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CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 10 - MCQExams.com
CBSE
Class 12 Medical Physics
Current Electricity
Quiz 10
Two cells of e.m.f. $$ E_1 = 8$$V and $$E_2 = 4V, $$ with internal resistance 1 ohm and 2 ohm respectively are connected so that they oppose each other. this combination of cells is connected to an external resistance of 5 ohm. the terminal potential difference across the cell $$ E_2 $$ is :
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0%
$$7.5V$$
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$$3.0V$$
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$$4.0V$$
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$$5.0V$$
The potential difference $$V$$ and the current $$I$$ flowing through an instrument in an ac circuit of frequency f are given by V = $$5$$ cos $$\omega t$$ volts and $$I =$$2$$ sin \omega t$$ amperes (where $$\omega=2\pi f$$). The power dissipated in the instrument is :-
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0%
Zero
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$$10 W$$
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$$5 W$$
0%
$$2.5 W$$
Explanation
Given,
$$I=2\,sin\,\omega t$$
$$V=5\,cos\,\omega t=5\,sin\,(\omega t+\dfrac{\pi}{2})$$
Power $$=V_{rms}\times I_{rms}\times cos\phi=0$$
$$($$ Since $$\phi=\dfrac{\pi}{2}$$, therefore, $$cos\phi=cos\dfrac{\pi}{2}=0$$ $$)$$
The diagram shows a circuit with two identical resistors. The battery has negligible resistance. Voltmeter and ammeter are ideal. It switch S is closed then which of the following statement is incorrect?
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Power dissipated across R in left branch increases
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Power dissipated across R in right branch increases
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Ammeter reading increases
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Voltmeter reading decreases
A source of e.m.f. E= 15 v and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increase with time as I = 1.2t +Then total charge will flow in first seconds will be:
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10C
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20C
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30C
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40C
Explanation
We knwo that $$i=\dfrac {dQ}{dt}$$
According to question:
$$\Rightarrow dQ=idt\Rightarrow Q=\displaystyle \int_{r_1}^{r_2}idt =\displaystyle \int_{0}^{5}(1.2t+3)dt$$
$$=\left[\dfrac {1.2t^2}{2}+3t \right]_0^5 =30\ C$$
Kirchhoff's first law at a junction is based on conservation of
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Angular momentum
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Mass
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Charge
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Energy
Kirchhoff's Loop rule is a direct consequence of law of conservation of
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Charge
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Momentum
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Angular momentum
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Energy
In the circuit shown in the figure below, the magnitude and direction of the current will be
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$$2.3 \ Amp$$ from A to B via E
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$$2.3 \ Amp$$
from B to A via E
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$$1.0 \ Amp$$ from B to A via E
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$$1.0 \ Amp$$ from A to B via E
Explanation
The voltage of the battery $$V_1=10 \ V$$ and that of the battery $$V_2=4 \ V$$ get added as they are connected in series with same polarity.
Net emf $$V_{net}=10\,V+4\,V=14\,V$$
The resistances $$1 \Omega, 3 \Omega$$ and $$2 \Omega$$ are connected in series.
Total resistance $$R_{series}=1+3+2=6\Omega$$
By Ohm's law,
$$V_{net}=IR_{series}$$
$$\Rightarrow I=\dfrac{V_{net}}{R_{series}}$$
Hence, Current $$I=\dfrac {14}{6}=2.3\,A$$
For the direction of current, we can see net emf is in the same sense of the individual batteries because $$10\,V $$ battery and $$4\,V $$ battery are connected back-to-back. So $$2.3\,A$$ current will flow in the direction $$A$$ to $$B$$ via $$E$$.
A lamp of $$ 600 \mathrm{W}-240 \mathrm{V} $$ is connected to 220 $$ \mathrm{V} $$ mains. Its resistance is
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0%
$$
96 \Omega
$$
0%
$$
84 \Omega
$$
0%
$$
90 \Omega
$$
0%
$$
64 \Omega
$$
When direct current passed through a spring then it :-
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0%
Contract
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Expand
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Vibrate
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Unchanged
For different values of resistance, R power consumptions in R are given. Then which of the following values are not possible?
a) 2 W b) 5 W
c) 8 W d) 4 W
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Only c
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b &c
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a,b,c
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All
The length of a potentiometer wire is $$l$$. A cell of emf E is balanced at a length $$\dfrac{l }{ 3}$$ from the positive end of the wire. If the length of the wire is increased by $$\dfrac{l} { 2}$$.
At what distance will the same cell give a balance point?
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0%
$$
\dfrac{2l}{3}
$$
0%
$$
\dfrac{l}{2}
$$
0%
$$
\dfrac{l}{6}
$$
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$$
\dfrac{4l}{3}
$$
A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns was to be quadrupled and the wire radius halved, the electric power dissipated would be
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Halved
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The same
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Doubled
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Quadrupled
Explanation
Power dissipated, $$P=\dfrac{e^2}{R}$$
where,
$$e=$$ Induced emf $$=-(\dfrac{d\phi}{dt})$$
and
$$\phi=NBA$$
Therefore,
$$e=-NA(\dfrac{dB}{dt})$$
Also,
$$R\propto \dfrac 1a\propto \dfrac{1}{r^2}$$
Therefore,
$$P=\dfrac{e^2}{R}\propto \dfrac{N^2A^2r^2}{l}\propto \dfrac{N^2r^4}{l}$$
As $$r$$ is halved, then $$N$$ is quadruples.
Therefore, $$P$$ remains the same.
Kirchhoff's first law is based on the law of conservation of.
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Charge
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Energy
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Momentum
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Sum of mass and energy
To increase sensitivity of a potentiometer, its
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area should be increased
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current should be decreased
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current should be increased
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length should be increased.
A battery is a parallel combination of cells.
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0%
True
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False
Graphs between electric current and potential difference across two conductors A and B are shown in the figure. Which of the following conductor has more resistance?
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B
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A
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Both have equal resistance
0%
None of these
Explanation
For comman voltage $$V$$ Current in conductor B is greter then currecnt in conductor A.
ie $$i_B>i_A$$ for same value of V
From ohms law,
$$R=\dfrac VI\\R\propto\dfrac1I$$
Hence,
If $$i_B>i_A$$ $$R_A>R_B$$
Option B
Four similar charges each of magnitude $$Q$$ are placed at the four comers of a square of side '$$a$$'. The intensity of the electric field at the intersection of the diagonal is
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0
0%
$$\dfrac { Q } { 4 \pi \varepsilon _ { 0 } a ^ { 2 } }$$
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$$\dfrac { 4 Q } { 4 \pi \varepsilon _ { 0 } a ^ { 2 } }$$
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$$\dfrac { Q } { 8 \pi \varepsilon _ { 0 } a ^ { 2 } }$$
Explanation
Four similar charges each of magnitude $$=Q$$
Square of side $$=a$$
intensity of electric field $$E=\dfrac { 4Q }{ 4\pi { \epsilon }_{ 0 }{ a }^{ 2 } } $$
This is the total charge $$=4Q$$.
If the radius of a potentiometer wire is increased four times, keeping its length constant, then the value of its potential gradient will become
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half
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two times
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four times
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unchanged
Calculate the value of the unknown potential $$V$$ for the given poteniometer circuit. The total length (400 cm) of the potentiometer wire has a resistance of $$10 \Omega$$ and the balance point is obtained at a length of $$240$$ cm .
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$$0.06V$$
0%
$$0.6V$$
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$$6V$$
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none of these
Explanation
$$\begin{array}{l} Resis\tan ce\, \, of\, \, 240\Omega \, \, cm\, of\, \, the\, \, wire, \\ =\dfrac { { 240 } }{ { 400 } } \times 10=6\Omega \, \\ So, \\ \dfrac { { { R_{ 2 } } } }{ { { R_{ 1 } } } } =\dfrac { V }{ 3 } \\ \Rightarrow \dfrac { 6 }{ { 290+10 } } =\dfrac { V }{ 3 } \\ \Rightarrow V=0.06V \end{array}$$
Among the following, the intensive property is
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molarity
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entropy
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resistance
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heat capacity
An infinite non-conducting sheet of charge has a surface charge density of $${ 10 }^{ -7 }C/{ m }^{ 2 }$$. The separation between two equipotential surfaces near the sheet whose potential difference by $$5V$$ is
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$$0.88cm$$
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$$08.8mm$$
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$$0.88m$$
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$$5\times { 10 }^{ -7 }m$$
Explanation
Given,
$$\sigma=10^{-7}\,C/m^2$$
$$V=5\,V$$
The electric field of the sheet, $$E=\dfrac{\sigma}{2\times \epsilon_0}$$
$$=\dfrac{1\times 10^{-7}}{2\times 8.854\times 10^{-12}}=5649.71\,C^2/Nm^2$$
The potential difference of any equipotential surface, $$V=E\Delta s$$
Then,
$$\Delta s=\dfrac VE$$
$$=\dfrac{5}{5649.72}=8.8\,mm$$
In metre bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is 'l'. If the temperature of water is increased, the new balancing length is
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>l
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none
0%
=0
0%
=l
Explanation
$$\begin{array}{l} \frac { { { R_{ unknown } } } }{ { { R_{ s\tan dard } } } } =\frac { l }{ { \left( { 1-l } \right) } } . \\ If\, \, temperature\, \, increases,\, \, resis\tan ce\, \, increases. \end{array}$$
Hence, Option $$A$$ is correct.
Find value of current in circuit diagram
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13 A
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11 A
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6 A
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14 A
In the circuit shown in the figure, the voltage across $$15\Omega$$ resistor is $$30$$V having the polarity as indicated. What is the value of R?
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$$10\Omega$$
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$$35\Omega$$
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$$17.5\Omega$$
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$$37.5\Omega$$
After switch is closed, current drawn from the battery is
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$$6A$$
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$$1.5A$$
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$$3A$$
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$$4A$$
Explanation
$$R_{eq} = 1W + \dfrac {6\times 3}{6 + 3} \Omega = 3\Omega$$
$$i = \dfrac {V}{R_{eq}} = \dfrac {9V}{3\Omega} = 3A$$.
A circuit connected to an ac source of emf $$e = e_0 \sin (100 t)$$ with t in seconds, gives a phase difference of $$\dfrac{\pi}{4}$$ between the emf e and current i. Which of the following circuits will exhibit this ?
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RC circuit with $$R = 1 k \Omega$$ and $$C = 1 \mu F$$
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RL circuit with $$R = 1 k \Omega $$ and $$L = 1 m H$$
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RL circuit with $$R = 1 k \Omega $$ and $$L = 10 mH$$
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RC circuit with $$R = 1 k \Omega $$ and $$C = 10 m F$$
Explanation
Given phase difference = $$\dfrac{\pi}{4}$$
and $$\omega = 100 rad/s$$
$$\Rightarrow $$ Reactance (X) = Resistance (R)
now by checking options
option (1)
$$R = 1000 \Omega$$ and $$X_C = \dfrac{1}{10^{-6} \times 100} = 10^4 \Omega$$
Option (2)
$$R = 10^3 \Omega$$ and $$X_L = 10^{-3} \times 100 = 10^{-1} \Omega$$
option (3)
$$R = 10^3 \Omega$$ and $$X_L = 10 \times 10^{-3} \times 100 = 1 \Omega$$
Option (4)
$$R = 10^3 \Omega$$ and $$X_C = \dfrac{1}{10 \times 10^{-6} \times 100} = 10^3 \Omega$$
Clear option (4) matches the given condition
The electric current in a wire may be calculated using the equation $$I=Anvq$$.
Which statement is not correct?
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n is the number of charge carriers per unit volume of the wire
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nA is the number of charge carriers per unit length of the wire
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q is the charge of each charge carrier
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v is the velocity of each charge carrier
Explanation
Given that,
Electric current in the wire , $$I=Anvq$$
Here $$A$$ is the cross-section of wire
$$n$$ is the number of charge carriers per unit volume of wire
$$v $$ is drift velocity of charge carriers
$$q$$ is the charge of each charge carier
Hence incorrect statement is $$(D)$$
Three equal resistors are connected in series across a source of emf together dissipate 10 walts of power.What would be the power dissipated if the same resistors are connected in parallel across the same source of emf ?
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60 watt
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90 watt
0%
100 watt
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30 watt
In measuring a resistant using metre bridge, the resistance in the gaps are interchanged to minimize error due to
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the yielding of the supports
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nonuniformity of the bridge wire
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the contact or end resistance
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Joule heating of the bridge wire.
An electrical conductor has a resistance of $$5.6k\Omega$$. A potential difference (p.d.) of $$9.0V$$ is applied
across its ends.
How many electrons pass a point in the conductor in one minute?
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$$6.0\times 10^{20}$$
0%
$$1.0\times 10^{19}$$
0%
$$6.0\times 10^{17}$$
0%
$$1.0\times 10^{16}$$
Explanation
We know that
$$V = IR$$
$$\implies I = \dfrac{V}{R}$$
$$\implies I = \dfrac{9}{5.6 \times 10^3}$$A
This means that no.of electron passes through a point in one second = $$\dfrac{9}{5.6 \times 10^3 \times 1.6 \times 10^{-19}}$$
$$\implies $$ no. of electron passes through a point in one minute = $$\dfrac{9}{5.6 \times 10^3 \times 1.6 \times 10^{-19}} \times 60$$
=$$6.0 \times 10^{17 } $$ electrons
A $$220 \ V$$ and $$100 \ W$$ lamp is connected to $$220 \ V$$ power supply. What is the power consumed?
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$$100$$W
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$$200$$W
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More than $$100$$W
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None of these
Explanation
The power consumed would be obviously $$100 \ W$$.
It is because the lamp is operated at its rated voltage $$V =220 \ V$$. So the power consumed will be equal to its power rating, i.e. $$100 \ W.$$
Let us solve it to prove it.
We know, power $$P=\dfrac{V^2}{R}$$
Putting $$P=100 \ W$$ and $$V =220 \ V $$
$$\Rightarrow R=\dfrac{(220)^2}{100}=484 \ \Omega$$
By Ohm's law,
Current in the circuit $$I=\dfrac{\text{power supply voltage V}}{\text{resistance R}}$$
$$\Rightarrow I=\dfrac{220}{484} \ A$$
Power consumed, P can be calculated as $$I^2R.$$
$$P={\left( \dfrac{220}{484}\right)}^2 \times 484=\dfrac{(220)^2}{484}=100 \ W$$
Hence, option (A) is correct.
The V-l graph for a conductor at temperature $$T_1$$ and $$T_2$$ are as shown in the figure, $$(T_2-T_1)$$ is proportional to?
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$$\dfrac{\cos 2\theta}{\sin^2\theta}$$
0%
$$\dfrac{\sin 2\theta}{\sin^2\theta}$$
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$$\dfrac{\cot 2\theta}{\sin^2\theta}$$
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$$\dfrac{\tan 2\theta}{\sin^2\theta}$$
Explanation
If the number of free electrons is $$5\times 10^{28}m^{-3}$$ then the drift velocity of electron in a conductor of area of cross-section $$10^{-4}m^2$$ for a current of $$1.2$$A is?
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$$1.5\times 10^{-2}$$ m/s
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$$1.5\times 10^{-3}$$ m/s
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$$1.5\times 10^{-4}$$ m/s
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$$1.5\times 10^{-6}$$ m/s
A copper wire with a cross-section area of $$2\times 10^{-6}m^2$$ has a free electon density equal to $$5\times 10^{22}/cm^2$$. If this wire carries a current of $$16A$$, the drift velocity of the electron is?
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$$1 m/s$$
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$$0.1 m/s$$
0%
$$0.01 m/s$$
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$$0.001 m/s$$
0%
$$0.00001 m/s$$
Explanation
Drift velocity, $$V_d=\dfrac{1}{n\in A}$$
Given, $$n=5\times 10^{22}/cm^3$$
$$=5\times 10^{22}\times 10^6/m^3$$
$$e=1.6\times 10^{-19}C$$
$$A=2\times 10^6m^2$$
$$I=16A$$
$$v_d=\dfrac{16}{5\times 10^{28}\times 16\times 10^{-19}\times 2\times 10^{-6}}$$
$$=\dfrac{1}{10^3}=0.001 m/s$$.
In the circuit shown, the current in $$3\Omega$$ resistance is?
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$$1$$A
0%
$$1/7$$A
0%
$$5/7$$A
0%
$$15/7$$A
Explanation
We know,
The voltage of ground is considered to be 0V,
Let the grounds be named as 1,2,3 respectively,
And let the current in $$1^{st}$$ $$2\Omega$$ be i, and the $$3\Omega$$ be $$i_1$$,
Applying Kirchhoff Voltage law , from ground 1 to ground 2,
$$0+10-2i-2(i-i_1)=0$$
$$2i-i_1=5$$ $$-(i)$$
Applying KVL from ground 1 to ground 3,
$$0+10-2i-3i_1-3i_1=0$$
$$2i+6i_1=10$$ $$-(ii)$$
Subtracting $$(i)$$ from $$(ii)$$,
$$7i_1=5$$
$$i_1=\dfrac{5}{7} A$$
Option $$\textbf C$$ is the correct answer.
Conductivity of water (in $$\Omega ^{-1}\,cm^{-1}$$) is
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0.4
0%
0.04
0%
0.0004
0%
0.00004
A resistance R carries a current I. The heat loss to the surroundings is $$\lambda(T-T_0)$$, where $$\lambda$$ is a constant, T is the temperature of the resistance and $$T_0$$ is the temperature of the atmosphere. If the coefficient of linear expansion is $$\alpha$$, the strain in the resistance is?
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0%
Proportional to the length of the resistance wire
0%
Equal to $$\dfrac{\alpha}{\lambda}I^2R$$
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Equal to $$\dfrac{1}{2}\dfrac{\alpha}{\lambda}I^2R$$
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Equal to $$\alpha \lambda(IR)$$
The power consumed by $$4$$V battery in the circuit as shown is?
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$$8$$W
0%
$$7$$W
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$$6$$W
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$$5$$W
Explanation
$$E_{\text{net}} = 10 - 4 = 6\ V $$
$$R = 3\ \Omega $$
$$i = \dfrac{V}{R} = \dfrac{6}{3} = 2A $$
$$\therefore $$ Power consumed by $$4V$$ battery
$$= VI = 4 \times 2 $$
$$ = 8 VA = 8W$$
In a meter bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If $$ X< Y $$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4 X against Y
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50 cm
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80 cm
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40 cm
0%
70 cm
Explanation
Case: 1
$$\dfrac{X}{Y}=\dfrac{20}{80}=\dfrac{1}{4}$$
or $$Y=4 X$$
Case: 2
When X become 4X
$$\dfrac{4X}{Y}=\dfrac{l}{100-l}$$
or $$\dfrac{4X}{4X}=\dfrac{l}{100-l}$$
or $$l=50$$ cm
When the switch is closed, the initial current through the $$1\Omega$$ resistor is?
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0%
$$12$$A
0%
$$4$$A
0%
$$3$$A
0%
$$\dfrac{10}{7}$$A
The temperature coefficient of resistance of a wire is $$0.00125$$ per $$^{0}C$$. At $$300 K$$, its resistance is $$1 \ ohm$$. This resistance of the wire will be $$2 \ ohm$$ at
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0%
$$1154 K$$
0%
$$1100 K$$
0%
$$1400 K$$
0%
$$1127 K$$
Explanation
Given : $$\alpha = 0.00125/^oC$$
Using $$R_T = R_{273} (1 + \alpha (T- 273) )$$
For $$T = 300$$, $$R_{300} = 1\Omega$$
$$\therefore$$
$$R_{300} = R_{273} (1 + \alpha (300- 273) )$$
OR
$$1 = R_{273} (1 + 0.00125 (300- 273) )$$ $$\implies R_{273} = 0.967\Omega$$
Let at temperature be $$T$$ when $$R_T = 2\Omega$$
$$\therefore$$
$$R_T = R_{273} (1 + \alpha (T- 273) )$$
OR
$$2 = 0.967(1 + 0.00125 (T- 273) )$$
OR $$T- 273 = 854.6$$
$$\implies$$ $$T = 273 + 854.6 = 1127.6 K$$
When temperature of the conductor is increased, then the collision frequency between current carries ( electrons ) and atoms increases. This results in
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decreased in the resistance of the conductor
0%
increase in the resistance of the conductor
0%
no change in the resistance
0%
decrease or increase in the resistance depending upon the type of the conductor
Explanation
When temperature increases,collision frequency increases, it means more number of collisions per time. Time will decrease the relaxation time. But $$R \propto \dfrac{1}{\tau }$$, so $$R$$ increases.
If two identical heaters each rated as ($$1000$$W-$$220$$V) are connected in parallel to $$220$$V, then the total power consumed is?
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0%
$$200$$W
0%
$$2500$$W
0%
$$250$$W
0%
$$2000$$W
Statement I: In a meter-bridge experiment, the null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance.
Statement II: Resistance of a metal increases with the increase in temperature.
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Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I.
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Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.
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Statement I is true, Statement II is false.
0%
Statement I is false, Statement II is true.
Express which of the following steups can be used to verify Ohm's law?
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0%
0%
0%
Explanation
In ohm's law, we check $$V = IR$$ where I is the current flowing through a resister and V is the potential difference across that resistor. Only option (a) fits the criteria. Remember that ammeter is connected in series with resistance and voltmeter in parallel with the resistance.
To verify Ohm's law, a student is provided with a test resistor $$R_T$$ a high resistance $$R_1$$, a small resistance $$R_2,$$ two identical galvanometer $$G_1$$ and $$G_2$$ and a variable voltage source V. The correct circuit to carry out the experiment is
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0%
0%
0%
0%
Which one of the following is not the unit of energy?
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joule
0%
newton meter
0%
kilowatt
0%
kilowatt hour
Explanation
$$(c)$$ kilowatt
Kirchhoff s junction rule is a reflection of
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conservation of current density vector.
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conservation of charge.
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the fact that the momentum with which a charged particle approaches a junction is
unchanged (as a vector) as the charged particle leaves the junction.
0%
the fact that there is no accumulation of charges at a junction.
Explanation
According to junction rule, the algebraic sum of current or charge flowing per unit time towards a junction in an electric network is zero, i.e., the law of conservation of charge verifies answer (b) and no any charges accumulate at junction as the sum of entering and out
going charge are equal, at any time interval. It verifies answer (d).
S.I. unit and commercial unit of electrical unit of electrical energy are same.
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0%
True
0%
False
Explanation
S.I. unit and commercial unit of electrical unit of electrical energy are $$joules$$ and $$kWh$$ respectively .
When a current I flows through a resistance R for time t, the electrical energy spent is given by
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$$IRt$$
0%
$$I^2Rt$$
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$$IR^2t$$
0%
$$I^2R/t$$
Explanation
$$\textbf{Step 1: Deriving expression for Energy spent}$$
$$\textbf{[Ref. Fig.]}$$
We know that, Power
$$P = i^2R$$
Since, $$\text{Energy} = \text{Power} \times \text{Time}$$
Therefore, in $$dt$$ time, the energy spent is given by:
$$dE = Pdt$$
$$dE = i^2 Rdt$$
Integrating both sides to get the Total Energy in time t:
$$E = \displaystyle \int dE = \int_0^t i^2 Rdt = i^2R \int_0^t dt = i^2 Rt$$
Hence, Option A is correct
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Practice Class 12 Medical Physics Quiz Questions and Answers
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