JEE physics MCQ

$\Rightarrow \frac{1}{4}=2 \pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}}$

$\Rightarrow \frac{1}{16}=4 \pi^2 \times \frac{9.8 \times 10^{-6}}{M \times 49 \times 10^{-3}}$

$\Rightarrow M=\frac{4 \pi^2 \times 9.8 \times 10^{-6}}{49 \times 10^{-3}} \times 16$

$\frac{4 \pi^2 \times 9.8 \times 16 \times 10^{-3}}{49}$

$\quad=12.8 \pi^2 \times 10^{-3} \times 10^{-2} \times 10^2$

$=1280 \pi^2 \times 10^{-5} \mathrm{Am}^2$

$\mathrm{A}_{1}=\frac{1}{2} \times \mathrm{a} \times \frac{\sqrt{3}}{2} \mathrm{a}$

$\mathrm{A}_{1}=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$

$\mu_{1}=\mathrm{N}_{1} \mathrm{IA}_{1}$

$\mu_{1}=\frac{4 \mathrm{I} \sqrt{3}}{4} \mathrm{a}^{2}$

$\mu_{1}=\sqrt{3} \mathrm{I} a^{2}$

$A_{2}=a^{2}$

$\mu_{2}=N_{2} I A_{2}$

$=3 \times I \times a^{2}$

$\mu_{2}=3 \mathrm{Ia}^{2}$

Radius of loop, $R=\frac{L}{2 \pi}$

Magnetic moment, $M =I A$

$=I\left(\frac{\pi L^{2}}{4 \pi^{2}}\right)$

$=\frac{I L^{2}}{4 \pi} A \;m^{2}$

$\mathrm{I}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{c}}$

Voltage sensitivity

$\mathrm{v}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{CR}_{\mathrm{G}}}$

So, resistance of galvanometer

$R_{G}=\frac{l_{s}}{V_{s}}=\frac{5 \times 1}{20 \times 10^{-3}}=\frac{5000}{20}=250\, \Omega$

Here $\overrightarrow{ A } \| \overrightarrow{ B }$

$\theta=0$

$\vec{\tau}= NIAB \sin \theta$

$\vec{\tau}=0$

$\mathrm{i}_{\mathrm{g}}=20 \mu \mathrm{A}$

$\mathrm{i}=\mathrm{i}_{\mathrm{g}}\left(\frac{\mathrm{G}}{\mathrm{S}}+1\right)$

$\Rightarrow 20 \times 10^{-3}=20 \times 10^{-6}\left(\frac{200}{\mathrm{~S}}+1\right)$

$\Rightarrow \frac{200}{\mathrm{~S}}=999$

$\Rightarrow \mathrm{S} \approx 0.2 \Omega$

$=140.4+\frac{240 \times 10}{240+10}$

$\text { Req }=140.4+\frac{2400}{250}$

$\text { Req. }=150 \Omega$

$\therefore$ Current in ammeter$   =\frac{24}{150}$

$=160 \mathrm{~mA}$

For ammeter

Let shunt resistance $=\mathrm{S}$

$i_g R=\left(i-i_g\right) S$

$20 \times 10^{-3} \times 100=10 S$

$S=20 \times 10^{-2} \Omega$

$\frac{2}{G+R}=K \theta$

$\Rightarrow \frac{1}{\theta}=\frac{(G+R) K}{2}=R\left(\frac{K}{2}\right)+\frac{K G}{2}$

$\text { Slope }=\frac{K}{2}=\frac{1}{4} \Rightarrow K=0.5=5 \times 10^{-1} A$

$\mathrm{C} \theta=\mathrm{BINA} \sin 90^{\circ}$

$\mathrm{C}=\frac{\mathrm{BINA}}{\theta}=\frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2 \times 10^{-4}}{0.05}$

$=4 \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad} .$

$\mathrm{x}=4$

$\frac{95}{100} \mathrm{IS}=\frac{5 \mathrm{I}}{100} \mathrm{G}$

$\mathrm{S}=\frac{\mathrm{G}}{19}$

$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}=\frac{\frac{\mathrm{G}^2}{19}}{\frac{20 \mathrm{G}}{19}}$

$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{G}}{20}$

$=20000-50$

$=19950 \Omega$

Now $5 \mathrm{x} \times \mathrm{G}=20 \mathrm{x} \times 24$

$ \mathrm{G}=4 \times 24 $

$ \mathrm{G}=96 \ \Omega$

$ \mathrm{I}_{\mathrm{g}}=3 \mathrm{~mA} $

$ \mathrm{I}=8 \mathrm{~A}$

In case of conversion of galvanometer into ammeter.

We have $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$

$ S=\frac{I_g G}{I-I_g} $

$ S=\frac{\left(3 \times 10^{-3}\right) 10}{8-0.003}=3.75 \times 10^{-5} \Omega$

$ \therefore \frac{i_2}{i_1}=\frac{\theta_2}{\theta_1} \Rightarrow \frac{i_2}{200 \mu \mathrm{A}}=\frac{\pi / 10}{\pi / 3}=\frac{3}{10}$

$ \Rightarrow \mathrm{i}_2=60\  \mu \mathrm{A}$

$\mathrm{W}=(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_{\mathrm{f}}-(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_i$

$=0+(\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_i$

$=\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J}$

$=1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{J}$

$\mathrm{W}_{\text {ext }} =\Delta \mathrm{U}+\Delta \mathrm{KE} \quad(\mathrm{P} . \mathrm{E} .=-\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}})$

$=-\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}_{\mathrm{f}}+\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}_{\mathrm{i}}+0$

$=-\mathrm{MB} \cos 90+\mathrm{MB} \cos 0$

$=\mathrm{MB}$

$=\mathrm{NIAB}$

$=200 \times 100 \times 10^{-6} \times \frac{5}{2} \times 10^{-4} \times 1=5 \mu$

$\mathrm{U}=-\mathrm{mB} \cos 0^{\circ}=-\mathrm{mB}$

At unstable equilibrium

$\mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB}$

$\mathrm{W}=\Delta \mathrm{U}$

$\text { W.D. }=2 \mathrm{mB}$

$=2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J}$

$=5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}})$

$=0.1(-\hat{\mathrm{i}})$

$\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}})$

$=2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m}$

$R =\frac{ V }{ I _{ g }}- G$

$=\frac{50}{10^{-3}}-54 \approx 50\,k \Omega$

For ammeter

$S=\frac{I_g G}{I-I_g}=\frac{10^{-3} \times 54}{(10-1) \times 10^{-3}}=6 \Omega( C )$

$i=\frac{25}{1050+R_G} \dots(ii)$

From $(i)$ and $(ii)$

$\frac{25}{1050+R_G}=\frac{5}{4\left(5+R_G\right)}$

$100\left(5+R_G\right)=1050 \times 5+R_G \times 5$

$95 R_G=4750$

$R_G=50\,\Omega$

$\Rightarrow V_{ s }=\frac{I_s}{G}$, If $G$ (galvanometer resistance) is constant, then $V _{ S } \propto I_S$

so percentage change in $V_S$ is also $25 \%$.

$\text { BiNA }= k \theta$

$\theta=\left(\frac{ BNA }{ k }\right) i ; \text { Current sensitive }=\frac{ BNA }{ k }$

So, if $N$ is doubled then current sensitivity is doubled.

Voltage sensitivity

$B \frac{ V }{ R } NA = k \theta$

$V =\frac{ BNA }{ Rk } \theta, \text { as } N \text { is doubled } R \text { is also doubled. }$

So, no change in voltage sensitivity.

$NiAB = K \theta$

$A =\frac{ K \theta}{ NiB }=\frac{4 \times 10^{-5} \times 0.05}{200 \times 10 \times 10^{-3} \times 0.01}$

On solving $A =10^{-4}\,m ^2=1\,cm ^2$

$=180\,\Omega$

$i =\frac{90}{180}=\frac{1}{2}\,A$

$\text { Reading }=\frac{1}{2} \times \frac{400}{500} \times 100$

$=40\,volt$

$\therefore \quad ILB = mg$

$\therefore \quad\left(\frac{ V }{ R }\right) LB = mg$

$\therefore \quad V =\frac{ mgR }{ LB }$

$=\frac{\left(1 \times 10^{-3}\,kg \right)\left(10\,m / s ^2\right)(10 \Omega)}{(0.1\,m )\left(10^3 \times 10^{-4}\,T \right)}=10\,V$

$=\left[\frac{ q ^{2} B ^{2} r _{0}^{2}}{2 m }\right]$

$=18\,MeV$

$\therefore \frac{ d \theta}{ di }=\frac{ NAB }{ K }$

$8=\frac{72}{\frac{I}{I_{g}}-1}$

$\frac{I}{I_{g}}-1=9$

$\frac{I}{I_{g}}=10 \Rightarrow \frac{I_{g}}{I}=\frac{1}{10} \quad \% I=\frac{I_{g}}{I} \times 100=10 \%$

$\frac{I_{g}}{I}=\frac{S}{S+G}$

$\Rightarrow \frac{1}{3}=\frac{S}{S+G} \Rightarrow S+G=3 S \Rightarrow G=2 S$

$I _{ g }= Kn$

$I =\frac{ I _{ g }}{ s }( G + S )$

$I =\frac{ nK }{ S }( G + S )$

$\text { Magnetic Moment }= IA$

$=14 \times \pi R ^{2}$

$= 14 \times(3.14) \times \frac{1}{4}$3

$=10.99 \approx 11.00$

$\overrightarrow{ M }=-7 \times \frac{22}{7}\left(\frac{25}{100}-\frac{9}{100}\right) \hat{ k }$

$=-22\left(\frac{16}{100}\right) \hat{ k }$

$\overrightarrow{ M }=-3.52 \hat{ k }\,Am ^{2}$

$=-\frac{7}{2} \hat{ k }\,Am ^{2}$

$\frac{\vec{\mu}}{\vec{L}}=\frac{q}{2 m}$

For $e^{-}$

$\vec{\mu}=-\frac{e}{2 m} \vec{L}$

Number of revolution $={n}$

${n}\left[2 \times {q}_{{P}} \times {V}\right]=\frac{1}{2} {m}_{{p}} \times {v}_{{P}}^{2}$

${n}\left[2 \times 1.6 \times 10^{-19} \times 12 \times 10^{3}\right.$

$=\frac{1}{2} \times 1.67 \times 10^{-27} \times\left[\frac{3 \times 10^{8}}{6}\right]^{2}$

${n}\left(38.4 \times 10^{-16}\right)=0.2087 \times 10^{-11}$

${n}=543.4$

${Rg}=245 \,\Omega$

$R=\frac{V}{{I}}=\frac{50\, {mV}}{25\, {mA}}=2\, \Omega$

In square ${N}_{{s}}=\frac{24 {a}}{4 {a}}=6$

$\frac{{M}_{{t}}}{{M}_{3}}=\frac{{N}_{{t}} {IA}_{{t}}}{{N}_{{s}} {I}_{{s}}}[{I}$ will be same in both $]$

$=\frac{8 \times \frac{\sqrt{3}}{4} \times a^{2}}{6 \times a^{2}}$

$\frac{{M}_{{t}}}{{M}_{{s}}}=\frac{1}{\sqrt{3}}$

${y}=3$

$=M B=\frac{i \sqrt{3} \ell^{2}}{4} B$

$=\sqrt{3} \times 10^{-5} {N}-{m}$

$\therefore$ Work by magnetic force on the point charge is zero.

Option $(1)$ is correct

Ammeter :- In series connection, the same current flows through all the components. It aims at measuring the curent flowing through the circuit and hence, it is connected in series.

Voltmeter :- A voltmeter measures voltage change between two points in a circuit, So we have to place the voltmeter in parallal with the circuit component.

$\Rightarrow 2= i _{ g }\left( R _{1}+ R _{2}+ G \right)$$...(2)$

$(1)/(2)$

$\Rightarrow \frac{1}{2}=\frac{G+R_{1}}{G+R_{1}+R_{2}}$

$G + R _{1}+ R _{2}=2 G +2 h _{1}$

$\left( R _{2}= G + R _{1}\right)$

$= C =\frac{6 \times 10^{-3}}{2}=3 \times 10^{-3} Am ^{2}$

$\mathrm{V}=\mathrm{i}_{\mathrm{g}}\left(\mathrm{R}+\mathrm{R}_{\mathrm{g}}\right)$

$10=1 \times 10^{-3}(\mathrm{R}+100)$

$\mathrm{R}=9.9 \mathrm{k} \Omega$

$F =\frac{\mu_{0} I ^{2} a }{\pi \sqrt{ b ^{2}+ a ^{2}}}$

$\tau= F \cos \theta \times 2 a$

$=\frac{\mu_{0} I^{2} a}{\pi \sqrt{b^{2}+a^{2}}} \times \frac{b}{\sqrt{b^{2}+a^{2}}} \times 2 a$

$\tau=\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}$

If $b \gg a$ then $\tau=\frac{2 \mu_{0} I ^{2} a ^{2}}{\pi b }$

But among the given options $(1)$ is most appropriate

$M=i A$

$M=\left(\frac{q}{T}\right) \times \pi r^{2}=\frac{q \pi r^{2}}{\left(\frac{2 \pi r}{v}\right)}=\frac{q v r}{2}$

$M=\frac{q v}{2} \times \frac{v m}{q B}$

$M =\frac{ mv ^{2}}{2 B }$

As we can see from the figure, direction of magnetic moment $(M)$ is opposite to magnetic field.

$\overrightarrow{ M }=-\frac{ mv ^{2}}{2 B } \hat{ B }$

$=-\frac{m v^{2}}{2 B^{2}} \vec{B}$

$=4 a ^{2} I \times \frac{\mu_{0} I }{2 \pi b }$

$M =20 Am ^{2}$

$U _{ i }+ K _{ i }= U _{ f }+ K _{ f }$

$0+0=- MB \cos 30^{\circ}+\frac{1}{2} I \omega^{2}$

$20 \times 4 \times \frac{\sqrt{3}}{2}=\frac{1}{2}(0.8) \omega^{2}$

$\omega=\sqrt{100 \sqrt{3}}=10(3)^{1 / 4}$