CBSE Class 10 Physics Electricity MCQs Set 11

Answer: (d) nature of the material

Explanation:
1. Resistivity is a characteristic property of the material.
2. It does not depend on the length, thickness or shape of the wire.
3. It depends only on the nature of the material (and the temperature), so (d) is correct.

Teacher's Note:
1. Resistance depends on length and area, but resistivity does not.
2. Resistivity is the resistance of a wire of unit length and unit area of cross-section.

Answer: (d) All the above

Explanation:
1. R = ρl/A, so it depends on the length and the area of cross-section.
2. The resistivity ρ also changes with temperature, so R depends on temperature too.

Teacher's Note:
1. Resistance depends on length, area, material and temperature.
2. All three given factors affect the resistance.

Answer: (d) density

Explanation:
1. R = ρl/A, where the resistivity ρ depends on the material and on the temperature.
2. So resistance depends on length, area and temperature but not on the density.

Teacher's Note:
1. Resistance depends on length, area of cross-section, material and temperature.
2. Density is not a factor.

Answer: (d) Pressure

Explanation:
1. Resistance of a conductor depends on its length, area of cross-section, nature of the material and temperature.
2. It does not depend on pressure.

Teacher's Note:
1. R = ρl/A, and the resistivity ρ changes with temperature.
2. Pressure is not a factor for the resistance of a conductor.

Answer: (c) the shape of the resistor is changed

Explanation:
1. Resistivity is a characteristic property of the material of the conductor.
2. It depends on the nature of the material and on the temperature only.
3. It does not depend on the shape, length or thickness of the resistor, so (c) is correct.

Teacher's Note:
1. Resistance changes with length and area, but resistivity does not.
2. Resistivity changes only when the material or the temperature changes.

Answer: (a) True

Explanation:
1. For metallic conductors, increase of temperature increases the collisions of electrons with the vibrating atoms.
2. So the resistivity (specific resistance) of a metallic conductor increases with temperature.
3. The statement is true.

Teacher's Note:
1. Specific resistance is another name for resistivity.
2. Alloys like constantan show only a small change, but in general resistivity of a conductor rises with temperature.

Answer: (b) increase

Explanation:
1. In a metallic conductor, the atoms vibrate more at a higher temperature, so the free electrons collide more often.
2. Hence the resistance of the conductor increases when its temperature increases.

Teacher's Note:
1. The resistance of metals increases with temperature.
2. The resistance of a semiconductor decreases with temperature, but this is outside this chapter.

Answer: (d) is high

Explanation:
1. R = ρl/A, so resistance is inversely proportional to the area of cross-section.
2. A thin wire has a smaller area, so its resistance is higher.

Teacher's Note:
1. The comparison is for the same material and the same length.
2. Thin wires offer more resistance than thick wires.

Answer: (c) greater when switched on

Explanation:
1. When the lamp is switched on, the filament becomes very hot (about 2500°C).
2. The resistance of a metal increases with temperature, so the resistance is greater when the lamp is switched on.

Teacher's Note:
1. A bulb filament is a non-ohmic conductor because its temperature changes.
2. The hot-filament resistance may be ten times the cold resistance.

Answer: (b) super conductors

Explanation:
1. Some materials lose all their resistance below a certain very low temperature.
2. Such materials are called superconductors.

Teacher's Note:
1. Superconductors carry current without any loss of energy as heat.
2. Mercury becomes a superconductor near 4 K.

Answer: (a) Resistance of C will be the least; C is the best conductor of electricity

Explanation:
1. For wires of equal length and area, R = ρl/A, so resistance is directly proportional to resistivity.
2. The smallest resistivity is 10⁻⁸ (wire C), so C has the least resistance.
3. A wire of least resistivity is the best conductor, so the correct row is (a).

Teacher's Note:
1. Resistivity is a property of the material, so it can be compared directly when length and area are equal.
2. Insulators have very large resistivity (10¹² to 10¹⁴), metals have very small values.

Answer: (a) 10⁻⁸ to 10⁻⁶ Ω m

Explanation:
1. Metals are good conductors with a very low resistivity.
2. Their resistivity lies in the range 10⁻⁸ to 10⁻⁶ Ω m.
3. Insulators have a very large resistivity of 10¹² to 10¹⁷ Ω m.

Teacher's Note:
1. Silver and copper have resistivities of the order of 10⁻⁸ Ω m.
2. Alloys have resistivity in the order of 10⁻⁶ Ω m.

Answer: (a) Higher resistivity

Explanation:
1. Alloys have a higher resistivity than the pure metals they are made from, so they produce more heat.
2. They also do not oxidise (burn) readily at high temperature, so they last long.
3. Among the given options the higher resistivity is the correct property.

Teacher's Note:
1. Nichrome is a common alloy used in heating elements.
2. A heating element needs high resistivity and a high melting point.

Answer: (b) very high

Explanation:
1. The resistivity of an alloy is generally higher than that of its constituent metals.
2. So among the given options the resistivity of alloys is described as very high compared with metals, but it is still far less than that of insulators.

Teacher's Note:
1. Alloys like nichrome have resistivity of about 10⁻⁶ Ω m.
2. Insulators have a resistivity of 10¹² Ω m or more.

Answer: (a) 5A

Explanation:
1. For the first conductor R = ρl/A, so ρ = RA/l.
2. For the second conductor 0.5R = ρ(2.5l)/A′.
3. A′ = ρ × 2.5l / 0.5R = (RA/l) × 2.5l / 0.5R = 5A.

Teacher's Note:
1. The two conductors are of the same material, so the resistivity ρ is the same.
2. Divide the two resistance equations to cancel ρ.

Answer: (c) 2A

Explanation:
1. For the same material the resistivity ρ is the same, so ρ = RA/l for the first conductor.
2. For the second conductor R = ρ(2l)/A′, so A′ = ρ(2l)/R = (RA/l)(2l)/R = 2A.

Teacher's Note:
1. Use the formula R = ρl/A for both conductors and eliminate ρ.
2. Doubling the length needs doubling of the area to keep the same resistance.

Answer: (b) thick, short and cool

Explanation:
1. R = ρl/A, so resistance is least for a small length and a large area.
2. Resistance also falls with a lower temperature.
3. So a thick, short and cool conductor has the least resistance.

Teacher's Note:
1. Resistance increases with length and temperature, and decreases with area.
2. Choose the option with the smallest l, the largest A and the lowest temperature.

Answer: (d) 5 × 10⁻⁷ Ω m

Explanation:
1. R = V/I = 2/4 = 0.5 Ω.
2. ρ = RA/l = (0.5 × 1 × 10⁻⁶)/1 = 5 × 10⁻⁷ Ω m.

Teacher's Note:
1. First find R using Ohm’s law, then find ρ from R = ρl/A.
2. Nichrome has resistivity of the order of 10⁻⁶ Ω m.

Answer: (a) 1.15 × 10⁻⁶ ohm m

Explanation:
1. l = 80 cm = 0.8 m, A = 1.00 mm² = 1 × 10⁻⁶ m², R = 0.92 Ω.
2. ρ = RA/l = (0.92 × 1 × 10⁻⁶)/0.8 = 1.15 × 10⁻⁶ Ω m.

Teacher's Note:
1. Convert cm to m and mm² to m² before substituting.
2. Resistivity is in ohm metre.

Answer: (b) 10 m

Explanation:
1. A = πr² = 3.14 × (10⁻³)² = 3.14 × 10⁻⁶ m².
2. l = RA/ρ = (0.1 × 3.14 × 10⁻⁶)/(3.14 × 10⁻⁸) = 10 m.

Teacher's Note:
1. Convert the radius to metres before finding the area.
2. Rearrange R = ρl/A to get l.

Answer: (d) 1.1 m

Explanation:
1. A = π(d/2)² = π(0.2 × 10⁻³)² = 1.257 × 10⁻⁷ m².
2. l = RA/ρ = (4.2 × 1.257 × 10⁻⁷)/(48 × 10⁻⁸) ≈ 1.1 m.

Teacher's Note:
1. Use the radius, 0.2 mm, to find the area.
2. Convert mm to m.

Answer: (b) 60 cm

Explanation:
1. A = 1 mm² = 1 × 10⁻⁶ m².
2. From R = ρl/A, l = RA/ρ = (1.56 × 10⁻² × 1 × 10⁻⁶)/(2.6 × 10⁻⁸).
3. l = 0.6 m = 60 cm.

Teacher's Note:
1. Convert the area from mm² to m² (1 mm² = 10⁻⁶ m²).
2. Rearrange R = ρl/A to find the length.

Answer: (c) 4R

Explanation:
1. The volume Al is constant, so when l becomes 2l the area becomes A/2.
2. R′ = ρ(2l)/(A/2) = 4ρl/A = 4R.

Teacher's Note:
1. When a wire is stretched at constant volume, R ∝ l².
2. Doubling the length makes R four times.

Answer: (d) 180 Ω

Explanation:
1. At constant volume, R ∝ l².
2. R′ = (3)² × 20 = 9 × 20 = 180 Ω.

Teacher's Note:
1. When the length becomes 3 times, the area becomes one third.
2. So the resistance becomes 3 × 3 = 9 times.

Answer: (d) 16R

Explanation:
1. Radius becomes r/2, so the area becomes A/4.
2. The volume is constant, so the length becomes 4l.
3. R′ = ρ(4l)/(A/4) = 16ρl/A = 16R.

Teacher's Note:
1. Area is proportional to r², so halving the radius makes the area one fourth.
2. Length increases four times to keep the volume.

Answer: (d) 2% increase

Explanation:
1. The volume of the wire is constant, so a 1% decrease in the area gives a 1% increase in length.
2. R ∝ l/A, so R increases by about 1% + 1% = 2%.

Teacher's Note:
1. For small changes, add the percentage change in length and the percentage change in area.
2. R ∝ l/A.

Answer: (b) the resistance will be halved and the specific resistance will remain unchanged

Explanation:
1. R′ = ρ(2l)/π(2r)² = ρ(2l)/(4πr²) = R/2.
2. Specific resistance is a property of the material, so it remains unchanged.

Teacher's Note:
1. Length doubling gives a factor of 2, radius doubling gives a factor of 1/4.
2. Resistivity remains the same for the same material and temperature.

Answer: (a) R/4

Explanation:
1. The circle has two semicircular halves between two diametrically opposite points.
2. Each half has resistance R/2 and they are in parallel.
3. R′ = (R/2 × R/2)/(R/2 + R/2) = R/4.

Teacher's Note:
1. Two equal resistors R/2 in parallel give R/4.
2. Diametrically opposite points divide the ring into two equal arcs.

Answer: (c) L/2 and 2A

Explanation:
1. The volume LA is fixed, so R = ρL/A = ρL²/(volume).
2. R is smallest for the smallest length.
3. Among the options, L/2 with 2A has the smallest length.

Teacher's Note:
1. For a fixed volume R ∝ L².
2. Compare L², 4L² and L²/4 for the three options.

Answer: (b) 0.25R

Explanation:
1. Doubling the diameter makes the area four times.
2. Resistance is inversely proportional to the area, so R′ = R/4 = 0.25R.

Teacher's Note:
1. Area is proportional to the square of the diameter.
2. Same length and material, so only the area changes.

Answer: (d) 625R

Explanation:
1. Diameter becomes d/5, so the area becomes A/25.
2. The volume is constant, so the length becomes 25l.
3. R′ = ρ(25l)/(A/25) = 625 ρl/A = 625R.

Teacher's Note:
1. Area ∝ d², and length increases to compensate for the volume.
2. R ∝ l/A = 25 × 25.

Answer: (d) R/n²

Explanation:
1. The volume is constant, so if the area becomes nA the length becomes l/n.
2. R′ = ρ(l/n)/(nA) = R/n².

Teacher's Note:
1. When the area increases n times, the length decreases n times.
2. So R changes by a factor of 1/n².

Answer: (a) R/4

Explanation:
1. On recasting, the volume stays the same: Al = A′l′.
2. New length l′ = l/2, so the new area A′ = 2A.
3. R′ = ρl′/A′ = ρ(l/2)/(2A) = R/4.

Teacher's Note:
1. The volume of the wire does not change on melting and recasting.
2. Halving the length doubles the area of cross-section.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Resistance R = ρl/A, so a thick wire (larger A) has less resistance.
2. With the same source voltage, the current I = V/R is greater in the thick wire, so A is true.
3. R is true and it is the reason why the thick wire has lower resistance, so R explains A.

Teacher's Note:
1. Resistance is inversely proportional to the area of cross-section.
2. A thick wire offers a wider path to electrons, so current flows more easily.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: a longer wire has more resistance because R ∝ l.
2. R is false: resistance is directly proportional, not inversely proportional, to the length.
3. So A is true but R is false.

Teacher's Note:
1. R = ρl/A, so resistance is directly proportional to length and inversely proportional to area.
2. Read the words directly and inversely carefully in such questions.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R = ρl/A, so R ∝ l for a wire of constant area (R is true).
2. When the length is doubled the resistance is doubled, so A is true and R explains A.

Teacher's Note:
1. Resistance is directly proportional to length and inversely proportional to area.
2. This holds when the area and the material stay the same.

Answer: (d) A is false, but R is true.

Explanation:
1. Conductivity is the reciprocal of resistivity, so resistivity × conductivity = 1 for every material.
2. Hence A is false: the product does not depend on the material.
3. R is true, because each of them depends on the material.

Teacher's Note:
1. Resistivity ρ and conductivity σ are related by σ = 1/ρ.
2. Their product is always 1.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: R = ρl/A, so the thinner wire A (smaller area) has the greater resistance.
2. R is false: both wires are of the same material, so their resistivity is the same.
3. So A is true but R is false.

Teacher's Note:
1. Resistivity depends only on the material and temperature.
2. The wires differ in the area of cross-section, not in the material.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R = ρl/A, so R is inversely proportional to the area (R is true).
2. If A becomes A/2 with the length constant, the resistance becomes 2R (A is true), and R explains A.

Teacher's Note:
1. Inverse proportionality means halving one quantity doubles the other.
2. Length and material are unchanged.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: bending changes neither the length nor the area of cross-section, so the resistance does not change.
2. R is true: resistance depends only on length, thickness, material and temperature.
3. R explains A, because bending is not one of these factors.

Teacher's Note:
1. Bending the wire does not change l, A, material or temperature.
2. So the resistance remains the same.