MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 12

A student was given the experiment to measure the emf of a unknown cell using potentiometer. He cheated result and wrote the observation arbitrarily and was caught. Which reading helped the teacher to arrive at the conclusion that he cheated?

S. No.	Value of rheostat	Null point
1.	$R_1$	50 cm
2.	$R_2$	60 cm
3.	$R_3$	70 cm
4.	$R_4$	80 cm

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

Explanation

Current at null point will be,

$i=\dfrac { 20 }{ 8+2+R } =\dfrac { 20 }{ 10+R }$ .

Also, at null point,

$\dfrac { 8l }{ 100 } i=10$

therefore,

$il=125$

From above equations,

$\dfrac { l }{ 10+R } =6.25$

$R=\dfrac { l }{ 6.25 } -10>0$

Hence,

$l>62.5$ .

First and second readings voilate the conditions.

In a Wheatstone bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances

$S_1$ and

$S_2$ connected in parallel. The condition for the bridge to be balanced will be

0%
$\displaystyle \frac{P}{Q} = \frac{R(S_1 + S_2)}{2S_1S_2}$
0%
$\displaystyle \frac{P}{Q} = \frac{R}{S_1 + S_2}$
0%
$\displaystyle \frac{P}{Q} = \frac{2R}{S_1 + S_2}$
0%
$\displaystyle \frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1S_2}$

Explanation

The balanced condition for Wheatstone bridge is

$\dfrac{P}{Q}=\dfrac{R}{S}$

Here

$S=\frac{S_1S_2}{S_1+S_2}$ (as

$S_1$ and

$S_2$ are connected in parallel in S arm)

so,

$\dfrac{P}{Q}=\dfrac{R(S_1+S_2)}{S_1S_2}$

If the temperature is increase, what will be the effect on the resistance of a conductor?

0%
Does not change
0%
Decreases
0%
Increases
0%
Cannot say

1 kWh is equal to

0%
$3.6 \times 10^6 MJ$
0%
$3.6 \times 10^5 MJ$
0%
$3.6 \times 10^2 MJ$
0%
$3.6 MJ$

Explanation

1 kilowatt hour is the energy produced by 1 kilowatt power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

$\implies 1kWh=3.6\times 10^6J$

$\implies 1kWh=3.6MJ$

Answer-(D)

How many positions are there on wire

$AB$ such that on touching

$K$ to those positions current from

$A$ to

$B$ remain same throughout if

$V>E$ and internal resistance of primary circuit battery is zero?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

An electric lamp whose resistance of 10 ohm, and a conductor of 2 ohm resistance are connected in series with a 6 V battery. The total current through the circuit and the potential difference across the electric lamp are :

0%
3.6 A, 6 V
0%
0.5 A, 5 V
0%
2.0 A, 0.2 V
0%
0.3 A, 3 V

If the filament of bulb D breaks then does bulb C give more light, less light, the same light or no light?

0%
more light
0%
less light
0%
same light
0%
no light

Explanation

It is assumed that every bulb is identical and has same resistance(

$R$ ).

So C and D are in parallel.And A , B and combination of (C and D) all these three are in series.

Initially resistance here was

$R/2$ (C and D both have

$R$ in parallel).So resistances in series are

$R$ ,

$R$ and

$R/2$ . Since resistance across third element is less than first two elements so less portion of cell voltage will be shared here.

Now bulb D breaks.So combination of C and D will reduced to only C.Now only C is there so net resistance here is

$R$ .So now A (

$R$ resistance) , B (

$R$ resistance) and C (

$R$ resistance) are in series and have equal resistances so voltage across C will be

$1/3$ of cell voltage.

In other words after breaking of D voltage across C has increased from before hence it will glow more brightly than before.

Identify the cases where the bulbs are connected totally in series.

Arrange the order of power dissipated in the given circuits, if the same current enters at point

$A$ in all the circuits and resistance of each resistor is

$R$ .

0%
$(ii) > (iii) > (iv) > (i)$
0%
$(iii) > (ii) > (iv) > (i)$
0%
$(iv) > (iii) > (ii) > (i)$
0%
$(i) > (ii) > (iii) > (iv)$

A cell of emf

$E$ and internal resistance

$r$ is connected in series with an external resistance

$nr$ . Then, the ratio fo the terminal potential difference to E.M.F is

0%
$1/n$
0%
$\cfrac { 1 }{ n+1 } \quad$
0%
$\quad \cfrac { n }{ n+1 }$
0%
$\cfrac { n+1 }{ n }$

Explanation

Applying voltage division rule, to find terminal voltage(v);

Voltage drop

$nR=E\cdot\cfrac{nR}{nR+R}=\cfrac{E\cdot nR}{R[n+1]}$

$\Rightarrow \cfrac{V}{E}=\cfrac{n}{[n+1]}$

A potentiometer is connected between A and B and the balance point is obtained at

$203.6\;cm$ . When the end of the potentiometer connected to B is shifted to C, then the balance point is obtained at

$24.6\;cm$ . If now the potentiometer be connected between B and C, the balance point will be at:

0%
$179.0\;cm$
0%
$197.2\;cm$
0%
$212.0\;cm$
0%
$228.0\;cm$

Explanation

Let the potential gradient of the potentiometer be

$x$ .

$x=\dfrac E{L}$

$x=\dfrac v{L}$

When potentiometer connected between

$AB$

$v=\epsilon_{ 1 }$ and

$l=20.6cm$

$\therefore x=\dfrac{\epsilon_{1}}{203.6}$

$\implies \epsilon_{1}=203.6x \rightarrow(1)$

When it is connected between

$AC$ then

$v=\epsilon_1-\epsilon_2$ and

$l=24.6x\rightarrow(2)$

Subtracting equation

$(1)$ and

$(2)$

$\epsilon_1=203.6x$

$v=\epsilon_{1}-\epsilon_{2}=24.6x$

$\epsilon_2=(203.6-24.6)x$

$\epsilon_2=179x$

$\implies x=\dfrac{\epsilon_2}{179}$

There when potentiometer is connected between

$B$ and

$C$ then the balance point will be

$179cm$

Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at

$40\ cm$ from left end. A

$30\Omega$ resistance is connected in series with the smaller of the two resistances, the null point shifts by

$20\ cm$ to the right end. The value of smaller resistance in

$\Omega$ is

0%
$12$
0%
$24$
0%
$36$
0%
$48$

Explanation

let the reistance at right be

$R$ and at left be

$r$

$\dfrac{r}{R}=\dfrac{l_x}{l_R}=\dfrac{40}{60}$ ,

$r$ is smaller

$\dfrac{r+30}{R}=\dfrac{60}{40}$

From above equation we get

$r=24\Omega$

Potentiometer wire of length

$1$ m is connected in series with

$490\Omega$ resistance and

$2$ V battery. If

$0.2$ mV/cm is the potential gradient, then resistance of the potentiometer wire is?

0%
$4.9\Omega$
0%
$7.9\Omega$
0%
$5.9\Omega$
0%
$6.9\Omega$

Explanation

$R_p$ is resistance of potentiometer.

Potential difference across potentiometer=potential gradient

$\times$ length

$=0.2\times 10^{-3}\times 100=0.02V$

Current in circuit

$=I=\dfrac{E}{R+R_p}=\dfrac{2}{490+R_p}$

Potential difference across potentiometer=

$IR_p$

$\implies \dfrac{2R_p}{490+R_p}=0.02$

$\implies 1.98R_p=9.8$

$\implies R_p=4.9\Omega$

Answer-(A)

The V-I graph for conductor at temperatures

$T_1$ and

$T_2$ are shown in the figure. (

$T_2$ -

$T_1$ ) is proportional to :

0%
$cos2\theta$
0%
$sin2\theta$
0%
$cot2\theta$
0%
$tan2\theta$

Explanation

We know that-

$R_1=R_o[1+(T_1-T_o)\alpha]$

$R_2=R_o[1+(T_2-T_o)\alpha]$

$\implies R_2-R_1=R_o\alpha(T_2-T_1)$

Now, we know that resistance is the slope of

$V-I$ curve.

Hence,

$R_2=\tan(90^{o}-\theta)=\cot\theta$

and

$R_1=\tan \theta$

Hence,

$\cot\theta-\tan\theta=R_o(T_2-T_1)\alpha$

$\implies R_o(T_2-T_1)\alpha=\dfrac{1-\tan^2\theta}{\tan\theta}=2 \dfrac{1-\tan^2\theta}{2\tan\theta}=\dfrac{2}{\tan2\theta}=2\cot2\theta$

$\implies T_2-T_1\propto \cot2\theta$

Answer-(C)

In the arrangement shown in figure, the current through

$5 \Omega$ resistor is

0%
$2 A$
0%
Zero
0%
$\dfrac{12}{7}A$
0%
$1 A$

Explanation

Using the Kirchhoff's voltage law

for loop ABCDA,

$2i_1+5(i_1+i_2)=12$ or

$7i_1+5i_2=12 ...(1)$

and for loop EBCFE,

$2i_2+5(i_1+i_2)=12$ or

$5i_1+7i_2=12 ...(2)$

Now,

$(1)\times 7 , (2)\times 5 \Rightarrow 49i_1+35i_2=84 ...(3)$ and

$25i_1+35i_2=60 ...(4)$

$(3)-(4) \Rightarrow 24i_1=24$ or

$i_1=1 A$

Substituting the value of

$i_1$ in

$(1)$ we get,

$i_2=\dfrac{12-7}{5}=1 A$

Thus the current through

$5 \Omega=i_1+i_2=1+1=2A$

A potentiometer (variable resistor) is connected in a simple circuit as shown below.
The student varies the resistance on the potentiometer and measures the corresponding current in the circuit. The table of data and the resistance vs. current graph created by the student is shown below.

$R$ (Ohm)

	$I$ (A)

0.5

5.99

1.0

3.02

1.5

2.07

2.0

1.56

2.5

1.19

3.0

1.04

3.5

0.83

4.0

0.72

4.5

0.70

5.0

0.62

5.5

0.53

6.0

0.49

6.5

0.45

8.5

0.36

9.0

0.33

9.5

0.31

10.00

0.29

Based on the experimental data collected by the student, determine the emf of the battery.

0%
0.33 V
0%
0.5 V
0%
1.0 V
0%
2.0 V
0%
3.0 V

Explanation

$E=IR$

For all the data, we see that

$E\approx 3V$

Answer-(E)

The temperature coefficient of resistance of a wire is

$\displaystyle { 0.00125 }/{ ^{ \circ }{ C } }$ . Its resistance is

$\displaystyle 1\Omega$ at

$300\ K$ . Its resistance will be

$\displaystyle 2\ \Omega$ at:

0%
$1127\ K$
0%
$1400\ K$
0%
$1154\ K$
0%
$1100\ K$

Explanation

Given :

$\alpha = 0.00125/^oC$

Using

$R_T = R_{273} (1 + \alpha (T- 273) )$

For

$T = 300$ ,

$R_{300} = 1\Omega$

$\therefore$

$R_{300} = R_{273} (1 + \alpha (300- 273) )$

$1 = R_{273} (1 + 0.00125 (300- 273) )$

$\implies R_{273} = 0.967\Omega$

Let at temperature be

$T$ when

$R_T = 2\Omega$

$\therefore$

$R_T = R_{273} (1 + \alpha (T- 273) )$

$2 = 0.967(1 + 0.00125 (T- 273) )$

$T- 273 = 854.6$

$\implies$

$T = 273 + 854.6 = 1127.6 K$

It is desired to make a long cylindrical conductor whose temperature coefficient of resistivity at

$20^{\circ}C$ will be close to zero. If such a conductor is made by assembling alternate disks of iron and carbon, Find the ratio of the thickness of a carbon disk to that an iron disk. (For carbon,

$\rho = 3500\times 10^{-8}\Omega m$ and

$\alpha = -0.50 \times 10^{-3}\ ^{\circ}C^{-1}$ for iron,

$\rho = 9.68\times 10^{-8}\Omega$ and

$\alpha = 6.5\times 10^{-3}\ ^{\circ}C^{-1})$ .

0%
$0.36$
0%
$0.036$
0%
$1.0$
0%
$2.0$

Explanation

Change in the resistance of the conductor on increasing its temperature,

$R - R_{0} = R_{0}\alpha_{av} (T - T_{0})$
The disks will be effectively in series, so we will add the resistances to get the total. Looking only at one pair of disk, we have

$R_{c} + R_{i} = R_{0c} (\alpha_{c}(T - T_{0}) + 1) + R_{0i} (\alpha_{i}(T - T_{0}) + 1)$

$= R_{0c} + R_{0i} + (R_{0c} \alpha_{c} + R_{0i}\alpha_{i})(T - T_{0})$ .
This equation will only be constant if the coefficient for the term

$(T - T_{0})$ vanishes.
Then

$R_{0c} \alpha_{c} + R_{0i}\alpha_{i} = 0$ ,
but

$R = \rho L/ A$ , and the disks have the same cross sectional area, so

$L_{c} \rho_{c} \alpha_{c} + L_{i}\rho_{i}\alpha_{i} = 0$

$\dfrac {L_{c}}{L_{i}} = \dfrac {\rho_{i}\alpha_{i}}{\rho_{c}\alpha_{c}} = -\dfrac {9.68\times 10^{-8}\times 6.5\times 10^{-3}}{3500\times 10^{-8}\times (-0.50\times 10^{-3})} = 0.036$ .

The Kirchhoff's first law

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

$\left(\displaystyle \sum iR=\displaystyle\sum E\right)$ , where the symbols have their usual meanings, are respectively based on

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

Explanation

$\sum i=0$ (Conservation of charge).

Charge can neither be created or destroyed.

$iR=\sum E$ (Conservation of energy).

The total potential around a loop

$=0$ or

Drop-in Potential energy=Energy dissipated.

According to Kirchhoff's first law, a junction can act neither as a source of charge nor as a sink of charge. This supports the law of conservation of charge. According to Kirchhoff's second law, the energy per unit charge transferred to the moving charges is equal to the energy per unit charge transferred from them. This supports the law of conservation of energy.

The resistance of a wire is

$5$ ohm at

$50^o$ C and

$6$ ohm at

$100^o$ C. The resistance of the wire at

$0^o$ C will be.

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

Two batteries of emf

${E}_{1}$ and

${E}_{2}$ a capacitor of capacitance

$C$ and resistor

$R$ are connected in a circuit as shown. When the switch S is shifted from position (1) to (2), then the charge flow and work done by battery

${E}_{2}$ are

0%
$\Delta q=C\left( { E }_{ 2 }-{ E }_{ 1 } \right) \quad$
0%
$\Delta W=C\left( { E }_{ 2 }-{ E }_{ 1 } \right) { E }_{ 2 }$
0%
$\Delta q=C\left( { E }_{ 2 }+{ E }_{ 1 } \right)$
0%
$\Delta W=\cfrac { { q }_{ 2 }^{ 2 } }{ 2c } -\cfrac { { q }_{ 1 }^{ 2 } }{ 2c }$

In the figure of a potentiometer arrangement,

$B$ is the main battery,

$C$ is the cell whose emf is to be determined,

$WW'$ is the potentiometer wire,

$G$ is the galvanometer and

$J$ is the jockey which may touch any point on the wire

$WW'$ . Choose the correct alternative/s for the potentiometer to work properly

0%
The emf of $B$ must be greater than the emf $C$
0%
Two batteries $B$ and $C$ must be joined so that the same polarity terminals are joined together
0%
The postive terminals of two cells must be together to end $W$
0%
The galvanometer resistance must be less than the resistnace of potentiometer to end $W$ .

Four equal resistance dissipated

$5W$ of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistance when connected in parallel across the same battery would be

0%
$125W$
0%
$80W$
0%
$20W$
0%
$5W$

Explanation

In series,

$R_s=4R$

Power

$P=\dfrac{V^2}{4R}$

$\implies 5=\dfrac{V^2}{4R}$

$\implies 20=\dfrac{V^2}{R}\rightarrow(1)$

In parallel,

$R+p=\dfrac{R}{4}$

Power consumed,

$P.\dfrac{v^2}{R_p}=4\dfrac{v^2}{R}$

$4\times 20$

$=80W$

$10m$ long potentiometer wire is connected to a battery having a steady voltage. A Leclanche cell is balanced at

$4m$ length of the wire. If the length is kept the same, but its cross-section is doubled, the null point will be obtained at

0%
$8m$
0%
$4m$
0%
$2m$
0%
None of these

The emf of the cell

${E}_{2}$ is:

0%
$1$ volt
0%
$2$ volt
0%
$3$ volt
0%
$4$ volt

The potential differences across the resistance are given in the following circuit as shown. If point

$Q$ is grounded, the potential at point

$S$ will be

0%
$20\ V$
0%
$-20\ V$
0%
$50\ V$
0%
$-50\ V$

In the circuit shown, the current in the

$1\ \Omega$ resistor is :

0%
$0.13\ A$ , from $Q$ to $P$
0%
$0.75\ A$ , from $P$ to $Q$
0%
$1.3\ A$ , from $P$ to $Q$
0%
$0\ A$

The internal resistance of the cell

${E}_{2}$ is:

0%
$4.5\Omega$
0%
$5.5\Omega$
0%
$6.5\Omega$
0%
$7.5\Omega$

Which of the following relation is wrong?

0%
1 ampere $\times$ 1 ohm = 1 Volt
0%
1 watt $\times$ 1 sec = 1 Joule
0%
1 newton per coulomb = 1 Volt per meter
0%
1 coulomb $\times$ 1 volt = 1 watt

If an alternate current main supply is given to be

$220 \ V$ . What would be the average electro motive force during a positive half cycle:

0%
$198 \ V$
0%
$386 \ V$
0%
$256 \ V$
0%
None of these

Explanation

$Vav= \dfrac{2}{\pi} Vo= \dfrac{2}{\pi} Vrms \times \sqrt{2}$

$= \dfrac{2\sqrt{2}}{\pi} Vrms$

$= \dfrac{2\sqrt{2}}{\pi} \times 220$

$198\ V$

A potentiometer wire has length $4$ m and resistance $8\Omega$ .What should be the resistance that must be connected in series with the wire and an accumulator of emf $2$ V, so as to get a potential gradient $1$ mV per cm on the wire?

0%
$32\Omega$
0%
$36\Omega$
0%
$40\Omega$
0%
$30\Omega$

Explanation

Given,

$l=4$ m

$R=$ Potentiometer wire resistance

$=8\Omega$

Potential gradient

$=\dfrac{dV}{dr}=1$ mVper cm

So, for

$400$ cm,

$\Delta V=400\times 1\times {10}^{-3}=0.4$ V

Let a resistor

${R}_{s}=$ connected in series, so as

$\Delta V=\dfrac{V}{R+{R}_{s}}\times R$

$0.4=\dfrac{2}{8+R}\times 8$

$\Rightarrow 8+R=\dfrac{16}{0.4}=40$

$\Rightarrow R=32\Omega$

The potential difference across the terminals of a battery is 8.4 volt when there is current of 1.50 A in the battery from negative to positive terminal. When the current 3.5 A in reverse direction, the potential difference becomes 9.4 volt. The internal resistance and e.m.f. battery are:

0%
0.2 ohm, 8.7 V
0%
0.4 ohm, 6.7 V
0%
0.5 ohm, 9.5 V
0%
0.2 ohm, 10.4 V

In a experiment for calibration of voltmeter a standard cell of e.m.f

$1.5 \ V$ is balanced at

$300 \ cm$ length of potentiometer wire. The potential difference across a resistance in the circuit is balanced at

$1.25 \ m$ . If a voltmeter is connected across the same resistance it reads

$0.65 \ V$ . The error in the volt meter is:

0%
$0.5 \ V$
0%
$0.025 \ V$
0%
$0.05 \ V$
0%
$0.25 \ V$

A uniform but time varying magnetic field is present in a circular region of radius R. The magnetic field is perpendicular and into the plane of the loop and the magnetic of field is increasing at a constant rate

$\alpha$ . There is a straight conducting rod of length 2R placed as shown in figure. The magnitude of induced emf across the rod is

0%
$60^\circ$
0%
$90^\circ$
0%
$30^\circ$
0%
$20^\circ$

Explanation

Angle =

${45^ \circ } + {45^ \circ } = {90^ \circ }$
1157229_1060187_ans_e9a29e10a4364cb7afc6e1a7d92b3741.JPG

Three identical conducting plates are arranged as shown in figure and they are given charges as indicated. When switch S is closed, the amount of charge flown through the switch is

0%
$20 \mu C$
0%
$60 \mu C$
0%
$120 \mu C$
0%
Zero

In a meter bridge an unknown resistance P is connected in the left gap and a

$50 \Omega$ resistance in the right gap. Null point is obtained at x cm from the left end. The unknown resistance now shunted with an equal resistance. Find the value of the resistance in the right gap so that the null point is not shifted.

0%
$60 \Omega$
0%
$38 \Omega$
0%
$25 \Omega$
0%
$50 \Omega$

A part of circuit with current is shown. The value of

$I$ is :

0%
$1A$
0%
$2A$
0%
$4A$
0%
Zero

The potential difference across the terminals of a battery is

$10V$ when there is a current of

$3A$ in the battery from the negative to the positive terminal. When the current is

$2A$ in the reverse direction, the potential difference becomes

$15V$ . The internal resistance of the battery is?

0%
$2.5\Omega$
0%
$3.2\Omega$
0%
$1.5\Omega$
0%
$4\Omega$

Two cells of electro motive force

$3 \ V$ and

$5 \ V$ and internal resistance

$r_1$ and

$r_2$ respectively are in series with an external resistance

$R$ . If the potential difference across 1st cell is zero, then

$R$ is:-

0%
$\dfrac{5 r_1 - 3 r_2}{3}$
0%
$\dfrac{2 r_1 - 3 r_2}{4}$
0%
$\dfrac{3 r_1 - 5 r_2}{3}$
0%
$\dfrac{4 r_1 - 5 r_2}{3}$

$AB$ is a uniform wire of meter bridge , across which an ideal

$20 volt$ cell is connected as shown. The resistor of

$1\Omega$ and

$X\Omega$ are inserted in slots of metre bridge. A cell of emf

$E volts$ and internal resistance

$r\Omega$ and a galvanometer is connected to jockey

$J$ as shown.

If

$E = 16 volts$ ,

$r = 4\Omega$ and distance of balance point

$P$ from end

$A$ is

$90 cm$ , then :

0%
the value of $X$ is $9\Omega$ .
0%
the current in $X$ is $2A$ .
0%
at the balanced conditionthe current through jockey is zero and the potential at $C$ and $P$ becomes equal.
0%
If we short the cell of emf $E$ by a thick wire then the current throughgalvanometer remains zero.

The electromotive force resistance of a single cell equivalent to a combination of

$6$ cells as shown in the figure will be:-

0%
$4 V, 1.5 \Omega$
0%
$6 V, 1.5 \Omega$
0%
$8 V, 1.5 \Omega$
0%
$0 V, 1.5 \Omega$

A horizontal straight conductor of length

$l$ placed along east-west direction falls under gravity from height

$h$ at a place where horizontal component of magnetic field is

$H$ and vertical component of magnetic field is

$V$ . Then :

0%
No electromotive force is induced along the length of the conductor
0%
An electromotive force is induced along the length which has maximum value $Hl\sqrt{2gh}$
0%
An electromotive force is induced along the length which has maximum value $Vl\sqrt{2gh}$
0%
An electromotive force is induced along the length which has maximum value $l\sqrt{2gh (V^2 + H^2)}$

If the current in the toroidal solenoid increases uniformly from zero to

$6.0 A$ in

$3.0\mu s$ .Self inductance of the toroidal solenoid is

$40 \mu H$ . The magnitude of self-induced emf is

0%
$24 V$
0%
$48 V$
0%
$80 V$
0%
$160 V$

In the given potentiometer arrangement, the null point

0%
Can be obtained for any value of V
0%
Can be obtained only if $V < V_0$
0%
Can be btained only if $V > V_0$
0%
Can never be obtained

In the determination of the internal resistance of a cell using a potentiometer, when the cell is shunted by a resistance

$"R"$ and connected in the secondary circuit, the balance length is found to be

$L_1$ . On doubling the shunt resistance, the balance length is found to increase to

$L_2$ . The value of the internal resistance is:

0%
$\dfrac{2 R(L_2 - L_1)}{(L_1 - 2 L_2)}$
0%
$\dfrac{2 R(L_2 - L_1)}{(2 L_1 - L_2)}$
0%
$\dfrac {R(L_2 - L_1)}{(L_1 - 2 L_2)}$
0%
$\dfrac{R(L_2 - L_1)}{(2 L_1 - L_2)}$

Three equal resistors are connected as shown in figure. the maximum power consumed by each resistor is 18 W. Then the maximum power consumed by the combination is :

0%
$18 W$
0%
$27 W$
0%
$36 W$
0%
$54 W$

Explanation

equivalent resistance from

$A\rightarrow B$

$\boxed{(R\parallel R)+R}$

$\Rightarrow \dfrac{1}{\dfrac{1}{R}+\dfrac{1}{R}}+R=\dfrac{1}{\dfrac{2}{R}}+R=\dfrac{R}{2}+R$

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

$\therefore$ equivalent resistance

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

Power across the single resistor

$=18\ W$ ( given)

$P=i^2R=18\ W.......(1)$

Power across the equivalent circuit is

$P_1=i^2\times \dfrac{3R}{2}=\dfrac{3i^2R}{2}\rightarrow .....(2)$

dividing

$(1)$ by

$(2)$

$\dfrac{P}{P_1}=\dfrac{18}{P_1}=\dfrac{i^2R}{\dfrac{3i^2R}{2}}=\dfrac{2}{3}$

$\Rightarrow P_1=\dfrac{3}{2}\times 18=27\ W$

$\therefore$ maximum power consumed by circuit

$=27\ W$

1457480_1101106_ans_e0622a5ac5384cd6b033a38c35ea73a6.png

A coil of area

$500{cm}^{2}$ having

$1000$ turns is held perpendicular to a uniform magnetic field

$0.4G$ . The induced emf in the coil if it is turned through angle of

${180}^{o}$ in

$\cfrac{1}{10}s$ is

0%
$0.01V$
0%
$0.02V$
0%
$0.03V$
0%
$0.04V$

Figure shows a potentiometer arrangement with

$R_{AB} = 10\ \Omega$ and rheostat of variable resistance

$x$ . For

$x= 0$ null deflection point is found at

$20$

$cm$ from

$A$ . For unknown value of a null deflection point was at

$30$

$cm$ from

$A$ . then the value of

$x$ is

0%
$10\ \Omega$
0%
$5\ \Omega$
0%
$2\ \Omega$
0%
$1\ \Omega$

In the potentionmeter circuit shown in figure the balance point with

$R = 10 W$ when

$S_1$ is closed and

$S_2$ open is 50 cm, While that when

$S_2$ is closed and

$S_1$ is open is 60 cm. What is the value of X?

0%
$1\Omega$
0%
$2\Omega$
0%
$3\Omega$
0%
$4\Omega$

Two resistors of resistances

$200 \Omega$ and

$1M\Omega$ , are connected with outer junctions maintained at potential of

$+3V$ and

$-15V$ as shown in figure. What is the potential at the junction

$X$ between the resistors?

0%
$0V$
0%
$+1V$
0%
$+0.6V$
0%
$-12V$

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

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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