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Question ID - 174386 | SaraNextGen Top Answer

The motion of free electrons in a conductor are continuous and random. They collide with positive metal ions and change direction during each collision. So thermal velocities are randomly distributed and average velocity is zero.

When a potential difference is applied across the ends of a conductor, electrons are drifted towards the positive terminal of the field, this velocity is called drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)$
$
\mathrm{v}_{\mathrm{d}}=-\frac{\mathrm{e} \overrightarrow{\mathrm{E}} \tau}{\mathrm{m}}=\frac{i}{n e A}
$

1.If $\mathrm{N}, \mathrm{e}, \tau$ and $\mathrm{m}$ are representing electron density, charge, relaxation time and mass of an electron respectively, then the resistance of wire of length $\ell$ and cross-sectional area A is given by
(a) $\frac{\mathrm{m} \ell}{\mathrm{Ne}^{2} \mathrm{~A}^{2} \tau}$
(b) $\frac{2 \mathrm{~m} \tau \mathrm{A}}{\mathrm{Ne}^{2} \ell}$
(c) $\frac{\mathrm{Ne}^{2} \tau \mathrm{A}}{2 \mathrm{~m} \ell}$
(d) $\frac{\mathrm{Ne}^{2} \mathrm{~A}}{2 \mathrm{~m} \tau \ell}$
2.When a current I is set up in a wire of radius $\mathrm{r}$, the drift velocity is $\mathrm{v}_{\mathrm{d}}$. If the same current is set up through a wire of radius $2 \mathrm{r}$, the drift velocity will be
(a) $4 \mathrm{v}_{\mathrm{d}}$
(b) $2 \mathrm{v}_{\mathrm{d}}$
(c) $\mathrm{v}_{\mathrm{d}} / 2$
(d) $\mathrm{v}_{\mathrm{d}} / 4$
3.A straight conductor of uniform cross-section carries a current I. If $\mathrm{s}$ is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is
(a) Is
(b) $\sqrt{\mathrm{I} / \mathrm{s}}$
(c) $\mathrm{I} / \mathrm{s}$
(d) $(\mathrm{I} / \mathrm{s})^{2}$

4.The resistance of a wire at room temperature $30^{\circ} \mathrm{C}$ is found to be $10 \Omega$. Now to increase the resistance by $10 \%$, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002$ per ${ }^{\circ} \mathrm{C}$ ]
(a) $36^{\circ} \mathrm{C}$
(b) $83^{\circ} \mathrm{C}$
(c) $63^{\circ} \mathrm{C}$
(d) $33^{\circ} \mathrm{C}$
5.The number of free electrons per $100 \mathrm{~mm}$ of ordinary copper wire is $2 \times 10^{21}$. Average drift speed of electrons is $0.25$ $\mathrm{mm} / \mathrm{s}$. The current flowing is
(a) $5 \mathrm{~A}$
(b) $80 \mathrm{~A}$
(c) $8 \mathrm{~A}$
(d) $0.8 \mathrm{~A}$

a)a,d,c,a,d   b)a,d,c,b,d   c)a,a,c,b,d   d)a,d,c,b,a

Verified Solutions by SaraNextGen


1 Answer
127 votes
Answer Key / Explanation : (b) -

(a) $\frac{\mathrm{m} \ell}{\mathrm{Ne}^{2} \mathrm{~A}^{2} \tau}$

(d) $\mathrm{I}=n A e v_{d}$ or $v_{d} \propto 1 / \pi r^{2}$

(c) $\mathrm{I} / \mathrm{s}$

(b) $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}(1+\alpha \mathrm{t})$
Initially, $\mathrm{R}_{0}(1+30 \alpha)=10 \Omega$
Finally, $\mathrm{R}_{0}(1+\alpha \mathrm{t})=11 \Omega$
$
\therefore \frac{11}{10}=\frac{1+\alpha t}{1+30 \alpha}
$
or, $10+(10 \times 0.002 \times t)=11+330 \times 0.002$
or, $0.02 \mathrm{t}=1+0.66=1.066$ or $\mathrm{t}=\frac{1.66}{0.02}=83^{\circ} \mathrm{C}$.
$
\text { (d) } \begin{aligned}
\mathrm{I} &=\mathrm{n} \mathrm{e} \mathrm{A} \mathrm{V}_{\mathrm{d}}=2 \times 10^{21} \times 1.6 \times 10^{-19} \times 10 \times 0.25 \times 10^{-3} \\
&=2 \times 1.6 \times 0.25=\frac{8}{10}=0.8 \mathrm{~A}
\end{aligned}
$


127 votes

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