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Science · 2024 · 5 marks
CBSE 2024 · Region 1 · Set 1 · Q36
(A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R). (ii) An electric oven is designed to work on the…
(i)
Define electric power. Express it in terms of potential difference (V) and resistance (R).
(ii)
An electric oven is designed to work on the mains voltage of $\displaystyle 220$ V. This oven consumes $\displaystyle 11$ units of electrical energy in $\displaystyle 5$ hours. Calculate :
(a)
power rating of the oven
(b)
current drawn by the oven
(c)
resistance of the oven when it is red hot
(i)
Write the relation between resistance R and electrical resistivity $\displaystyle \rho$ of the material of a conductor in the shape of cylinder of length $\displaystyle l$ and area of cross-section A. Hence derive the SI unit of electrical resistivity.
(ii)
The resistance of a metal wire of length $\displaystyle 3$ m is $\displaystyle 60 \Omega$. If the area of cross-section of the wire is $\displaystyle 4 \times 10^{-7} \mathrm{~m}^{2}$, calculate the electrical resistivity of the wire.
(iii)
State how would electrical resistivity be affected if the wire (of part 'ii') is stretched so that its length is doubled. Justify your answer.
Marking-scheme solution
(A) (i)
Electric power : Rate at which electrical energy is dissipated or consumed / Rate of supplying energy to maintain the flow of current through a circuit.
\(\displaystyle \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
(ii) (a) \(\displaystyle (1\) unit \(\displaystyle =1 \mathrm{kWh})\)
Power, \(\displaystyle \mathrm{P}=\frac{\text { Electrical energy consumed }}{\text { Time }}\)
\[=\frac{11 \mathrm{kWh}}{5 \mathrm{~h}}=2.2 \mathrm{~kW} \text { or } 2200 \mathrm{~W}
\]
(b) \(\displaystyle I=\frac{P}{V}\)
\[=\frac{2200}{220}=10 \mathrm{~A}
\]
(c) \(\displaystyle \mathrm{R}=\frac{V^{2}}{P}\)
\[=\frac{(220)^{2}}{2200}=22 \Omega
\]
(B)(i)
\(\displaystyle R=\rho \frac{l}{A}\)
\[\begin{aligned}
& \rho=\frac{\mathrm{R} \times \mathrm{A}}{\mathrm{l}} \\
= & \text { Ohm } \times \frac{(\text { metre })^{2}}{\text { metre }} \\
= & \text { ohm metre } / \Omega \mathrm{m}
\end{aligned}
\]
(ii)
Here \(\displaystyle l=3 \mathrm{~m}, \mathrm{~A}=4 \times 10^{-7} \mathrm{~m}^{2}, \mathrm{R}=60 \Omega\)
\[\begin{aligned}
\rho & =\frac{R \times A}{l} \\
& =\frac{60 \times 4 \times 10^{-7}}{3} \\
& =80 \times 10^{-7} \Omega \mathrm{~m}
\end{aligned}
\]
(iii)
Resistivity will not change.
because Resistivity does not depend on the dimension of the conductor / It only depends on the nature of the material.
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