Physics Ncert Solutions Class 12th

3.22 Constant emf of the standard cell, $E_1$ = 1.02 V

Balance point on the wire, $l_1$ = 67.3 cm

A cell of unknown emf, $?$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is, $\frac{E_1}{l_1}$ = $\frac{?}{l}$

Hence, $?$ = $\frac{l}{l_1}*E_1$ = $\frac{82.3}{67.3}*1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

The method would not work if the emf of the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{eff}$ is maximum. $R_{eff}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{eff}$ is minimum. $R_{eff}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{nR}{\frac{R}{n}}$ = $n^2$

Let us assume, $R_1$ = 1 Ω, $R_2$ = 2 Ω, $R_3$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$Ω

From the circuit the equivalent resistance is given by

R = $\frac{2*1}{2+1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{2*3}{2+3}$ + 1 = $\frac{6}{5}$ + 1 = $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

Required equivalent resistance, R = $\frac{6}{11}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{1*2*3}{1*2+2*3+3*1}$ = $\frac{6}{11}$ Ω

(a)

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2 $\Omega$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R = $\frac{2*4}{2+4}$ = $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is $\frac{4}{3}$ Ω $*4=\frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5 $*R=5R$

Q.3.18  (a) When a steady current flows in a metallic conductor of non-uniform cross section, only the current flowing is constant. Current density, Electric field and Drift speed are inversely proportional to the cross section area, hence not constant.

(b) Ohm's law is not applicable to all conductors, vacuum diode semi-conductor is a non-ohmic conductor.

(c) According to ohm's law V = IR, voltage is directly proportional to current, hence to draw high current from a low voltage source, internal resistance ®, needs to be low.

(d) To prevent the drawing of extra current, which can cause short circuit, the internal resistance for a high voltage system needs to be high.

3.16 Given, resistivity of aluminium, ${\rho }_1$ = 2.63 $*{10}^{-8}$ Ωm

Resistivity of copper, ${\rho }_2$ = 1.72 $*{10}^{-8}$ Ωm

Relative density of aluminium = 2.7

Relative density of copper = 8.9

For Aluminium wire: Let

$l_1$ = length, $m_1$ = mass, $R_1$ = resistance, $A_1$ = cross-section area, ${\rho }_1$ = density

For Copper wire: Let

$l_2$ = length, $m_2$ = mass, $R_2$ = resistance, $A_2$ = cross-section area, ${\rho }_2$ = density

From the relation R = $\frac{\rho l}{A}$ , we get

$R_1=\frac{{\rho }_1{l}_1}{A_1}$ ………(1), and

$R_2=\frac{{\rho }_2{l}_2}{A_2}$ ………(2)

It is given $R_1$ = $R_2$ and $l_1$ = $l_2$ . Hence

$\frac{A_1}{A_2}$ = $\frac{{\rho }_1}{{\rho }_2}$ = $\frac{2.63}{1.72}$ ( ${10}^{-8}$ cancel each other)

From the relation mass = volume $*\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}$ = length $*$ cross sectional area $*$ relative density, we get

$m_1=l_1{A}_1{d}_1$ and $m_2=l_2{A}_2{d}_2$ , where $d_1$ = relative density of aluminium and $d_2$ = relative density of copper

$\frac{m_1}{m_2}=\frac{l_1{A}_1{d}_1}{l_2{A}_2{d}_2}$ = $\frac{A_1{d}_1}{A_2{d}_2}$ = $\frac{2.63}{1.72}*\frac{2.7}{8.9}$ = 0.464

It can be inferred from this ratio that $m_1$ is less than $m_2$ . Hence, aluminium is preferred as overhead power cables over copper.