4 Marks Questions

Question 14 Marks

The number of free electrons in a unit volume copper conductor estimated in Example 8.1 is 8.5 x 10²⁸per m³.How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 x 10^-6 m²and it is carrying a current of 3.0 A.

Answer

Given that :
Number density of electrons,
$n=8.5 \times 10^{28} m^3$
Length of wire, $l=3 m$
Area of cross-section of wire,
$A =2.0 \times 10^{-0} m^2$
Current in the wire, $I =3.0 A$
Current on electron, $e=1.6 \times 10^{-19} C$
Let time taken by the electron in moving from one end of the wire to other end $=t$
t = ?
Using the relation,
current, $\quad I =n e A v_d$
$\therefore \quad v_d=\frac{ I }{n e A}$
Putting value$v_d=\frac{3}{8.5 \times 10^{28} \times 1.6 \times 10^{-9} \times 2.0 \times 10^{-6}}$
$=\frac{3}{27.2} \times 10^{-3} ms^{-1}=1.1 \times 10^{-4} ms^{-1}$
Time taken in the draft
$\begin{aligned}
t & =\frac{l}{v_d}=\frac{3}{1.1 \times 10^{-4}} \\&
=2.727 \times 10^4 s \text { or } 7.5 \text { hours }\end{aligned}$

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Question 24 Marks

In the given figure, a part ABC of electric circuit is shown. The magnitudes of potentials at points $A, B$ and $C$ are and respectively, then find the value of potential at point $O$.

Answer

Sol. By junction rule, at point O
$-i_1-i_2-i_3=0$ or $i_1+i_2+i_3=0$
If magnitude of potential at point O is, then by Ohm 's law, magnitudes of current in different branches will be
$i_1=\frac{V_0-V_{ A }}{ R _1}, i_2=\frac{V_0-V_{ B }}{ R _2}$ and $i_3=\frac{V_0-V_{ C }}{ R _3}$
Putting the value of $i_1, i_2$ and $i_3$
$\frac{V_0-V_A}{R_1}+\frac{V_0-V_B}{R_2}+\frac{V_0-V_C}{R_3}=0$
or $V _0\left(\frac{1}{ R _1}+\frac{ l }{ R _2}+\frac{1}{ R _3}\right)=\frac{ V _{ A }}{ R _1}+\frac{ V _{ B }}{ R _2}+\frac{ V _{ C }}{ R _3}$
$V _0=\left(\frac{ V _{ A }}{ R _1}+\frac{ V _{ B }}{ R _2}+\frac{ V _{ C }}{ R _3}\right)\left(\frac{1}{ R _1}+\frac{1}{ R _2}+\frac{1}{ R _3}\right)^{-1}$

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Question 34 Marks

A thick wire of $15 \Omega$ is stretched to three times its original length. Assuming that the density of the wire remains unchanged, calculate the resistance of the new wire.

Answer

Let original length of the wire be $l_1$ and original area of cross-section be $A _1$. Let the new area of cross-section be $A_2$ and new length be $l_2$.
Since density of the wire remains unchanged, so the volume of wire before stretching is equal to the volume of the wire after stretching. $ \begin{aligned} A_1 l_1 & =A_2 l_2 \\ A_2 & =\frac{A_1 l_1}{l_2} \\ l_2 & =3 l_1 \end{aligned} $ or Since $\therefore$ Area of wire after stretching.
$A _1 l_1= A _2 l_2$
or $A _2=\frac{ A _1 l_1}{l_2}$
Since $l_2=3 l_1$
$\therefore$ Area of wire after stretching
$A _2=\frac{ A _1 l_1}{3 l_1}=\frac{1}{3} A_1$
Given that, $R =\frac{\rho l}{A}$
$R _1=\rho\left(\frac{l_1}{A_1}\right)$ and $R _2=\rho\left(\frac{l_2}{A_2}\right)=\frac{\rho \times 3 l_1}{A_2}$
$\therefore \quad \frac{ R _2}{ R _1}=\frac{3 \rho l_1}{A_2} \times \frac{ A _1}{\rho l_1}=3\left(\frac{A_1}{A_2}\right)$
$\Rightarrow \quad \frac{ R _2}{ R _1}=\frac{3 A_{\perp}}{\frac{1}{3} A_1}=9$
$\Rightarrow \quad \frac{ R _2}{ R _1}=9$
$R_2=9 R_1$
But given that, $R _1=15 \Omega$
$\therefore \quad R _2=9 \times 15=135 \Omega$ Ans.

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Question 44 Marks

A wire is stretched to reduce its diameter to half. What will be the magnitude of its new resistance?

Answer

Let original length be $l_1$ and original diameter be $d_1$, new length be $l_2$ and new diameter be $d_2$.
$A _1 l_1= A _2 l_2$
$\frac{\pi d_1^2}{4} l_1=\frac{\pi d_2^2}{4} l_2$
$l_2=\left(\frac{d_1}{d_2}\right)^2 l_1$
$\because \quad d_2=\frac{1}{2} d_1$
$\therefore \quad \frac{d_1}{d_2}=\frac{2}{1}=2$
$l_2=(2)^2 l_1=4 l_1$
New length $l_2=4 l_1 $
$\therefore \quad R _1=\frac{\rho \times 4 l_1}{\pi d_1^2}$
and $R _2=\frac{4 l_2 \rho}{\pi d_2^2}$
$\therefore \quad \frac{ R _2}{ R _1}=\frac{l_2 d_1^2}{l_1 d_1^2}$
$=\frac{4 l_1 d_1^2 \times 4}{l_1 \times d_1^2}=16$
New resistance $R _2=16 R _1$
Resistance will become 16 times of its original value.

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