JEE physics MCQ

so, it is minimum for deutron.

$\mu_0 \varepsilon_0=\frac{1}{ c ^2}$

$\frac{E}{B}=c$

$\frac{ E ^2 \mu_0 \varepsilon_0}{ B ^2}=1$

Dimension of $1=\left[ M ^0 L ^0 T ^0\right]$

Option $B-$It does not oxidise easily.

Option $C-$It has low torsional constant.

Option $D-$It is non-magnetic.

Hase,Option $A$ Is Correct Answer.

$S_{i_1}=\frac{N i A B}{k i}=\frac{N A B}{k}$

$S_{i_2}=\frac{1.2 N A B}{k}$

$S_{v_1}=\frac{N A B}{k R}=\frac{S_{i_1}}{R}$

$S_{v_2}=\frac{S_{i_2}}{2 R}=\frac{1.2(N A B)}{k(2 R)}$

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{8}}-\mathrm{G}$

$\mathrm{I}_{\mathrm{R}}$ is the current in the galvenometer for full scale deflection.

For the choice $( 1)$

${\rm{R}} = \frac{{50}}{{50 \times {{10}^{ - 6}}}} - 100 \ne 10{\rm{k}}\Omega $

For the choice $(2)$

$\mathrm{R}=\frac{10}{50 \times 10^{-6}} -100=2 \times 10^{5}-100 $

$ \simeq 2 \times {10^5}\Omega $

$ \simeq 200{\rm{k}}\Omega $

Hence, the choice $( 2)$ is conrect.

Further, the galvanometer may be converted inte an ammeter if a small resistance $'s'$ is connectec (or shunted) in parallel with the galvenometer.

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

$I$ is the range of the annmeter.

For the choice $( 3)$

$S = \frac{{50 \times {{10}^{ - 6}}}}{{5 \times {{10}^{ - 3}} - 50 \times {{10}^{ - 6}}}} \times 100$

$ \simeq \frac{{50 \times {{10}^{ - 6}}}}{{5 \times {{10}^{ - 3}}}} \times 100$

$ \simeq 1\Omega $

Hence. the choices $( 3)$ and $( 4)$ are not contect.

$= \frac{2 \mathrm{NAB}_{\mathrm{s}}}{\mathrm{R}} $

$=\frac{2 \times 100 \times \pi \times\left(1 \times 10^{-2}\right)^{2} \times \mu_{0} \mathrm{nI}}{20}$

$=2 \times 10^{-4} \mathrm{\,C}$

${\frac{I}{5} \times G=\frac{4}{5} I \times 4} $

${G=16\, \Omega}$

When additional shunt of $4\, \Omega$ used

${\rm{I'}} \times 16 = 2({\rm{I}} - {\rm{I'}})$

$16{\rm{I}} = 2{\rm{I}} - 2{\rm{I'}}$

$18{\rm{I'}} = 2{\rm{I}}$

${\rm{I'}} = \frac{{\rm{I}}}{9}$

$\therefore S=\frac{I_{g} \cdot G}{\left(I-I_{g}\right)}$

Given $\mathrm{I}=5 \mathrm{\,A}, \mathrm{I}_{\mathrm{g}}=1 \mathrm{\,Amp} ., \mathrm{G}=60 \mathrm{\,M}$

$\therefore \quad S=\frac{60 \times 1}{4}=15\, \Omega$ in parallel

$\mathrm{R}_{\mathrm{g}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}_{\mathrm{g}}+\mathrm{R}_{\mathrm{H}}}$

$\Rightarrow \mathrm{R}_{\mathrm{H}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{R}_{\mathrm{g}}$

$\Rightarrow \mathrm{R}_{\mathrm{H}}=5988\, \Omega$

$\mathrm{V}=\frac{\mathrm{NAB}}{\mathrm{kR}}$

$\therefore \frac{V_{s_{2}}}{V_{s_{1}}}=\frac{n_{2} B_{2} A_{2} \cdot k_{1} R_{1}}{k_{2} R_{2} n_{1} B_{1} A_{1}}$

$=\frac{42 \times 0.50 \times 1.8 \times 10^{-3} \times \mathrm{k} \times 10}{\mathrm{k} \times 14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1$

Maximum current,

$\mathrm{I}=\frac{\mathrm{R}_{\mathrm{G}}+\mathrm{R}_{\mathrm{S}}}{\mathrm{R}_{\mathrm{G}}} \times \mathrm{I}_{\mathrm{G}}=\frac{(40+2) \times 10}{2}=210 \mathrm{\,mA}$

$\mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$

$\mathrm{I}_{\mathrm{g}}=\left(\frac{\mathrm{I}}{\theta}\right) \times \mathrm{N}=\left(\frac{2 \times 10^{-6}}{2}\right) \times 30$

$\Rightarrow \mathrm{I}_{\mathrm{g}}=30 \times 10^{-6} \mathrm{\,A}$

$\therefore \mathrm{S}=\frac{30 \times 10^{-6} \times 50}{5-30 \times 10^{-6}} \cong 3 \times 10^{-4}\, \Omega$

$\therefore {{\rm{I}}_a}:{{\rm{I}}_g} = \frac{9}{1}$

$\frac{9}{10}$ th current goes through shunt

$25 \times {10^{ - 3}} = 25{{\rm{R}}_{\rm{s}}}$

$ \Rightarrow {{\rm{R}}_{\rm{s}}} = 0.001$

$ \Rightarrow {\rm{N}}i{\rm{AB}} = {\rm{C}}\frac{\pi }{2} \Rightarrow {\rm{C}} = \frac{{2{\rm{N}}i{\rm{AB}}}}{\pi }$         ........$(1)$

When $Q$ is passed,

$\int \tau  dt = \int {Ni} ABdt$

$\mathrm{I} \omega=\mathrm{N} \mathrm{A} \mathrm{B} \mathrm{Q}$       ........$(2)$

Also, maximum deflection happens when entire $\frac{1}{2} \mathrm{I} \omega^{2}$ converts to $\frac{1}{2} \mathrm{C} \theta_{\mathrm{max}}^{2}$

So ${\theta _{\max }} = \sqrt {\frac{{{\rm{I}}{\omega ^2}}}{{\rm{C}}}}  = \frac{1}{{\sqrt {{\rm{IC}}} }}{\rm{I}}\omega  = \frac{{{\rm{NABQ}}}}{{\sqrt {{\rm{IC}}} }}$        (from $(1)$)

$\frac{20 \times 50}{1000}=i_{S} \times 0.01$

$\mathrm{i}_{\mathrm{S}}=100 \mathrm{\,A}$

$C=\frac{2 N i A B}{\pi}$

$\mathrm{I} \omega=\mathrm{NABQ} \Rightarrow \frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{C} \quad \theta_{\max }^{2}$

$\phi=\frac{\pi}{6}=30^{\circ}$

$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}} \Rightarrow \frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{E}}{\mathrm{R}}=\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right) \frac{1}{\mathrm{R}}$

$\Rightarrow \mathrm{q}=\frac{\Delta \phi}{\mathrm{R}}=\frac{\mathrm{NBA}}{\mathrm{R}}$

Total charge $q=\frac{500 \times 0.25 \times 0.04}{25}=0.2 \mathrm{\,C}$

$I = i _{ g +} I _{ S }$

$I =2$,

$V$ across $A B=0.50 v$.

$Ig =0.5 / 25=0.02\,V$

$I_{ S }=2-0.02=1.98\,v$

$Rs =0.5 / 1.98=0.252\,V$

$\mathrm{i}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{i}-\mathrm{i}_{\mathrm{g}}\right) \mathrm{R}_{\mathrm{S}}$

$50 \times 10^{-3} \mathrm{r}^{3}=(1-0.05) \times \mathrm{R}$

$ S=\left(\frac{50 \times 10^{-6}}{95 \times 10^{-50} \times 10^{-60}}\right)(100) \approx 1 \Omega$

To change it in voltmeter a high resistance $R$ is put in series, where $R$ is given by $R=\frac{V}{i_{g}}-G$

$R=\frac{10}{50 \times 10^{-6}}-100 \approx 200 k \Omega$

$v=I_{g} R_{x}=I \times 10^{-3} \times R_{x}$

As $R_{x}=$ very large voltmeter, $v_{0}$ range will be more.

Maximum voltage may be of range $10 \mathrm{v}$.

Given least count of voltmeter $=0.1 V$ i.e. $0.1$ volts$/$division

hence, $V=\frac{\text { Total no. of divisions }}{\text { least count }}$

$1000=\frac{\text { Total no. of divisions }}{0.1}$

Total no. of divisions $=100$

Hence, $I_{g}=\frac{0.2}{2}=0.01 A$

To convert galvanometer into a $10 \mathrm{A}$ ammeter, the galvanometer coil should have to connect a shunt resistance in parallel, given by

$S=\frac{R_{g} I_{g}}{I-I_{g}}$

$S=\frac{(20)(0.01)}{10-0.01}$

$S=\frac{0.2}{9.99}=0.02 A$

$G \times 1 / 5=4 \times(4 / 5)=16 \Omega$

$\left(i_{g}\right)_{2}=\frac{4 / 3}{16+(4 / 3)} i=\frac{1}{13} i$

$20+R_{g}=40 \Rightarrow R_{g}=20 \Omega$

$==>$ ${N_2} = 60$.

$\therefore $ ${i_g}G = (0.03 - {i_g})4r$ ..... $(i)$

and ${i_g}G = (0.06 - {i_g})r$  ..... $(ii)$

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

$0.12 - 4{i_g} = 0.06 - {i_g}$ $ \Rightarrow {i_g} = 0.02A$.