CBSE Class 10 Physics HOTs Electricity Set 05

This CBSE Class 10 Physics HOTS set on Electricity has 24 higher-order questions with step-by-step answers, simple-words summaries and teacher's notes. It tests how resistance changes when wires are stretched, folded or cut, why bulbs and heaters behave as they do in series and parallel circuits, and Joule heating in everyday appliances.

Very Short Answer Questions (HOTS)

Question 1. How will the resistivity of a conductor change when its length is tripled by stretching it?
Answer:
Resistivity is a property of the material (and temperature) only.
It does not depend on length or area, so it will remain the same. The resistance, however, becomes 9 times.
In simple words: Stretching changes resistance, not resistivity.

Teacher's Note:
1. Separate the two ideas: resistance depends on dimensions, resistivity does not.
2. Volume stays constant on stretching, so A becomes one-third and R becomes 9 times.

Question 2. A given length of a wire is doubled on itself. By what factor does the resistance of the wire change?
Answer:
When the wire is doubled on itself, its new length is L/2 and its new area of cross-section is 2A.
R′ = ρ(L/2)/(2A) = (1/4)(ρL/A) = R/4.
The resistance becomes one-fourth of its original value.
In simple words: Folding a wire in two gives half the length and twice the area.

Teacher's Note:
1. Folding is the opposite of stretching.
2. Both the length factor and the area factor reduce R.

Question 3. A given length of wire is doubled on itself and the process is repeated once again. By what factor does the resistance of wire will change?
Answer:
Doubling the wire on itself makes its length half and its area twice. Repeating the process once again, the length becomes 1/4 of the original and the area becomes 4 times.
R′ = ρ(l/4)/(4A) = (1/16)(ρl/A), so the resistance becomes 1/16 of the original value (it decreases 16 times).
In simple words: Folding twice gives one-sixteenth of the resistance.

Teacher's Note:
1. Each folding multiplies R by 1/4.
2. The source answer said 16 times; the wire's resistance actually decreases.

Question 4. The electric bulb glows immediately when switch is on. Why?
Answer:
When the switch is closed, the electric field is set up throughout the wire almost at the speed of light, so all the free electrons in the wire and the filament begin to move together at once. The bulb therefore glows immediately, even though the electrons drift slowly.
In simple words: The whole wire starts carrying current at the same instant.

Teacher's Note:
1. The signal travels much faster than the electrons drift.
2. Electrons are already present in the filament.

Question 5. A heater joined in parallel with a 60 W bulb is replaced by a 100 W bulb. Will the rate of heat produced by the heater be more, less or remain the same? Why?
Answer:
The rate of heat produced by the heater remains the same.
In a parallel connection each branch has the same potential difference, so the heater's current and power (P = V²/R) do not depend on what is connected in the other branch.
In simple words: A branch in parallel does not affect the other branches.

Teacher's Note:
1. The voltage across the heater is unchanged.
2. P = V²/R depends only on that branch.

Question 6. What happens to the value of resistance when each of the length and the area of cross section of a conductor are tripled?
Answer:
R′ = ρ(3l)/(3A) = ρl/A = R. The resistance remains the same.
In simple words: Tripling length and area leaves R unchanged.

Teacher's Note:
1. Length and area effects cancel.
2. R = ρl/A.

Question 7. If the radius of a current carrying conductor is halved, how does current through it change?
Answer:
Halving the radius makes the area one-fourth, so the resistance becomes 4 times (R ∝ 1/r²).
At the same potential difference, I = V/R, so the current becomes one-fourth.
In simple words: A thinner conductor carries less current.

Teacher's Note:
1. Area depends on r².
2. I ∝ 1/R at constant voltage.

Question 8. Two unequal resistances are connected in parallel. If you are not provided with any other parameters (e.g. numerical values of I and R), what can be said about the voltage drop across the two resistors?
Answer:
In a parallel combination the voltage drop (potential difference) is the same across both resistors.
In simple words: Parallel branches share the same voltage.

Teacher's Note:
1. Voltage is common in parallel; current divides.
2. This holds whatever the resistances are.

Short Answer Questions (HOTS)

Question 9. Though the same current flows through line wires or the filament of a bulb, yet only the latter glows. Why?
Answer:
The filament has a very high resistance, whereas the line wires have a negligible resistance.
Heat produced H = I²Rt is proportional to the resistance when the current is the same, so the filament produces much more heat and starts glowing.
In simple words: A high-resistance filament becomes white hot; the thick wires stay cool.

Teacher's Note:
1. Use H = I²Rt in the answer.
2. Same current, different resistance: that is the whole idea.

Question 10. Which has higher resistance: a 50 W lamp bulb or a 25 W lamp bulb and how many times?
Answer:
For bulbs on the same voltage, R = V²/P, so R is inversely proportional to P.
The 25 W bulb has the higher resistance, and it is 2 times the resistance of the 50 W bulb.
In simple words: Lower wattage on the same voltage means higher resistance.

Teacher's Note:
1. Compare the bulbs at the same rated voltage.
2. R₂₅/R₅₀ = 50/25 = 2.

Question 11. Two wires P and Q are made of copper. The wire P is long and thin, while the wire Q is short and thick. Which will have more specific resistance? Give a reason for your answer.
Answer:
Both wires will have the same specific resistance, since both are made of the same material (copper) and are at the same temperature. Resistivity does not depend on length or thickness.
(The long, thin wire P has the greater resistance.)
In simple words: Same material, same resistivity; only the resistance differs.

Teacher's Note:
1. Distinguish clearly between resistance and resistivity.
2. R = ρl/A: P has larger l and smaller A.

Question 12. How does the resistance of a metallic wire depend on its temperature? Explain the reason to your answer.
Answer:
The resistance of a metallic wire increases with increase in temperature.
On heating, the atoms (ions) of the metal vibrate with larger amplitude and frequency, so the free electrons collide with them more often. This increases the opposition to the flow of electrons, i.e. the resistance.
In simple words: A hotter metal wire has more atomic vibrations, so more resistance.

Teacher's Note:
1. Give the reason in terms of collisions.
2. The resistance of metals falls when they are cooled.

Question 13. The electric power consumed by a device may be calculated by either of the two expressions P = I²R or P = V²/R. The first expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained?
Answer:
The expression P = I²R is used when the current is the same for different resistances (as in series), so P ∝ R.
The expression P = V²/R is used when the voltage is the same across different resistances (as in parallel), so P ∝ 1/R.
There is no contradiction: the proportionality holds only for the quantity (I or V) that is kept constant.
In simple words: Power rises with R at constant current and falls with R at constant voltage.

Teacher's Note:
1. State which quantity is constant in each case.
2. Series: same I. Parallel: same V.

Question 14. A bulb gets dimmer when an electric iron or geyser is switched on, why?
Answer:
An electric iron or geyser draws a heavy current from the supply line. The heavy current causes a larger voltage drop in the connecting wires, so the voltage across the bulb falls and the bulb glows less brightly (dimmer).
In simple words: A heavy load lowers the voltage available to the bulb.

Teacher's Note:
1. Think of the voltage drop in the supply wires.
2. Brightness depends on the voltage across the bulb.

Question 15. A fuse wire melts at 5 A. If it is desired that the fuse wire of same material melt at 10 A, then whether the new fuse wire should be of smaller or larger radius than the earlier one? Give reasons for your answer.
Answer:
The new fuse wire should have a larger radius.
A wire carrying a larger current produces more heat (H = I²Rt) and loses heat from its surface. To let 10 A pass without melting, the wire must have a smaller resistance and a larger surface, i.e. a larger area of cross-section (thicker wire). A thinner wire would melt at a smaller current.
In simple words: A thicker fuse wire can carry more current before melting.

Teacher's Note:
1. Resistance falls as the radius increases.
2. Thicker wire, higher rating.

Question 16. A battery E is connected to three identical lamps P, Q and R as shown in figure. Initially, the switch S is kept open and the lamps P and Q are observed to glow with same brightness. Then, switch S is closed. How will the brightness of glow of bulbs P and Q change? Justify your answer.

Answer:
On closing S, the lamps Q and R are in parallel and this combination is in series with lamp P.
The total resistance of the circuit decreases, so the current in the circuit (and in P) increases. The glow of bulb P increases.
The current in P divides between Q and R, so a smaller current flows through Q. The glow of bulb Q decreases.
In simple words: Adding a parallel branch lowers the resistance, brightening P and dimming Q.

Teacher's Note:
1. Draw the circuit before and after closing S.
2. Brightness depends on the current through the lamp.

Long Answer and Numerical Questions (HOTS)

Question 17. Two identical resistors of 24 Ω each are connected to a battery of 6 V. Calculate the ratio of the power consumed by the resulting combinations with (a) minimum resistance and (b) maximum resistance.
Answer:
(a) Minimum resistance (parallel): R_p = (24 × 24)/(24 + 24) = 12 Ω.
(b) Maximum resistance (series): R_s = 24 + 24 = 48 Ω.
P = V²/R, so P_p/P_s = R_s/R_p = 48/12.
Ratio = 4 : 1
In simple words: For the same battery, the smaller resistance dissipates more power.

Teacher's Note:
1. Same V, so power is inversely proportional to resistance.
2. Write the ratio in the order asked.

Question 18. A wire of resistance 6 Ω is bent to form a closed square. What is the resistance across a diagonal of the square?
Answer:
Each side of the square has a resistance 6/4 = 1.5 Ω.
Between opposite corners there are two parallel paths, each with two sides in series: 3 Ω each.
R = (3 × 3)/(3 + 3) = 1.5 Ω
In simple words: Two equal 3 Ω paths in parallel give 1.5 Ω.

Teacher's Note:
1. Split the wire into four equal sides first.
2. Series inside each path, then parallel.

Question 19. Explain how the resistance of the conductor varies if (i) area is halved (ii) length is doubled (iii) area is doubled (iv) both area and length are doubled.
Answer:
Use R = ρl/A, with the original resistance R.
(i) Area halved: R′ = ρl/(A/2) = 2R.
(ii) Length doubled: R′ = ρ(2l)/A = 2R.
(iii) Area doubled: R′ = ρl/(2A) = R/2.
(iv) Both doubled: R′ = ρ(2l)/(2A) = R (no change).
In simple words: Resistance goes up with length and down with area.

Teacher's Note:
1. Change one factor at a time.
2. In (iv) the two effects cancel.

Question 20. A wire has a resistance of 16 Ω. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance?
Answer:
The volume stays constant, so when the length becomes l/2 the area becomes 2A.
R′ = ρ(l/2)/(2A) = R/4 = 16/4 = 4 Ω.
Change in resistance = 16 − 4 = 12 Ω, i.e. (12/16) × 100 = 75% decrease.
In simple words: Half the length and double the area gives one-fourth of the resistance.

Teacher's Note:
1. Volume (length × area) is constant when a wire is melted and redrawn.
2. Percentage change = change ÷ original × 100.

Question 21. If 5 equal resistances are connected in series and this combination dissipates 40 watt of power across a source of emf, the power dissipated if the resistors are connected in parallel across the same source of emf is ________.
Answer:
Let each resistance be R and the source voltage V.
Series: R_s = 5R, so P_s = V²/5R = 40 W.
Parallel: R_p = R/5, so P_p = 5V²/R = 25 × (V²/5R) = 25 × 40 = 1000 W.
In simple words: Parallel dissipates n² times the series power, here 25 × 40 W.

Teacher's Note:
1. P = V²/R for the same source.
2. R_s/R_p = 25.

Question 22. A piece of wire of resistance 20 Ω is drawn out of a material so that its length is increased to twice of its original length. Calculate the resistance in the new situation.
Answer:
When the wire is drawn out, its volume stays constant: the length becomes 2l and the area becomes A/2.
R′ = ρ(2l)/(A/2) = 4ρl/A = 4R = 4 × 20 = 80 Ω.
In simple words: Doubling the length by drawing out makes the resistance four times.

Teacher's Note:
1. Do not forget that the area also halves.
2. The source answer of 40 Ω ignored the change of area.

Question 23. When a potential difference of 2 volt is applied across the ends of a wire of 5 m length, a current of 1 A is found to flow through it. Calculate: (i) the resistance per unit length of the wire; (ii) the resistance of 2 m length of this wire; (iii) the resistance across the ends of the wire if it is doubled on itself.
Answer:
Resistance of the whole wire R = V/I = 2/1 = 2 Ω.
(i) Resistance per unit length = 2/5 = 0.4 Ω per metre.
(ii) R(2 m) = 2 × 0.4 = 0.8 Ω.
(iii) Doubled on itself, the length becomes 2.5 m and the area doubles: R′ = 2/(2 × 2) = 0.5 Ω.
In simple words: Folding the wire in two makes its resistance one-fourth.

Teacher's Note:
1. Resistance is proportional to length for the same wire.
2. Folding halves the length and doubles the area.

Question 24. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source. (i) Will the bulbs in the two circuits glow with the same brightness? Justify your answer. (ii) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
Answer:
(i) No. In the series circuit the source voltage V is shared, so each bulb gets V/3 and its power is only (1/3)² = 1/9 of its rated power: dim. In the parallel circuit each bulb gets the full voltage V and glows at its rated power: bright.
(ii) In the series circuit, the circuit is broken and the other bulbs will not glow. In the parallel circuit the other bulbs continue to glow, because each is on its own branch.
In simple words: Series: dim and all go off together. Parallel: bright and independent.

Teacher's Note:
1. Compare the voltage across each bulb.
2. Power is proportional to V² for a fixed resistance.