<img src="data:image/png;base64,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" alt=""/> Potentiometer wire Pq of length 1m is connected to a standard cell $$ E_1 $$ Another cell of E2 of emf1.02v/s connected as shown in the circuit diagram with a resistance 'r' and with a switch 's' . With the switch S open, null position is obtained at a distance of 51 cm from p,Calculate

When switch S is closed, will null point moves towards p or Q?

If $${i_1} = 3\sin \omega t\,and\,{i_2} = 4\cos \omega t$$, then $$i_3$$ is

$$5\sin \left( {\omega t + {{53}^ \circ }} \right)$$

$$5\sin \left( {\omega t + {{37}^ \circ }} \right)$$

$$5\sin \left( {\omega t + {{45}^ \circ }} \right)$$

$$5\cos \left( {\omega t - {{53}^ \circ }} \right)$$

MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 8

The supply voltage to a room is

$120\ V$ . The resistance of the lead wires is

$6\ \Omega$ . A

$60W,120V$ bulb is already switched on. What is the decrease of voltage across the bulb, when a

$240 W,120V$ heater is switched on in parallel to the bulb?

0%
$2.9$ $volt$
0%
$13.3$ $volt$
0%
$10.04$ $volt$
0%
$zero\ volt$

Explanation

HINT:

$\bullet$ Calculate the equivalent resistance of the bulb and heater first.

$\bullet$ Apply Ohm's Law.

STEP 1: Draw the circuit diagram

STEP 2: Resistance of bulb $(R_b)$

Using

$P = \dfrac{V^2}R\;\;\;or\;\;\;R = \dfrac{V^2}P$

we have,

$R_b = \dfrac {V^2}P = \dfrac{120^2}{60} = 240\;\Omega$

STEP 3: Resistance for heater $(R_h)$

$R_h = \dfrac{V^2}P = \dfrac{120^2}{240} = 60\;\Omega$

STEP 4: When bulb and heater are connected in parallel

$R_{eq} = \dfrac{R_bR_h}{R_b + R_h} = \dfrac{240\times 60}{240+60} = \dfrac{14400}{300} = 48\;\Omega$

STEP 5: Voltage drop across bulb before heater was connected

Apply Ohm's Law V=i R

Where V is potential difference

and i is current

$i=\dfrac{120}{R_b+6}$

$V_2 = \dfrac{120}{R_b +6} \times R_b = \dfrac{120}{240+6}\times 240 = 117.7\;V$

STEP 6: Voltage drop after heater is connected

$i=\dfrac{120}{R+6}$

$V_1 = \dfrac{120}{R+6} \times R = \dfrac{120}{48+6} \times 48 = 106.66\;V$

So,

$V_2 - V_1 = 10.04\;V$

Thus, decrease in voltage across bulb is 10.04 V.

Thus, option C is correct.

$220$ volt,

$1000$ watt bulb is connected across a

$110$ volt mains supply. The power consumed will be.

0%
$750$ watt
0%
$500$ watt
0%
$250$ watt
0%
$1000$ watt

The resistance of a

$10m$ long potentiometer wire is

$50\Omega$ . It is connected in series with a

$3V$ battery and

$10\Omega$ resistor. The potential difference between two points separated by distance

$40cm$ is equal to ______

0%
$0.02V$
0%
$0.1V$
0%
$0.06V$
0%
$1.2V$

Explanation

$I=\cfrac { 3 }{ 60 } =\cfrac { 1 }{ 20 } A\\ \therefore { V }_{ p }=\cfrac { 1 }{ 20 } \times 50=2.5V\\ 10m\longrightarrow 50\Omega \\ 1m\longrightarrow \cfrac { 50 }{ 10 } \Omega \\ 1cm\longrightarrow \cfrac { 5 }{ 100 } \Omega \\ 40cm\longrightarrow \cfrac { 5 }{ 100 } \times 40\Omega \\ 40cm\longrightarrow 2\Omega \\ \therefore V=2\Omega \times \cfrac { 1 }{ 20 } =0.1V$
1180233_961258_ans_273d4b152f2d4738bee6a983243a0c4b.png

1180233_961258_ans_273d4b152f2d4738bee6a983243a0c4b.png

If the figure shows a part of an electric circuit, then the current

$I$ is

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

Explanation

Outgoing Current

$=3+2+5=10A$

Incoming Current

$=6+1+I=(7+I)A$

By comparing ,we get

$\Rightarrow\,10=7+I$

$\Rightarrow I=3$

1103163_1016741_ans_bd402b0a4f224bdf90713b84aec7e6d1.png

${i}_{1}=3\sin{\omega t}$ and

${i}_{2}=4\cos{\omega t}$ , then

${i}_{3}$ is -

0%
$5\sin{(\omega t+{53}^{o})}$
0%
$5\sin{(\omega t+{37}^{o})}$
0%
$5\sin{(\omega t+{45}^{o})}$
0%
$5\cos{(\omega t+{53}^{o})}$

Explanation

Usin KCL at the node

$i_1+i_2=i_3$

$i_3= 3sin\ \omega t+4 cos\ \omega t$

$i_3= 5(3/5 sin\ \omega t+4/5 cos\ \omega t)$

$i_3= 5\ sin(\omega t+\ 53^o)$

A solid spherical conducting shell has inner radius

$a$ and outer radius

$2a$ . At the centre of the shell a point charge

$+Q$ is located. What must be the charge of the shell be in order for the charge density on the inner and outer surfaces of the shell to be exactly equal?

0%
$-5Q$
0%
$+3Q$
0%
$-4Q$
0%
$+4Q$

Explanation

Inner radius

$=a$

outer radius

$=2a$

point charge

$=+Q$

The total charge enclosed for the inner sphere is

$q$ . Hence, the surface charge density for the inner sphere is

${ \sigma }_{ inner }=\dfrac { q }{ 4\pi { R }_{ 1 }^{ 2 } } =\dfrac { q }{ 4\pi { a }^{ 2 } }$

The total charge enclosed for the outer sphere is

$=Q+q$

Hence, the surface change density for the outer sphere is -

${ \sigma }_{ inner }=\dfrac { q }{ 4\pi { \left( a \right) }^{ 2 } }$

The electric field at a point at a distance

$x$ from the sphere is given by Gauss's law,

by imaging a sphere of radius

$2a$

$E\times \left( 4\pi .{ 4a }^{ 2 } \right) =\dfrac { { Q }_{ enclosed } }{ { \epsilon }_{ 0 } }$

or,

$E=\dfrac { Q+q }{ 4\pi { \epsilon }_{ 0 }\times { 4a }^{ 2 } } =\dfrac { Q+q }{ 16\pi { \epsilon }_{ 0 }{ a }^{ 2 } } =-5Q$

1241190_1026888_ans_eabd77f3ae90462180c8378816120f84.png

Consider a cube as shown in the fig-1; with uniformly distributed charge within its volume. The potential at one of its vertex P is ${V_0}$ .A cubical portion of half the size (half edge length) of the original cube is cut and removed as shown in the fig-Find the modulus of potential at the point P in the new structure.

0%
${{7} \over 8}{V_0}$
0%
${{{V_0}} \over 2}$
0%
${{3{V_0}} \over 4}$
0%
${{{V_0}} \over 4}$

Explanation

Potential at one vertex

$=V-0$

When one of the vertex out of

$8$ is removed

Portion of vertex

$=\cfrac{1}{8}$

Potential of the vertex

$=\cfrac{1}{8}(V_0)$

Potential of remaining cube

$=(V_0-\cfrac{1}{8}V_0)=\cfrac{7}{8}V_0$

The number of turns in th coil of an

$ac$ generator is

$5000$ and the area of the coil is

$0.25\ {m}^{2}$ . The coil is rotated at the rate of

$100$ cycles/s in a magnetic field of

$0.2\ T$ . The peak value of emf generated is nearly:

0%
$786\ kV$
0%
$440\ kV$
0%
$220\ kV$
0%
$157.1\ kV$

Explanation

As given,

$N=5000 turns$

$B=0.2 w/m^{2}$

$A=0.25m^{2}$

$V=1000sec^{-1}$

$E_{0}=NBAW$

$=5\times { 10 }^{ 3 }\times 2.5\times 0.25\times 100\times 2\times 3.14$

$=3.14\times 0.5\times 10^{6}V$

$E_{0}=157kV$

A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:

0%
0.2 V
0%
0.4 V
0%
0.08 V
0%
0.16 V

Explanation

The potentiometer has

$10\Omega$ resistance at

$100cm$ length . at

$40cm$ length , it'll have resistance

$R_{1}=\dfrac{10}{100}\times 40=4\Omega$ Drives cell has emf

$E_{1}$ connected to a secondary circuit having a cell of emf

$E_{2}=2V$ and series resistance

$R_{2}=40\Omega$ At balance point,current through both primary and secondary circuits would be same.

$\Rightarrow I=\dfrac{E_{1}}{R_{1}}=\dfrac{E_{2}}{R_{2}}$ (follows from ohm's law

$V=1R$ )

$\Rightarrow \dfrac{E_{1}}{4}=\dfrac{E_{2}\times 2}{40}\Rightarrow E_{2}=\dfrac{2}{40}\times 4=0.2V$

Refer to teh circuit shown. What will be the total power dissipation in the circuit if

$P$ is the power dissipated in

${R}_{1}$ ? It is given that

${R}_{2}=4{R}_{1}$ and

${R}_{3}=12{R}_{1}$

0%
$4P$
0%
$7P$
0%
$13P$
0%
$17P$

Explanation

Total equivalent resistance=

$\frac { { R }_{ 2 }{ R }_{ 3 } }{ { R }_{ 2 }{ +R }_{ 3 } } +{ R }_{ 1 }\\ =\frac { 48{ { R }_{ 1 } }^{ 2 } }{ 16{ R }_{ 1 } } +{ R }_{ 1 }\\ =3{ R }_{ 1 }+{ R }_{ 1 }=4{ R }_{ 1 }$

Power dissipation at

$R_1=I^2R_1=P$

$\therefore$ Power dissipation in the total circuit=

$I^24R_1$

$=4I^2R_1$

$=4P$

A potentiometer wire of

$10\ m$ length and having a resistance of

$1\ ohm/m$ is connected to an accumulator of emf

$2.2$ volt and a high resistance box. To obtain a potential gradient of

$2.2mV/m$ , the value of resistance used from the resistance box is :-

0%
$790\ ohm$
0%
$810\ ohm$
0%
$990\ ohm$
0%
$1000\ ohm$

The temperature coefficient of resistance of wire is

$12.5 \times 10^{-4} / C^0$ . At 300K the resistance of the wire is 1 ohm. The temperature at which resistance will be 2 ohm is

0%
1154K
0%
1100K
0%
1400K
0%
1127K

Explanation

linear temperature coefficient

$\alpha =\dfrac{\triangle R}{R\times \triangle \theta}$

$\triangle R=2\Omega -1\Omega$

$\triangle \theta =T-27$

$\alpha =0.00125$

$0.00125=\dfrac{1}{1\times (T-27)}\implies T-27=\dfrac{1}{0.00125}=800\implies T=800+27=827C^0=827+273=1100k$

For the following circuits, the potential difference between
X and Y in volt is $\left( {{V_x} - {V_y}} \right)$

0%
1
0%
-1
0%
2
0%
-2

Explanation

$V_x+2\times 1-3\times 1=V_y$

$V_x-V_y=1$ volt

The correct option is 'A'.

2077950_1043163_ans_d2f83f15ce844e9f95e155389779662f.png

Write T against true and F against false in the statements.
A charged glass road attract a charged plastic straw

0%
True
0%
False

If current

$I_1 = 3A$ sin

$\omega t$ and

$I_2 = 4A$ cos

$\omega t$ , then

$I_3$ is :

0%
$5A sin (\omega t + 53^0)$
0%
$5A sin (\omega t + 37^0)$
0%
$5A sin (\omega t + 45^0)$
0%
$5A sin (\omega t + 30^0)$

Explanation

Given,

$I_{1}=3Asin\omega t$ and

$I_{2}=4Acos\omega t$

$I_{3}=I_{1}+I_{2}$

$\Rightarrow I_{3}=5\times(\dfrac{3}{5}sin\omega t+\dfrac{4}{5}cos\omega t)=5Asin(\omega t+53^0)$ .

Hence the correct option is A

In the circuit shown, the current in the

$1 ohm$ resistor is:

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

In the diagram shown

$P$ is a point negative charge. It's weight is balanced by the electric force due to the fixed very long wire. The equilibrium of the particle is

0%
Stable for vertical displacements
0%
Neutral for vertical displacements
0%
Stable for horizontal displacements (parallel to the wire)
0%
Neutral for horizontal displacements (parallel to the wire)

Find the potentials of points A and B:-

0%
$V_A = + 10 \ V; V_B = 0 \ V$
0%
$V_A = + 7.5 \ V; V_B = - 2.5 \ V$
0%
$V_A = + 2.5 \ V; V_B = - 7.5 \ V$
0%
$V_A = + 0 \ V; V_B = - 10 \ V$

Explanation

(Refer to Image)

Finding the potentials of point A and B

$V_A-I \times 4R-V_B+10=0$

or,

$V_A-V_B=-10/4$ or,

$V_B=+2.5$

when,

$V_A=-7.5$ due to,

$C$ will be ground.

1228958_1066581_ans_2c1eb56c4812408e9413ed9a65851a96.png

If the resistance in the circuit is increased four times by keeping the potential difference the same, the current in the circuit is ___________.

0%
Remains same
0%
Four times
0%
One fourth
0%
Half

Explanation

According to ohm's law-

$V = IR$

$I = \dfrac{ V }{ R }$

$V$ : Applied potential difference

$I$ : Current in the circuit

$R$ : Resistance of the circuit

Now, new resistance becomes four times of initial

$R' = 4R$

So new current will be-

$I' = \dfrac{ V }{ 4R }$

$V$ , remains same

$\Rightarrow I' = \dfrac{ I }{ 4 }$

So, the current becomes one-fourth of the initial current when resistance becomes four times.

$AB$ is part of a circuit as shown, that absorbs energy at a rate of

$50 \ W$ .

$E$ is an electromotive force device that has no internal resistance.

0%
Potential difference across $AB$ is $48 V$
0%
Electromotive force device is $48 V$ .
0%
Point $B$ is connected to the positive terminal of $E$ .
0%
Rat of conversion from electrical to chemical energy is $48 W$ in device $E$ .

Explanation

$\begin{array}{l} { i^{ 2 } }\left( 2 \right) +i.E=50 \\ \Rightarrow 2+E=50 \\ \Rightarrow E=48V \\ Hence, \\ option\, \, B\, is\, \, correct\, \, answer. \end{array}$

Charges

$25Q,9Q$ and

$Q$ are placed at point

$ABC$ such that

$AB=4m$ ,

$BC=3m$ and angle between

$AB$ and

$BC$ is

$90^\circ$ . then force on the charge

$C$ is:

0%
Zero
0%
$\frac{q^2}{{\pi { \in _0}\sqrt 5 }}$
0%
$\frac{{2{Q^2}}}{{\pi { \in _0}}}$
0%
$\frac{{5{Q^2}}}{{4\pi { \in _0}}}$

Explanation

R.E.F image

Let us define,

$q_{1} = 25; q_{2} = 9q; q_{3} = q$ placed at A, B and C respectively, as shown in figure.

The total force on

$q_{3}$ will be the vector sum of force on

$q_{3}$ due to

$q_{2}$ (ie

$F_{32}$ ) and force on

$q_{3}$ due to

$q_{1}$ (ie

$F_{31}$ ).

$F_{32} = \dfrac{q_{3}q_{2}}{4\pi \epsilon _{0}(BC)^{2}} = \dfrac{9q^{2}}{4\pi \epsilon _{0}\times 9} = \dfrac{q^{2}}{4\pi \epsilon _{0}}$

$F_{31} = \dfrac{q_{3}q_{1}}{4\pi \epsilon _{0}((AB)^{2}+(BC)^{2})} = \dfrac{25q^{2}}{4\pi \epsilon _{0}\times 25} = \dfrac{q^{2}}{4\pi \epsilon _{0}}$

$F_{total} = \sqrt{F_{32}^{2}+F_{31}^{2}+2F_{32}F_{31}cos\theta }$

here,

$cos\theta = \dfrac{3}{5}$

$\Rightarrow F_{total} = \dfrac{q^{2}}{4\pi \epsilon _{0}} \sqrt{1+1+2\times \dfrac{3}{5}}$

$\Rightarrow F_{total} = \dfrac{q^{2}}{\sqrt{5}\pi \epsilon _{0}}$

1131882_1071647_ans_b394eb4bb7364ea68174f42c15809141.PNG

The

$V-I$ graph for a conductor at temperature

$T_{1}$ and

$T_{2}$ are as shown in figure. The term

$(T_{2}-T_{1})$ is proportional to:

0%
$\cos 2\theta$
0%
$\sin 2\theta$
0%
$\cot 2\theta$
0%
$\tan 2\theta$

Explanation

$\textbf{Step1: Slope of $$V-I$$ graph}$

Ohm's Law states that $V=IR$

From the graph

Slope of

$V-I$ graph will be resistance

So, $\dfrac{{{V}_{1}}}{{{I}_{1}}}\ =\tan \theta \ = \ {{R}_{1}}$

$\&$ $\dfrac{{{V}_{2}}}{{{I}_{2}}}\ =\tan (90-\theta )\ =\cot \theta \ =\ {{R}_{2}}$

$\textbf{Step2: Calculation of}$ $T_2 -T_1$

Temperature dependence of resistance is given by:

$R_1 = R_0(1+\alpha T_1)$

$R_2 = R_0(1+\alpha T_2)$

Subtracting above equations, we get

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

$\therefore T_2-T_1$ is proportional to $R_2-R_1$

$\Rightarrow$ ${{T}_{2}}-{{T}_{1}}\ \propto \ \left( \cot \theta -\tan \theta \right)=\dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\sin \theta \cos \theta }$

${{T}_{2}}-{{T}_{1}}\ \propto \ \dfrac{2\ \cos 2\theta }{\sin 2\theta }\ =\ 2\cot 2\theta$

Hence, ${{T}_{2}}-{{T}_{1}}\ \propto \ \cot 2\theta$

Therefore, Option C is correct

An electric motor operates on a

$50 \ V$ supply and a current of

$1 \ A$ . If the efficiency of the motion is

$30 \ \%$ , what is the resistance of winding of motion ?

0%
$35 \ \Omega$
0%
$4 \ \Omega$
0%
$2.9 \ \Omega$
0%
$3.1\ \Omega$

Potentiometer wire Pq of length 1m is connected to a standard cell

$E_1$ Another cell of E2 of emf1.02v/s connected as shown in the circuit diagram with a resistance 'r' and with a switch 's' . With the switch S open, null position is obtained at a distance of 51 cm from p,Calculate

0%
emf of the cell.
0%
Potential gradient of the wire.
0%
When switch S is closed, will null point moves towards p or Q?
0%
insufficient data

Explanation

Length of potentiometer wire $l=1\,m$

Emf $e={{E}_{1}}$

Emf of cell ${{E}_{2}}=1.02\,v/s$

Null point $=51\,cm=0.51\,m$

(a)Potential gradient of the wire

$V=\dfrac{1.02}{0.51}$

$V=2\,V/m$

(b) Emf of cell

${{V}_{PQ}}={{E}_{1}}$

${{E}_{1}}=2\,V$

Potentiometer is based on

0%
Deflection method
0%
Zero deflection method
0%
Both (a) and (b)
0%
None of these

Two cells each of electromotive force

$E$ and internal resistance

$r$ are concerned in parallel across the resistance

$R$ . The maximum energy given to the resistor per second if:

0%
$R = \dfrac{r}{2}$
0%
$R = r$
0%
$R = 2r$
0%
$R = 0$

Explanation

${R_{eq}} = \dfrac{r}{2} + R = \dfrac{{r + 2R}}{2}$

$I = \dfrac{{2E}}{{r + 2R}}$

For max. power consumption

$.$ I should be max. so denominator should be min. for that

$r + 2R = {\left( {\sqrt r - \sqrt {2R} } \right)^2} + 2\sqrt r \sqrt {2R}$

$\Rightarrow \sqrt r - \sqrt {2R} = 0$

$\Rightarrow R = \dfrac{r}{2}$

Hence,

option

$(A)$ is correct answer.

The wire of potentiometer has resistance

$4 \Omega$ and length

$1 m$ . It is connected to a coil of electromotive force

$2$ volt and internal resistance

$1 \Omega$ . If a cell of electromotive force

$1.2$ volt is balanced by it, the balancing length will be:

0%
$90 \ cm$
0%
$60 \ cm$
0%
$50 \ cm$
0%
$75 \ cm$

For comparing the electromotive force of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wire. Which of the following possibilities would make the experiment unsuccessful?

0%
The electromotive force system of the standard cell is larger than the $E$ electromotive force system of the two cells.
0%
The diameter of the wires is the same and uniform throughout
0%
The number of wires is ten
0%
The electromotive force of the standard cell is smaller than the electromotive force of the two cells.

Two identical batteries, each of electromotive force

$2 V$ and internal resistance

$r = 1 \Omega$ are connected as shown. The maximum power that can be developed across

$R$ using these batteries is:

0%
$2 W$
0%
$3.2 W$
0%
$8,2 W$
0%
$4 W$

$n$ identical cells each of electromotive force

$e$ and internal resistance

$r$ are connected in series of this combination. The current through

$V$ is:

0%
$\frac{{2e}}{n}$
0%
$\dfrac{n \space e}{nR + r}$
0%
$\dfrac{e}{R + nr}$
0%
$\dfrac{n \space e}{R + r}$

Explanation

$I = \left( {\dfrac{{ne - 2e}}{{nr}}} \right)$

$I = \dfrac{{\left( {n - 2} \right)e}}{{nr}}$

$V = e - Ir$

$V = e - \dfrac{{\left( {n - 2} \right)e}}{n}$

$V = \dfrac{{\left( {ne - ne + 2e} \right)}}{n}$

$V = \dfrac{{2e}}{n}$

Hence,

option

$(A)$ is correct answer.

A resistor of resistance

$1000 \Omega$ is connected to an AC source

$E=220 V sin(100 \pi rad/s$ . the power consumed by resistor is

0%
$48.4 W$
0%
$24.2 W$
0%
$12.1 W$
0%
ZERO

Explanation

Given,

$R=1000\Omega$

$E=220V sin(100\pi rad/s )$

$E_0=220V$

The power consumed by the resistance will be given by

$P=\dfrac{E_0^2}{R}$

$P=\dfrac{220\times 220}{1000}=48.4W$

The correct option is A.

For an AC Circuit, the potential difference and current are given by

$V=10\sqrt { 2 } \sin { \omega t }$ (inV) and

$I=2\sqrt { 2 } \cos { \omega t }$ (in A) respectively.The power dissipated in the instrument is:

0%
$20W$
0%
$40W$
0%
$40\sqrt { 2 }W$
0%
Zero

Explanation

Given,

$V=10\sqrt{2}sin\omega t$

$I=2\sqrt{2}cos\omega t=2\sqrt{2}sin(\omega t+\dfrac{\pi}{2})$

The power dissipated in the instrument is,

$P=V_{rms}.I_{rms}.cos\phi$

$P=\dfrac{V_0}{\sqrt{2}}.\dfrac{I_0}{\sqrt{2}}.cos\phi$

$P=\dfrac{10\sqrt{2}}{\sqrt{2}}\times \dfrac{2\sqrt{2}}{\sqrt{2}}\times cos\dfrac{\pi}{2}$

$P=10\times 2\times 0=0W$

The correct option is D.

Eight resistances each of resistance

$5\Omega$ are connected in the circuit as shown in figure. The equivalent resistance between

$A$ and

$B$ is

0%
$\dfrac{8}{3}\Omega$
0%
$\dfrac{16}{3}\Omega$
0%
$\dfrac{15}{7}\Omega$
0%
$\dfrac{19}{2}\Omega$

Two resistances of

$400\Omega$ and

$800\Omega$ are connected in series with a

$6$ volt battery of negligible internal resistance. A voltmeter of resistance

$10,000\Omega$ is used to measure the potential difference across

$400\Omega$ . The error in the measurement of the potential difference in volts approximately is :

0%
$0.01$
0%
$0.02$
0%
$0.03$
0%
$0.05$

${i_1} = 3\sin \omega t\,and\,{i_2} = 4\cos \omega t$ , then

$i_3$ is

0%
$5\sin \left( {\omega t + {{53}^ \circ }} \right)$
0%
$5\sin \left( {\omega t + {{37}^ \circ }} \right)$
0%
$5\sin \left( {\omega t + {{45}^ \circ }} \right)$
0%
$5\cos \left( {\omega t - {{53}^ \circ }} \right)$

In wheatstone's bridge

$P = 9$ ohm,

$R = 4$ ohm and

$S = 6$ ohm. How much resistance must be put in parallel to the resistance S to balance the bridge?

0%
$24$ ohm
0%
$13.5$ ohm
0%
$7.2$ ohm
0%
$18.7$ ohm

Explanation

ACC, to what stone's bridge

At balance condition

$PS=QR$

$\dfrac{PS}{R}=Q$

$\dfrac{9\times 6}{4}=Q=\dfrac{54}{4}$

$\boxed {Q=13.5\Omega}$

1055533_1110868_ans_f4069add8ef843709da6ca973a33581a.png

A circular coil of area $8{m^2}$ and number of turns 20 is placed in a magnetic field of 2T with its plane perpendicular to it. It is rotated with an angular velocity of 20rev/s about its natural axis. The emf induced is

0%
$400 V$
0%
$800\pi V$
0%
$Zero$
0%
$400\pi V$

In the following diagram, the deflection in the galvanometer in a potentiometer
circuit is zero, then

0%
${E_1} > {E_2}$
0%
${E_2} > {E_1}$
0%
${E_1} = {E_2}$
0%
${E_1} + {E_2} = E$

Explanation

Correct Answer: Option B
Step 1: Relate cell potential with balancing length
Refer figure

$1$

Let the potential drop per unit length of the potentiometer wire be

$k$

As seen in the figure

$l_1$ &

$l_2$ are the balancing length of the cells

$E_1$ and

$E_2$ respectively.

From the figure

$l_1 < l_2$

And we know,

$E_1 = kl_1$ ...........

$(I)$

$E_2 = kl_2$ .........

$(II)$

So,

$E_2 > E_1$

Hence, the correct answer is option B.

2192162_1145121_ans_75c1db0e4b454f78beca4a54aa6a550f.png

At what temperature will the resistance of a copper wire become three times its value at

$0^{0}$ C? [Temperature coefficient of resistance for copper =

$4\times 10^{-3}per^{0}C$ ]:-

0%
$500^{0}C$
0%
$450^{0}C$
0%
$600^{0}C$
0%
None of these

Explanation

Given,

Resistance become $R=3{{R}_{o}}$

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

$3{{R}_{o}}={{R}_{o}}\left( 1+\alpha \Delta T \right)$

$\Delta T=\dfrac{2}{\alpha }=\dfrac{2}{4\times {{10}^{-3}}}=500{{\,}^{o}}C$

Hence, when rise in temperature is $500{{\,}^{o}}C$

Taking the electric charge 'e' and permittivity

$\epsilon$ .use dimensional analysis to determine the correct expression for Wp. Where Wp

$\rightarrow$ angular frequency N

$\rightarrow$ number of density of free electron, m

$\rightarrow$ mass

0%
$\sqrt{\dfrac{Ne}{m\epsilon}}$
0%
$\sqrt{\dfrac{me}{N\epsilon}}$
0%
$\sqrt{\dfrac{Ne^2}{m\epsilon}}$
0%
$\sqrt{\dfrac{Ne^3}{m\epsilon}}$

Wattless current in the circuit is

0%
3.75 A
0%
5 A
0%
2.1 A
0%
zero

Explanation

Given,

$R=4\Omega ,{ X }_{ L }=7\Omega ,\quad { X }_{ C }=4\Omega$

And

${ E }_{ m }=25v$

$Impedance(Z)=\sqrt { { R }^{ 2 }+{ \left( { { X }_{ L }-{ X }_{ C } } \right) }^{ 2 } }$

Substitute all value in above equation

$Impedance(Z)=\sqrt { { 4 }^{ 2 }+{ \left( 7-4 \right) }^{ 2 } } =\sqrt { 16+9 } =5\Omega$

Then,

${ I }_{ m }=\frac { 25 }{ 5 } =5A$

$sin\theta =\frac { { X }_{ L } }{ Z } =\frac { 3 }{ 5 }$

$Wattless\quad current={ I }_{ RMS }\times sin\theta \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 5 }{ \sqrt { 2 } } \times \frac { 3 }{ 5 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 3 }{ \sqrt { 2 } } =2.1A$

Three resistances

$R_1,R_2$ &

$R_3$

$(R_1> R_2> R_3)$ are connected in series. If current

$I_1,I_2$ &

$I_3$ respectively is flowing through them, the correct relation will be

0%
$I_1=I_2=I_3$
0%
$I_1> I_2> I_3$
0%
$I_1< I_2< I_3$
0%
$I_1> I_2< I_3$

Explanation

In series, same amount of current will flow through all the resistances.

Current

$I_1=I_2=I_3=I=\dfrac{V}{R_{eq}}=\dfrac{V}{R_1+R_2+R_3}$

Option A is correct.

1935723_1130018_ans_05714a7879634f9c8f0afca430e4af69.jpeg

The inductance of a coil is

$L = 10 H$ and resistance

$R = 5 \Omega$ . If applied voltage of battery is

$10 V$ and it switches off in

$1$ millisecond, find induced electromotive force of inductor.

0%
$2 \times 10^4 V$
0%
$1.2 \times 10^4 V$
0%
$2 \times 10^{-4} V$
0%
None of these

Explanation

According to question.....................

$\begin{array}{l} \varphi =Li \\ or,\, \, \, e-\frac { { d\varphi } }{ { dt } } =-\frac { d }{ { dt } } (Li) \\ or,\, \, \, e=-L\frac { { di } }{ { dt } } \\ Induced\, current\, =\frac { V }{ R } =\frac { { 10 } }{ 5 } =2A \\ Circuit\, \, switches\, \, off\, in\, \, 1\, milli{ { second } } \\ or,\, \, \, dt=1\times { 10^{ -3 } }s \\ and,\, \, \, L=10H \\ \therefore \, \, \, \, Induced\, emp\, in\, \, inductor\, is\, \left| e \right| =10\times \frac { 2 }{ { 1\times { { 10 }^{ -3 } } } } =2\times { 10^{ 4 } }V \\ so\, the\, \, option\, \, is\, \, A.\, \end{array}$

If the circuit shown in the figure, the internal resistance of the battery is

$1.5\Omega$ and

${V}_{p}$ and

${V}_{Q}$ are potentials at

$P$ and

$Q$ respectively. What is the potential difference between the points

$P$ and

$Q$

0%
$Zero$
0%
$4\ volt\ \left({V}_{p}>{V}_{Q}\right)$
0%
$4\ volt\ \left({V}_{Q}>{V}_{P}\right)$
0%
$2.5\ volt\ \left({V}_{Q}>{V}_{P}\right)$

In the given figure, if the heat flowing through the 5

$ohm$ resistance is 10 calorie then how much heat is flowing through the 4

$ohm$ resistance ?

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

Explanation

We know that Heat =

${ I }^{ 2 }Rt$

Here,

${ H }_{ 5 }=10cal$

In given circuit

$5\Omega \parallel (4\Omega \& 6\Omega )$

Therefore,

Voltage across both terminal will be same and current following through 4 and 6 ohm will be same as both are connected in series

Current following through

$4\Omega$ will be half of the current following through

$5\Omega$ Because resistance is inversely proportional to current.

${ H }_{ 5 }={ I }^{ 2 }{ R }_{ 5 }t$

Substitute all value in above equation

$10={ I }^{ 2 }5t\\ { I }^{ 2 }t=2$

Similarly,

${ H }_{ 4 }={ \left( \frac { I }{ 2 } \right) }^{ 2 }4t\\ { H }_{ 4 }=\frac { { I }^{ 2 } }{ 4 } \times 4t\\ \quad \quad ={ I }^{ 2 }t=2cal/sec$

Hence,option (B) is correct option

An electron is rotating around an infinite positive linear charge in a circle of radius 0.1 m, if the linear charge density is

$1 \mu C/m,$ then the velocity of electron in m/s will be

0%
$0.562 \times 10^7$
0%
$5.62 \times 10^7$
0%
$562 \times 10^7$
0%
$0.0562 \times 10^7$

Explanation

REF.Image.

solution

Consider a Gaussian surface

which is cylindrical

symmetrical about the damage

so, Net flux

$= \frac{ginclosed}{\epsilon _{0}}$

$\Rightarrow E(2\pi rl) = \frac{\lambda l}{e_{0}}$

$\Rightarrow E = \frac{\lambda }{2\pi r\epsilon _{0}}$

Here, Electrostatic force = centripetal force

$\Rightarrow (\frac{\lambda }{2\pi r\epsilon} _{0})e = \frac{mv^{2}}{r}$

$\Rightarrow v = \sqrt{\frac{\lambda _{0}}{2\pi \epsilon _{0}\pi }} = \sqrt{2\times\frac{9}{2}\times10^{9}\times10^{-6+3}\times 1.759\times 10^{8}}$

$v = 5.62\times 10^{7}$ option - B is correct.

1152178_1174962_ans_156ca3b173124e4bbd00a79031c5763c.jpg

$3hp$ motors requires

$2.4 kw$ to drive it; its efficiency is about:

0%
$90$ %
0%
$75$ %
0%
$60$ %
0%
$50$ %

Explanation

We know,

Efficiency

$=\dfrac{\textrm{Output}}{\textrm{Input}}$

$=\dfrac{3\times 745}{2400}=0.93$

Which is

$93%$ , close to

$90%$ in the option pallette.

Option

$\textbf A$ is the correct answer

For which of the following substances the resistance decreases with the decrease in temperature:

0%
Germanium
0%
Silicon
0%
Silver
0%
None

The temperature coefficient of resistance of a wire is 0.00125 per degree celcius. At 300 K its resistance of the wire is 2 ohms. The temperature when resistance doubles is :-

0%
1154 K
0%
1100 K
0%
600 K
0%
1400 K

The name of the instrument which records electrical energy consumed by customer is

0%
Ampere-hour meter
0%
Watt-hour meter
0%
Var-hour meter
0%
Volt-ampere meter

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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