Science — CBSE Class 10 board question

(i) Definition of 1 Volt:

The potential difference between two points is said to be 1 volt when 1 joule of work is done in moving a charge of 1 coulomb from one point to the other.

$$1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}$$

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(ii) Finding the length of the wire:

Given: diameter $d = 0.2 \text{ mm} = 2 \times 10^{-4}$ m, $\rho = 1.6 \times 10^{-8}$ Ω m, $R = 14$ Ω

Area of cross-section:
$$A = \frac{\pi d^2}{4} = \frac{\pi \times (2 \times 10^{-4})^2}{4} = \pi \times 10^{-8} \text{ m}^2$$

Using $R = \rho \dfrac{l}{A}$:

$$l = \frac{RA}{\rho} = \frac{14 \times \pi \times 10^{-8}}{1.6 \times 10^{-8}} = \frac{14\pi}{1.6} \approx 27.5 \text{ m}$$

Effect of doubling the diameter:

If diameter is doubled, area becomes $4A$. Since $R \propto \dfrac{1}{A}$, resistance becomes $\dfrac{14}{4} = 3.5$ Ω.

Resistance decreases by 10.5 Ω (from 14 Ω to 3.5 Ω).

Source: Chapter 11 – Electricity, Section 11.4 & 11.5

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• (i): Examiners expect the definition in terms of work done per unit charge, with the formula. Simply writing "1 J/C" without context loses marks.
• (ii): Use $R = \rho l/A$ and $A = \pi d^2/4$. The key insight for the second part is that doubling diameter → area becomes 4 times → resistance becomes ¼ of original. Always state the change in resistance, not just the new value. Show formula, substitution, and result clearly for full 3 marks.