-sectional areas are in the ratio 3:1, the resistance of the thinner wire will be 3 times that of the thicker wire. Thus, the resistance of the thinner wire is 30Ω. In series, the total resistance is the sum of individual resistances: 10Ω + 30Ω = 40Ω.

Question 12

The charge flowing through a resistance R varies with time t as Q = at − bt², where a and b are positive constants. The total heat produced in R is:

Accepted Answer

a³R / 6b. The current is the time derivative of charge, I = dQ/dt = a - 2bt. The heat produced is given by H = ∫(I²R) dt. Substituting I and integrating over the time interval where current flows (from t = 0 to t = a/2b, where I becomes zero), the result is H = a³R / 6b.

Question 13

Three resistances each of 4Ω are connected to form a triangle. The resistance between any two terminals is

Accepted Answer

6Ω. When three resistances of 4Ω each are connected in a triangle, the equivalent resistance between any two terminals is calculated by considering two resistances in parallel and one in series. The equivalent resistance is 6Ω.

Question 14

Two wires are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω. The power loss in the wires is:

Accepted Answer

19.2 kW. The total resistance of the wire is R = 0.5 Ω/km × 150 km = 75 Ω. The total voltage drop is V = 8 V/km × 150 km = 1200 V. Power loss is given by P = V²/R = (1200)²/75 = 19.2 kW.

Question 15

If voltage across a bulb rated 220 Volt-100 Watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is:

Accepted Answer

5%. The power of a bulb is proportional to the square of the voltage (P ∝ V²). A 2.5% drop in voltage corresponds to a decrease in power by (2 × 2.5%) = 5%.

Question 16

A 5–ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is:

Accepted Answer

0.2 ohm. The power dissipated in the fuse wire is given by P = I²R. Substituting P = 1 W and I = 5 A, we get R = P/I² = 1/(5²) = 0.2 ohm.

Question 17

When three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be

Accepted Answer

20 watt. When connected in series, the total resistance of the bulbs is 3 times the resistance of one bulb. The resistance of one bulb is R = V²/P = (200)²/60 = 666.67 Ω. Total resistance = 3R = 2000 Ω. The current through the circuit is I = V/R = 200/2000 = 0.1 A. Power drawn = I²R = (0.1)² × 2000 = 20 W.

Question 18

Ten identical cells connected in series are needed to heat a wire of length one meter and radius ‘r’ by 10°C in time ‘T’. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time ‘T’?

Accepted Answer

20. The heat generated is proportional to the resistance of the wire and the square of the current. Resistance is proportional to the length of the wire. Doubling the length doubles the resistance, so to maintain the same heating in the same time, the voltage (and hence the number of cells) must also double. Thus, 20 cells are required.

Question 19

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be

Accepted Answer

50 watt, 200 watt. In the series connection, the total resistance doubles, so the power is halved to 50 W. In the parallel connection, the total resistance is halved, so the power is doubled to 200 W.

Question 20

A battery of 10 V and internal resistance 0.5Ω is connected across a variable resistance R. The value of R for which the power delivered is maximum is equal to

Accepted Answer

0.5Ω. The power delivered to the external resistance R is maximum when R equals the internal resistance of the battery, according to the maximum power transfer theorem. Here, the internal resistance is 0.5Ω, so R = 0.5Ω.

NEET Physics: Current Electricity questions with solutions

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