Physics M.C.Q (1 Marks)

$\frac{I_{2}}{I_{1}}=\frac{A_{2}}{A_{1}} \times \frac{v_{d_{2}}}{v_{d_{1}}}=\frac{1}{4} \times \frac{2 v}{v} \Rightarrow I_{2}=\frac{I}{2}$

$\frac{4 I}{\pi d^{2}}=n e v$

$\frac{161}{\pi d^{2}}=n e v^{\prime}$

From equaiton $( I )$ and $( II )$

$\frac{4 I}{16 I}=\frac{v^{\prime}}{v^{\prime}} \Rightarrow v^{\prime}=4 v$

$v_{P}>v_{Q}$

$I=n e A v$

Where I = current

$\mathrm{n}=$ charge carrier current density

$\mathrm{e}=$ charge on the carriers

$A=$ cross sectional area

$\mathrm{v}=$ drift velocity

In the above example

$l=$ current $=\mathrm{i}$

$\mathrm{n}=$ charge carrier current density $=\mathrm{p}$

$\mathrm{e}=$ charge on the carriers $=\mathrm{q}$

$A=$ cross sectional area $=s$

$\mathrm{v}=$ drift velocity $=\mathrm{v}$

hence $i=p q s v$

$\therefore v=\frac{i}{p q{s}}$

$I=J A$

As $A -cross$ section of wire changes, $J -current$ density changes and $I$ current remains same.

The free electron density is a property of the material and doesnot change when current passes.

As the cross section changes, current density changes and hence the drift velocity changes.

$t=\frac{1}{4} \sec$

$\mathrm{H}-\int_{0}^{1 / 4}(3-12 \mathrm{t})^{2} \times \mathrm{Rdt}$

$=\left.\frac{(3-12 t)^{2}}{3 x-12}\right|_{0} ^{1 / 4} R$

$=\frac{+1}{36}[27]=\frac{3 \mathrm{R}}{4}$

$=\frac{1}{2}\left(I_{0}\right)(\Delta t)$

$I_{0}=\frac{2 q}{\Delta t}$

also, $\mathrm{i}=-\left(\frac{\mathrm{I}_{0}}{\Delta \mathrm{t}}\right) \mathrm{t}+\mathrm{I}_{0}$

heat loss $=\int i^{2} \mathrm{Rdt}=\frac{4}{3} \frac{\mathrm{q}^{2} \mathrm{R}}{\Delta \mathrm{t}}$

$i=J A$

$\mathrm{ev}_{\mathrm{d}} \mathrm{B}=\mathrm{e} \mathrm{E}$

$\operatorname{ne} \mathrm{A} \mathrm{v}_{\mathrm{d}}=\mathrm{JA}$

$\mathrm{E}=\frac{\mathrm{JB}}{\mathrm{ne}}$

Constant of proporionality $=\frac{1}{\mathrm{ne}}=\frac{\mathrm{m}^{3}}{\mathrm{C}}$

$\sigma=\frac{\mathrm{J}}{\mathrm{E}}=\frac{\mathrm{neVd}}{\mathrm{E}}=\mathrm{ne}\left(\frac{\mathrm{Vd}}{\mathrm{E}}\right)$

$\sigma=$ $\mathrm{ne}$ (mobility)

$6.4 \times 100\, r^{-1} m^{-1}$

$=n\left(1.6 \times 10^{-19}\right)\left(4 \times 10^{3} \times 10^{-1}\right)$

$\Rightarrow n=10^{22} \mathrm{\,m}^{-3}$

$\mathrm{J}=\mathrm{neVd}$

$\mathrm{i}=$ $\mathrm{Ane}$ $\mathrm{Vd}$

$\mathrm{i}=$ constant

$\mathrm{AVd}=$ Constant

$ \Rightarrow \frac{{{{\rm{n}}_{\rm{e}}} \times {\rm{e}}}}{{\rm{t}}} = {10^{20}} \times 1.6 \times {10^{ - 19}} \times 0.25 \times 1.5$

$= 16 \times \frac{1}{4} \times \frac{3}{2}=6 $

$ \Rightarrow  \frac{\mathrm{n}_{e}}{\mathrm{t}}=\frac{6}{1.6 \times 10^{-19}} $

$ \Rightarrow \frac{{{{\rm{n}}_{\rm{p}}}}}{{\rm{t}}} = 6.25 \times {10^{19}}\,$   [ Given ${{\rm{ }}\frac{{{{\rm{n}}_{\rm{p}}}}}{{\rm{t}}} \times 60\%  = \frac{{{{\rm{n}}_{\rm{e}}}}}{{\rm{t}}}}$]

$\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{4 \rho \ell}{\pi \mathrm{d}^{2}}$

$\therefore {{\rm{V}}_{{\rm{drift}}}} = \frac{{\rm{V}}}{{{\rm{RneA}}}} = \frac{{\rm{V}}}{{\frac{{\rho \ell }}{{\rm{A}}} \times {\rm{neA}}}} = \frac{{\rm{V}}}{{\rho \ell {\rm{ne}}}} \propto \frac{{\rm{V}}}{\ell }$

$\mathrm{q}=$ area under curve $=\frac{1}{2} \times 150 \times 0.2=\frac{30}{2} \mathrm{\,C}$

$\mathrm{v}=\mathrm{E} \cdot \mathrm{d}=400 \times 10^{3} \times 1500=60 \times 10^{7}$

energy released $=60 \times 10^{7} \times 15=900 \times 10^{7}$

$=9 \times 10^{9} \mathrm{\,J}$

$\times$ momentum of each electron

$=\mathrm{Nmv}_{\mathrm{d}}=\mathrm{nA} \ell \mathrm{mv}_{\mathrm{d}}$

$=\frac{\mathrm{I} \ell}{\mathrm{e}} \mathrm{m}=\frac{\mathrm{I} \ell}{\mathrm{S}}$

is same $2(\mathrm{n})$ e $\mathrm{A} \mathrm{v}_{\mathrm{L}}=\mathrm{n}$ e $(2 \mathrm{A}) \mathrm{v}_{\mathrm{R}} \Rightarrow \frac{\mathrm{v}_{\mathrm{L}}}{\mathrm{v}_{\mathrm{R}}}=1$

$\therefore $ total current $\mathrm{I}=(2 \mathrm{e}+\mathrm{e}+\mathrm{e}) \times 10^{19}$

${=4 \times 1.6 \times 10^{-19} \times 10^{19}} $

${=6.4 \mathrm{\,A}}$

$\mathrm{J}=$ current density

$=\frac{\mathrm{I}}{\pi \mathrm{r}^{2}}=\frac{1.7}{\pi \times\left(0.51 \times 10^{-3}\right)^{2}}=\mathrm{nev}_{\mathrm{d}}$

$=8.5 \times 10^{27} \times\left(1.6 \times 10^{-19}\right) \times v_{d}$

$\therefore $ $\mathrm{v}_{\mathrm{d}}=\frac{1.7}{\pi \times\left(0.51 \times 10^{-3}\right)^{2} \times 8.5 \times 10^{27} \times 1.6 \times 10^{-19}}$

$=1.5 \times 10^{-3} \mathrm{\,m} / \mathrm{sec} .=1.5 \mathrm{\,mm} / \mathrm{sec}$

$=\frac{\mathrm{J}_{0} \mathrm{r}}{\mathrm{R}} \times 2 \pi \mathrm{dr}$

$\mathrm{I}=\frac{\mathrm{J}_{0} 2 \pi}{\mathrm{R}} \int_{0}^{\mathrm{R}} \mathrm{r}^{2} \mathrm{dr}=\frac{2 \pi \mathrm{J}_{0}}{\mathrm{R}} \times \frac{\mathrm{R}^{3}}{3}=\frac{2}{3} \mathrm{J}_{0} \cdot \pi \mathrm{R}^{2}$

$=\frac{2}{3} J_{0} A$

$\mathrm{I}=$ constant

$\mathrm{V}_{\mathrm{d}} \propto \frac{1}{\mathrm{A}}$

$\left(\mathrm{V}_{\mathrm{d}}\right)_{\mathrm{B}}>\left(\mathrm{V}_{\mathrm{d}}\right)_{\mathrm{A}}$

$\frac{I_{2}}{I_{1}}=\frac{A_{2}}{A_{1}} \times \frac{v_{d_{2}}}{v_{d_{1}}}=\frac{1}{4} \times \frac{2 v}{v} \Rightarrow I_{2}=\frac{I}{2}$

Molecular mass of copper $M=63 g m=0.063 k g$

Number of copper atoms per unit volume $n=\frac{\rho N_{A}}{M}$

$\therefore n=\frac{9 \times 10^{3} \times 6.023 \times 10^{23}}{0.063}=8.60 \times 10^{28} \mathrm{m}^{-3}$

As one copper atom gives one free electron, thus number of electrons per unit volume is $8.6 \times 10^{28} \mathrm{m}^{-3}$

Cross-section area of wire $A=\frac{\pi d^{4}}{4}=\frac{3.14 \times(0.001)^{2}}{4}=7.85 \times 10^{-7} \mathrm{m}^{2}$

Drift velocity $v_{d}=\frac{I}{A n e}$

$\therefore v_{d}=\frac{1.1}{\left(7.85 \times 10^{-7}\right)\left(8.6 \times 10^{28}\right)\left(1.6 \times 10^{-19}\right)}=0.01 \times 10^{-2} \mathrm{m} / \mathrm{s}=0.1 \mathrm{mm} / \mathrm{s}$

$\int \mathrm{d} q=\int i d t$

$\mathrm{q}=\int_{0}^{60} 2+3 \mathrm{t} \mathrm{dt}\left(\times 10^{-3}\right)$

$\mathrm{q}=\left[2 \mathrm{t}+\frac{3 \mathrm{t}^{2}}{2}\right]_{0}^{60} \times 10^{-3}$

$=[120+5400] \times 10^{-3}=5.52$ $\mathrm{coulamb}.$

$(ii)$ $\mathrm{t} \uparrow \mathrm{I} \downarrow$

$(iii)$ For $\mathrm{I}=0 ; \mathrm{a}-2 \mathrm{bt}=0 ; \mathrm{t}=\mathrm{a} / 2 \mathrm{b}$

$(iv)$ $\frac{\mathrm{dI}}{\mathrm{dt}}=-2 \mathrm{b}$

Hence option $(4)$ is incorrect.

Here $\mathrm{I}_{\mathrm{p}}=1 \mathrm{\,A}, \mathrm{C}=2 \times 10^{-6} \mathrm{\,F}$

$4 i= neA V_D$

$i=n e(4 A) V_D^{\prime}$

$4=\frac{V_D}{4 V_D^a}$

$\frac{V_D}{V_D^{\prime}}=16: 1$

$\Delta q=$ Area $( l / t )$

$\Delta q=\frac{1}{2} \cdot 10 \times 15=75 \,C$

$i_{\text {avg }}=\frac{75}{15}=5 \,A$

$V =\int_{ a }^{ b } E \cdot d \ell=\frac{\lambda}{2 \pi \varepsilon_0 r } \ln \left(\frac{ b }{ a }\right)$

Now, I $=\sigma \int \overrightarrow{ E } \cdot \overrightarrow{ dA }=\sigma \int \frac{\lambda}{2 \pi \varepsilon_0 r } \cdot 2 \pi dr =\frac{\sigma \lambda}{\varepsilon_0}$

From $(1)$: $I =\frac{2 \sigma \pi \varepsilon_0}{\varepsilon_0 \ln ( b / a )}=\frac{2 \pi \sigma}{\ln ( b / a )} v$

Area of cross-section $A = 1 \,cm \times 1 \,cm= 1\, cm^2 = 10^{-4} \,m^2$

Resistance $  R = 3 \times 10^{-7}  \times \frac{{{1}}}{{{{10}^{ - 4}}}}= 3 \times 10^{-3}\, ohm$

$l = 100 \,cm = 1\,m$,  $A = 1\,cm^2 = 10^{-4} \,m^2$, so resistance $R = 3 \times 10^{-7}  \times \frac{1}{{{{10}^{ - 4}}}}= 3 \times 10^{-3} \,\Omega $

Hence new resistance $R' = 10 + 10 $ of $20\%$

$ = 10 + \frac{{20}}{{100}} \times 10 = 12\,\Omega .$

$ \Rightarrow $  ${R_{100}} = {R_{50}}[1 + \alpha (100 - 50)]$ 

$ \Rightarrow $ $7 = 5\,[1 + (\alpha \times 50)]$ $ \Rightarrow $ $ \propto \, = \,\frac{{(7 - 5)}}{{250}} = 0.008\,{^o}C$

$ \Rightarrow t = 167\,^o C$

Since wires have same material so $r$ and $d$ is same for both.

Also they have same mass $ \Rightarrow $ $Al$ $=$ constant $ \Rightarrow $ $l \propto \frac{1}{A}$

$ \Rightarrow $ $\frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}} \times \frac{{{A_2}}}{{{A_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^2} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4}$

$ \Rightarrow $ $\frac{{34}}{{{R_2}}} = {\left( {\frac{r}{{2r}}} \right)^4}$ $ \Rightarrow $ ${R_2} = 544\,\Omega $

Again by ${R_t} = {R_0}(1 + \alpha t)$

$ \Rightarrow \,\,5 = {R_0}\left( {1 + \frac{1}{{200}} \times 50} \right) \Rightarrow {R_0} = 4\,\Omega .$

$3 \times {R_0} = {R_0}(1 + 4 \times {10^{ - 3}}t)$$ \Rightarrow \,\,t = {500\,^o}C$.

$ \Rightarrow \,\,\,\,\,{R_1}:{R_2}:{R_3} = \frac{9}{1}:\frac{4}{2}:\frac{1}{3} = 27:6:1$.

$ \Rightarrow $${R_2} = 4{R_1}$. Change in resistance $ = {R_2} - {R_1} = 3{R_1}$

Now, $\frac{{{\rm{Change}}\,\,{\rm{in}}\,\,{\rm{resistance}}}}{{{\rm{Original}}\,\,{\rm{resistance}}}} = \frac{{3{R_1}}}{{{R_1}}} = \frac{3}{1}$

$ \Rightarrow $  $\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{100}}{{110}}} \right)^2}$ $ \Rightarrow $ ${R_2} = 1.21\,{R_1}$

$\%$ change $\frac{{{R_2} - {R_1}}}{{{R_1}}} \times 100 = 21\% $

In series ${R_{eq}} = {R_1} + {R_2}$

$\frac{{{\rho _{eq.}}({l_1} + {l_2})}}{A} = \frac{{{\rho _1}{l_1}}}{A} + \frac{{{\rho _2}{l_2}}}{A}$ $ \Rightarrow \,\,{\rho _{eq}} = \frac{{{\rho _1}{l_1} + {\rho _2}{l_2}}}{{{l_1} + {l_2}}}$.

$V = iR$

$ \Rightarrow $ $5 = \left( {\frac{{110}}{{20 \times {{10}^3} + R}}} \right) \times 20 \times {10^3}$

$ \Rightarrow $ ${10^5} + 5R = 22 \times {10^5}$

$ \Rightarrow $ $R = 21 \times \frac{{{{10}^5}}}{5} = 420\,K\Omega $

When $R_A$ and $R_B $ are connected in parallel then equivalent resistance

${R_{eq}} = \frac{{{R_A}{R_B}}}{{({R_A} + {R_B})}} = \frac{{16}}{{17}}{R_A}$

If ${R_A} = 4.25\,\Omega $ then ${R_{eq}} = 4\,\Omega $ i.e. option $(a)$ is correct.

$\frac{{10}}{{{R_2}}} = \frac{{(1 + 5 \times {{10}^{ - 3}} \times 20)}}{{(1 + 5 \times {{10}^{ - 3}} \times 120)}}$

${R_2} \approx 15\,\Omega $

Also $\frac{{{i_1}}}{{{i_2}}} = \frac{{{R_2}}}{{{R_1}}}$

$\frac{{30}}{{{i_2}}} = \frac{{15}}{{10}}$

${i_2} = 20\,mA$