Physics M.C.Q (1 Marks)

Similarly area of cross-section of $B=$ Area of cross-section of $C = a^2$

Hence according to formula $R = \rho \frac{l}{A};$   resistances of all the conductors are equal i.e.$ R_A = R_B = R_C$

So resistance $=$ slope $= \tan \,(90 -\theta) = \cot\,\theta$.

Resistance for contact $A-A$

${R_{AA}} = \rho \frac{x}{{2x \times 4x}} = \frac{\rho }{{8x}}$

Similar for contacts $B-B$ and $C-C$ are respectively

${R_{BB}} = \rho .\frac{{2x}}{{x \times 4x}} = \frac{\rho }{{2x}} = \frac{{4\rho }}{{8x}}$
and ${R_{CC}} = \rho \frac{{4x}}{{x \times 2x}} = \frac{{2\rho }}{x} = \frac{{16\rho }}{{8x}}$

It is clear maximum resistance will be for contact $C-C$.

and conductance $G = \frac{1}{R}$

$ \Rightarrow \,\,GR = 1$  .....  $(ii)$

From equation $(i)$ and $(ii)$ $\sigma = \frac{{GR}}{\rho }$

For conductor $B$, ${R_B} = \frac{{\rho \,l}}{{\pi (r_2^2 - r_1^2)}}$

$ \Rightarrow $   $\frac{{{R_A}}}{{{R_B}}} = \frac{{r_2^2 - r_1^2}}{{r_1^2}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} - 1 = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} - 1 = {\left( {\frac{2}{1}} \right)^2} - 1 = 3$

Hence,

$\frac{\rho_{B} l_{B}}{A}\left(\alpha_{B} \Delta T\right)=\frac{\rho_{C} l_{C}}{A}\left(\alpha_{C} \Delta T\right)$

$\therefore \frac{l_{B}}{l_{C}}=\frac{\rho_{C} \alpha_{C}}{\rho_{B} \alpha_{B}}$

$\rho=\frac{m}{n e^{2} \tau}$

where, $m$ is mass of an electron, $e$ is charge on electron, $n$ is number of free electrons per unit volume in conductor and $\tau$ is relaxation time.

$\Rightarrow \rho \propto \frac{1}{\tau}$$.............(1)$

Also, the conductivity is reciprocal of resistivity i.e.

$\sigma=\frac{1}{\rho}$$...............(2)$

$\Rightarrow \sigma \propto \tau$$............(3)……….$ from $(1)$ and $(2)$

$\frac{\rho}{\sigma} \propto \frac{1}{\tau^{2}} ........$ from $(1)$ and $(3)$

When temperature increases relaxation time will decrease and hence, ratio of resistivity and conductivity will increase.

Where $R=$ resistance $V=$ Voltage $I=$ Current

Also $R$ is directly proportional to temperature $T$ and inversely proportional to the slope.

$\therefore$ from the above example

$R_{1} \propto T_{1}$ and $R_{2} \propto T_{2}$

but from given figure $R_{1} < R_{2}$

$\therefore T_{1} < T_{2}$

$\mathrm{E}_{2} \mathrm{A}-\mathrm{E}_{1} \mathrm{A}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_{0}}$

$\mathrm{JA}\left(\frac{1}{\sigma_{2}}-\frac{1}{\sigma_{1}}\right)=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_{0}}$

${\varepsilon _0}I\left( {\frac{1}{{{\sigma _2}}} - \frac{1}{{{\sigma _1}}}} \right) = {q_{{\rm{in}}}}$

The resistance per unit length $\mathrm{r}$ is given by $x=\rho \frac{1}{A}$

where $\rho$ is the resistivity and $\mathrm{A}$ is the cross-sectional area of the conductor $\mathrm{PQ}$.

Thus, $\mathrm{r}$ is inversely proportional to the cross-sectional area $\mathrm{A}$ of the conductor.

$\frac{1 \Delta \mathrm{R}}{\mathrm{R}}=\frac{1 \Delta \ell}{\ell}+\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \rho}{\rho}$

Maximum error is least count of an instrument

$\therefore \quad 1.5 \%$

$\frac{\mathrm{A}_{1}+\mathrm{A}_{2}}{\rho_{\mathrm{eq}} \ell}=\frac{\mathrm{A}_{1}}{\rho_{1} \ell}+\frac{\mathrm{A}_{2}}{\rho_{2} \ell}$

$\rho_{\mathrm{eq}}=\frac{\rho_{1} \rho_{2}\left(\mathrm{A}_{1}+\mathrm{A}_{2}\right)}{\rho_{2} \mathrm{A}_{1}+\rho_{1} \mathrm{A}_{2}}$

$\frac{V_{A}}{V_{B}}=\frac{R_{A}}{R_{B}}=\frac{\ell_{A} / r_{A}^{2}}{\ell_{B} / r_{B}^{2}}$

$\Rightarrow \frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}=\sqrt{\frac{\mathrm{V}_{\mathrm{A}} \ell_{\mathrm{B}}}{\mathrm{V}_{\mathrm{B}} \ell_{\mathrm{A}}}}=\frac{1}{2}$

$\Rightarrow \mathrm{R}_{\mathrm{A}}=\mathrm{R}_{\mathrm{B}}$

$R_1 \alpha_1+R_2 \alpha_2=0$

$(100)\left(4.5 \times 10^{-3}\right)=R\left(5 \times 10^{-2}\right)$

$0.9 \times 10=R$

$R=9 r$

Violet $= 7$, yellow $= 4$, brown $= 1$ and gold $= 5\,\ %$ (tolerance)

So $R = 74 \times {10^1} \,\pm 5\% $ $ = 740\, \pm 5\% $

So its value will be nearest to $741\,\Omega $.

$\Rightarrow \frac{n R T}{V} v^{2 /3}=$ const.

$\Rightarrow T V^{-1 / 3}=$ const.

$\Rightarrow T=$ conts. $\times V^{1 / 3}$

$\therefore$ On increasing volume, temperature increase

For minimum thickness increment $2 \mu_{1} \Delta \mathrm{t}=\frac{\lambda}{2}$

$\Rightarrow\left(t_{2}-t_{1}\right)=\frac{\lambda}{4 \mu_{1}}=\frac{9.6 \times 10^{-7}}{4 \times 1.2}=2 \times 10^{-7} \mathrm{\,m}$

$56 \mathrm{\,k} \Omega \pm 5 \%=56 \times 10^{3}\, \Omega \pm 5 \%$

Sequence : Green, Blue, Orange, Golden

$\mathrm{R}_{\mathrm{f}}=\frac{\mathrm{R}_{\mathrm{i}}}{2}=12 \times 10^{\circ}\, \Omega$

After molding $\mathrm{R}_{f}=4 \mathrm{R}_{i}=96\, \Omega$

For identical resistances, potential difference distributes equally among all. Hence potential difference across each resistance is $\frac{2}{3}\,V,$ and potential difference between $A$ and $B$ is $\frac{4}{3}\,V.$

Let equivalent resistance be ${r_{R,}}$

then $\frac{1}{{{r_R}}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r}$ ........... $10\;{\rm{times}}$

$\therefore \frac{1}{{{r_R}}} = \frac{{10}}{r} = \frac{{10}}{{R/10}} = \frac{{100}}{R} \Rightarrow {r_R} = \frac{R}{{100}} = 0.01\,R$

This is in series with $\frac{2}{3}\,\Omega $ resistor.

$\therefore $Total resistance $ = \frac{2}{3} + \frac{1}{3} = \frac{3}{3}\,\Omega = 1\,\Omega $

Total resistance

$ = 2 + \frac{2}{3} = \frac{8}{3}\,\Omega $

Current through ammeter

$ = \frac{2}{{8/3}} = \frac{6}{8} = \frac{3}{4}\,A$

$ = \frac{{(2 + 2) \times 2}}{{2 + 2 + 2}} = \frac{8}{6} = \frac{4}{3}\, \Omega $

In series, $R + R + R$ .... $n$ times $=$ $nR = n (nx) = n^2x$

${R_{AB}} = \frac{{9 \times 6}}{{9 + 6}} = \frac{{9 \times 6}}{{15}} = \frac{{18}}{5} = 3.6\, \Omega $

So the equivalent resistance between points $A$ and $B$ is equal to $R = \frac{{6 \times 3}}{{6 + 3}} = 2\, \Omega $

Suppose ${R_2}$ is broken then ${R_1} = 2\, \Omega $ ….. $ (ii)$

On solving equations $(i)$ and $(ii) $ we get ${R_2} = 6/5\,\Omega $

So potential difference across $3$ will be $V = \frac{{3 \times 1}}{5} = 0.6\,V$

${R_1} - {R_2} = \sqrt {{{({R_1} + {R_2})}^2} - 4{R_1}{R_2}} = \sqrt {81 - 72} = 3$

${R_1} = 6\,\Omega ,\,{R_2} = 3\,\Omega $

$\frac{{{i_1}}}{{{i_2}}} = \frac{3}{3}$ $ \Rightarrow $ ${i_1} = {i_2}$

${i_2} = 0.5\,A = {i_1}$

$ \Rightarrow $ ${R_{eq}} = 9\,\Omega $

$\frac{1}{{{R_1}}} = \left( {\frac{4}{R}} \right)4 \Rightarrow {R_1} = \frac{R}{{16}}$

So the current in the circuit $ = \frac{{20}}{5} = 4\,ampere$

Hence the current flowing through each resistance $= 2\,ampere$.