In-text Questions — Heating Effect — Electricity

1. 12 marksNCERT Cl-10 Science, In-text Qs after §11.7, Q1

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

   Hint. Both carry the same current — so the difference must come from somewhere else in the heat formula.

   Step 1 — Recall the heat formula. H = I²Rt.

   Step 2 — Compare the resistances. The heating element (e.g. nichrome coil) has a much higher resistance than the cord (usually copper, a good conductor).

   Step 3 — Apply this with the same current. Since both carry the same current (they're in series) but H depends on R, the high-resistance element produces far more heat — enough to glow — while the low-resistance cord stays comparatively cool.

   ✦ Answer: Both carry the same current, but the heating element's much higher resistance means it produces far more heat (H = I²Rt), while the low-resistance cord stays cool and doesn't glow.

   Where students slip. Assuming the current must be different in the cord and the element — they're in series, so the current through both is identical; it's the resistance that differs dramatically.

2. 22 marksNCERT Cl-10 Science, In-text Qs after §11.7, Q2

   Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

   Hint. You already have the total charge, so you don't need to bring time into this calculation at all.

   Step 1 — Recall the energy formula in terms of charge. W = VQ (work done per unit charge, extended to the whole charge moved).

   Step 2 — Substitute the given values. W = 50 × 96000 = 4,800,000 J = 4.8 × 10⁶ J, since the charge here is already a total rather than a rate.

   ✦ Answer: 4.8 × 10⁶ J of heat is generated.

   Where students slip. Trying to bring the '1 hour' into the calculation via H = I²Rt — since the total charge is already given directly, W = VQ gives the answer immediately, without needing current, resistance or time separately.

3. 32 marksNCERT Cl-10 Science, In-text Qs after §11.7, Q3

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

   Hint. Apply Joule's law directly — all three quantities you need are already given.

   Step 1 — Apply H = I²Rt. H = 5² × 20 × 30.

   Step 2 — Compute. H = 25 × 20 × 30 = 15000 J = 1.5 × 10⁴ J.

   ✦ Answer: 1.5 × 10⁴ J (15,000 J) of heat is developed.

   Where students slip. Squaring the resistance instead of the current (H = IR²t) — Joule's law squares the current specifically, not the resistance.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc111.pdf) — seven in-text question sets (23 questions total, not 16 as some older manifests claim) plus one end-of-chapter Exercise (18 questions, correctly counted). Unchanged by rationalisation. Table 11.2's resistivity values (used for several answers) were read directly from the book, including nichrome's 100 × 10⁻⁶ Ω·m.. Questions are referenced from the NCERT textbook for identification.