JEE physics MCQ

Current through the galvanometer,

$I_{G}=0.2 \% \text { of } I=\frac{0.2}{100} I=\frac{1}{500} I$

$\therefore$ Current through the shunt,

$I_{S}=I-I_{G}=I-\frac{1}{500} I=\frac{499}{500} I$

As shunt and galvanometer are in parallel

$\therefore \quad I_{G} G=I_{S} S$

$\left(\frac{1}{500} I\right) G=\left(\frac{499}{500}\right) S \text { or } S=\frac{G}{499}$

Resistance of the ammeter $R_{A}$ is

${\frac{1}{R_{A}}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{1}{G}=\frac{500}{G}}$

${R_{A}=\frac{1}{500} \,G}$

where, $\vec{M}$ is the magnetic moment of the loop and $\vec{B}$ is the magnetic field.

or $\quad \tau=M B \sin \theta$ where $\theta$ is angle between $M$ and $B$ When $\vec{M}$ and $\vec{B}$ are parallel (i.e. $\theta=0^{\circ}$ ) the equilibrium is stable and when they are antiparallel (i.e. $\theta=\pi$ ) the equilibrium is unstable.

or $\quad B=\frac{2 \pi m v}{e}.........(i)$

As $\frac{m v^{2}}{R}=e v B$

${\text{or }}v = \frac{{eBR}}{m} = \frac{{e2\pi mvR}}{{me}}\,\,\,\,\,(U\sin g\,(i))$

$ = 2\pi vR{\text{ }}.........{\text{(ii)}}$

${\text{Kinetic energy, }}K = \frac{1}{2}m{v^2} = \frac{1}{2}m{(2\pi vR)^2}{\text{ }}$

$=2m\pi ^2v^2R^2$

Neglecting $I_{z}$

$\therefore \quad S=\frac{V_{g}}{I}=\frac{25 \times 10^{-3} \,\mathrm{V}}{25 \,\mathrm{A}}=0.001\, \Omega$

$\therefore \frac{G S}{G+S}+R=G$

$R=G-\frac{G S}{G+S} \Rightarrow R=\frac{G^{2}+G S-G S}{G+S}$

$R=\frac{G^{2}}{G+S}$

Current for full scale deflection, $I_{g}=30 \,\mathrm{mA}$

$=30 \times 10^{-3}\, \mathrm{A}$

Range of voltmeter, $V=30\, \mathrm{V}$

To convert the galvanometer into an voltmeter of a given range, a resistance $R$ is connected in series with it as shown in the figure.

From figure, $V=I_{g}(G+R)$

$ \text { or }  R=\frac{V}{I_{8}}-G =\frac{30}{30 \times 10^{-3}}-100\, \Omega $

$=1000-100=900\, \Omega$

$ M=N I A=2000 \times 2 \times 1.5 \times 10^{-4}=0.6 \,\mathrm{J} / \mathrm{T} $

$\text { Torque } \tau =M B \sin 30^{\circ} $

$=0.6 \times 5 \times 10^{-2} \times \frac{1}{2}=1.5 \times 10^{-2}\, \mathrm{N} \mathrm{m}$

$S=\frac{1.0 \times 60}{4}=15 \,\Omega $ in parallel.

$\varepsilon=3\, \mathrm{V}$

$\therefore$ Current $=\frac{3\, \mathrm{V}}{3000 \,\Omega}=1 \times 10^{-3} \,\mathrm{mA}$

If the deflection has to be reduced to $20$ divisions, current $i=1\, \mathrm{mA} \times \frac{2}{3}$ as the full deflection scale for

$1\,\mathrm{mA}=30$ divisions.

$3\, \mathrm{V}=3000\, \Omega \times 1 \,\mathrm{mA}=x\, \Omega \times \frac{2}{3} \,\mathrm{mA}$

$\Rightarrow x=3000 \times 1 \times \frac{3}{2}=4500\, \Omega$

But the galvanometer resistance $=50\, \Omega$

Therefore the resistance to be added

$=(4500-50) \,\Omega=4450\, \Omega$

Given: $I=750\, \mathrm{A}$

$I_{\mathrm{g}}=100\, \mathrm{A}, R_{G}=13\, \Omega$

From the figure,

$I_{g} R_{G}=\left(I-I_{g}\right) S$

or $100 \times 13=[750-100] S$

or $1300=650\, S$

$\therefore  S=1300 / 650=2\, \Omega$

$\tau=\operatorname{BiN} A \sin \theta$

$\tau=\operatorname{BiN} A \sin 90^{\circ}$

$\tau= Bi \times \frac{\sqrt{3}}{4} I ^{2} \times 1$

$\left[\because\right.$ Area of equilateral triangle $=\frac{\sqrt{3}}{4} I ^{2}$ and $\left. N =1\right]$

$\Rightarrow I ^{2}=\frac{4 \tau}{\sqrt{3} Bi } \Rightarrow I =2\left[\frac{\tau}{ Bi \sqrt{3}}\right]^{1 / 2}$

Note$:$The resistance of an ideal voltmeter should be infinite.

${i_g}(S + G) = iG$

$\frac{{{i_g}}}{i} = \frac{G}{{S + G}} = \frac{9}{{2 + 9}} \approx 0.8$

$A=0.01 \mathrm{~m}^2, \mathrm{I}=10 \mathrm{~A}, \mathrm{~B}=0.1 \mathrm{~T},$

$\theta=0^{\circ}$

To find:

Torque acting on the loop, $\tau$

Solution:

Magnetic moment, $M=(I)(A)$

The torque acting on the loop is given as, $\vec{\tau}=\vec{M} \times \vec{B}$

$\tau=M B \sin \theta $

$\tau=1 A B \sin 0$

$\tau=0\;\mathrm{Nm}$

But current carrying coil experience torque inside magnetic field, which is cross product of magnetic moment of the coil and the magnetic field. $\vec{\tau}=\vec{M} \times \vec{B}$

Option $A$ is correct.                                    

$M=i \times \pi r^{2}$

$=\frac{2 e}{2 \pi r} \times v \times \pi r^{2}$

$=\frac{2 e v r}{2}$

$=e v r$

$\tau=N L A B \sin \theta$

$=N I\left(\pi R^{2}\right) B \sin 30^{\circ}$

$=100 \times \pi \times(2)^{2} \times \frac{1}{\pi} \times \frac{1}{2}$

$=200 N - m$

Where, $G=$ voltmeter (resistance)

Given, $V_{A B}=120 V , I=2 A, R=75 \Omega$

$\Rightarrow 120=\left(2-I_{g}\right) 75 \Rightarrow I_{g}=0.4 A$

Now, $V_{A B}=I_{g} G \Rightarrow G=\frac{120}{0.4}=300 \Omega$

$R_{G}=60 \Omega$

When the galvanometer is shunted by a resistance $r$, its effective resistance

$R_{P}=\frac{R_{G} r}{R_{G}+r}=\frac{60 \times 0.02}{60+0.02} \approx 0.02 \Omega$

Total resistance of the circuit $=R+R_{P}=R+0.02$

Current, $I=\frac{5}{R+0.02} \Rightarrow 1=\frac{5}{R+0.02}$

$R+0.02=5 \Rightarrow R=5-0.02=4.98 \approx 5 \Omega$

For a voltmeter, in which current $I_g$ passes,

when range is lower, $5= G \times I_g$

when range is higher, $20=(G+R) I_g$

where $G=$ resistance of voltmeter $=20000\,\Omega$

$\therefore \quad \operatorname{From}(i)$ and $(i i)$

$\frac{5}{20}=\frac{G}{G+R}$ or $\frac{1}{4}=\frac{G}{G+R}$

or $R=3 G=3 \times 20000=60000$

or extra resistance $=60000\,\Omega$ in series.

$s\, = \,\frac{\theta }{i} \simeq \frac{{\tan \,\theta }}{i} = \frac{{{\mu _0}N}}{{2R{B_H}}}.$

If a magnetic material is placed inside coil of galvanometer, then

$s\,'\, = \,\frac{{{\mu _r}{\mu _0}N}}{{2R{B_H}}}$

When a voltmeter of resistance $150\,\Omega$ is connected across $A$ and $B$, the resistance across $A$ and $B$ will be

$R _{ ab }=\frac{150 \times 100}{150+100}=60\,\Omega$

Now the equivalent resistance of the circuit will be $R_{a c}=R_{a b}+R_{b c}=60+100=$ $160\,\Omega$

The current in the circuit will be $I=50 / 160=0.31\,A$

Thus, Potential drop across B and C will be $V _{ bc }= IR _{ bc }=(0.31)(100)=31\,V$

Shunt, $\mathrm{S}=\left(\frac{\mathrm{I}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}\right) \mathrm{G}$

$\Rightarrow $   $\mathrm{S}=\left(\frac{100 \times 10^{-6}}{10-100 \times 10^{-6}}\right) \times 50$

$=\frac{10^{-4}}{10} \times 50=5 \times 10^{-4}\,\Omega$

$ \Rightarrow $ $S = 5 \times {10^{ - 4}}\,\Omega $

We know that time period of a particle

$\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{0}} ; \frac{\mathrm{mv}_{0}^{2}}{\mathrm{r}}=\mathrm{Bqv}_{0} \Rightarrow \mathrm{v}_{0}=\frac{\mathrm{Bqr}}{\mathrm{m}}$

$\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{Bqr}} \times \mathrm{m}=\frac{2 \pi}{\mathrm{Bq}} \mathrm{m}$

Frequency $=\frac{1}{\mathrm{T}}=\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}$

This does not dpeend upon velocity so Reason is incorrect.

$dI =\frac{ dq }{ dt }$ $=\frac{2 \pi \cdot \sigma \cdot x \cdot dx }{ dt }$

$\frac{2 \pi}{ dt }=\omega$

$\therefore dI =\omega \cdot \sigma \cdot x \cdot dx$

$dM = dI \times \pi \times x ^{2}$

$=\omega \sigma x d x \cdot \pi \cdot x^{2}$

$=\omega \sigma x ^{3} dx$

$M=\int_{0}^{R} d M$

$=\int_{0}^{R} \omega \sigma \pi x^{3} d x$

$=\omega \sigma \pi \int_{0}^{ R } x ^{3} \cdot dx$

$=\omega \cdot \sigma \cdot \pi\left(\frac{ x ^{4}}{4}\right)_{0}^{ R }$

$=\omega \cdot \sigma \cdot \pi\left(\frac{ R ^{4}}{4}\right)$

$=\frac{1}{4} \pi R ^{4} \cdot \sigma w$

$ \Rightarrow $   $G = \frac{{10}}{2} = 5\,\Omega $.

Here ${i_g} = $ Full scale deflection current $ = \frac{{150}}{{10}} = 15\,mA$.

$V =$ Voltage to be measured $= 150 × 1 = 150 \,V.$

Hence $R = \frac{V}{{{i_g}}} - G = \frac{{150}}{{15 \times {{10}^{ - 3}}}} - 5 = 9995\,\Omega $.

$ \Rightarrow $ $\frac{{10}}{1} = 1 + \frac{{0.81}}{S}$

$ \Rightarrow S = 0.09\,\Omega $.

$E$. $2 \pi r=\pi r^{2} \frac{d B}{d t}$

$\mathrm{E} .2 \pi \mathrm{r}=\pi \mathrm{r}^{2} \mathrm{x}$

$\mathrm{E}=\frac{\pi \mathrm{R}^{2}}{2 \pi \mathrm{r}} \mathrm{B}_{0}=\frac{\mathrm{R}^{2} \mathrm{B}_{0}}{2 \mathrm{r}}$

$a=\frac{q E}{m}=\frac{q}{m} \frac{R^{2} B_{0}}{2 r}=\frac{q B_{0} R^{2}}{2 m r}$

Energy increased in each revolution $=2 \times 100 \times 10^3 \,eV$

$=2 \times 10^5 \,eV$

Now for energy $E=2 \times 10^7 \,eV$

Number of revolution $=\frac{2 \times 10^7 \,eV }{2 \times 10^5 \,eV }=100$

$f=\frac{q B}{2 \pi m} \quad\left(\because T =\frac{2 \pi m}{q B}\right)$

$\Rightarrow 10 \times 10^6=\frac{1.6 \times 10^{-19} \times B}{2 \pi\left(1.6 \times 10^{-27}\right)}$

$B=0.656 \,T$

$\therefore v=2 \pi r f=2 \times 3.14 x 50 \times 10^{-2} \times 10 \times 10^6$

$=3.14 \times 10^7\,ms ^{-1}$

${\mathrm{f}=\frac{\mathrm{q} \mathrm{B}}{2 \pi \mathrm{m}}=\frac{1.6 \times 10^{-19} \times 1}{2 \times 3.14 \times 9.1 \times 10^{-31}}} $

${=28 \times 10^{9} \mathrm{\,Hz}=28 \mathrm{\,GHz}}$

$E_{k}=\frac{q^{2} B^{2} r^{2}}{2 m}$

But frequency, $f=\frac{\mathrm{qB}}{2 \pi \mathrm{m}}$

$\therefore \mathrm{E}_{\mathrm{k}}=\frac{(2 \pi \mathrm{mf})^{2} \mathrm{r}^{2}}{2 \mathrm{m}}=2 \pi^{2} \mathrm{mf}^{2} \mathrm{r}^{2}$

or $\quad \mathrm{E}_{\mathrm{k}}=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times\left(10 \times 10^{6}\right)^{2} \times(0.5)^{2}$

$=8.23 \times 10^{-13} \,\mathrm{J}$

$\therefore $ $\mathrm{E}_{\mathrm{k}}=\frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}}=5.1 \times 10^{6} \mathrm{eV}=5.1\, \mathrm{MeV}$