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ScienceCBSE Class 10 · Board exam 2024, set 31/2/1 · 4 marks · Case-based

CBSE Class 10 Science 2024: question 39(a)

Set in the CBSE Class 10 board exam of 2024, paper 31/2/1, section E, question 39(a), for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q39(a) · 4 marks · Case-basedCompetency-basedElectricity

When electric current flows in a purely resistive circuit electrical energy gets fully converted into heat energy. The amount of heat produced (H) in the circuit is found to be directly proportional to (i) the square of current () (ii) the resistance (R) of the conductor and (iii) the time (t) for which current flows. In other words . Electrical devices such an electric fuse, electric heater, electric iron etc. are all based on this effect called heating effect of electric current.

  1. (a)

    List two properties of heating elements.

    1 mark
  2. (b)

    List two properties of electric fuse.

    1 mark
  3. (c)

    Name the principle on which an electric fuse works. Explain how a fuse wire is capable of saving electrical appliances from getting damaged due to accidently produced high currents.

    2 marks
  4. (c)

    The power of an electric heater is 1100 W. If the potential difference between the two terminals of the heater is 220 V, find the current flowing in the circuit. What will happen to an electric fuse of rating 5 A connected in this circuit ?

    2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 31/2/1.

Hints

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

Hint 1

A heating element must get very hot without melting or burning.

Hint 2

A fuse must be the weakest point in the circuit, so it melts before the appliance is harmed.

Hint 3

Use I = P/V and compare the result with the fuse rating.

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) Heating elements (for example, nichrome in heaters) should have: (1) high resistivity, so more heat is produced ().

    ½ mark
  2. (a) (2) A high melting point, and they should not oxidise (burn) readily at high temperatures. For this reason alloys are used.

    ½ mark
  3. (b) An electric fuse wire should have: (1) high resistivity, so it heats up quickly.

    ½ mark
  4. (b) (2) A low melting point, so it melts as soon as the current goes above its rating.

    ½ mark
  5. (c)(a) Principle: an electric fuse works on the heating effect of electric current ().

    1 mark
  6. (c)(a) The fuse is connected in series with the appliance. If an unduly high current flows by accident, for example in a short circuit or overload, the heat produced in the fuse wire rises sharply. The wire melts and breaks the circuit, which stops the current and saves the appliance from damage.

    1 mark
  7. (c)(b) [OR] Given: W, V. Use .

    ½ mark
  8. (c)(b)

    ½ mark
  9. (c)(b) The current (5 A) does not exceed the fuse rating (5 A). So the fuse will not melt; it has no effect and the heater works normally.

    1 mark

Answer: (a) High resistivity; high melting point / does not oxidise at high temperature. (b) High resistivity; low melting point. (c)(a) Heating effect of current; the wire melts on high current and breaks the circuit. (c)(b) I = 5 A; the 5 A fuse is not affected (does not blow)

Where the marks are

  • Two properties of heating elements: high resistivity, high melting point/no oxidation (½ + ½ marks)
  • Two properties of fuse wire: high resistivity, low melting point (½ + ½ marks)
  • Names the heating effect of current (1 mark)
  • Explains that the fuse melts on high current and breaks the circuit (1 mark)
  • OR: uses P = VI (½ mark), gets I = 5 A (½ mark), states no effect on the 5 A fuse (1 mark)

Revise the chapter

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.