MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 3

SI unit of electrical energy is

0%
Joule
0%
Kilowatt hour
0%
KiloJoule
0%
Watt hour

Explanation

SI unit of electrical energy is Joule.

$1 Joule = 1 Volt \times 1 ampere \times 1 second$

When does two elements are said to be in series?

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When same current physically flows through both elements
0%
When different current physically flows through both elements
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When no current physically flows through both elements
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None of these

Electromotive Force (EMF) may be defined as:

0%
work done per coulomb on the charge.
0%
drift velocity of electrons.
0%
flow of electron.
0%
the number of coulombs of charge per second.

Consider the following two statements:
1) Kirchoff's junction law follows from the conservation of charge
2) Kirchoff's loop law follows from the conservation of energy.
Which of the following is correct?

0%
both 1 and 2 are wrong
0%
1 is correct and 2 is wrong
0%
1 is wrong and 2 is correct
0%
both 1 and 2 are correct

Which of the following is not a correct statement with respect to the given circuit?

0%
Switch $S$ is closed
0%
Current is flowing in the circuit
0%
$B$ and $C$ are connected in series
0%
$A$ is connected in series to $B$ and $C$

Your are given three equal resistances. In how many combinations can they be arranged?

0%
Three
0%
Four
0%
Five
0%
Two

Which of the following component is connected in parallel in the circuit?

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Rheostat
0%
Capacitor
0%
Bulb
0%
LED

Which of the following methods does not help to minimize the error due to contact resistance in Wheatstone's meter bridge experiment:

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The wire used must be uniform
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The wire used must be non-uniform
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Null point is obtained near the middle of the wire
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Interchange the positions of unknown resistance (X) and known resistance(R)

For a metallic wire, the ratio

$\displaystyle \frac{V}{i}$ (where, V = applied potential difference and i = current flowing)

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is independent of temperature
0%
increases as the temperature rises
0%
decreases as the temperature rises
0%
increases or decreases as temperature rises depending upon the metal.

Kirchhoff's law of junction,

$\displaystyle \sum { I } =0$ , is based on

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Conservation of energy
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Conservation of charge
0%
Conservation of energy as well as charge
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Conservation of momentum

Potentiometer measures potential more accurately because

0%
it measures potential in the open circuit .
0%
it uses sensitive galvanometer for null deflection .
0%
it uses high resistance potentiometer wire .
0%
it measures potential in the closed circuit .

Explanation

A potentiometer has very large resistance. To read load voltage or voltage across any circuit element, potentiometer is connected parallel across it. Thus, the load voltage or circuit voltage gets applied across the potentiometer but as the resistance is very high compared to load resistance, you can safely assume their is almost no current flowing through potentiometer. As power=

$VI$ , the power loss in potentiometer is almost equal to zero as current through it remains zero.

Therefore it measures the potential in the open circuit.

A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is /are

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current , electric current and drift velocity
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drift speed only
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current and drift speed only
0%
current only

Explanation

Current does not depends on the area of conductor , hence it remains a constant.

Current density is inversely a constant.

Current density is inversely proportional to area (i.e.,)

$J\propto \dfrac 1A$ electric field drift speed is also inversely proportional to area (i.e.,)

$E\propto\dfrac 1A , V_d\propto \dfrac 1A$

Hence current is constant along the conductor.

Which of the following statements are true?

0%
During electrolysis, charge flows through electrolytic solution via electrons
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The randomly moving electrons in a metal wire will start moving in a particular direction when a potential difference is applied across it
0%
A negatively charged particle has higher electric potential than a positively charged particle
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Charge flows only through negative charge carriers like electrons

State whether given statement is True or False
In conductors, the innermost electrons are free to move :

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True
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False

What happens when negative terminal of cell is connected to other negative terminal of cell in a particular circuit?

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Current will not flow in circuit
0%
Current will flow in circuit
0%
temperature of cell will increase
0%
cell gets damaged

Explanation

$1.$ Current means flow of electrons and electrons are negative charged and so are attracted to the positive end of the battery and repelled by the negative end.

$2.$ When battery is connected in a circuit that lets the electron flow through it, they flow from negative to positive.

$3.$ When negative terminal of cell is connected to other negative terminal of the cell in a particular circuit then, current will not flow in circuit as electrons cannot flow from negative to negative terminal.

Kirchoff's junction law is equivalent to ___________.

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Conservation of energy
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Conservation of charge
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Conservation of electric potential
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Conservation of electric flux

Materials that allow electrons to flow through them are known as

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Insulator
0%
Conductors
0%
Electrolytes
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Bases

State whether given statement is True or False
The wire of potentiometer is made up of Aluminium.

0%
True
0%
False

Why is electrical wiring usually made from copper?

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Because copper is shiny.
0%
Because copper conducts electricity.
0%
Because copper is not magnetic
0%
none of these

Which of the following has negative temperature coefficient of resistance?

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Copper
0%
Aluminium
0%
Iron
0%
Germanium

A potentiometer can be used to measure

0%
emf of newly desined cell
0%
internal resistance of a cell
0%
unknown resistance
0%
potential difference across a resistor

Identify the correct way to connect cells to make battery.

0%
0%
0%
0%
None of the above

If the length of potentiometer wire is increased, then the length of the previously obtained balance point will

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Increase
0%
Decrease
0%
Remains unchanged
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Becomes two times

The value of

$I$ in the figure shown below is

0%
$19 A$
0%
$21 A$
0%
$8 A$
0%
$4 A$

Explanation

Krichhoff's 1st law states that the total current flowing towards a junction is equal to the total current flowing away from that junction.

At junction a :

$I_1 = 20 - 5 = 15$ A

At junction b :

$I_3=I_1 + 4$

$\therefore$

$I_3 =15 + 4 = 19$ A

At junction c :

$I_2 = 5-3 =2$ A

At junction d :

$I_4 = I_3 + I_2$

$\therefore$

$I_4 = 19+2 =21$ A

A cell of emf

$6\ V$ and resistance

$0.5\ ohm$ is short circuited. The current in the cell is

0%
$3\ amp$
0%
$12\ amp$
0%
$24\ amp$
0%
$6\ amp$

Explanation

$i=\dfrac Er=\dfrac{6}{0.5}=12\ amp$

The current in a car headlamp is

$3.0$ A when connected to a

$12$ V battery. What is the resistance of the lamp when it is lit?

0%
$0.25\Omega$
0%
$4.0\Omega$
0%
$15\Omega$
0%
$36\Omega$

Explanation

Given that,

Current in a car headlamp ,

$I=3.0\ A$

Voltage applied across it ,

$V= 12\ V$

Resistance of the lamp ,

$R=\dfrac VI= \dfrac{12}{3}= 4.0\ \Omega$

Write true or false for the following statements :
Electric power is also called BOTU.

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True
0%
False

The given circuit is part of a certain circuit. The current through resistors are shown. The potential difference

$V_P-V_Q$ is

0%
$-2V$
0%
$19V$
0%
$22V$
0%
$-19V$

The Kirchhoff's first law (

$\sum$ i=0) and second law (

$\sum$ iR=

$\sum$ E) , where the symbols have their usual meanings, are respectively based on :

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Conservation of charge, conservation of momentum
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Conservation of energy, conservation of charge
0%
Conservation of momentum, conservation of charge
0%
Conservation of charge, conservation of energy

A wire has a diameter of

$0.2 mm$ and a length of

$50 cm$ . The specific resistance of it's material is

$40\times 10^{-6}$

$ohm cm$ . The current through it, when a potential difference of

$2 V$ is applied across it, is

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$3.14 A$
0%
$31.4 A$
0%
$0.314 A$
0%
$0.0314 A$

Explanation

Given:

$l = 50$ cm,

$\rho = 40 \times 10^{-6} \Omega$ cm,

$r = 0.1$ mm

$= 0.01$ cm
The resistance R is given by,

$R = \dfrac{\rho l}{A} = \dfrac{40 \times 10^{-6} \times 50}{\pi \times (0.01)^2} \approx 6.37 \Omega$
Now the current through it when a potential difference of 2V is applied across it is:

$i=\dfrac{V}{R} = \dfrac{2}{6.37} =0.314$ A.

In a potentiometer whose wir e resistance is 10

$\Omega$ the potential fall per cm is V volts. To reduce it to V/4 Volt cm

$^{-1}$ , the resistance that must be connected in series with the potentiometer wire is

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40 $\Omega$
0%
30 $\Omega$
0%
20 $\Omega$
0%
10 $\Omega$

Explanation

Let the length and resistance of the wire be L and

$10 \Omega$

$\dfrac{\Delta V}{l} = \dfrac{V}{1}$

$\Delta V = Vl$

Emf of cell

$= Vl/4$

Later

$\dfrac{\Delta V}{l} = \dfrac{V}{4}$

of four parts of potential, 1 part is across wire and other 3 parts are at R.

R must be 3 times of Resistance of wire

$R = 3 \times 10 = 30 \Omega$

The resistance of a wire is 5 ohm at

$50^{\circ}$ C and 6 ohm at

$100^{\circ}$ C. The resistance of the wire at

$0^{\circ}$ C will be:

0%
2 ohm
0%
1 ohm
0%
4 ohm
0%
3 ohm

Explanation

$R={ R }_{ 0 }(1+\alpha \Delta T)$

$R$ at

${ 100 }^{ 0} C=6\Omega$

$R$ at

${ 50}^{ 0 } c =5\Omega$

$\Rightarrow \dfrac{5}{(1+\alpha (50))}=\dfrac{6}{(1+\alpha (100))}$

$\Rightarrow 6(1+\alpha (50))=5(1+\alpha (100))$

$\Rightarrow \alpha =\dfrac { 1 }{ 200}$

$R={ R }_{0 } (1+\alpha \Delta T)$

$R$ at

${ 50 }^{ 0 } C =5\Omega$

${ R }_{0 }$ at

${ 0}^{ 0 } C ={ R }_{ 0}$

$\Rightarrow R={R }_{0 }(1+\dfrac { 1}{ 200} \times 50)$

$\Rightarrow 5={R }_{0 }(1+\dfrac { 1}{ 200} \times 50)$

$\Rightarrow R_0 = 4\Omega$

A carbon filament has resistance of 120

$\Omega$ at

$0^{0}C$ . The resistance of a copper filament connected in series with carbon so that the combination has same resistance at all temperatures must be

$\alpha$ of carbon

$= -7\times 10^{-4}/^{0}C$ ;

$\alpha$ of copper

$=4\times 10^{-3}/^{0}C$

0%
120 $\Omega$
0%
21 $\Omega$
0%
60 $\Omega$
0%
210 $\Omega$

Explanation

$R_{ca} = R^{0}_{ca}(1+\alpha_{ca} \Delta T)$

$R_{cu} = R^{0}_{cu}(1+\alpha_{cu} \Delta T)$

$R_{ca} + R_{cu} = R^{0}_{ca} + R^{0}_{cu} + \Delta T(R^{0}_{ca}\alpha_{ca} + R^{0}_{cu}\alpha_{cu})$

$\Rightarrow R_{ca}+R_{cu}$ is independent of temperature

$\Rightarrow R^{0}_{ca}\alpha_{ca} + R^{0}_{cu}\alpha_{cu} = 0$

$120 \times 7\times 10^{-4} = R^{0}_{cu} \times 4 \times 10^{-3}$

$R^{0}_{cu} = 21 \Omega$

$n$ identical cells, each of internal resistance

$(r)$ are first connected in parallel and then connected in series across a resistance

$( R)$ . If the current through

$R$ is same in both the cases then:

0%
$R = r/2$
0%
$r = R/2$
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$R = r$
0%
$r = 0$

Explanation

If the emf of each cell be E, then for parallel connection, equivalent emf is

$E_{eq}= \dfrac{\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}+....}{\dfrac{1}{r_1}+\dfrac{1}{r_2}+....}= \dfrac{\dfrac{E}{r}+\dfrac{E}{r}+....}{\dfrac{1}{r}+\dfrac{1}{r}+.....}=E$

And equivalent internal resistance of the cells,

$r_{eq}=\dfrac{r}{n}$

So net current in parallel connection is

$I_p=\dfrac{E_{eq}}{r_{eq}+R}=\dfrac{E}{\dfrac{r}{n}+R}$

And for series connection of cells, eqv emf ,

$E_{eq}= E_1+E_2+...=nE$

and equivalent internal resistance for series,

$r_{eq}= r_1+r_2+....=nr$

So current ,

$I_s=\dfrac{E_{eq}}{r_{eq}+R}=\dfrac{nE}{nr+R}$

Now according to question, both the currents are equal.

So,

$I_p=I_s$ reduces to

$\dfrac{E}{\dfrac{r}{n}+R}=\dfrac{nE}{nr+R}$ which gives

$R=r$

Assertion : The equivalent resistance between the points X and Y in the figure is 10.
Reason : According to Wheatstone bridge, points A and C have the same potential.

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Both (A) and (R) are true and (R) is the correct explanation of A
0%
Both (A) and (R) are true but (R) is not the correct explanation of A
0%
(A) is true but (R) is false
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(A) is false but (R) is true

Explanation

Simplified circuit:

${ R }_{xy }=\dfrac {20 \times 20}{2 \times 20} =10\Omega$

According to Wheatstone condition, A and C are of same potential.

But solving for

${ R }_{xy }$ and Wheatstone condition are not related.

The resistance of a bulb filament is 100

$\Omega$ at a temperature of 100

$^{0}$ C. If its temperature coefficient of resistance be 0.005 per

$^{0}$ C , its resistance will become 200

$\Omega$ at a temperature of :

0%
300 $^{0}$ C
0%
400 $^{0}$ C
0%
500 $^{0}$ C
0%
200 $^{0}$ C

Explanation

Given

${ R }_{ 0} = 100\ \Omega\ at\ { 100}^{ 0 }c.$

$\alpha =0.005 ; R=200\ \Omega\ at\ T=?$

$\Rightarrow (R-{ R }_{ 0 })={ R }_{ 0 }\alpha (T-{ T }_{ 0 })$

$\Rightarrow 100=100\times \dfrac { 5 }{ 1000 } (T-{ T }_{ 0 })$

$\Rightarrow T=200+{ T }_{ 0 }={ 300 }^{ 0 }c$

In a potentiometer of ten wires, the balance point is obtained on the sixth wire. To shift the balance point to eighth wire, we should

0%
increase resistance in the primary circuit.
0%
decrease resistance in the primary circuit.
0%
decrease resistance in series with the cell whose e.m.f. has to be measured.
0%
increase resistance in series with the cell whose e.m.f. has to be measure.

Explanation

By increasing resistance in the primary circuit the P.G of the wire decreases thereby it requires more length to measure the emf of secondary cell.

$\therefore$ By increasing the resistance in series circuit we can move balancing point from sixth to eight wire.

A potentiometer wire 10 m long has a resistance of 40

$\Omega$ . It is connected in series with a resistance box and a 2V storage cell. If the potential gradient along the wire is 0.1 mV/cm , the resistance in the box is

0%
760 $\Omega$
0%
260 $\Omega$
0%
1060 $\Omega$
0%
960 $\Omega$

Explanation

$= 0.1 \dfrac {mV}{cm}$

$\Rightarrow \dfrac {\triangle V}{l} = \dfrac {0.1 \times 10^{-3}}{1 \times 10^{-2}}$

$\triangle V = l \times 0.01 = 0.1$

$\Rightarrow \dfrac {0.1}{2 - 0.1} = \dfrac {40}{R}$

$\Rightarrow R = 760 \Omega$

The current that passes through 20 ohm resistance when it is connected in parallel with a 30 ohm resistance and this set is connected to a battery of 2V is

0%
0.2A
0%
0.3A
0%
0.1A
0%
0.016A

Explanation

Given,

$R_1=20\Omega\\R_2=30\Omega\\E=2\ V$

According to question, both resistance are in parallel and it is connected with

$2\ V$ battery.

Potential difference across each resistance

$2\Omega$

Current in

$20\Omega$ resistance

$I_{20\Omega}=\dfrac{\Delta V}{R}=\dfrac{2}{20}=0.1\ A$

The temperature coefficient resistivity of a material is

$0.0004/K$ . When the temperature of the material is increased by

$50^{o}C$ , its resistivity increases by

$2\times 10^{-8}\ ohm-meter$ .The initial resistivity of the material of the resistance

0%
$50\times 10^{-8}$
0%
$90\times 10^{-8}$
0%
$100\times 10^{-8}$
0%
$200\times 10^{-8}$

Explanation

$(\rho-\rho_{0}) = \rho_{0} \propto \Delta T$ (using standard result)

$\Rightarrow 2 \times10^{-8} = \rho_{0} \times4 \times 10^{-4} \times50$

$\Rightarrow 2\times 10^{-8} = \rho_{0}\times2 \times 10^{-2}$

$\Rightarrow \rho_{0} = 100\times10^{-8} ohm-meter$

The potentiometer is more appropriate for measuring potential difference than a voltmeter because

0%
the resistance of voltmeter is high
0%
the sensitivity of a potentiometer is higher than that of voltmeter
0%
the resistance of potentiometer wire is very low
0%
the potentiometer does not draw any current from the unknown source of emf

Resistance of a resistor at temperature

$t^{0}C$ is

$R_{t}=R_{0}(1+\alpha t+\beta t^{2})$ . Here

$R_{0}$ is the resistance at

$0^{0}C$ . The temperature coefficient of resistance at temperature

$t^{0}C$ is

0%
$\dfrac{(1+\alpha t+\beta t^{2})}{\alpha +2\beta t}$
0%
$\alpha +2\beta t$
0%
$\dfrac{\alpha +2\beta t}{(1+\alpha t+\beta t^{2})}$
0%
$\dfrac{\alpha +2\beta t}{2(1+\alpha t+\beta t^{2})}$

Explanation

Resistance of a resistor at temperature

$t^{0}C$ is

$R_{t}=R_{0}(1+\alpha t+\beta t^{2})$ .

The temperature coefficient of resistance is given by:

Temperature coefficient of resistance

$= \dfrac{1}{R_t} \dfrac{dR}{dt}$ .

Temperature coefficient of resistance

$=\dfrac{R_0}{R_0(1+\alpha t+\beta t^{2})} \dfrac {d}{dt}(1+\alpha t+\beta t^{2})$

Temperature coefficient of resistance

$=\dfrac{\alpha +2\beta t}{(1+\alpha t+\beta t^{2})}$ .

Four conductors of resistance 4, 3, 9 and 6 ohm are connected in AB,BC, CD and DA arms of a Wheatstone bridge. The bridge can be balanced by connecting.

0%
1) 6 ohm in series with 3 ohm conductor
0%
2) 4 ohm in parallel with 6 ohm conductor
0%
3) 3 ohm in series with 3 ohm conductor
0%
4) 5 ohm in series with 6 ohm conductor

Explanation

To make it a wheatstone bridge

$\dfrac { { R }_{ AB } }{ { R }_{ AD } } =\dfrac { { R }_{ BC } }{ { R }_{ CD } }$

Verification of options.

$\Rightarrow { R }_{ BC }=9\Omega$

$\Rightarrow \dfrac { 2 }{ 3 } \neq 1$

$\Rightarrow { R }_{ AD }=\dfrac { 6\times 4 }{ 10 } =2-4$

$\Rightarrow \dfrac { 1 }{ 0.6} \neq 1/3$

$\Rightarrow { R }_{ BC }=6$

$\Rightarrow \dfrac { 2 }{ 3 } =\dfrac { 2 }{ 3 }$

$=$ 2/3

$\therefore$ Option C is correct

Aluminium

$\left ( \alpha =4\times 10^{-3}K^{-1} \right )$ resistance of 60

$\Omega$ and carbon

$\left ( \alpha =0.5\times 10^{-3}K^{-1} \right )$ resistance 40

$\Omega$ are connected in parallel. The combination is heated. The effective resistance is

0%
Greater than 24 $\Omega$
0%
Less than 24 $\Omega$
0%
Greater than 40 $\Omega$
0%
Greater than 100 $\Omega$

Explanation

$R_{eff} = \dfrac{R_{A} R_{C}}{R_{A}+R_{C}}$

Before heating

$= \dfrac{60 \times 40}{100} = 24 \Omega$

$\dfrac{\Delta R_{eff}}{R_{eff}} = \dfrac{-\Delta R_{A}}{R_{A}}- \dfrac{\Delta R_{B}}{R_{B}}$

$\alpha_{eff} \Delta T = \Delta T(-\alpha_{A}-\alpha_{B})$

$\Rightarrow \alpha_{eff} = -(\alpha_{A}+\alpha_{B})$

If in a Wheatstone bridge the battery and Galvanometer are interchanged, the condition for balance

0%
is disturbed
0%
is not disturbed
0%
depends on the internal resistance of the bridge
0%
depends on the values of the resistances in the bridge

Explanation

Case 1

$\dfrac{R_{1}}{R_{2}} = \dfrac{R_{3}}{R_{4}}$

Case 3

Simplified

$\dfrac{R_{1}}{R_{3}} = \dfrac{R_{2}}{R_{4}}$

$\dfrac{R_{1}}{R_{2}} = \dfrac{R_{3}}{R_{4}}$

Condition hasn't changed.

The resistance of a metallic conductor increases with temperature due to:

0%
change in carrier density
0%
change in dimension of the conductor
0%
increase in the number of collisions among the carriers
0%
increase in the rate of collisions between the carriers and the vibrating atoms of the conductor.

Consider a thin square sheet of side

$L$ and thickness

$t$ , made of a material of resistivity

$\rho$ . The resistance between two opposite faces, shown by the shaded areas in the figure is____

0%
directly proportional to L
0%
directly proportional to t
0%
independent of L
0%
independent of t

Explanation

$R= \rho \dfrac{l}{a}$

Therefore

$R=\rho \dfrac{l}{l\times t}=\dfrac{\rho}{t}$

Consider a long conductor, the middle of which is earthed. If the potential difference across the two ends of the conductor is

$220\ V$ , then what is the potential at the ends and the middle point?

0%
$220\ V$ all over the conductor
0%
$110\ V$ and $-110\ V$ at ends and $0\ V$ at the mid point
0%
$0\ V$ at ends and $220\ V$ at midpoint
0%
$-220\ V$ all over the conductor

Explanation

Let the potential at the ends be

$V_1\,and\, V_2$ .

Since the middle point is earthed, its potential needs to be zero.

Also,

$V_1 - V_2 = 220$

$\Rightarrow V_1 = 110\ V \,and\, V_2 = -110\ V$

The length of a potentiometer wire is l. A cell of emf E is balanced at a length L/3 from the
positive end of the wire. If the length of the wire is increased by L/At what distance will the
same cell give a balance point.

0%
$\dfrac{2L}{3}$
0%
$\dfrac{L}{2}$
0%
$\dfrac{L}{6}$
0%
$\dfrac{4L}{3}$

A wheel having mass m has charges

$+q$ and

$q$ on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E

$=$

0%
$\dfrac{\mu g}{q}$
0%
$\dfrac{\mu g}{2q}$
0%
$\dfrac{\mu g\tan\theta }{2q}$
0%
None of these

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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