(a) The resistance of a wire of 0.01 cm radius and 1.0 cm length is $7\Omega$ . Calculate its resistivity. (b) An electric heater is rated 220 V; 11 A. Calculate the power consumed if the heater is operated at 200 V.

Electricity

Resistivity and Electric Power

Official Answer

(a) Resistivity $\rho \approx 2.2 \times 10^{-5}\ \Omega\,\text{m}$ .

(b) Power consumed at 200 V is $2000\ \text{W} = 2\ \text{kW}$ .

resistivityresistancearea of cross-sectionOhm's lawelectric powerP=V^2/R

Hint

First find the resistance of the heater from its rated V and I using $R=V/I$ ; this R stays constant. For resistivity, use $R = \rho l/A$ with $A=\pi r^2$ .

Quick Oral Answer

Resistivity depends only on the material and temperature, not on the wire's dimensions. Power of a resistive device varies as the square of the applied voltage for a fixed resistance, so a small drop in voltage causes a proportionally larger drop in power.

Analysis & Explanation

This is a two-part numerical based on Section 11.5 (resistivity, Eq. 11.10, $R=\rho l/A$ ) and Section 11.8 (electric power, Eq. 11.22, $P=V^2/R$ ), both illustrated by NCERT Examples 11.5 and 11.3/11.10.

Part (a): The resistance of a uniform conductor depends on its length, area of cross-section and the resistivity of its material: $R = \rho \dfrac{l}{A}$ . Given the radius and length of the wire and its resistance, the area is found from $A = \pi r^2$ , and $\rho$ is obtained by rearranging the formula. This mirrors NCERT Example 11.5, which computes resistivity from a wire's resistance, length and diameter using $\rho = \dfrac{RA}{l} = \dfrac{R\pi d^2}{4l}$ .

Part (b): The heater's resistance is a fixed property of its filament (assuming resistance does not change with the small change in operating voltage). It is calculated from the rated values using Ohm's law, $R = V/I$ , and then the power at the new (lower) voltage is found using $P = V^2/R$ (Eq. 11.22). Since $R$ stays the same while $V$ decreases, power consumed decreases — consistent with the chapter's point that a device draws different currents/powers at different applied voltages for the same resistance (cf. Example 11.4).

Common Mistakes

1Forgetting to convert radius and length from cm to m before substituting into SI-unit formulas, giving a resistivity off by several orders of magnitude.
2Using diameter instead of radius in $A=\pi r^2$ without adjusting the formula.
3Assuming the heater's power scales linearly with voltage instead of using $P=V^2/R$ (quadratic dependence), leading to an incorrect answer of $2000\times(200/220)=1818\ \text{W}$ instead of the correct $V^2/R$ calculation.
4Recomputing a 'new' resistance at 200 V instead of recognizing R is a fixed property of the heater filament.

Interesting Facts

A resistivity of about $2.2\times10^{-5}\ \Omega\,\text{m}$ falls in the range typical of alloys such as manganin ( $44\times10^{-6}\ \Omega\,\text{m}$ ) or constantan ( $49\times10^{-6}\ \Omega\,\text{m}$ ) rather than a pure metal like copper, whose resistivity is far lower ( $1.62\times10^{-8}\ \Omega\,\text{m}$ ).

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2026?

This question carries 4 marks in the CBSE Class 10 Science 2026 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Electricity" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Resistivity and Electric Power" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2026 paper?

This is a Short Answer question from Section C in the CBSE Class 10 Science 2026 paper.