CBSE Class 10 Science sample paper 2025-26: question 36

Hints

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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. (i) The wire has a circular cross-section, so $A=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2$.

   Diameter $d=0.80\text{ mm}=0.80\times 10^{-3}\text{ m}$, so radius $r=0.40\times 10^{-3}\text{ m}$.

   ½ mark

2. $A=3.14\times (0.40\times 10^{-3}{)}^2=3.14\times 0.16\times 10^{-6}\approx 5.0\times 10^{-7}{\text{m}}^2$

   ½ mark

3. (ii) Use $R=\frac{\rho l}{A}$ and rearrange for length:

   $l=\frac{RA}{\rho }$

   ½ mark

4. Substitute $R=0.12\Omega$, $A=5\times 10^{-7}{\text{m}}^2$ and $\rho =1.8\times 10^{-8}\Omega \text{m}$:

   $l=\frac{0.12\times 5\times 10^{-7}}{1.8\times 10^{-8}}=\frac{6\times 10^{-8}}{1.8\times 10^{-8}}$

   1 mark

5. $l\approx 3.33\text{m}$ (about 3.3 m of the copper wire is needed).

   ½ mark

Answer: (i) A ≈ 5 × 10⁻⁷ m²; (ii) l ≈ 3.33 m

Revise the chapter

• Chapter: Electricity (notes, formulas, where marks are lost and every past question)
• Topic: Resistance and Resistivity