CBSE Class 10 Science 2024: question 36(B)

Hints

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Step-by-step solution

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1. (i) Resistance of a cylindrical conductor: $R=\rho \frac{l}{A}$, where $\rho$ is the resistivity of the material.

   1 mark

2. Rearranging, $\rho =\frac{RA}{l}$. SI unit $=\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre).

   1 mark

3. (ii) Given $l=3$ m, $R=60\Omega$, $A=4\times 10^{-7}{\text{m}}^2$.
   $\rho =\frac{RA}{l}=\frac{60\times 4\times 10^{-7}}{3}=80\times 10^{-7}=8\times 10^{-6}\Omega \text{m}$.

   1 mark

4. (iii) The resistivity will not change when the wire is stretched to double its length.

   1 mark

5. Justification: resistivity depends only on the nature of the material (and temperature), not on the length or area of cross-section of the conductor. (Only the resistance changes on stretching.)

   1 mark

Answer: (i) $R=\rho l/A$; SI unit of ρ = Ω m. (ii) $\rho =8\times 10^{-6}\Omega$ m. (iii) No change, as resistivity depends only on the material.

Revise the chapter

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