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Physics · 2025 · 2 marks
CBSE 2025 · Region 4 · Set 1 · Q22
(a) (i) Derive an expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time. (ii) The figure…
(a)(i)
Derive an expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.
(ii)
The figure shows the plot of current through a cross-section of wire over two different time intervals. Compare the charges ( $\displaystyle \mathrm{Q}_{1}$ and $\displaystyle \mathrm{Q}_{2}$ ) that pass through the cross-section during these time intervals.
(b)(i)
A battery of emf E and internal resistance r is connected to a variable external resistance $\displaystyle \mathrm{R}$.
(I)
Obtain the expression for current I in the circuit and the value of maximum current the battery can supply.
(II)
Obtain the terminal voltage V across the battery and its maximum possible value.
(ii)
The above battery sends a current $\displaystyle \mathrm{I}_{1}$ when $\displaystyle \mathrm{R}=\mathrm{R}_{1}$ and a current $\displaystyle \mathrm{I}_{2}$ when $\displaystyle \mathrm{R}=\mathrm{R}_{2}$. Obtain the internal resistance of the battery in terms of $\displaystyle \mathrm{I}_{1}, \mathrm{I}_{2}, \mathrm{R}_{1}$ and $\displaystyle \mathrm{R}_{2}$.
Marking-scheme solution
(a)(i)
Total charge transported along E is:
$\displaystyle I \Delta t=\frac{e^{2} A}{m} \tau n \Delta t E$
$\displaystyle \frac{I}{A}=\frac{n e^{2}}{m} \tau E$
$\displaystyle J=\frac{1}{\rho} E$
$\displaystyle \rho=\frac{m}{n e^{2} \tau}$
OR
Current in the conductor:
$\displaystyle I=n e A v_{d}$
$\displaystyle \frac{I}{A}=n e \frac{e E}{m} \tau$
$\displaystyle J=\frac{n e^{2} \tau}{m} E$
$\displaystyle J=\frac{1}{\rho} E$
$\displaystyle \rho=\frac{m}{n e^{2} \tau}$
(ii)
From given graph:
$\displaystyle \frac{Q_{1}}{Q_{2}}=\frac{A_{1} \text { (Area of rectangle) }}{A_{2} \text { (Area of triangle) }}=\frac{3}{2}$
$\displaystyle Q_{1}>Q_{2}$
OR
(b)(i)
(I)
$\displaystyle V=E-I r$
$\displaystyle I R=E-I r$
$\displaystyle I=\frac{E}{R+r}$
For maximum value of current $\displaystyle R=0$:
$\displaystyle I_{\max }=\frac{E}{r}$
(II)
$\displaystyle V=E-I r$
$\displaystyle V_{\max }=E$, when $\displaystyle I=0$
(ii)
$\displaystyle I_{1} R_{1}+I_{1} r=I_{2} R_{2}+I_{2} r$
$\displaystyle r=\frac{I_{2} R_{2}-I_{1} R_{1}}{I_{1}-I_{2}}$
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