Physics M.C.Q (1 Marks)

$R^{\prime}  =\frac{\rho l}{10 A}=\frac{R}{10}$

$R_S =5 \times \frac{R}{10} \quad \text { [series] }$

$R_S =50$

$R_p =\frac{K}{50} \quad \text { [parallel] }$

$R_{e q} =R_S+R_p$

$=52 \Omega$

$\text { Terminal voltage } =E-i R$

$=10-2 \times 1=8 \mathrm{~V}$

$\frac{P_A}{P_B}=\frac{R_B}{R_A}$

$R_A=2 R_B$

For Series Combination

$P_S=\frac{V^2}{3 R_B}$

For Parallel Combination

$P_P=\frac{3 V^2}{2 R_B}$

$\frac{P_S}{P_P}=\frac{2}{9}$

$6.8=2[1+\alpha(80-0)]$

$\alpha=\frac{2.4}{80}=0.03 /{ }^{\circ}\,C =3 \times 10^{-2} /{ }^{\circ}\,C$

$\frac{10-2}{400}=\frac{2}{R} \Rightarrow R=\frac{2 \times 400}{8}=100\,\Omega$

$=0.5\,A$

from $A$ to $B$ through $E$.

$I_P=\frac{E}{R / 10}=\frac{10 E}{R}$

$n=\frac{I_P}{I_S}=100 \Rightarrow n=100$

Acc. to color code Third Band

$\rightarrow$ Orange (color code for digit $3$ is orange)

Cross sectional area $A =\pi r ^{2}=10^{-4} m ^{2}$ $j =\frac{ i }{ A }=\left(\frac{ V }{ R }\right) \cdot \frac{1}{ A }=\frac{ E \ell}{ RA } \quad R =\frac{\rho \ell}{ A }$ $j =\frac{10 \times 10}{10 \times 10^{-4}}=10^{5} A / m ^{2}$

$J=\sigma E \Rightarrow \frac{ E }{\rho}=\frac{ E \ell}{ RA }=\frac{10 \times 10 \times \pi}{10 \times 10^{-4} \times \pi}$

$\Rightarrow 10^{5} A / m ^{2}$

So, $V_{A C}=\frac{R_{A C}}{R_{A C}+R_{C B}} V_0$

$=\frac{\frac{ RR _0}{4 R + R _0}}{\frac{ RR _0}{4 R + R _0}+\frac{3 R _0}{4}} \times V _0=\frac{4 RV _0}{16 R +3 R _0}$

$P =\frac{ V ^{2}}{ R } \Rightarrow P \propto \frac{1}{ R }$

$\frac{ P _{1}}{ P _{2}}=\frac{ R _{2}}{ R _{1}}=\frac{200}{100}=\frac{2}{1}$

$=2: 1$

$\qquad =4-0.25$

TPD $=3.75 \text { volt }$

$\Rightarrow x=1+\frac{x \times 1}{x+1}+1$

$\Rightarrow(x-2)=\frac{x}{x+1}$

$\Rightarrow x^2-x-2=x$

$\Rightarrow x^2-2 x-2=0$

$\text { So, } x=\frac{2 \pm \sqrt{12}}{2}$

$=1 \pm \sqrt{3}\,\Omega$

neglecting negative value, $x =1+\sqrt{3}\,\Omega$

$V_{d}=\frac{e E}{m} I ; \rho=\frac{m}{n e^{2} \tau}$ or $\rho=\frac{E}{J}, J=n e v_{d}$

$A \rightarrow R \quad B \rightarrow S \quad D \rightarrow Q$

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

$C \rightarrow P$

$E_{\text {unknown }}=\phi I_{b} \Rightarrow E_{\text {unknown }} \propto I_{b}$

$\frac{E_{1}}{E_{2}}=\frac{I_{1}}{I_{2}} \Rightarrow \frac{1.5}{2.5}=\frac{36}{x}$

$x=\frac{36 \times 5}{3}=60\, \mathrm{~cm}$

$\mathrm{R}=1$

$\mathrm{R}_{\mathrm{s}}=4 \mathrm{R}=4\, \Omega$

$\frac{I_{3}}{I_{1}}=\frac{I_{1} r_{2}}{\left(r_{2}+r_{3}\right) I_{1}}=\frac{r_{2}}{r_{2}+r_{3}}$

$R =470 \Omega, 5 \%$

$V_{A}-1 \times 4-1 \times 1+4-1 \times 1+2=V_{A}$

$-61+6=0$

$1=1 A$

$v _{ a }- i _{2} R _{2}- E _{2}+ E _{3}+ i _{3} R _{1}= v _{ B }$

$i _{2} R _{2}+ E _{2}- E _{3}- i _{3} R _{1}=0$

$\frac{ R }{10}=\frac{3}{2}$

$R=15 \Omega$

Length of $15 \Omega$ resistance wire is $1.5 m$

length of $1 \Omega$ resistance wire $=\frac{1.5}{15}=0.1$

$=1.0 \times 10^{-1} m$

$=2.5 \times 10^{6}$

$R=\frac{\rho l}{A}$

$\therefore \quad A=\frac{\rho l}{R}$

$\Rightarrow \frac{A_{1}}{A_{2}}=\frac{\rho_{1}}{\rho_{2}} \times \frac{L_{1}}{I_{2}}\left(\frac{R_{2}}{R_{1}}\right)$

$\Rightarrow \frac{A_{1}}{A_{2}}=1 \quad\left[\because R_{1}=R_{2}, I_{1}=l_{2}\right.$ and for same

material $\left.\rho_{1}=\rho_{2}\right]$

$\mathrm{I}_{1}=\frac{10}{10}=1 \;\mathrm{A}$

$\mathrm{I}_{2}=\frac{10}{10}=1\; \mathrm{A}$

$V_1=10\;V$

$V_{2}=10\;V$

$\mathrm{R}_{\mathrm{eq}_{2}}=\mathrm{R} / 2+\mathrm{R}=\frac{3 \mathrm{R}}{2}$

$\mathrm{P}_{\mathrm{eq}_1}=\frac{\mathrm{E}^{2}}{2 \mathrm{R} / 3}=\frac{3 \mathrm{P}}{2}$

$\mathrm{P}_{\mathrm{eq}_1}=\frac{\mathrm{E}^{2}}{3 \mathrm{R} / 2}=\frac{2 \mathrm{P}}{3}$

$P_{eq_{1}}: P_{eq_{2}}=9: 4$

$I=\frac{E}{n R+R}$        .....$(i)$

Current drawn from same battery when $n$ resistors are connected in parallel is

$10 I=\frac{E}{R / n+R}$        ......$(ii)$

On dividing eqn. $(ii)$ by $(i),$ $10=\frac{(n+1) R}{(1 / n+1) R}$

After solving the equation, $n=10$.

cell is

$I=\frac{n \varepsilon}{n r}=\frac{\varepsilon}{r}$

$\mathrm{So}, I$ is independent of $n$ and $I$ is constant.

Yellow - Violet - Orange - Silver

$A$ and resistivity $\rho$ is given as

$R=\frac{\rho l}{A}$

Given, $l'=n l$

As the volume of the wire remains constant

$\therefore A^{\prime} l^{\prime}=A l$

$A^{\prime}=\frac{A l}{l^{\prime}}=\frac{A l}{n l} \text { or } A^{\prime}=\frac{A}{n}$

$ \therefore \quad R^{\prime} =\frac{\rho l^{\prime}}{A^{\prime}} $

$ R^{\prime} =\frac{\rho n l}{\frac{A}{n}}=\frac{n^{2} \rho l}{A}=n^{2} R $

$I=\frac{10}{25}$

$I=0.4$

$I_{(20\;\Omega)}=\frac{30 \times \frac{4}{10}}{20+30}=\frac{12}{50}=\frac{6}{25} A$

$\left(\text { also } \varepsilon_{1}>\varepsilon_{2}\right)$

Potential difference per unit length of the potentiometer wire $=k$ (say)

When $\varepsilon_{1}$ and $\varepsilon_{2}$ are in series and supporteach other then

$\varepsilon_{1}+\varepsilon_{2}=50\, \times k$       .....$(i)$

When $\varepsilon_{1}$ and $\varepsilon_{2}$ are in opposite direction

$\varepsilon_{1}-\varepsilon_{2}=10 \times k$     ....$(ii)$

On adding eqn. $(i)$ and eqn. $(ii)$

$2 \varepsilon_{1}=60\, k \Rightarrow \varepsilon_{1}=30 \,k$ and $\varepsilon_{2}=50\, k-30 \,k=20\, k$

$\therefore \quad \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{30 \,k}{20 \,k}=\frac{3}{2}$

Power of the bulb in the circuit,

$P =V I $

$I =\frac{P}{V_{B}} $

$=\frac{500}{100}=5\, \mathrm{A}$

$V_{R} =I R \Rightarrow(230-100)=5 \times R $

$\therefore \quad R =26 \,\Omega$

$\therefore \quad I=\frac{d Q}{d t}=a-2 b t$

At $t=0, Q=0 \Rightarrow I=0$

Also, $I=0$ at $t=a / 2 b$

$\therefore \quad$ Total heat produced in resistance $R$

$H = \int\limits_0^{a/2b} {{I^2}Rdt = R\int\limits_0^{a/2b} {{{(a - 2bt)}^2}dt} } $

${ = R\int\limits_0^{a/2b} {\left( {{a^2} + 4{b^2}{t^2} - 4abt} \right)dt} }$

$=R\left[a^{2} t+4 b^{2} \frac{t^{3}}{3}-4 a b \frac{t^{2}}{2}\right]_{0}^{a / 2 b} $

$=R\left[a^{2} \times \frac{a}{2 b}+\frac{4 b^{2}}{3} \times \frac{a^{3}}{8 b^{3}}-\frac{4 a b}{2} \times \frac{a^{2}}{4 b^{2}}\right]$

$=\frac{a^{3} R}{b}\left[\frac{1}{2}+\frac{1}{6}-\frac{1}{2}\right]=\frac{a^{3} R}{6}$

$ = \frac{{{n_{( + )}}{q_{( + )}}}}{t} + \frac{{{n_{( - )}}{q_{( - )}}}}{t}$

$ = \frac{{{n_{( + )}}}}{t} \times 2e + \frac{{{n_{( - )}}}}{t} \times e$

$= 3.2 \times {10^{18}} \times 2 \times 1.6 \times 10^{-19} + 3.6 \times {10^{18}} \times 1.6 \times 10^{-19}$

$= 1.6\, A$ (towards right)

$ \Rightarrow $  $i = \frac{{({n_ + })}}{t} \times e + \frac{{({n_ - })}}{t} \times e$

$ = 2.9 \times {10^{18}} \times 1.6 \times {10^{ - 19}} + 1.2 \times {10^{18}} \times 1.6 \times {10^{ - 19}}$

$ \Rightarrow $  $i = \,0.66\,A$

$ \Rightarrow \,\frac{{{v_P}}}{{{v_Q}}} = {\left( {\frac{{{d_Q}}}{{{d_P}}}} \right)^2} = {\left( {\frac{{d/2}}{d}} \right)^2} = \frac{1}{4} \Rightarrow {v_P} = \frac{1}{4}{v_Q}$.

$ = \left[ {\frac{{1.2{t^2}}}{2} + 3t} \right]_0^5 = 30\,C$

$\Rightarrow $ $Q = \int_{t = 2}^{t = 3} {\,Idt} = $ $\left[ {2\int\limits_2^3 {tdt} + 3\int\limits_2^3 {{t^2}dt} } \right]$

$= \left[ {{t^2}} \right]_2^3 + \left[ {{t^3}} \right]_2^3= (9 -4) + (27 -8) = 5 + 19 = 24\,C$.

$A$ be the area of cross section,

$e$ is the charge on electron and

$I$ is the current flowing through both the wires

Hence, $I=n_{1} e A V_{d_{1}}=n_{1} e A V_{d_{2}}$

$\therefore n_{1} V_{d_{1}}=n_{2} V_{d_{2}}$

$\therefore \frac{V_{d 1}}{V_{d_{2}}}=\frac{n_{2}}{n_{1}}$

$\therefore \frac{V_{d 1}}{V_{d_{2}}}=\frac{4}{1}$