Figure given below shows uniformly positively charged, thin rod of length $$L$$ and four points $$A, B, C$$ and $$D$$ at the same distance $$d$$ from the rod, with position as marked. If $$V_{A}, V_{B}, V_{C}$$ and $$V_{D}$$ are their respective potentials then.

$$V_{B} > V_{A} > V_{C} > V_{D}$$

$$V_{B} > V_{A} > V_{C} = V_{D}$$

$$V_{D} > V_{B} > V_{A} > V_{C}$$

MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 9

If a wire is stretched , so that its length is 20% more than its initial length, the percentage increase in the resistance of the wire is

0%
40%
0%
10%
0%
44%
0%
25%

The electrical energy consumed by a

30 W

$30\ W$ bulb in

5

$5$ minutes is ______.

0%
$\text{9000 kJ}$
0%
$\text{9000 MJ}$
0%
$\text{9 KJ}$
0%
$\text{9 MJ}$

Explanation

Given,

Power,

P = 30 W

$P=30W$

Time,

t = 5 m i n = 5 \times 60 = 300 s e c

$t=5min=5\times 60=300sec$

The electrical energy consumed,

E = P t

$E=Pt$

E = 30 \times 300 = 9000 J

$E=30\times 300=9000J$

E = 9 K J

$E=9KJ$

The correct option is C.

A wire has a non-uniform cross-section as shown in figure. A steady current flows through it. The drift speed of electrons at points P and Q is

v_{P}

$v_P$ and

v_{Q}

$v_Q$ , then?

0%
$v_P=v_Q$
0%
$v_P < v_Q$
0%
$v_P > v_Q$
0%
Data is insufficient

Explanation

Answer C:

v_{P} > v_{Q}

${ v }_{ P }>{ v }_{ Q }$

Because $I=neAvd$ and as $A$ increases $vd$ must decrease to keep $I$ constant.

Two positive charges

q

$q$ and

n q

$nq$ are placed on a straight line. A charge

Q

$Q$ is placed at the neutral point so as to make the entire system in equilibrium. The charge

Q

$Q$ is

0%
$\dfrac{nq}{(1-\sqrt n)^2}$
0%
$\dfrac{-nq}{(1+\sqrt n)^3}$
0%
$\dfrac{-nq}{(1+\sqrt n)^2}$
0%
$\dfrac{q}{4}$

The post office box works on the principle of :

0%
Potentiometer
0%
Principle of wheat stone bridge
0%
Matter waves
0%
Ampere's law

The colour code on a resistor is red, red, black, the value of its resistance is

0%
$22 \omega$ exact
0%
$22\pm 2.2\ \omega$
0%
$\left(22\pm 4.4\right)\omega$
0%
$22 \pm 0.44\ \omega$

An electric heater rated as 2 kW is used to heat 200 kg of water from

10^{o} C

$10^o C$ to

70^{o} C

$70^o C$ . Assuming no heat losses, the time taken is :

0%
$25.2 s$
0%
$6 \times 10^3 s$
0%
$25.2 \times 10^3 s$
0%
$25.2 \times 10^6 s$

Explanation

Power of water

= 2 k w = 2000 w

$= 2 \ kw = 2000w$

Mass of water

= 200 k g

$= 200kg$

difference in temperature

Δ T = 70 - 10 = 60^{o} C

$\Delta T = 70 - 10 = 60^oC$

Concept

energy required to heat the water

=

$=$ energy given by water in time

t = p t

$t=pt$

energy required to increase tempeature of water by

60^{o} C, Q = m s Δ T

$60^oC, Q = ms\Delta T$

S =

$S=$ specific heat

= 4200 J / k g^{o} C

$= 4200J/kg^oC$

p t = m s Δ T

$pt = ms\Delta T$

2000 \times t = 200 \times 4200 \times 60

$2000\times t = 200\times 4200 \times 60$

t = 25200

$t = 25200$

t = 25.2 \times 10^{3} s e c

$t=25.2\times 10^3sec$ .

A long straight wire of radius a carries a steady current

I .

$I.$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields

B

$B$ and

B^{'},

$B',$ at radial distances

\frac{a}{2}

$\dfrac{a}{2}$ and

2 a

$2a$ respectively, from the axis of the wire is

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$4$

Explanation

Consider two amperian loops of radius

a / 2

$a/2$ and

2 a

$2a$ as shown in the diagram

.

$.$

Applying ampere's circuital law for these loops

,

$,$ we get

\oint B . d L = μ_{0} I_{e n c l o s e d}

$\oint {B.dL = } {\mu _0}{I_{enclosed}}$

For the smaller loop

,

$,$

\Rightarrow B \times 2 π \frac{a}{2} = μ_{0} \times \frac{1}{π a^{2}} \times {(\frac{a}{2})}^{2}

$\Rightarrow B \times 2\pi \dfrac{a}{2} = {\mu _0} \times \dfrac{1}{{\pi {a^2}}} \times {\left( {\dfrac{a}{2}} \right)^2}$

= μ_{0} I \times \frac{1}{4} = \frac{μ_{0} I}{4}

$= {\mu _0}I \times \dfrac{1}{4} = \dfrac{{{\mu _0}I}}{4}$

\Rightarrow B_{1} = \frac{μ_{0} I}{4 π a}

$\Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi a}}$

B^{'} \times 2 π (2 a) = μ_{0} I

$B' \times 2\pi \left( {2a} \right) = {\mu _0}I$

\frac{B}{B^{'}} = \frac{μ_{0} I}{4 π a} \times \frac{4 π a}{μ_{0} I} = 1

$\dfrac{B}{{B'}} = \dfrac{{{\mu _0}I}}{{4\pi a}} \times \dfrac{{4\pi a}}{{{\mu _0}I}} = 1$

Hence,

option

(C)

$(C)$ is correct answer.

The Wheatstone bridge shown in the above figure is balanced. If the positions of the cell C and the galvanometer G are now interchanged, G will show zero deflection

0%
in all cases
0%
only if all the resistance are equal
0%
only if $R_{1}=R_{3}and R_{2}=R_{4}$
0%
only $R_{1}/R_{3}and R_{2}/R_{4}$

Find the value of I ?

0%
-1.4 A
0%
-2.1 A
0%
2 A
0%
-4.2 A

Explanation

\begin{array}{l} i_{1} + i_{2} + I = 0 \\ \Rightarrow \frac{10 - 2}{5} + \frac{5 - 2}{6} + I = 0 \\ \Rightarrow \frac{8}{5} + \frac{3}{6} + I = 0 \\ ∴ I = - 2.1 A \end{array}

$\begin{array}{l} { i_{ 1 } }+{ i_{ 2 } }+I=0 \\ \Rightarrow \dfrac { { 10-2 } }{ 5 } +\dfrac { { 5-2 } }{ 6 } +I=0 \\ \Rightarrow \dfrac { 8 }{ 5 } +\dfrac { 3 }{ 6 } +I=0 \\ \therefore I=-2.1A \end{array}$

∴

$\therefore$ Option

B

$B$ is correct.

1211357_1262224_ans_473f76d60a18471ab10d0b2bb0e10b1f.png

When resistor are connected in series what remains same.?

0%
Current
0%
Voltage
0%
Resistance
0%
power

Explanation

From the figure we can see

V_{1} = I R_{1}

${ V }_{ 1 }=I{ R }_{ 1 }$

V_{2} = I R_{2}

${ V }_{ 2 }=I{ R }_{ 2 }$

V_{3} = I R_{3}

${ V }_{ 3 }=I{ R }_{ 3 }$

Hence, current is constant when resistor are connected in series.

1380546_1242930_ans_3324fd6b714d43e08db376697ec5cd0d.jpg

The magnetic field of

2 \times 10^{2} T

$2\times { 10 }^{ 2 } T$ acts at right angles to a coil of area

100 c m^{2}

$100 cm^{2}$ with

50

$50$ turns. The average emf induced in the coil is

0.1 V

$0.1 V$ , When it is removed from the field in time

L

$L$ .The value of

t

$t$ is

0%
$0.1 s$
0%
$1 s$
0%
$0.01 s$
0%
$20 s$

A room has AC run hours a day at a voltage of

220

$220$ V. The wirng of the room consists of Cu of

1

$1$ mm radius and a length of

10

$10$ m. Power consumption per day is

10

$10$ commercial units . The fraction of it goes in the joule heating in wires is:

ρ_{c u} = 1.7 \times 10^{- 8} Ω m

$\rho_{cu}=1.7\times10^{-8}\Omega m$

0%
$0.32\%$
0%
$0.2\%$
0%
$2\%$
0%
$3.2\%$

Explanation

Power consumption

= \frac{10 u n i t s}{5 h o u r} = \frac{2 u n i t s}{h o u r}

$=\dfrac{10\;units}{5\;hour}=\dfrac{2\;units}{hour}$

= 2 k w

$=2kw$

= 2000 J / S

$=2000J/S$

So,

I = \frac{P}{V}

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

= \frac{2000}{220}

$=\dfrac{2000}{220}$

= 9.09 A

$=9.09A$

Power loss in copper wire

= I^{R}

$=I^R$

= I^{2} \frac{P L}{A}

$=I^2\dfrac{PL}{A}$

Power loss in copper wire

= 9.09 \times 9.09 \times \frac{1.7 \times 10^{- 8} \times 10}{17 \times {(10^{- 3})}^{2}}

$=9.09\times 9.09\times \dfrac{1.7\times 10^{-8}\times 10}{17\times \left( 10^{-3}\right)^2}$

= 4.47 J

$=4.47J$

= \frac{4.47}{2000} \times 100 %

$=\dfrac{4.47}{2000}\times 100\%$

= 0.22 %

$=0.22\%$

Power loss in aluminium wire

= I^{2} R = I^{2} \frac{P L}{A}

$=I^2R=I^2\dfrac{PL}{A}$

Power loss in copper wire

= 9.09 \times 9.09 \times \frac{2.7 \times 10^{- 8} \times 10}{17 \times {(10^{- 3})}^{2}}

$=9.09\times 9.09\times \dfrac{2.7\times 10^{-8}\times 10}{17\times \left( 10^{-3}\right)^2}$

= 7.1 J

$=7.1J$

= \frac{7.1}{2000} \times 100 %

$=\dfrac{7.1}{2000}\times 100\%$

= 0.355 %

$=0.355\%$

Hence, the answer is

0.355 % .

$0.355\%.$

i = t^{2} 0 < t < T

$i=t^2 0 < t < T$ then r.m.s value of current is

0%
$\dfrac {T^2} {\sqrt2}$
0%
$\dfrac {T^2} {2}$
0%
$\dfrac {T^2} {\sqrt5}$
0%
none of these

Explanation

\begin{array}{l} i = t^{2} \\ ⟨ r . m . s ⟩ = ⟨ l^{2} ⟩ = \sqrt{\frac{{(t^{2})}^{2} d t}{d t}} \\ = \frac{t^{5}}{4 t} = {[\int_{0}^{1} \frac{t^{4}}{4}]}^{\frac{1}{2}} \\ ⟨ r . m . s ⟩ = {[\frac{t^{4}}{4}]}^{\frac{1}{2}} = \frac{T^{2}}{2} \\ ∴ l_{r . m . s} = \frac{T^{2}}{2} \end{array}

$\begin{array}{l} i={ t^{ 2 } } \\ \left\langle { r.m.s } \right\rangle =\left\langle { { l^{ 2 } } } \right\rangle =\sqrt { \dfrac { { { { \left( { { t^{ 2 } } } \right) }^{ 2 } }dt } }{ { dt } } } \\ =\dfrac { { { t^{ 5 } } } }{ { 4t } } ={ \left[ { \int _{ 0 }^{ 1 }{ \dfrac { { { t^{ 4 } } } }{ 4 } } } \right] ^{ \dfrac { 1 }{ 2 } } } \\ \left\langle { r.m.s } \right\rangle ={ \left[ { \dfrac { { { t^{ 4 } } } }{ 4 } } \right] ^{ \dfrac { 1 }{ 2 } } }=\dfrac { { { T^{ 2 } } } }{ 2 } \\ \therefore { l_{ r.m.s } }=\dfrac { { { T^{ 2 } } } }{ 2 } \end{array}$

Hence, option

B

$B$ is the correct answer.

If the emf of the battery in primary circuit of a potetimeter is doubled the value of potential gradient will:-

0%
Remain Uncharged
0%
Be halved
0%
Be doubled
0%
None of these

Explanation

\begin{array}{l} P . G = (\frac{E}{R + R_{w}}) \frac{R_{w}}{l_{w}} \\ T h e n, \\ I f E i s d o u b l e t h e n p o t e n t i a l g r a d i e n t i s a l s o d o u b l e . \\ H e n c e, o p t i o n C i s t h e c o r r e c t a n s w e r . \end{array}

$\begin{array}{l} P.G=\left( { \frac { E }{ { R+{ R_{ w } } } } } \right) \frac { { { R_{ w } } } }{ { { l_{ w } } } } \\ Then, \\ If\, E\, is\, double\, then\, potential\, gradient\, is\, also\, double. \\ Hence,\, option\, \, C\, is\, the\, correct\, answer. \end{array}$

Power dissipated in the circuit shown in the figure 30 watt . The value of R is

0%
5 $\Omega$
0%
15 $\Omega$
0%
10 $\Omega$
0%
20 $\Omega$

Explanation

R_{1} = R_{2} = ?

${R_1} = {R_2} = ?$

R_{2} = 10 Ω, V = 15 V

${R_2} = 10\Omega ,\,V = 15V$

and

P = 30 W

$P = 30W$

Hence

,

$,$

p = \frac{V^{2}}{R_{1}} + \frac{V^{2}}{R_{2}}

$p = \dfrac{{{V^2}}}{{{R_1}}} + \dfrac{{{V^2}}}{{{R_2}}}$

\frac{{(10)}^{2}}{R} = 30 - \frac{{(10)}^{2}}{10}

$\dfrac{{{{\left( {10} \right)}^2}}}{R} = 30 - \dfrac{{{{\left( {10} \right)}^2}}}{{10}}$

\frac{100}{R} = 30 - 10

$\dfrac{{100}}{R} = 30 - 10$

R = 20 Ω

$R = 20\Omega$

Hence,

option

(D)

$(D)$ is correct answer.

All of the following can induce an emf in a coil of wire EXCEPT

0%
moving a magnet through the coil
0%
placing a stationary coil of wire in an increasing magnetic field
0%
Placing a stationary coil of wire in a stationary magnetic field.
0%
Placing a stationary coil of wire in a decreasing magnetic field.

To increase sensitivity of a potentiometer, its:

0%
Area should be increase
0%
Current should increase
0%
length should increase
0%
Current should decrease

Figure given below shows uniformly positively charged, thin rod of length

L

$L$ and four points

A, B, C

$A, B, C$ and

D

$D$ at the same distance

d

$d$ from the rod, with position as marked. If

V_{A}, V_{B}, V_{C}

$V_{A}, V_{B}, V_{C}$ and

V_{D}

$V_{D}$ are their respective potentials then.

0%
$V_{B} > V_{A} > V_{C} > V_{D}$
0%
$V_{B} > V_{A} > V_{C} = V_{D}$
0%
$V_{A} = V_{B} > V_{C} = V_{D}$
0%
$V_{D} > V_{B} > V_{A} > V_{C}$

Explanation

V α \frac{1}{r}

$V\alpha \dfrac { 1 }{ r }$

none radius is less potential.

hence,

V_{A} = V_{B} > V_{C} = V_{D}

${ V }_{ A }={ V }_{ B }>{ V }_{ C }={ V }_{ D }$

1240072_1278688_ans_8d4b67b86b434a719eb266d12da9409e.jpg

For circuit shown, the value of emf of battery 'E' is.

0%
15 V.
0%
8 v
0%
200v/7
0%
65v/7.

At what moment when the potentiometer is balanced ,

0%
Current flows only in the primary circuit.
0%
Current flows only in the secondary circuit.
0%
current flows in both primary and secondary circuits .
0%
current doesnot flow in any circuit

An electric bulb marked

40 W - 220 V

$40W-220V$ is connected with an electric supply of

110 v

$110v$ . Its elctric power is

0%
$100\ W$
0%
$40\ W$
0%
$20\ W$
0%
$10\ W$

Explanation

Formula,

P = \frac{V^{2}}{R}

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

\to R = \frac{V^{2}}{P}

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

= \frac{220^{2}}{40}

$=\dfrac{220^2}{40}$

= 1210 Ω

$=1210\Omega$

P = \frac{V^{2}}{R}

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

= \frac{110^{2}}{1210}

$=\dfrac{110^2}{1210}$

= 10 W

$=10W$

Potential difference across

A B

$AB$ shown in the figure is:

0%
10 V
0%
12 V
0%
8 V
0%
0 V

Explanation

In the given figure, $A$ and $B$ are on the same potential as there is a direct connection between $A$ and $B$ without any other element (via $D$ )

V_{A} = V_{B}

$V_A=V_B$

Potential difference

= V_{B} - V_{A} = 0

$=V_B-V_A=0$

Alternatively,

we can see that

Voltage between

A

$A$ and

C

$C$ i.e.

V_{A C} = V_{A} - V_{C} = 10 V

$V_{AC}=V_A-V_C= 10 \ V$ . . . (i)

And,

Voltage between

C

$C$ and

B

$B$ i.e.

V_{C B} = V_{C} - V_{B} = - 10 V

$V_{CB}=V_C-V_B=-10 \ V$ . . . (ii)

Adding

(i)

$(i)$ and

(i i)

$(ii)$ ,

(V_{A} - V_{C}) + (V_{C} - V_{B}) = 10 + (- 10)

$(V_A-V_C)+(V_C-V_B)=10+(-10)$

\Rightarrow V_{A} - V_{B} = 0

$\Rightarrow V_A-V_B=0$

So Voltage between

A

$A$ and

B

$B$ i.e.

V_{A B} = V_{A} - V_{B} = 0 V

$V_{AB}=V_A-V_B= 0 \ V$

Thus, option D is correct.

2050624_1309413_ans_0e25510e8ebc4b058cab8662cb0cbf9c.png

See the diagram. Area of each plate is

2.0 m^{2} a n d d = 2 \times 10^{- 3} m

$2.0 m^2 and\ d = 2 \times 10^{-3} m$ A charge of

8.85 \times 10^{- 8} C

$8.85 \times 10^{-8} C$ is given to Q. Then the potential of Q becomes :

0%
13 V
0%
10 V
0%
66.7 V
0%
8.825 V

2 V

$2V$ battery, a

990 Ω

$990\Omega$ resistor and a potentiometer of

2 m

$2m$ length, all are connected in series of the resistance of potentiometer wire is

10 Ω

$10\Omega$ , then the potential gradient of the potentiometer wire is

0%
$0.05Vm^{-1}$
0%
$0.5Vm^{-1}$
0%
$0.01Vm^{-1}$
0%
$0.1Vm^{-1}$

In a meter bridge setup, which of the following should be the properties of the one meter long wire?

0%
High resistivity and low temperature coefficient
0%
Low resistivity and low temperature coefficient
0%
Low resistivity and high temperature coefficient
0%
High resistivity and high temperature coefficient

\begin{array}{l} Two electric charges 12 μ C and - 6 μ C are placed 20 c m apart in air.Net \\ electric field intensity at zero potential point on the line joining the charges \\ and outside the region between them \end{array}

$\begin{array} { l } { \text { Two electric charges } 12 \mu C \text { and } - 6 \mu C } \ { \text { are placed } 20 \mathrm { cm } \text { apart in air.Net } }\\ { \text { electric field intensity at zero potential } } \ { \text { point on the line joining the charges } } \\ { \text { and outside the region between them } } \end{array}$

0%
$1.25 \times 10 ^ { 6 } N / C$
0%
$3.75 \times 10 ^ { 5 } N / C$
0%
$4.5 \times 10 ^ { 6 } N / C$
0%
$6.75 \times 10 ^ { 5 } N + C$

Explanation

Hence, the option

C

$C$ is the correct answer.
1380137_1309358_ans_0a3fff8fc11e4b2f8335703e3fcee290.png

Six identical cells, each of emf of

6 V

$6V$ , are connected in parallel. The net emf across the battery is

0%
$6V$
0%
$36V$
0%
$0V$
0%
Between $6V$ and $36V$

Kirchoff's voltage law and current law are respectively in accordance with the conservation of

0%
Charge and momentum
0%
Charge and energy
0%
Energy and charge
0%
Energy and momentum

For different values of resistance, R power consumption in R are given. Then which of the following values are not possible?
(a)

2 W

$2W$
(b)

5 W

$5W$
(c)

8 W

$8W$
(d)

4 W

$4W$

0%
Only c
0%
b and c
0%
a, b, c
0%
All

In which of the cases no emf is induced in conducting wire?

0%
open circuit
0%
conductor at rest
0%
L perpendicular to B and v
0%
None of the above

The figure shows three circuits I, II and III which are connected to a 3V battery. if the powers dissipated by the configurations I,II and III are

P_{1}, P_{2}

$P_1,P_2$ and

P_{3}

$P_3$ respectively, then :

0%
$P_1 > P_2 > P_3$
0%
$P_1 > P_3 > P_2$
0%
$P_2 > P_1 > P_3$
0%
$P_3 > P_2 > P_1$

A capacitor C is connected to a battery of potential V. the distance between its plates is reduced to half, keeping the battery connected . then the work done by the battery is

0%
$\frac {CV^2}{4}$
0%
$\frac {CV^2}{2}$
0%
$\frac {3CV^2}{2}$
0%
$CV^2$

Terminal potential difference across the cell of emf 5v is:

0%
$V_a-V_b=10/3 V$
0%
$V_a-V_b=5/3 V$
0%
$V_a-V_b=5 V$
0%
$V_a-V_b=8/3 V$

Four charges

+ Q, - Q, + Q, - Q

$+Q , -Q, +Q, -Q$ are placed at the corners of a square taken order. At the centre of the square :

0%
$E = 0,$ $V = 0$
0%
$E = 0$ , $V \neq 0$
0%
$E \neq 0$ , $V = 0$
0%
$E \neq 0$ , $V \neq 0$

Explanation

Four charges

+ Q, - Q, + Q, - Q

$+Q,-Q,+Q,-Q$ .

In this context

E = 0

$E=0$ as well

V = 0

$V=0$

because, at the centre there is needed all cancel out because

θ

$\theta$ . If

E = 0

$E=0$ and

V = 0

$V=0$

1244373_1342261_ans_ff1e14ecdf3d483d9111363c50d4ebb5.jpg

An electric bulb is rated

220 V

$220 V$ and

100 W

$100 W$ . When it is operated om

110 V

$110 V$ ,the power consumed will be-

0%
$100 W$
0%
$75 W$
0%
$50 W$
0%
$25 W$

Explanation

Given,

P_{1} = 100 W

$P_1=100W$

V_{1} = 220 V

$V_1=220V$

P_{2} = ?

$P_2=?$

V_{2} = 110 V

$V_2=110V$

Power,

P = \frac{V^{2}}{R}

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

\frac{P_{2}}{P_{1}} = (\frac{V_{2}}{V_{1}})^{2}

$\dfrac{P_2}{P_1}=(\dfrac{V_2}{V_1})^2$

\frac{P_{2}}{P_{1}} = (\frac{110}{220})^{2} = \frac{1}{4}

$\dfrac{P_2}{P_1}=(\dfrac{110}{220})^2=\dfrac{1}{4}$

P_{2} = \frac{P_{1}}{4} = \frac{100}{4}

$P_2=\dfrac{P_1}{4}=\dfrac{100}{4}$

P_{2} = 25 W

$P_2=25W$

The correct option is D.

Which unit is not used in either the definition of the coulomb or the definition of the volt?

0%
ampere
0%
joule
0%
ohm
0%
second

When the switch

S

$S$ , in the circuit shown, is closed, then the value of current

i

$i$ will be :

0%
$3A$
0%
$5A$
0%
$4A$
0%
$2A$

Explanation

Let voltage at

C = x v

$C = xv$

K C L : i_{1} + i_{2} = i

$KCL : i_1 + i_2 = i$

\frac{20 - x}{2} + \frac{10 - x}{4} = \frac{x - 0}{2}

$\dfrac{20-x}{2} + \dfrac{10-x}{4} = \dfrac{x-0}{2}$

\Rightarrow x = 10

$\Rightarrow x = 10$

and

i = 5 A m p

$i = 5Amp$ .

1142064_1329657_ans_869ea608ed4d41bea2ba6b1f756a0ae8.png

The magnitude and direction of current I (in A) indicated in the following circuit is

0%
14 $\rightarrow$
0%
8 $\rightarrow$
0%
$\leftarrow$ 4
0%
$\leftarrow$ 6

Current l in the network shown in figure is :-

0%
16 A
0%
3 A
0%
19 A
0%
12 A

In the situation shown, the currents through

3 Ω

$3\Omega$ and

2 Ω

$2\Omega$ resistances are in the ratio

0%
$3:2$
0%
$7:4$
0%
$4:3$
0%
$3:1$

Kirchhoff's junction law is equivalent to

0%
conservation of energy
0%
conservation of charge
0%
conservation of electric potential
0%
conservation of electric flux

A cell of e.m.f. E is connected with an external resistance R; p.d. across the cell is v. the internal resistance of the cell will be:

0%
$( E-V) R/E$
0%
$( E-V) R/V$
0%
$(V-E)R/V$
0%
$(V-E)R/E$

Explanation

From ohm's law:

Let the current in the circuit

= i = \frac{V}{R}

$=i=\dfrac VR$

P.D Across the cell,

E = V + i r

$E=V+ir$

\Rightarrow r = \frac{E - V}{i} = \frac{E - V}{V / R} = (\frac{E - V}{V}) R

$\Rightarrow r=\dfrac{E-V}{i}=\dfrac{E-V}{V/R}=\left( \dfrac{E-V}{V}\right)R$

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10

Ω

$\Omega$ Its internal resistance is -

0%
2.0 ohm
0%
zero
0%
1.0 ohm
0%
0.5 ohm

Explanation

Hence, the option

C

$C$ is the correct answer
1380135_1361721_ans_f3162bf1df80461c994d11872afb4543.png

In the given circuit diagram , the currents ,

I_{1} = - 0.3 A, I_{4} = 0.8 A, a n d I_{5} = 0.4 A,

${ I }_{ 1 }=-0.3\quad A,{ I }_{ 4 }=0.8\quad A,and\quad { I }_{ 5 }=0.4\quad A,$ are flowing as shown . the currents

I_{2}

${ I }_{ 2 }$ ,

I_{3}

${ I }_{ 3 }$

0%
0.4 A, 1.1 A,0.4 A
0%
1.1 A, 0.4 A, 0.4 A
0%
1.1 A, -0.4 A, 0.4 A
0%
-0.4 A, 0.4 A , 1.1 A

The peak value of an alternating e.m.f. which is given by

E = E_{0}

$E={ E }_{ 0 }$

ω

$\omega$ is 10 volts and its frequency is 50 Hz. At time

t = \frac{1}{600}

$t=\dfrac { 1 }{ 600 }$ s, The instantaneous e.m.f.

0%
10 V
0%
$5\sqrt { 3 } V$
0%
5 V
0%
1 V

A potential gradient is created in the wire by standard cell for the comparision of emf's of two cells in a potentiometer experiment. Which possibility of the following will cause failure of the experiment.

0%
the emf of the standard cell is higher than that of the other cells.
0%
the diameter of the wires is equal and similar,
0%
3) the number of wires is ten.
0%
the emf of the standard cell is smaller than those of both the cells.

Four concentric spherical shell is

A, B, C

$A,B,C$ AND

D

$D$ has radius

a, 2 a, 3 a

$a, 2a, 3a$ and

4 a

$4a$ shell

B

$B$ and

D

$D$ has charge

+ q

$+q$ and

- q

$-q$ .The shell

^{'} C^{'}

$'C'$ is earthed. Find

V_{A} - V_{C}

$V_A - V_C$

0%
$\dfrac{kq}{2a}$
0%
$\dfrac{kq}{3a}$
0%
$\dfrac{kq}{4a}$
0%
$\dfrac{kq}{6a}$

Explanation

V_{C} = \frac{K q}{3 a} + \frac{{K q}^{1}}{3 a} - \frac{K q}{4 a}

${ V }_{ C }=\dfrac { Kq }{ 3a } +\dfrac { { Kq }^{ 1 } }{ 3a } -\dfrac { Kq }{ 4a }$

V_{C} = 0

${ V }_{ C }=0$

\frac{K q}{12 a} + \frac{{K q}^{1}}{3 a} = 0

$\dfrac { Kq }{ 12a } +\dfrac { { Kq }^{ 1 } }{ 3a } =0$

q^{1} = - \frac{q}{4}

${ q }^{ 1 }=-\dfrac { q }{ 4 }$

V_{A} = \frac{K q}{2 a} - \frac{\frac{K q}{4}}{3 a} - \frac{K q}{4 a}

${ V }_{ A }=\dfrac { Kq }{ 2a } -\dfrac { \dfrac { Kq }{ 4 } }{ 3a } -\dfrac { Kq }{ 4a }$

V_{A} = \frac{K q}{4 a} - \frac{K q}{12 a} = \frac{K q}{6 a}

${ V }_{ A }=\dfrac { Kq }{ 4a } -\dfrac { Kq }{ 12a } =\dfrac { Kq }{ 6a }$

V_{A} - V_{C} = \frac{K q}{6 a}

${ V }_{ A }-{ V }_{ C }=\dfrac { Kq }{ 6a }$

1244435_1377839_ans_67ca1cbaafbf4d7eac7a1449d62178fc.jpg

The potential difference

V

$V$ and the current

I

$I$ flowing an instrument in an ac circuit of frequency f are given by

V = 5 c o s ω t

$V = 5 cos \ \omega t$ volts and

I = 2 s i n ω t

$I = 2 sin \ \omega t$ amperes (where

ω = 2 π f

$\omega= 2 \pi f$ ). The power dissipated in the instrument is :-

0%
Zero
0%
10 W
0%
5 W
0%
2.5 W

Explanation

V = 5 c o s ω t = 5 s i n (ω t + \frac{π}{2})

$V=5\,cos\omega t=5\,sin(\omega t+\dfrac{\pi}{2})$ and

I = 2 s i n ω t

$I=2\,sin\omega t$

P o w e r = V_{r m s} \times I_{r m s} \times c o s ϕ = 0

$Power=V_{rms}\times I_{rms}\times cos\,\phi=0$

(Since

ϕ = \frac{π}{2}

$\phi=\dfrac{\pi}{2}$ , therefore

c o s ϕ = c o s \frac{π}{2} = 0

$cos\,\phi=cos\dfrac{\pi}{2}=0$ )

Current in a circuit falls from

5.0

$5.0$ A to

0.0

$0.0$ A in

0.1

$0.1$ S. If an average emf of

200

$200$ V induced, given an estimate of the self-inductance of the circuit.

0%
$4$ H
0%
$40$ H
0%
$0.4$ H
0%
$0.04$ H

Explanation

Change in current

d I = 5 - 0 = 5

$dI=5-0=5$ A

d t = 0.1

$dt=0.1$ s

E_{a v} = 200

${E}_{av}=200$ V

e = L \frac{d I}{d t}

$e=L\dfrac{dI}{dt}$

∴ 200 = L (\frac{5}{0.1})

$\therefore 200=L\left(\dfrac{5}{0.1}\right)$

\Rightarrow L = \frac{200}{50} = 4

$\Rightarrow L=\dfrac{200}{50}=4$ H

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 9 - 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 9 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

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