Physics NCERT Exemplar Solutions Class 12th Chapter Six Electromagnetic Induction

This is a short answer type question as classified in NCERT Exemplar

Mutual inductance of coil A with respect to B

M₂₁=N₂ $\varnothing$ ₂/I₁ = $\frac{{10}^{-2}}{2}$ = 5mH

N₁ $\varnothing$ ₁= M₁₂I₂= 5mH (1A)= 5mWb

This is a short answer type question as classified in NCERT Exemplar

 $\varnothing =B.A=BAcos\theta$

$\varnothing =B_o\pi a^{2}coswt$

But according to faraday’s law e= B_{o $\pi a^{2}wsinwt$}

So the current will be I=B_{0 $\frac{\pi a^{2}wsinwt}{R}$}

For t= $\frac{\pi }{2w}$

I= $\frac{B_o\pi a^{2}w}{R}$ along j

Sinwt= sin(w $\frac{\pi }{2w}$ ) = sin $\frac{\pi }{2}$ =1

T= $\frac{\pi }{w}$ , I= $\frac{B\left(\pi a^2\right)w}{R}$

Sinwt=sinw $\frac{\pi }{w}$  = sin $\pi =0$

T= $\frac{3\pi }{2w}$

I= $\frac{B_o\pi a^{2}w}{R}$

So it become sin $\frac{3\pi }{2w}$ =-1

This is a short answer type question as classified in NCERT Exemplar

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let dφ = BA represents magnetic lines in an area to. By the concept of continuity of lines cannot end or start in space, therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂.Therefore, in both the cases we gets the same answer for flux.

This is a short answer type question as classified in NCERT Exemplar

The motional emf along PQ = length PQ × field along PQ

= length PQ ×  vBsin $\mathrm{\theta }$

= $\frac{d}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\theta }\mathrm{}}\times$ vB sin $\mathrm{\theta }$ = vBd

So emf make the flow of current in the circuit with resistance R

I= dvB/R

This is a short answer type question as classified in NCERT Exemplar

The back emf in solenoid force in solenoid is U a maximum rate of change of current . so maximum back emf will be obtained between 5s

Since the back emf at t = 3s also the rate of change of current at t= 3s, s= slope of OA from t=0s to t= 5s=1/5 A/sec

So we have if u= L1/5 (for t= 3s, dI/dt=1/5) (L is a constant). Applying e=-LdI/dt

For 5s

At t= 7s, u₁=-3e

For 10s

For t>30s, u₂=0

Thus back emf at t=7s,15s and 40s are -3e, e/2 and 0 respectively.

This is a long answer type question as classified in NCERT Exemplar

Let us assume that parallel wires at are y=0 i.e along x-axis and y=d.at t=0, AB has x=0 i.e along y-axis and moves with a velocity v

Motional emf across AB is = (B₀sinwt)vd (-j)

Emf due to change in field along OBAC = -B₀wcoswt x (t)d

Total emf in the circuit = emf due to change in field (along OBAC)+ the motional emf across AB= - B₀d {wx coswt+v sinwt}

Electric current in clockwise direction is given by =   $\frac{B_{0}d}{R}$   (wxcoswt+vsinwt)

The force acting on the conductor is given by F= iLbsin90=iLb

Substituting the values

F=   $\frac{B_{0}d}{R}$   (wx coswt+v sinwt) × d × B₀sinwt

  =  $\frac{B_0^2{d}^2}{R}$ (wx coswt+v sinwt)sinwt

This is a long answer type question as classified in NCERT Exemplar

Let us assume that the parallel wires at are y= 0, i.e., along x-axis and y= l. At t= 0, XY has x = 0 i.e., along y-axis.

(i) Let the wire be at x x = (t) at time t.

The magnetic flux linked with the loop is given by

φ = B.A= Bacos0= BA

magnetic flux = B(t)(lx(t))

total emf = emf due to change in field along XYAC+ the motional emf across XY

E= - $\frac{d\varnothing }{dt}$  = - $\frac{dB\left(t\right)}{dt}$ lx(t)- B(t)lv(t)

Electric current in clockwise direction is I= E/r

So force is F= ilBsin90=ilB

Force = $\frac{IB\left(t\right)}{R}$ [- $\frac{dB\left(t\right)}{dt}Ix\left(t\right)-B\left(t\right)Iv\left(t\right)$ ]i

Applying newton 2^nd law

m $\frac{d^{2}x}{d{t}^2}$ = $\frac{I^{2}B\left(t\right)dB}{Rdt}x\left(t\right)$ - $\frac{I^2{B}^2\left(t\right)dx}{Rdt}$

(ii) $\frac{dB}{dt}$ =0

Substituting in eqn 1

$\frac{d^{2}x}{d{t}^2}$ + $\frac{I^2{B}^2\left(t\right)dx}{mRdt}$ =0

$\frac{dv}{dt}+\frac{I^2{B}^{2}v}{mR}$ =o

V= Aexp $(\frac{I^2{B}^{2}v}{mR}$ )=0

(iii) p= I²R= $\frac{I^2{B}^2{v}^{2}R}{R^2}=\frac{B^2{I}^2}{R}{u}_0^2$ exp(-2I²B²tlmR)

This proves decrease in kinetic energy.

This is a long answer type question as classified in NCERT Exemplar

At t=o t=T/8= $\pi /4w$  the rod will make contact with side BD.at any time 0 $\pi /4w$

So magnetic flux through area ODQ is $\varnothing =B\times areaOQD=B\times \frac{1}{2QD}\times OD$

=B $\times \frac{1ltan\theta }{2}\times l$

$\varnothing =\frac{1}{2}B{L}^{2}wtan\theta =\frac{1}{2}B{L}^{2}w{sec}^{2}wt$

I=e/R

And R= $\lambda x=\frac{\lambda l}{coswt}$  so I= $\frac{1}{2}\frac{BLw}{coswt\lambda }$

Same result for other side too

This is a long answer type question as classified in NCERT Exemplar

B= $\frac{{?}_{0}I}{2R}$ (ouside the paper)

Flux= $\frac{{?}_{0}I}{2R}l{?}_{x_0}^x\frac{dr}{r}$ = $\frac{{?}_{0}I}{2R}$ in $\frac{x}{x_0}$

Induced emf= e/R=I

Induced current = $\frac{1}{R}\frac{d?}{dt}$ = $\frac{e}{R}$ = $\frac{{?}_{0}I?}{2?R}$ in $\frac{x}{x_0}$

This is a long answer type question as classified in NCERT Exemplar

applying ohm’s law

I=E/R=d $\varnothing /Rdt$

I= dQ/dt or dQ/dt= d $\varnothing /Rdt$

Q (t₁)-Q (t₂)= $\frac{1}{R}$ [ $\varnothing$ t₁- $\varnothing$ t₂]

$\varnothing$ t₁= L_{1 $\frac{{\mu }_0}{2\pi }{\int }_x^{L_2+x}\frac{dx}{x}I$} t₁

= $\frac{{\mu }_0{L}_1}{2\pi }$ I1in [ $\frac{L_2+x}{x}$ ]

This is a long answer type question as classified in NCERT Exemplar

Since, the magnetic field is brought to zero in time? t, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in ring by the

phenomenon of EMI. The induces emf causes the electric field E generation around the

ring.

The induced emf= electric field E (2 πb) (Because V =E d  ) . (i)

By Faraday'slaw of EMI

The induced emf= rate of change of magnetic flux

=rate of change of magnetic field × area= $\frac{B\pi a^2}{dt}$ ……. (ii)

From (i) and (ii)

2 $\pi bE$ =emf= $\frac{B\pi a^2}{dt}$

Since, the charged ring experienced a electric force= QE

This force try to rotate the coil, and the torque is given by

Torque =  b (Force)

= Qeb= Q { $\frac{B\pi a^2}{2\pi bdt}$ }b

= Q $\frac{B\pi a^2}{2dt}$

dL= torque (dt)= Q $\frac{B{a}^2}{2}$

as intital momentum is zero

so final momentum is = mb²w= QBa²/2

w= QBa²/2mb²

This is a long answer type question as classified in NCERT Exemplar

The component of magnetic field perpendicular the plane = Bcosθ

But when there is a motional emf then it is e = Blv=vBcosθd

And current Is I= $\frac{\mathrm{v}\mathrm{B}\mathrm{c}\mathrm{o}\mathrm{s}\theta d}{R}$ but when discuss about force F= llB

And component of force which only act is fcosθ

So force is IBd(cosθ) = $\frac{\mathrm{v}\mathrm{B}\mathrm{c}\mathrm{o}\mathrm{s}\theta d}{R}$  where v=dx/dt

Also component of weight is mg cosθ

So by applying newton second law of motion

m $\frac{d^{2}x}{d{t}^2}=mgsin\theta -\frac{Bcos\theta d}{R}\left(\frac{dx}{dt}\right)\times \left(Bdcos\theta \right)$

$\frac{dv}{dt}=gsin\theta -\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}v$

$\frac{dv}{dt}+\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}v=gsin\theta$

On solving this equation we get

V= $\frac{gsin\theta }{\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^2}+Aexp(-\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}t)$

This is a long answer type question as classified in NCERT Exemplar

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vB d.

Since, switch S is closed at time t= 0. capacitor is charged by this potential difference. Let

Q ( t) is charge on the capacitor and current flows from A to B.

Now, the induced current I= $\frac{vBd}{R}$ - $\frac{Q}{RC}$

On rearranging  = $\frac{Q}{RC}$ + dQ/dt = $\frac{vBd}{R}$

On solving differential equation we get I= $\frac{vBd}{R}$ e^-t/RC

This is a long answer type question as classified in NCERT Exemplar

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vBd.

Since, switch S is closed at time t= 0. current start growing in inductor by the potential

Difference due to motional emf.

Applying Kirchhoff’s voltage rule, we have

-L $\frac{dI}{dt}$ +vBd=IR

On solving differential equation we get I= $\frac{vBd}{R}$ + Ae^-Rt/2

At t=0 I=0

A= -vBd/R

I= $\frac{vBd}{R}$ (1-e^-Rt/L)

This is a long answer type question as classified in NCERT Exemplar

$?=B_z\left(\pi l^2\right)=B_0(1+\lambda z)\left(\pi l^2\right)$

According to faraday’s law e= - $\frac{d\varnothing }{dt}$ and also ohm’s law V=IR

So by equating these two

We get B_o( $\pi l^2$ ) $\lambda \frac{dz}{dt}=IR$

I= $\frac{\pi l^2{B}_o\lambda v}{R}$

And energy lost/second = I²R= $\frac{{\left(\pi l^2\lambda \right)}^2{B}_o^2{v}^2}{R}$

As change in PE=mg $\frac{dz}{dt}=mgv$ but if we compare both energy then

Mgv= $\frac{{\left(\pi l^2\lambda \right)}^2{B}_o^2{v}^2}{R}$  or v=

This is a long answer type question as classified in NCERT Exemplar

This is a multiple choice answer as classified in NCERT Exemplar

(c) $\varnothing$ =B.A= B₀ (2i+3j+4k)= 4B₀L²wb

This is a multiple choice answer as classified in NCERT Exemplar

$\varnothing =B.A$

And A= A₁+A₂= (L²k+L²i)

B= B_o (i+k)T

$\varnothing =B.A$= B_o (i+k) (L²k+L²i)

= 2B_oL²Wb

This is a multiple choice answer as classified in NCERT Exemplar

(b) When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A.

This is a multiple choice answer as classified in NCERT Exemplar

(b) When the A stops moving the current in B become zero, it is possible only if the current in A is constant. If the current in A would be variable, there must be an induced emf (current) in B even if the A stops moving.

This is a multiple choice answer as classified in NCERT Exemplar

(a) When the current in B (at t= 0) is counter-clockwise and the coil A is considered above to it. The counterclockwise flow of the current in B is equivalent to north pole of magnet and magnetic field lines are emanating upward to coil A. When coil A start rotating at (t= 0), the current in A is constant along clockwise direction by Lenz's rule.

This is a multiple choice answer as classified in NCERT Exemplar

(b) The self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length, filled the inside of the solenoid with a material of relative

permeability (e g. ., soft iron, which has a high value of relative permeability) is given by the relation L = ${\mu }_r{\mu }_o$ n²A/l

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d) A metal plate is getting heated when a DC or AC current is passed through the plate, known as heating effect of current. Also, when metal plate is subjected to time varying magnetic field, the magnetic flux linked with the plate changes and eddy currents comes into existence which make the plate hot.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) Here, magnetic flux linked with the isolated coil change when the coil being in a time varying magnetic field, the coil moving in a constant magnetic field or in time varying magnetic field.

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) Mutual inductance of a coil increases when they came closer also the relation for the same will be given by

M₂₁= ${\mu }_0$ n₁n₂ r₁²l= M₁₂

M₂₁= M₁₂=M

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably in such a way that the cross product of magnetic field and surface area of plane of coil remain constant at every instant.

This is a short answer type question as classified in NCERT Exemplar

When the switch is thrown from the off position (open circuit) to the on position (closed circuit), then neither B, nor A nor the angle between B and A change. Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently no current will flow in the circuit.

This is a short answer type question as classified in NCERT Exemplar

When the coil is stretched so that there are gaps between successive elements of the spiral coil i.e., the wires are pulled apart which lead to the flux leak through the gaps.

According to Lenz's law, the emf produced must oppose this decrease, which can be done by an increase in current. So, the current will increase.

This is a short answer type question as classified in NCERT Exemplar

When the iron core is inserted in the current carrying solenoid, the magnetic field increase due to the magnetisation of iron core and consequently the flux increases.

According to Lenz's law, the emf produced must oppose this increase in flux, which can be done by making decrease in current. So, the current will decrease.

This is a short answer type question as classified in NCERT Exemplar

When the current is switched on, magnetic flux is linked through the ring. Thus, increase in flux takes place. According to Lenz's law, this increase in flux will be opposed and it can happen if the ring moves away from the solenoid.

This happen because the flux increases will cause a counter clockwise current (as seen from the top in the ring in figure.) i.e., opposite direction to that in the solenoid. This makes the same sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming same magnetic pole in front of each other. Hence, they will repel each other and the ring will move upward.

This is a short answer type question as classified in NCERT Exemplar

When the current is switched on, magnetic flux is linked through the ring. Thus, increase influx takes place. According to Lenz's law, this increase in flux will be opposed and it can happen if the ring moves away from the solenoid.

This happen because the flux increases will cause a counter clockwise current (as seen From the top in the ring in figure.) i.e., opposite direction to that in the solenoid. This makes the same sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming same magnetic pole in front of each other. Hence, they will repel each other and the ring will move upward.

This is a short answer type question as classified in NCERT Exemplar

When cylindrical bar magnet of radius 0.8 cm is dropped through the metallic pipe with an inner radius of 1 cm, flux linked with the cylinder changes and consequently eddy currents are produced in the metallic pipe. According to Lenz's law, these currents will oppose the (cause) motion of the magnet.

Therefore, magnet's downward acceleration will be less than the acceleration due to gravity g. On the other hand, an non magnetised iron bar will not produce eddy currents and will fall with an acceleration due to gravity g.

Thus, the magnet will take more time to come down than it takes for a similar unmagnetized Cylindrical iron bar dropped through the metallic pipe.

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

= dimensionless of

= MLT²

By conservation of Angular momentum

L_i = L_f

$M{R}^2\omega =\left(m{R}^2+2m{R}^2\right)\omega \text{'}$

$\frac{2M}{M+2m}=\omega \text{'}$

$\Rightarrow$ $\omega \text{'}=\frac{2M}{M+2m}$

$g\propto r0\le r\le R$

$g\propto \frac{1}{r^2}r>R$

After switch ‘S’ is closed

$Q_1+Q_2=C_{1}V$ - (1)

Using KVL

$\frac{Q_1}{C_1}-\frac{Q_2}{C_2}=0$

$Q_1=Q_2\frac{C_1}{C_2}$ - (2)

from (1) & (2)

$\Rightarrow Q_2\left[\frac{C_1+C_2}{C_2}\right]=C_{1}V\Rightarrow Q_2=\left(\frac{C_1{C}_2}{C_1+C_2}\right)V$

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

Since

$\overrightarrow{E}.\overrightarrow{B}=0\left(\overrightarrow{E}\perp \overrightarrow{B}\right)$

$\&\frac{E_0}{B_0}=C=3\times 10^{8}m/sec$

for option D

$\frac{E_0}{B_0}=\frac{\sqrt[]{301.6^2+452.4^2}}{10^{-6}\sqrt[]{1.5^2+1^2}}\begin{array}{cc}\hfill =301.6\times 10^6\hfill & \hfill =3.01\times 10^{8}m/sec\hfill \end{array}$

 ${\lambda }_e=\frac{h}{mv}=\frac{h}{p_e}$ - (1)

${\lambda }_{Ph}=\frac{h}{P_{Ph}}$ - (2)

A/C to question

${\lambda }_e={\lambda }_{Ph}$

$P_e=P_{Ph}\Rightarrow \frac{P_P}{P_{Ph}}=1$ - (3)

$k.E_e=E_e=\frac{1}{2}m{v}^2$

$=\frac{1}{2}Pev$ - (4)

$k.E_{Ph}=E_{Ph}=m{c}^2$

= mc c

q = P_Ph c- (5)

$\left(4\right)÷\left(5\right)$

$\frac{E_e}{E_{Ph}}=\frac{v}{2c}$

Dynamic Resistance at 2v

= $\frac{\Delta v}{\Delta I}=\frac{2.1-2}{\left(10-5\right)\times 10^{-3}}=\frac{0.1}{5}\times 10^3$

Dynamic Resistance at 4 v

= $\frac{\Delta v}{\Delta I}=\frac{4.2-4}{\left(250-200\right)}=\frac{0.2}{50\times 10^{-3}}=\frac{0.2}{50}\times 10^3$

$\frac{R_{2v}}{R_{4v}}=\frac{0.1\times 10^3\times 50}{5\times 0.2\times 10^3}=5:1$

Let ‘h’ be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h=ut+\frac{1}{2}a{t}^2$

$=0\times 1+\frac{1}{2}\times 10\times 1\times 1$

= 5m

Decay of current in Inductor is given by,

$i=i_0{e}^{-t/\tau }\left[Where\tau =\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$R$}\right.;i_0=\frac{v}{R}=\frac{20}{10}=2\right]$

At t = 100 $\mu s$

i = 0

i.e. i = i0 $e^{-100/\tau }=0$ -(1)

e.m.f induced

$e=\frac{-Ldi}{dt}=-L\frac{d}{dt}\left[i_0{e}^{-t/\tau }\right]$

= $=-L{i}_0\frac{-1}{2}{e}^{-t/\tau }$

$e=\frac{L{i}_0}{\tau }{e}^{-t/\tau }$

$e_{avg}=\frac{{\displaystyle \underset{0}{\overset{200\mu s}{\int }}edt}}{{\displaystyle \underset{0}{\overset{200\mu s}{\int }}dt}}=\frac{{\displaystyle \underset{0}{\overset{200\mu s}{\int }}\frac{L{i}_0}{\tau }{e}^{-t/\tau }dt}}{{\displaystyle \underset{0}{\overset{200\mu s}{\int }}dt}}$

$=\frac{L{i}_0}{200\times 10^{-6}sec}=\frac{20\times 10^{-3}\times 2}{200\times 10^{-6}}=400$

${\delta }_{total}=360°-2\theta$

= 360 – 2 × 75° = 210°

Total deviation = $\delta =150°$

Let ‘x’ he the value of one division of main scale

$x=\frac{1}{20}cm=0.05cm$                

Let y be value of one division on venire scale given

10 y = 9 x

$y=\frac{9x}{10}$                

Least count = $x-\frac{9x}{10}=\frac{x}{10}=\frac{0.05}{10}$  

= 0.005 cm

= 5 × 10^-2 mm

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$=\left(l_L+l_C\right)$

Therefore current through R circuit at resonance will be zero

$\rho =800kg/m^3$

P₁ – P₂ = 4100 Pa

Bernoulli’s question b/w (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+z_2$

$\frac{P_1-P_2}{\rho g}+\frac{v_1^2}{2g}+1=\frac{v_2^2}{2g}+0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁-(2)

from (1) & (2)

$\Rightarrow \frac{P_1-P_2}{\rho g}+\frac{v_1^2}{2g}+1=\frac{{\left(2v_1\right)}^2}{2g}$

$\frac{4100}{800\times 10}+\frac{v_1^2}{2g}+1=\frac{4v_1^2}{2g}$

$\frac{12100}{8000}=\frac{3v_1^2}{2g}$

$\Rightarrow v_1^2=\frac{2\times 10}{3}\times \frac{12100}{8000}$

$=\frac{2}{3}\times \frac{121}{8}$

x = 363

F.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss.

At maximum height velocity (v) = 0

So, momentum = mv = m × 0 = 0

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$\eta =\left(1-\frac{T_2}{T_1}\right)\times 100=\left(1-\frac{273}{373}\right)\times 100=26.81\mathrm{\%}$

$g_{eff}=g+a$

$=g+\frac{g}{6}$

$=\frac{7g}{6}$

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$T\text{'}=\sqrt[]{\frac{6}{7}}T$

$KE=\Delta U$

$\Delta U=\frac{1}{2}m{v}^2$

$n{C}_v\Delta T=\frac{m{v}^2}{2}$

$\Rightarrow \frac{m}{M}\frac{R}{y-1}\Delta T=\frac{m{v}^2}{2}$

$\Delta T=\frac{0.4}{2}\frac{M{v}^2}{R}$

$\Delta T=\frac{M{v}^2}{5R}$

 $V=\frac{-d?}{dt}=-\left(15t^2+8t+2\right)$

at t = 2 sec, V = (15 × 4 + 8 × 2 + 2)

= 78 V

$I=\frac{V}{R}=\frac{78}{5}=15.6A$

Volume = constant

$A\mathcal{l}=C$

$Ad\mathcal{l}+\mathcal{l}dA=0$

$\frac{d\mathcal{l}}{\mathcal{l}}+\frac{dA}{A}=0$

$\frac{dR}{R}\times 100=\left(\frac{d\mathcal{l}}{\mathcal{l}}-\frac{dA}{A}\right)\times 100$

= 0.4 (0.4)

= 0.8 %

 $r=\frac{mv}{qB}$

$r_{\alpha }=\frac{m_{\alpha }v}{q_{\alpha }B}$

$r_P=\frac{m_{P}v}{q_{P}B}$

$\frac{r_{\alpha }}{r_P}=\frac{m_{\alpha }{q}_P}{m_P{q}_{\alpha }}=\frac{m_{\alpha }}{m_P}\left(\frac{q_P}{q_{\alpha }}\right)$

$=\frac{4m}{m}\left(\frac{q}{2q}\right)=2:1$

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

B fighter jet

A anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos = 200

$\begin{array}{l}cos\theta =\frac{1}{2}\\ \theta =60°\end{array}$

 $\Delta L$  (change in length) $=\frac{F\mathcal{l}}{AE}=\frac{mg\mathcal{l}}{AE}$

$\Delta \mathcal{l}\propto g$

$\frac{\Delta {\mathcal{l}}_1}{g_1}=\frac{\Delta {\mathcal{l}}_2}{g_2}\Rightarrow \frac{10^{-4}}{10}=\frac{6\times 10^{-5}}{g_1}$

$g_1=\frac{6\times 10^{-4}}{10^{-4}}=$ 6 m/sec²

$\begin{array}{l}{D}_1\\ D_3\end{array}\}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I=\frac{v}{R}=\frac{10}{10}=1Amp$

$Tsin\theta =\frac{k{q}^2}{r^2}$

$Tcos\theta =mg$

$tan\theta =\frac{k{q}^2}{mg{r}^2}$

Using the above equation find the value of ‘q’

$P=\sqrt[]{2Km}$

$\frac{n}{2}=\sqrt[]{\frac{m_1}{m_2}}=\frac{1}{2}$

n = 1

Efficiency in the Carnot engine is defined as :

$\eta =\frac{work}{heat}=\frac{1}{4}$

$\eta =1-\frac{T_c}{T_h},2\eta =1-\frac{T_c-52}{T_h}\Rightarrow T_h=208K$

i₂ = 2i₁

i₁ + i₂ = 6

i₁ = 2A.

$v=\sqrt[]{\frac{T}{\mu }}$

new v’ = $\sqrt[]{\frac{1.04T}{\mu }}$

% change = $\left(\sqrt[]{1.04}-1\right)\times 100$

= 2%

Energy corresponding to a particle is mc²

$\frac{hc}{\lambda }=m{c}^2\Rightarrow \frac{hc}{\lambda }=\left(\frac{x}{3}h\right)c^2\Rightarrow x=\frac{3}{\lambda c}=\frac{3}{10\times 10^{-10}\times 3\times 10^8}$ = $\frac{1}{10^{-1}}=10$

Using conservation of energy :

$\frac{1}{2}m{v}_i^2-\frac{Gmm}{R}=0-\frac{Gmm}{11R}$

$v_i^2=\frac{20}{11}\frac{Gm}{R}$

Escape velocity is defined as $\sqrt[]{\frac{2GM}{R}}$

$v_i=\sqrt[]{\frac{10}{11}{v}_e}$

If angle between them is 180° then

$\overrightarrow{P}\times \overrightarrow{Q}=\overrightarrow{q}\times \overrightarrow{P}=0$

After contact, charges an each sphere will be

$\frac{q_1+q_2}{2}=1nc\Rightarrow F=\frac{k{q}_1{q}_2}{r^2}=36\times 10^{-9}N$

$l=\frac{1}{2}c{\epsilon }_0{E}_0^2\Rightarrow \frac{8}{4\pi {\left(10\right)}^2}\times \frac{1}{2}=\frac{1}{4}\times c\times \frac{1}{c^2{\mu }_0}\times E_0^2\Rightarrow E_0=\frac{2}{10}\sqrt[]{\frac{{\mu }_{0}c}{\pi }}\Rightarrow x=2$