MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 2

Change in number of magnetic field lines induces

0%
current in coil
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EMF in the coil
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frequency in coil
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both A and C

Which of the following statement regarding transformer is incorrect?

0%
A transformer makes use of Faradays Law.
0%
The ferromagnetic properties of an iron core is used by it to efficiently raise or lower AC voltage
0%
It cannot increase power so that if the voltage is raised, the current is proportionally lowered and vice versa.
0%
The current induced in primary circuit is of same nature as in secondary circuit with differences in amplitude only.

The induction coil works on the principle of.

0%
Self-induction
0%
Mutual induction
0%
Ampere's rule
0%
Fleming's right hand rule

Which of the following best describe the electromagnetic induction?

0%
The ability of a changing magnetic field to induce a voltage in a conductor
0%
The ability of a conductor to generate a magnetic field
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The ability of a static magnetic field to induce a voltage in a conductor
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The ability of a permanent magnet to induce a voltage in a coil
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The ability of a conductor to induce a magnetic field

Which of the following are not units of self inductance?

0%
$Weber/m^2$
0%
$Ohm-second$
0%
$Joule-ampere$
0%
$Joule-ampere^-2$

What is the unit for EMF?

0%
Ampere
0%
Potential
0%
Watt
0%
Volt

Explanation

EMF is defined as the voltage developed by the electrical energy source.

For example : Battery

Thus EMF has the unit of volt denoted by

$V$ .

The dimensions of self-inductance L are :

0%
$[ML^2T^{-2}A^{-2}]$
0%
$[ML^2T^{-1}A^{-2}]$
0%
$[ML^2T^{-1}A^{-1}]$
0%
$[ML^{-2}T^{-2}A^{-2}]$

Explanation

Self-inductance L

$= \dfrac {e}{\Delta i/\Delta t}=\dfrac{e\Delta t}{\Delta i}$

$\therefore$ Unit of L

$=\dfrac{Volt-second}{Ampere}$

$=\dfrac{(Joule/Coulomb)second}{Ampere}$

$=\dfrac{Newton-metre-second}{Coulomb-ampere}$

$\Rightarrow$ Unit of L

$= \dfrac{Newton-metere}{Ampere^2}$

$\therefore [L]=\dfrac{[MLT^2][L]}{[A^2]}=[ML^2T^{-2}A^{-2}]$

A coil carrying electric current is placed in uniform magnetic field, then :

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Torque is formed
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emf is induced
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Both (a) and (b) are correct
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None of the above

Explanation

As magnetic field is uniform so its rate of change with time is zero. By Faraday's law of induction, no emf is induced.

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.

Given below are the symbols of a few electronic components. Which of these components denote a variable inductor ?

The induced electromotive force increases when a magnet is moved fast in a stationary coil of wire because

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Magnetic field increases
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Rate of change of magnetic field increases
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Rate of change of magnetic field decreases
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Magnetic field decreases

.................. gave the principle of Electromagnetic Induction.

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Volta
0%
Ampere
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Faraday
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Orested

The self inductance of a straight conductor is.

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Zero
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Infinity
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Very large
0%
Very small

If a magnet is plugged into a coil, then the magnitude of induced emf does not depend upon:

0%
the number of turns in the coil
0%
the medium of the core of the coil
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the insertion speed of the magnet
0%
the strength of the magnet
0%
the resistance of the coil

Explanation

The magnitude of induced emf is given by

$e=\displaystyle\frac{Nd\phi}{dt}=NB\displaystyle\frac{ds}{dt}$
Which depends upon number of turns(N), magnetic field intensity(B) which depends upon medium, rate of change of area

$(ds/dt)$ which depends upon insertion speed of magnet. It is independent of resistance of coil.

The self-inductance of a long solenoid carrying current is independent of:

0%
its length
0%
the current
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its cross-sectional area
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magnetic permeability of the core
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the number of turns

Explanation

Self-inductance of a solenoid is given by

$L=\displaystyle\frac{\mu_0N^2A}{l}$
where,

$N=$ number of turns in solenoid

$l=$ length of the solenoid

$A=$ area of one turn of solenoid.
The self-inductance of solenoid does not depend on current passing through it.

An electromagnetic field exists only when there is

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an increasing current
0%
decreasing current
0%
voltage
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current

Alternative current generator is basically based upon :

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Amperes law
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Lenz's law
0%
Faradays law
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coulombs law

Flux density under trailing pole tips in case of generator will :

0%
increase
0%
decrease
0%
either increase or decrease
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none of the above

In one second, a current of 10 A changes through a coil. The induced emf is 10V, then, self-inductance of the coil is

0%
2H
0%
5H
0%
1H
0%
0.5H

Explanation

From Faradays law the induced emf is

$e= L \frac {di}{dt}$
Given,

$di = 10A, dt= 1s, e=10V$

$\therefore 10 = L\frac {10}{1}$

$\Rightarrow L = 1H$

Multiple Correct Answers Type
The SI unit of inductance, henry, can be written as

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Weber/ampere
0%
Volt-second / ampere
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$Joule / (ampere)^2$
0%
Ohm-second

Explanation

$L=\cfrac{\phi}{i}$

$L=\cfrac{weber}{Ampere}$

$V=L\cfrac{Ldi}{dt}$

$L\to$ volt.second/ampere

$E=\cfrac{1}{2}Li^2$

$L\to$ joule/(ampere

$)^2$

$\omega L=X_L$

$L\to$ ohm.second

Hence all are correct.

An emf can be induced between the two ends of straight copper wire when it is moved through a uniform magnetic field.

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True
0%
False

The magnitude of induced emf produced in a coil when a magnet is inserted in it , does not depend upon the

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magnetic movement of the magnet
0%
speed of approach of the magnet
0%
number of turns in the coil
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resistance of the coil

An inductor is connected to a direct voltage source through a switch. Now

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Very large emf is induced in inductor when switch is closed
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Larger emf is induced when switch is opened
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Large emf is induced whether switch is closed or opened
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No emf is induced whether switch is closed or opened

Explanation

As direct voltage doesn't change with time so the flux of inductor will not change with time. So there will not be $any$ emf induced.
Above statement is true even switch is closed or not.

So in a direct voltage circuit, the inductor behaves like a resistor.

The current in a coil of 0.4 mH coil increases by 250 mA in 0.1 seconds. The induced emf will be:-

0%
+1 V
0%
-1 V
0%
+ 1 mV
0%
-1 mV

Explanation

Fig: Inductor circuit

The emf induced in inductor i,

$\varepsilon= -L\dfrac{dI}{dt}$

Given,

$L=0.4mH$

$dI=250mA$

$dt=0.1sec$

$\varepsilon=-\dfrac{0.4\times 10^{-3}\times 250\times10^{-3}}{0.1}$

$\varepsilon=-1mV$

The correct Option is

$D$

1417745_1049518_ans_f2b057916acf4278ab3ca0b6b54071d2.jpeg

A generator works on the principle of electromagnetic induction.

0%
True
0%
False

Two different loops are concentric and lie in the same plane. The current in the outer loop is clockwise and increasing with time. The induced current in the inner loop then is

0%
clockwise
0%
zero
0%
counter clock wise
0%
in a direction that depends on the ration of the loop radii

Explanation

Current will induced in inner loop in counter clockwise direction. According to Lenz law induced emf always gives rise to a current whose magnetic field opposes the change in magnetic flux.

Emf is given by

$e=\dfrac{-d\phi }{dt}$

$IR=\dfrac{-d\phi }{dt}$

$I=\dfrac{1}{R}\dfrac{-d\phi }{dt}$

Which of the following does not have the same dimensions as the Henry?

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$\dfrac{\text {joule}}{(\text{ampere})^2}$
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$\dfrac{\text {tesla} - m^2}{(\text{ampere})^2}$
0%
$\text{ohm-second}$
0%
$\dfrac{1}{\text{Farad-second}}$

Explanation

Option

$D$ is does not have the same dimensions as the Henry.

So, Option

$D$ is correct.

Which device is used to produce electricity? Describe with a neat diagram.

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Electric motor
0%
Galvanometer
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Electric Generator (DC)
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Voltmeter

Explanation

The device used for producing electricity is Electric generator

$(DC)$ . It is based on the phenomenon of electromagnetic induction.

1746563_1823105_ans_0a369748677d450fa37f63896046b5a7.png

When a wire loop is rotated in a magnetic field, the direction of induced e.m.f. changes once in each

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$1/4$ revolution
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$1/2$ revolution
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$1$ revolution
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$2$ revolution

Explanation

Flux of the magnetic field through the loop

$\phi =B.A\cos \omega t$

Emf= $-\dfrac{d\phi }{dt}=B.A\omega \sin \omega t$

$\omega t=0\,to\,\pi \,\,,\,\,\,\,\,\sin \omega t=+ve$

$\omega t=\pi \,to\,2\pi ,\,\,\,\,\,\,\,\sin \omega t=-ve$

Hence, the direction of induced e.m.f. changes once in $\dfrac{1}{2}$ revolution.

A coil carries a current in a magnetic field. The coil experiences a turning effect. Which device uses this effect?

0%
A d.c motor
0%
An electromagnet
0%
A relay
0%
A transformer

In faraday's experiment current in not produced in

0%
the coil is moved and magnetic is stationary
0%
the magnetic is moved the coil is stationary
0%
both coil and magnetic are moved in same direction with same speed
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both coil and magnet stationary

An induced emf is produced when a magnet is plunged into a coil. The strength of the induced emf is independent of :

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the strength of the magnet
0%
number of turns of coil
0%
the resistivity of the wire of the coil
0%
speed with which the magnet is moved

Explanation

$EMF\quad =-\dfrac { nd\phi }{ dt } =\quad \dfrac { -d\left( nBA \right) }{ dt }$
Where

$n$ is number of turns in coil
So, emf is dependent on number of turns of coil, strength of magnet and speed with which magnet is moved. emf is independent of resistivity of wire of the coil.

To measure the field

$B$ between the poles of an electromagnet, a small test loop of area

$1$

$cm^{2}$ , resistance

$10$

$\Omega$ and

$20$ turns is pulled out of it. A galvanometer shows that a total charge of

$2 \mu C$ passed through the loop. The value of

$B$ is

0%
$0.001 T$
0%
$0.01 T$
0%
$0.1 T$
0%
$1.0 T$

Explanation

$emf=\dfrac { BAn }{ t }$

$i=\dfrac { emf }{ R } =\dfrac { BAn }{ tR }$

$q=it=\dfrac { BAn }{ tR } =\dfrac { BAn }{ R }$

$\therefore 2\times { 10 }^{ -6 }=\dfrac { B\times { 10 }^{ -4 }\times 20 }{ 10 }$

$B={ 10 }^{ -2 }T$

$B= 0.01T$

An average emf of

$32 V$ is induced in a coil in which the current drops from

$10 A$ to

$2 A$ in

$0.1 s$ . The inductance of the coil is:

0%
$0.32 H$
0%
$0.4 H$
0%
$4 H$
0%
$0.04 H$

Explanation

$emf=L\dfrac { di }{ dt }$

$emf=L\dfrac { \triangle i }{ \triangle t }$

So,

$L=emf\times \dfrac { \triangle t }{ \triangle i }$
=

$32\times \dfrac { 0.1 }{ 8 }$

$= 0.4H$

Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45 to the vertical JUST after release. Then X is nearly

0%
$1\times {10 }^{-5 }V$
0%
$1\times {10 }^{-7 }V$
0%
$1\times {10 }^{-9 }V$
0%
$1\times {10 }^{-10 }V$

Explanation

As we know that F = qE also f = mg due to mass and gravitation

there fore

$qE =mg$

$q(V/d)=mg$

$V=mgd/q$

now putting the values

$V = \dfrac{1.67\times 10^{-27}\times10\times10^{-2}}{1.6\times10^{-19}}$

we get V =

$10^{-9}$ v hence the option (c) is correct

When rate of change of current in a circuit is unity, the induced emf is equal to

0%
total flux linked with the coil.
0%
induced charge.
0%
number of turns in the circle.
0%
coefficient of self induction.

Explanation

Induced emf is equal to

$emf =-\dfrac { d(N\emptyset ) }{ dt } = \dfrac { -d\left( Li \right) }{ dt } =-L\dfrac { di }{ dt }$

So, when rate of change of current is equal to unity, emf is equal to coefficent of self inductance.

A current of

$2A$ is passed through a coil of

$1000$ turns to produce a flux of

$0.5\mu Wb$ . Self inductance of the coil

0%
$2.5\times 10^{-4}H$
0%
$2.5\times 10^{-5}H$
0%
$2.5\times 10^{-6}H$
0%
$2.5\times 10^{-7}H$

Explanation

$N\phi =Li$

$L=\dfrac { N\phi }{ L }$
=

$\dfrac { 1000\times 0.5\times { 10 }^{ -6 } }{ 2 }$

$= 2.5 \times 10^{-4} H$

A coil has

$100$ turns. When a current of

$5A$ produces a magnetic flux of

$1\mu Wb$ , the coefficient of self induction is :

0%
$10\mu H$
0%
$20\mu H$
0%
$30\mu H$
0%
$40\mu H$

Explanation

$N\phi =Li$

$L=\dfrac { N\phi }{ i }$
=

$\dfrac { 100\times 1\times { 10 }^{ -6 } }{ 5 }$
=

$20\times { 10 }^{ -6 }H$
=

$20\mu H$

The rate of change of current needed to induce an emf of

$8 V$ in

$0.1 H$ coil is

0%
$0.8 A/s$
0%
$0.125 A/s$
0%
$80 A/s$
0%
$8 A/s$

Explanation

$emf=L\dfrac { di }{ dt }$

$\dfrac { di }{ dt } =\dfrac { emf }{ L }$

$=\dfrac { 8 }{ 0.1 }$

$=80A/s$

A magnet is moved towards the coil
(i) quickly in one case, and
(ii) slowly in another case.

Then the induced emf is :

0%
larger in case (i)
0%
smaller in case (i)
0%
equal in both
0%
larger or smaller depending upon the radius of the coil

Explanation

$EMF\quad =-\dfrac { d\phi}{ dt } \quad$
So, induced emf is inversely proportional to time, so, when time is less (i.e quickly) induced emf will be more.

Assertion : The induced emf and current will be same in two identical loops of copper and aluminium, when rotated with same speed in the same magnetic field.
Reason : Mutual induction does not depend on the orientation of the coils.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
0%
Assertion is true but Reason is false
0%
Assertion and Reason both are false

In an inductance coil the current increases from zero to

$6A$ in

$0.3$ second by which an induced e.m.f. of

$60V$ is produced in it. The value of coefficient of self-induction of coil is :

0%
$1H$
0%
$0.5H$
0%
$2H$
0%
$3H$

Explanation

$emf=L\dfrac { di }{ dt }$

$L=\dfrac { emf }{ (\dfrac { di }{ dt } ) }$

$=\dfrac { 60 }{ (\dfrac { 6 }{ 0.3 } ) }$

$= 3H$

Which of the following units denotes the dimensions

${M}{L}^{2}/{Q}^{2}$ , where

${Q}$ denotes the electric charge?

0%
$Weber$ ( $Wb$ )
0%
$Wb$ $/\mathrm{m}^{2}$
0%
$Henry$ ( $H$ )
0%
$\mathrm{H}/\mathrm{m}^{2}$

Explanation

Weber

$=ML^2T^{-2}I^{-1}$

$=ML^2T^{-2}Q^{-1}T=ML^2T^{-1}Q^{-1}$ (

$I=QT^{-1}$ )
Henry H is SI unit of inductance.

$H=ML^2T^{-2}I^{-2}$ also

$I=QT^{-1}$
so

$H=ML^2T^{-2}Q^{-2}T^2=ML^2Q^{-2}$

An equilateral triangular loop ADC of some finite magnetic field

$\vec{B}$ as shown in
the figure. At time t = 0, side DC of loop is at edge of the magnetic field. Magnetic
field is perpendicular to the paper inwards (or perpendicular to the plane of the
coil). The induced current versus time graph will be as

Assertion : When two coils are wound on each other, the mutual induction between the coils is maximum.
Reason : Mutual induction does not depend on the orientation of the coils.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true

Assertion : When number of turns in a coil is doubled, coefficient of self inductance of the coil becomes 4 times.
Reason : This is because

$L\propto N^{2}$ .

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true

Explanation

Both are true

$L\ \alpha \ N^2$

The area of a coil is

$500cm^{2}$ and the number of turns in it is

$2000$ . It is kept perpendicular to a magnetic field of induction

$4 \times10^{-5}Wb /m^{2}$ . The coil is rotated through

$180$ in

$0.1$ second. If the resistance of the total circuit is

$20\Omega$ , then the value of the induced charge flowing in the circuit will be :

0%
$1 \times10^{-4}C$
0%
$2 \times10^{-4}C$
0%
$3 \times10^{-4}C$
0%
$4 \times10^{-4}C$

Explanation

Answer is B

Initial flux through the coil,

$= BA \cos \theta$

$= BA \cos 0^o$ , as the coil is perpendicular to magnetic field

$= BA \ \ Wb$

Final flux after the rotation,

$f = BA \cos 180^o$ , the coil is rotated through

$180^o$

$= -BA \ \ Wb$

Therefore, estimated value of the induced emf (

$E$ ) as per Faraday's law is,

$E = \dfrac{N}{t} = \dfrac{N (BA-(-BA))}{ t} = \dfrac{2NBA}{ t}$ -------(i)

Further,

$E= IR= \dfrac{QR}{ t}$ -------(ii)

From (i) & (ii)

$Q=\dfrac{2NBA}{R} = \dfrac{2\times2000\times4\times10^{-5}\times500\times10^{-4}}{20} = 4\times10^{-4}C$

A long straight wire is placed along the axis of a circular ring of radius R. The mutual inductance of this system is

0%
$\frac{\mu _{0}R}{2}$
0%
$\frac{\mu _{0}\pi R}{2}$
0%
$\frac{\mu _{0}}{2}$
0%
$0$

Explanation

Magnetic field due to wire at the ring is

$\begin{array}{l} B=\frac { { \mu oi } }{ { 4\pi R } } ,\therefore then\, \, \emptyset = \\ \emptyset =\overrightarrow { B } .\overrightarrow { A } \Rightarrow \left| B \right| \left| A \right| \cos \theta \\ \theta =90^{ \circ } \\ \emptyset =O,as\cos 90^{ \circ }=0 \\ \therefore mutual\, \, induction=0 \end{array}$

Select correct alternative :
Statement -1 : The COM of a rod is at its geometrical centre. This means the rod is uniform.
Statement -2 : The total potential energy of an electrostatic system may be positive.
Statement -3 : The line integral of electric field over a closed loop drawn in a region having only a time varying magnetic field is zero.

0%
FFF
0%
FTF
0%
TFF
0%
TTT

Which of the following statements is/are correct regarding the electric field produced by changing magnetic field due to the motion of charge q?

0%
It is non-conservative in nature
0%
It is non-electrostatic in nature
0%
Potential can be defined corresponding to this field
0%
The lines of this field are closed curves

Two ends of a conducting rod of varying cross-section are maintained at

$200^\circ C$ and

$0^\circ C$ respectively, figure. In steady state

0%
Temperature difference across AB and CD are equal
0%
Temperature difference across AB is greater than that of CD
0%
Temperature difference across AB is less than that across CD
0%
Temperature difference may be equal of different depending on the thermal conductivity of the rod

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 2 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 2 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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