CBSE 2025 · Region 7 · Set 1 · Q29 · 4 marks

In a metallic conductor, an electron, moving due to thermal motion, suffers collisions with the heavy fixed ions but after collision, it will emerge out with the same speed but in random directions. If we consider all the electrons, their average velocity will be zero. When an electric field is applied, electrons move with an average velocity, known as drift velocity ( $\displaystyle \mathrm{v}_{\mathrm{d}}$ ). The average time between successive collisions is known as relaxation time $\displaystyle (\tau)$. The magnitude of drift velocity per unit electric field is called mobility ( $\displaystyle \mu$ ). An expression for current through the conductor can be obtained in terms of drift velocity, number of electrons per unit volume ( n ), electronic charge (- e), and the cross-sectional area (A) of the conductor. This expression leads to an expression between current density ( $\displaystyle \hat{j}$ ) and the electric field ( $\displaystyle \overrightarrow{\mathrm{E}}$ ). Hence, an expression for resistivity ( $\displaystyle \rho$ ) of a metal is obtained. This expression helps us to understand increase in resistivity of a metal with increase in its temperature, in terms of change in the relaxation time $\displaystyle (\tau)$ and change in the number density of electrons $\displaystyle (\mathrm{n})$.

(i)

Consider two cylindrical conductors A and B, made of the same metal connected in series to a battery. The length and the radius of B are twice that of A. If $\displaystyle \mu_{\mathrm{A}}$ and $\displaystyle \mu_{\mathrm{B}}$ are the mobility of electrons in A and $\displaystyle \mathrm{B}$ respectively, then $\displaystyle \frac{\mu_{\mathrm{A}}}{\mu_{\mathrm{B}}}$ is:

(A)

$\displaystyle \frac{1}{2}$

(B)

$\displaystyle \frac{1}{4}$

(C)

$\displaystyle 2$ (D) $\displaystyle 1$ (ii) A wire of length $\displaystyle 0.5$ m and cross-sectional area $\displaystyle 1.0 \times 10^{-7} \mathrm{~m}^{2}$ is connected to a battery of $\displaystyle 2$ V that maintains a current of $\displaystyle 1.5$ A in it. The conductivity of the material of the wire (in $\displaystyle \Omega^{-1} \mathrm{~m}^{-1}$ ) is:

(A)

$\displaystyle 2.5 \times 10^{4}$

(B)

$\displaystyle 3.0 \times 10^{5}$

(C)

$\displaystyle 3.75 \times 10^{6}$

(D)

$\displaystyle 5.0 \times 10^{7}$

(iii)

The temperature coefficient of resistance of nichrome is $\displaystyle 1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. In order to increase resistance of a nichrome wire by $\displaystyle 8.5 \%$, the temperature of the wire should be increased by:

(A)

$\displaystyle 250^{\circ} \mathrm{C}$

(B)

$\displaystyle 500^{\circ} \mathrm{C}$

(C)

$\displaystyle 850^{\circ} \mathrm{C}$

(D)

$\displaystyle 1000^{\circ} \mathrm{C}$

(iv)

(a)OR(b)

Consider the contribution of the following two factors I and II in resistivity of a metal: I. Relaxation time of electrons II. Number of electrons per unit volume The resistivity of a metal increases with increase in its temperature because:

(A)

I decreases and II increases.

(B)

I increases and II is almost constant.

(C)

Both I and II increase.

(D)

I decreases and II is almost constant.

CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.