Q: Match Column I (physical terms for flow of current through a metallic conductor) with Column II (mathematical relations): Column I: A. Drift velocity; B. Electrical resistivity; C. Relaxation period; D. Current density Column II: 1. \frac{m}{ne^2\rho}; 2. nev_d; 3. \frac{eE}{m}\tau; 4. \frac{E}{J}

A→3, B→4, C→1, D→2. Drift velocity v_d = \frac{eE}{m}\tau (3); resistivity \rho = \frac{E}{J} (4); relaxation period \tau = \frac{m}{ne^2\rho} (1); current density J = nev_d (2).

Q: A charged particle having drift velocity of 7.5 \times 10^{-4} ms⁻¹ in an electric field of 3 \times 10^{-10} Vm⁻¹, has a mobility (in m²V⁻¹s⁻¹) of

2.5 \times 10^6. Mobility \mu = \frac{v_d}{E} = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^6 m²V⁻¹s⁻¹

Question 1

Match Column I (physical terms for flow of current through a metallic conductor) with Column II (mathematical relations):

Column I: A. Drift velocity; B. Electrical resistivity; C. Relaxation period; D. Current density

Column II: 1. \frac{m}{ne^2ho}; 2. nev_d; 3. \frac{eE}{m}	au; 4. \frac{E}{J}

Accepted Answer

A→3, B→4, C→1, D→2. Drift velocity v_d = \frac{eE}{m}	au (3); resistivity ho = \frac{E}{J} (4); relaxation period 	au = \frac{m}{ne^2ho} (1); current density J = nev_d (2).

Question 2

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

Accepted Answer

4 Ω. Parallel: \frac{1}{0.25} = \frac{4}{R} \Rightarrow R = 1 Ω per wire. Series: R_{eq} = 4R = 4 Ω

Question 3

Three resistors r_1, r_2 and r_3 are connected in a circuit where the current i_1 enters through r_1 and then splits into i_2 (through r_2) and i_3 (through r_3), with r_2 and r_3 in parallel. The ratio \frac{i_3}{i_1} in terms of the resistances is

Accepted Answer

\frac{r_2}{r_2 + r_3}. For the parallel branches, i_2 r_2 = i_3 r_3. With i_1 = i_2 + i_3: i_1 = \frac{i_3 r_3}{r_2} + i_3, giving \frac{i_3}{i_1} = \frac{r_2}{r_2 + r_3}

Question 4

Two solid conductors are made up of same material, have same length and same resistance. One of them has a circular cross-section of area A_1 and the other one has a square cross-section of area A_2. The ratio A_1/A_2 is

Accepted Answer

1. R = \frac{\rho l}{A}. Same material, length and resistance means same A, so \frac{A_1}{A_2} = 1

Question 5

A charged particle having drift velocity of 7.5 	imes 10^{-4} ms⁻¹ in an electric field of 3 	imes 10^{-10} Vm⁻¹, has a mobility (in m²V⁻¹s⁻¹) of

Accepted Answer

2.5 	imes 10^6. Mobility \mu = \frac{v_d}{E} = \frac{7.5 	imes 10^{-4}}{3 	imes 10^{-10}} = 2.5 	imes 10^6 m²V⁻¹s⁻¹

Question 6

The colour code of a resistance is given as (in order): Yellow, Violet, Brown, Gold. The values of resistance and tolerance respectively, are

Accepted Answer

470 Ω, 5%. Yellow = 4, Violet = 7, Brown = ×10¹, Gold = ±5%. So R = 47 × 10¹ = 470 Ω, ±5%

Question 7

Which of the following acts as a circuit protection device?

Accepted Answer

Fuse. A fuse has low melting point and low resistivity; it melts and breaks the circuit when excessive current flows, protecting it.

Question 8

A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be

Accepted Answer

Yellow - Violet - Orange - Silver. 47 kΩ = 47 × 10³ Ω with ±10% tolerance. 4 = Yellow, 7 = Violet, 3 (×10³) = Orange, ±10% = Silver → Yellow-Violet-Orange-Silver.

Question 9

The resistance of a wire is R ohm. If it is melted and stretched to n times its original length, its new resistance will be

Accepted Answer

n^2R. Volume is constant, so stretching to n times the length reduces area to \frac{A}{n}. R' = \frac{\rho(nl)}{A/n} = n^2\frac{\rho l}{A} = n^2R

Question 10

A, B and C are voltmeters of resistance R, 1.5R and 3R respectively, connected so that A is in series with the parallel combination of B and C, across a potential difference applied between X and Y. When some potential difference is applied, the voltmeter readings are V_A, V_B and V_C respectively. Then,

Accepted Answer

V_A = V_B = V_C. B (1.5R) and C (3R) in parallel give \frac{1.5R 	imes 3R}{1.5R + 3R} = R, equal to A. The same current I flows through A and through the parallel section, so V_A = IR and V_B = V_C = IR — all three readings are equal.

Question 11

Across a metallic conductor of non-uniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is

Accepted Answer

current. Charge is conserved, so the current is the same through every cross-section. Current density, drift velocity and electric field all vary because the area varies.

Question 12

Two metal wires of identical dimensions are connected in series. If \sigma_1 and \sigma_2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is

Accepted Answer

\frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}. Series resistances add: \frac{2L}{\sigma A} = \frac{L}{\sigma_1 A} + \frac{L}{\sigma_2 A}, giving \frac{2}{\sigma} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2}, so \sigma = \frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}

Question 13

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 Ω all connected in series. If the ammeter has a coil of resistance 480 Ω and a shunt of 20 Ω, then the reading in the ammeter will be

Accepted Answer

0.5 A. Ammeter resistance (coil ∥ shunt) = \frac{480 	imes 20}{500} = 19.2 Ω. Total = 40.8 + 19.2 = 60 Ω.

I = \frac{30}{60} = 0.5 A

Question 14

A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be

Accepted Answer

16 Ω. Stretching to n times length gives R' = n^2R = 2^2 	imes 4 = 16 Ω

Question 15

The mean free path of electrons in a metal is 4 	imes 10^{-8} m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units of Vm⁻¹

Accepted Answer

5 	imes 10^7. Energy gained = eE 	imes (mean free path). E = \frac{2	ext{ eV}}{e 	imes 4 	imes 10^{-8}} = \frac{2}{4 	imes 10^{-8}} = 5 	imes 10^7 V/m

Question 16

A wire of resistance 12 Ω/m is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points A and B is

Accepted Answer

0.6π Ω. Circumference = 2\pi(0.1) = 0.2\pi m; total resistance = 12 	imes 0.2\pi = 2.4\pi Ω. Two semicircles each 1.2\pi Ω in parallel: \frac{1.2\pi}{2} = 0.6\pi Ω

Question 17

A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively

Accepted Answer

1.21 times, same. Specific resistance (a material property) is unchanged. R \propto l^2 (constant volume), so R_2 = (1.1)^2 R_1 = 1.21R_1

Question 18

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

Accepted Answer

\frac{R}{4}. Two semicircular halves, each \frac{R}{2}, in parallel: \frac{(R/2)(R/2)}{R/2 + R/2} = \frac{R}{4}

Question 19

n 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

n^2R. Parallel: R = \frac{r}{n}, so r = nR. Series: R_s = nr = n(nR) = n^2R

Question 20

The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then

Accepted Answer

the resistance will be halved and the specific resistance will remain unchanged. R = \frac{\rho l}{\pi r^2}. Doubling l and r: R' = \frac{\rho(2l)}{\pi(2r)^2} = \frac{2\rho l}{4\pi r^2} = \frac{R}{2}. Specific resistance (material property) is unchanged.

Ohm's Law and Resistance — NEET Physics PYQ MCQs with Solutions

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