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NCERT Class 10 Science Electricity: In-text questions (page 190) Solutions

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of In-text questions (page 190) in NCERT's Class 10 Science book, chapter “Electricity”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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  1. In-text (page 190) Q1

    Why does the cord of an electric heater not glow while the heating element does?

    1. The heating element is made of a material with very high electrical resistance.
    2. The cord is made of a material like copper which has very low electrical resistance.
    3. According to Joule's law, the heat produced in a conductor is directly proportional to its resistance.
      why?

      Joule's law of heating: the heat produced in a resistor is H = I²Rt.

    4. Since the resistance of the heating element is much higher than that of the cord, it produces significantly more heat.
    5. The heating element reaches a very high temperature and glows, while the cord stays cool and does not glow.

    Answer: The heating element has high resistance and produces enough heat to glow, whereas the cord has low resistance and does not produce sufficient heat to glow.

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  2. In-text (page 190) Q2

    Compute the heat produced when 96000 C of charge is transferred in one hour through a potential difference of 50 V.

    1. The heat produced is equal to the work done in moving the charge.
    2. The formula for work done is W = VQ, where V is potential difference and Q is charge.
    3. Given V = 50 V and Q = 96000 C.
    4. Substitute the values into the formula: W = 50 × 96000.
    5. Calculate the product: 50 × 96000 = 4800000 J.
    6. Express the result in scientific notation: 4.8 × 10⁶ J.

    Answer: 4.8 × 10⁶ J

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  3. In-text (page 190) Q3

    An electric iron of resistance 20 Ω takes 5 A. Calculate the heat produced in 30 s.

    1. Given resistance R = 20 Ω, current I = 5 A, and time t = 30 s.
    2. Use the formula for heat produced: H = I²Rt.
      why?

      Joule's law of heating: the heat produced in a resistor is H = I²Rt.

    3. Substitute the values: H = (5)² × 20 × 30.
    4. Calculate the square of the current: 5² = 25.
    5. Multiply the terms: 25 × 20 = 500.
    6. Multiply by time: 500 × 30 = 15000 J.
    7. Express the result in scientific notation: 1.5 × 10⁴ J.

    Answer: 1.5 × 10⁴ J

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Other exercises in this chapter

← Electricity: notes, practice and all exercises

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