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Answer the following. (Class 10 Science 2025, Q25)

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ScienceCBSE Class 10 · Board exam 2025, set 31/5/1 · 2 marks · Very short answer

CBSE Class 10 Science 2025: question 25

Set in the CBSE Class 10 board exam of 2025, paper 31/5/1, section B, question 25, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q25 · 2 marks · Very short answerElectricity

Answer the following.

  1. (A)

    State two applications of Joule's heating in domestic electric circuit.

    0 marks
  2. (B)(a)

    Establish the relationship between the commercial unit of electric energy and the SI unit of electric energy.

    0 marks

  3. OR

    (B)(b)

    Determine the total resistance of the parallel combination of three resistances of 2 Ω, 4 Ω and 6 Ω.

    0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Which household devices work because a wire gets hot when current flows through it?

Hint 2

1 kW = 1000 W and 1 h = 3600 s; 1 W s = 1 J.

Hint 3

For parallel resistors, add the reciprocals using the LCM 12, then take the reciprocal.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (A) First application of Joule's heating at home: the electric iron / electric heater / room heater, where the heating element becomes hot when current flows.

    ½ mark

  2. (A) Second application: the electric fuse, whose wire melts from the heat produced and breaks the circuit when the current is too large. The filament of an electric bulb also glows because of heating.

    ½ mark

  3. (B)(a)

    ½ mark

  4. (B)(a) . So joules.

    ½ mark

  5. (B)(b) [OR alternative to (B)(a), ½ + ½] For resistors in parallel:

  6. (B)(b) , so .

Answer: (A) Electric iron/heater and electric fuse (or bulb); (B)(a) 1 kWh = 3.6 × 10⁶ J; (B)(b) Rp = 12/11 Ω ≈ 1.09 Ω

Where the marks are

  • Any two applications of heating effect, e.g. iron, heater, fuse, bulb (½ + ½ mark)
  • Writes 1 kWh = 1000 W × 3600 s (½ mark)
  • Concludes 1 kWh = 3.6 × 10⁶ J (½ mark)
  • OR: correct parallel formula with substitution (½ mark) and Rp = 12/11 Ω ≈ 1.09 Ω (½ mark)

Revise the chapter

  • Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 31/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.