MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 13

Find the potential differences across the

$24\Omega$

0%
$48 volts$
0%
$2 volts$
0%
$4 volts$
0%
$1 volts$

Explanation

Let the equivalent resistance consisting of six resistances in parallel be

$R_{eq}.$

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}+\dfrac{1}{R_6}$

$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{1}{48}+\dfrac{1}{48}+\dfrac{1}{24}+\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1+1+2+8+12+24}{48}$

$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{48}{48}$

$\Rightarrow R_{eq}=1 \Omega$

Let the voltage across

$R_{eq}$ be

$V_{eq}.$

Applying Ohm's law for

$R_{eq}$

$V_{eq}=IR_{eq}$

$V_{eq}=(2 \ A) \times (1 \ \Omega)=2 \ V$

The voltage

$V_{eq}=2 \ V$ also appears across

$R_1$ ,

$R_2, R_3 . . . R_6.$

$\therefore$ Voltage across

$24 \Omega$ is

$V_{24 \ \Omega}=V_{eq}=2 \ V$

Answer is option (B).

A potentiometer wire has a resistance 40

$\Omega$ and its length is 10 m. It is connected to a resistance of 760

$\Omega$ in series. If emf of battery is 2 V then potential gradient is:-

0%
$0.5\times 10^{-6}V/m$
0%
$1\times 10^{-6}V/m$
0%
$1\times 10^{-2}V/m$
0%
$2\times 10^{-6}V/m$

A thin uniform ring of radius R carrying uniformly distributed charge Q and mass M rotates about its axis with angular velocity . The ratio of its magnetic moment and angular momentum is .

0%
$\frac{Q}{M}$
0%
$\frac{M}{Q}$
0%
$\frac{1}{2}\frac{Q}{M}$
0%
$\frac{1}{2}\frac{M}{Q}$

An aluminium (Al) rod with area of cross-section

$4\times 10^{-6}m^2$ has a current of 5A. Flowing through it. Find the drift velocity of electron in the rod. Density of

$Al=2.7\times 10^3kgm^{-3}$ and atomic wt.=Assume that each Al atom provides one electron.

0%
$8.6\times 10^{-4}ms^{-1}$
0%
$6.2\times 10^{-4}ms^{-1}$
0%
$1.2\times 10^{-4}ms^{-1}$
0%
$3.8\times 10^{-3}ms^{-1}$

Explanation

Given,

Cross sectional area is

$4\times { 10 }^{ -6 }{ m }^{ 2 }$

Current in Wire is

$5A$

Atomic Mass of copper is

$27g$

Density of copper is

$2.7\times { 10 }^{ 3 }kg{ m }^{ -3 }$

${ N }_{ A }=6.0\times { 10 }^{ 23 }Per gram mole$

As given one conduction electron per atom therefore Conduction electron density is calculated by,

$\frac { { N }_{ A } }{ mass\quad of\quad copper }\times density$

Substituting all value in equation

$=6.23\times { 10 }^{ 23 }\frac { electrons }{ mole } \times \frac { 1mole }{ 27g } \times \frac { 2.7 }{ 1c{ m }^{ 3 } } \times { \left( \frac { 100c{ m } }{ 1m } \right) }^{ 3 }\\ =6.23\times { 10 }^{ 28 }\frac { electrons }{ { m }^{ 3 } }$

We know that Current density is

$J=\frac { I }{ A } =ne{ v }_{ d }$

Here

${ v }_{ d }$ is current density

${ v }_{ d }=\frac { I }{ neA }$

Substituting all value in above equation

${ v }_{ d }=\frac { 5 }{ 6.23\times { 10 }^{ 28 }\times 1.61\times { 10 }^{ -19 }\times 4\times { 10 }^{ -6 } } =\frac { 5 }{ 40.12 } \times { 10 }^{ -3 }=1.2\times { 10 }^{ -4 }m/s$

Hence, Option (c) is correct option.

The emf of a celll is

$E$ , and its internal resistance is

$1\ \Omega$ . A resistance of

$4\Omega$ is joined to battery in parallel this is connected in secondary circuit of potentio meter the balancing length is

$160\ cm$ . If

$1V$ cell balances for

$100\ cm$ of potentio meter wire, the emf of cell

$E$ is

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

Pick up the correct statement.

0%
Charge on surface of inner sphere is non-uniformly distributed.
0%
Charge on surface of outer shell is non-uniformly distributed.
0%
Charge on surface of inner shell is non-uniformly distributed.
0%
All the above statements are false .

A battery of emf

$E_0 = 12 V is$ connected across a 4 m long uniform wire resistance

$4 \Omega m$ . The cells of small emfs

$E_1 = 2 V and E_2 = 4 V$ having internal resistance

$2 \Omega and 6 \Omega$ respectively, are connected as shown in the figure, If galvanometer of point N from the point A is is equal to

0%
2.5 m
0%
3 m
0%
1.25 m
0%
0.75 m

In he circuit shown, the current in the

$1\Omega$ resistor is :

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

A clock face has negative point charges

$-q, -2q, -3q.............-12q$ fixed at position of corresponding numerals, The clock hands donot perturb the net field due to the point charges. At what time does the hour hand point in the same direction as the field vector at the centre of dial.

0%
$6 : 0$
0%
$9 : 30$
0%
$9 : 0$
0%
$10 : 30$

An AC supply is connected to a resistor. When the peak value of the

$e.m.f.$ of the supply is

$V_0$ and the frequency is

$f$ , the mean power disspated in the resistor is

$P$ . The supply frequency is then changed to

$2f$ , the peak value of the

$e.m.f.$ remaining as

$V_0$ what is now the mean power in the resistor?

0%
P
0%
$\sqrt2$ P
0%
$2P$
0%
$4P$

A conducting loop is placed in a uniform magnetic field of induction

$B$ tesla with its plane normal to the field. Now, the radius of the loop starts shirking at the rate

$(\dfrac{dr}{dt})$ .
Then the induced emf at the instant when the radius is

$r$ is

0%
$\pi rB \left(\dfrac{dr}{dt} \right)$
0%
$2 \pi rB \left(\dfrac{dr}{dt} \right)$
0%
$\pi r^2 \left(\dfrac{dB}{dt} \right)$
0%
$\left(\dfrac{\pi r^2}{2}\right)^2 B\left(\dfrac{dr}{dt}\right)$

$n$ cells of each of EMF

$E$ and internal resistance

$r$ send the same current

$R$ whether the cells are connected in series or parallel, then :

0%
$R = nr$
0%
$R = r$
0%
$r = nR$
0%
$R = \sqrt {n} \dfrac{E}{r}$

The diagram shows three capacitor with their capacitances with breakdown voltages. what should be the maximum value of external emf of source such that no capacitor breaks down :-

0%
$\frac{ 33 }{ 2 }$ volt
0%
$\frac{ 11 }{ 3 }$ volt
0%
$\frac{ 13 }{ 3 }$ volt
0%
$\frac{ 11 }{ 2 }$ volt

Explanation

Maximum charge on

$1\mu F$ ,

$Q_1 = CV = 1\times 2= 2\mu C$

Maximum charge on

$2\mu F$ ,

$Q_2 = CV = 2\times 1= 2\mu C$

Maximum charge on

$3\mu F$ ,

$Q_3 = CV = 3\times 3= 9\mu C$

So, maximum charge on the system is

$Q = 2\mu C$

Equivalent capacitance

$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$

So,

$\dfrac{1}{C_{eq} }=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}$

$\implies \ C_{eq} = \dfrac{6}{11} \ \mu F$

External emf

$V = \dfrac{Q}{C_{eq}} = \dfrac{2}{6/11} = \dfrac{11}{3} \$ volt

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

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

A sphere of

$4$ cm radius is suspended within a hollow sphere of

$6$ cm radius. The inner sphere is charged to a potential

$3$ e.s.u. When the outer sphere is earthed, the charge on the inner sphere is:

0%
$54$ e.s.u.
0%
$\dfrac{1}{4}$ e.s.u.
0%
$30$ e.s.u.
0%
$36$ e.s.u.

Explanation

The sphere of radius

$=4cm$

$=4\times { 10 }^{ -2 }m$

hollow sphere

$=6cm$

$=6\times { 10 }^{ -2 }m$

inner sphere charged to a potential

$3$ e.s.u.

$1esu=300V$

$1esu=3\times { 10 }^{ -10 }C$

Let, Charge on inner sphere is

$+Q$ , then,

$-Q$ is induced in the surface of outer sphere.

Now, potential on inner sphere

$=\dfrac { { Q }_{ 1 } }{ { r }_{ 1 } } +\dfrac { { Q }_{ 2 } }{ { r }_{ 2 } }$

Here,

${ Q }_{ 1 }$ is the charge on inner sphere.

${ r }_{ 1 }$ is the radius of inner sphere.

${ Q }_{ 2 }$ is the charge on outer sphere.

${ r }_{ 2 }$

$=$ the radius of the outer sphere.

Given, Potential on inner sphere

$=3esu$ of potential

${ r }_{ 1 }=4cm$

${ r }_{ 2 }=6cm$

$3esu=\dfrac { Q }{ 4 } -\dfrac { Q }{ 6 }$

$\Rightarrow Q\left[ \dfrac { 1 }{ 4 } -\dfrac { 1 }{ 6 } \right] =3$

or,

$\dfrac { Q }{ 12 } =3$

or,

$Q=36 esu$

Hence, charge

$=36esu$ of charge.

1240454_1221072_ans_794a2c51475246e79f73c48652f8bbaf.jpg

The drift velocity of free electrons in a conductor is

$v$ , when a current.

$I$ is flowing in it if both the radius and current are doubled, then drift velocity will be.

0%
$v/4$
0%
$v/2$
0%
$2v$
0%
$4v$

Explanation

$\textbf{Step1 : Relation between current and drift velocity }$

We know that,

$I=neAv$

where,

$A$ is the cross-sectional area

$n$ is the no. of electrons.

$e$ is the charge of electrons.

$\therefore v=\dfrac{I}{neA}$

$=\dfrac{I}{n e\pi r^2}$

$\Rightarrow v\propto \dfrac{I}{r^2}$

$......(1)$

$\textbf{Step-2: New drift velocity}$

Let, New drift velocity

$=v'$ ,

New radius,

$r'=2r$

New current,

$I'=2I$

From equation

$(1)$

$\dfrac{v'}{v}=\dfrac{(I'/r'^2)}{(I/r^2)}=\dfrac{I'}{I}\times \left(\dfrac{r}{r'}\right)^2$

$\Rightarrow\ \dfrac{v'}{v}=\dfrac{2I}{I}\times \left(\dfrac{r}{2r}\right)^2$

$\therefore\ \ v'=\dfrac{v}{2}$

$\therefore$ New drift velocity is

$\dfrac{v}{2}$

Hence, Option B is correct.

If a variable resistance is connected to a cell of constant emf then, which graph represent the relation between current

$I$ and resistance

$R$ ?

In the figure the magnet is moved along the axis of coil from one position to another position in

${10^{ - 3}}$ sec.Now magnet is at rest for 2 sec, in its new position.The duration of induced emf in the coil is:

0%
${10^{ - 3}}$ sec
0%
2 sec
0%
2x ${10^{ - 3}}$ sec
0%
0.5x2x ${10^{ - 3}}$ sec sec

Consider a cell of emf

$E$ and internal resistance

$r$ . The current drawn (i)from cell can be varied . Output power from cell is same when current drawn is

${i_1} = 2A$ and

${i_2} = 8A$ . For what value what value of current the output power will be maximum.

0%
$4A$
0%
$6A$
0%
$2A$
0%
$5A$

Calculate the value of current I in given circuit .

0%
2 A
0%
1 A
0%
1 .5 A
0%
5 A

Explanation

REF.Image

in the given figure inclined wire is shorting the circuit,

and due to this shorting No current will flow through

Resister B,D,E,F (as marked in figure)

so for analysis parpose we can remove this resistor and

modified figure would be :

So Total resistance =

$5+5= 10\Omega$

so current

$(I)=\frac{V}{R}= \frac{15}{10}= 1.5A$

1195076_1275780_ans_dd277b2680194e9ab0b1a4efd82bd508.jpg

$8$ identical spherical conducing droplets are charged to the same potential such that energy stored in the electrical field of each is

$E_0$ . When they are mixed form a large drop, the new energy stored in the electrical field is

0%
$8E_0$
0%
$16E_0$
0%
$32E_0$
0%
$64E_0$

Dimension of electromotive force (e.m.f) is -

0%
$M{L}^{2}{T}^{-2}$
0%
$M{L}^{2}{T}^{-2}{I}^{-3}$
0%
$ML{T}^{-2}$
0%
$M{L}^{2}{T}^{-3}{I}^{-1}$

EMF of a battery, when compared to its potential difference, is

0%
less
0%
more
0%
equal
0%
none of these

Kirchhoff's equation is

0%
$log \frac{K_2}{K_1} = \frac{E_a}{2.303 R}[\frac{1}{T_1} - \frac{1}{T_2}]$
0%
$log \frac{P_2}{P_1} = \frac{\Delta H_v}{2.303 R}[\frac{T_2 - T_1}{T_1 T_2}]$
0%
$\Delta C_p = \frac{\Delta H_2 - \Delta H_1}{T_2 - T_1}$
0%
$log \frac{K_2}{K_1} = \frac{\Delta H_v}{2.303 R}[\frac{1}{T_1} - \frac{1}{T_2}]$

Diagram shows three capacitors with capacitance and breakdown voltage mentioned. What should be maximum value of the external emf of source such that no capacitor breaks down?

0%
1 V
0%
2 V
0%
1.5 V
0%
4 V

Explanation

We know that

$q=cv$

Total charge of two parallel circuit is ,

${ q }_{ T }=CV+2CV\\ \quad \quad q_T=3CV$

Maximum charge stored by 6C capacitor

$q=6C\times V=6CV$

Hence q < q

$_T$ So Maximum charge flow through the battery

$3CV$

And

${ C }_{ eq }=\dfrac { 3\times 6 }{ 3+6 } =2C$

$Source\quad voltage=\dfrac { 3C }{ 2C } =1.5V$

CORRECT OPTION IS C

Which of the following statement is false?

0%
Wheatstone bridge is the most sensitive, when all the four resistance are of the same order of magnitude.
0%
In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.
0%
A rheostat can be used as a potential divider.
0%
Kirchhoff's second law represents energy conservation

The current i in the circuit (see figure) is :

0%
$\dfrac {1}{45} A$
0%
$\dfrac {1}{15} A$
0%
$\dfrac {1}{10} A$
0%
$\dfrac {1}{5} A$

Explanation

R.E.F.Image.

Apply KVL Law from ODAB

$2-30i_{1}=0$

$i_{1}= 2/30 = 1/15 AMP$

Now apply KVL in ODA CBO

$2-30(i-i_{1})-30(i-i_{1})=0$

$2 = 60 (i-i_{1})$

$\frac{1}{30}=(i-i_{1})$

$i = \frac{1}{15} = \frac{1}{30}$

$i = \frac{1}{30}+\frac{1}{15}= \frac{3}{30}$

$i = \frac{1}{10}AMP$

1222094_1280116_ans_32cd5330bfae4cfe8d5399ffa6ce46ff.png

A bar magnet is paced on the axis in a circular coil, close to the coil. Both of them are moved in the same direction. The coil covers

$1\ m$ in

$0.5 s$ and the magnet travels,

$2\ m$ in

$1s$ then the introduced emf produced in the coil is _______ V.

0%
$0.5$
0%
$0$
0%
$2$
0%
$1$

In the circuit diagram given below, a cell of

$9V$ and internal resistance

$0.5 \Omega$ is connected across a resistor A of

$2\Omega$ in series and two resistors

$2\Omega$ and

$6 \Omega$ which are in parallel.
a) What will be the values of the total resistance

b) the total current

c) the current in the

$6\Omega$ resistor and

d) the potential difference across the terminals of the cell respectively:

0%
$4\Omega, 2.5 A, 0.5625 A, 7.875 V$
0%
$4\Omega, 3 A, 0.5625 A, 7.875 V$
0%
$4\Omega, 2.5 A, 0.5625 A, 8 V$
0%
$4\Omega, 2.5 A, 0.75 A, 7.875 V$

A wire of length 1 m and radius

$10^{-3}$ m is carrying a heavy current and is assumed to radiate as a black body. At equilibrium, its temperature is 900 K while that of surrounding is 300 K . The resistivity of the material of the wire at 300 K is

$\pi^2 \times 10^{-8}$ ohm-m and its temperature coefficient of resistance is

$7.8 \times 10^{-3}/^oC$ ( Stefan's constant

$\sigma = 5.68 \times 10^{-8} W/m^2K^4 )$

The resistivity of wire at 900 K is nearly

0%
$2.4 \times 10^7 ohm-m$
0%
$2.4 \times 10^{-7} ohm-m$
0%
$1.2 \times 10^{-7} ohm-m$
0%
$1.2 \times 10^7 ohm-m$

Two cells of e.m.f. 10

$\mathrm { V } \& 15 \mathrm { V }$ are connected in parallel to each other between points A

$\& \mathrm { B }$ .The cell of e.m.f. 10

$\mathrm { V }$ is ideal but the cell of e.m.f. 15

$\mathrm { V }$ has internal resistance 1

$\mathrm { \Omega }$ . The equivalent e.m.f. between

$\mathrm { A }$ and

$\mathrm { B }$ is:

0%
$\frac { 25 } { 2 } \mathrm { V }$
0%
not defined
0%
15 $\mathrm { V }$
0%
10 $\mathrm { V }$

In the given circuit of potentiometer length of the wire is 6 m then current through the galvanometer when jockey touches the wire at 5.6 m.

0%
1.5 from negative terminal to jockey
0%
1.5 A from jockey to negative terminal
0%
3A from jockey to negative terminal
0%
3A from negative terminal to jockey

A voltmeter of 998 ohm resistance is connected to a cell of emf 2 volt, having internal resistance of 2 ohms The error in measuring emf will b

0%
$4 \times 10 ^ { - 1 } \mathrm { V }$
0%
$2 \times 10 ^ { - 3 } \mathrm { V }$
0%
$4 \times 10 ^ { - 3 } \mathrm { V }$
0%
$2 \times 10 ^ { - 1 } \mathrm { V }$

Explanation

$\begin{array}{l} V=iR=\dfrac { E }{ { r+R } } \times R \\ =\dfrac { 2 }{ { 2+998 } } \times 998=1.996\, V \\ \therefore \Delta V=2-1.996=0.004=4\times { 10^{ -3 } }\, V \end{array}$

Hence,

option

$(C)$ is correct answer.

1237708_1371202_ans_de27a91f88aa45578ad51f361e4e597e.JPG

The emf of the battery shown in figure is

0%
$12$ $V$
0%
$16$ $V$
0%
$18$ $V$
0%
$15$ $V$

For different values of resistance, R power consumption in R are given. Then which of the following values are possible ?
(a) 2 W
(b) 5 W
(c) 8 W
(d) 4 W

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

A bulb rated at (100W-200V) is used on a 100 V line. The current in the bulb is

0%
1/4 amp
0%
4 amp
0%
1/2 amp
0%
2 amp

Explanation

Given,

$P = 100\ W$ ; power of the bulb

$V = 200\ V$ ; voltage rating

So value of resistance-

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

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

$\Rightarrow R = \dfrac{200^2}{100}$

$\Rightarrow R = 400\ \Omega$

Now applied voltage changes to

$100\ V$ , so current passing through the resistance-

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

$\Rightarrow I = \dfrac{100}{400}$

$\Rightarrow I = \dfrac{1}{4} A$

A metre bridge cannot be used to determine

0%
resistance of a wire.
0%
specific resistance,
0%
conductivity.
0%
e.m.f of a cell.

Potential gradient remains constant until

0%
current in potentiometer wire remains constant
0%
potential difference across the potentiometer wire remains constant
0%
none of the above
0%
both

In the given figure if

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

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

$i _ { 3 }$

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

$V \rightarrow I$ graph is given above, observation of object

$P$ is done at

$T _ { 1 }$ temperature and observation of object

$Q$ is done at

$T _ { 2 }$ temperature. What we can conclude for these two temperature?

0%
$T _ { 1 } > T _ { 2 }$
0%
$T _ { 1 } < T _ { 2 }$
0%
$T _ { 1 } = T _ { 2 }$
0%
We can not predict from the graph

Figures (38-E5) shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. the front edge entres the 20 cm wide magnetic field at t=0 . find the emf induced in the loop at

0%
$t=2s,$ $3\times 10^{-4}$ V
0%
$t=10 s,$ $3\times 10^{-4}$ V
0%
$t= 22 s,$ $0$ V
0%
$t= 30s,$ $0$ V

An external variable load resistance R is connected with a battery of emf E and internal resistance r. The output power across R is maximum when

0%
$R=r$
0%
$R=\frac{r}{2}$
0%
$R=2r$
0%
$R=\frac{r}{4}$

A cell of constant emf first connect to a resistance

$R_1$ and then to connected to the resistance

$R_2.$ If power delivered in both cases in the same then internal resistance of the cell

0%
$\dfrac { R _ { 1 } - R _ { 2 } } { 2 }$
0%
${ R _ { 1 } + R _ { 2 } }$
0%
$\sqrt { R _ { 1 } R _ { 2 } }$
0%
$\sqrt { R _ { 1 } + R _ { 2 } / 2 }$

Explanation

The cell has constant emf $,$

Resistance is $R_1$ and $R_2$

$P = V_2 / R_1 + r$

$P = V_2 / R_2 - r$

$( R_1- R_2 ) / 2$

The internal resistance of the cell is $( R_1 - R_2 ) / 2$

Hence,

option $(A)$ is correct answer.

In which of the following cases,potential at point O is not zero?

For a

$DC$ circuit, Kirchoff's rules yield the following equations.

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

$10 = 3I_{1}-2I_{2}$

$50=2I_{2}+9.6I_{3}$
What is the current

$I_{2}$ (Amps)?

0%
$0.131$
0%
$-1.37$
0%
$0.245$
0%
$-3.5$
0%
$1.00$

A train of mass

$100$ tons (

$1$ ton

$= 1000 { kg }$ ) runs on a meter gauge track (distance between the two rails is

$1{ m } .$ The coefficient of friction between the rails and the train is

$0.045 .$ The train is powered by an electric engine of

$90 \%$ efficiency. The train is moving with uniform speed of

$72$ Kmph at its highest speed limit. Horizontal and vertical component of earth's magnetic field are

$B _ { H } = 10 ^ { - 5 } T$ and

$B _ { V } = 2 \times 10 ^ { - 5 } T.$ Assume the body of the train and rails to be perfectly conducting.
If now a resistor of

$10 ^ { - 3 } \Omega$ is attached of between the two rails, the extra units of energy (electricity) consumed during a

$324 { km }$ run of the train is (

$1$ unit of power

$= 1 kW$ hour ) (assume the speed of train to remain unchanged)

0%
$8 \times 10 ^ { - 4 } { KW }$ hour
0%
$8 \times 10 ^ { - 5 } { KW }$ hour
0%
$8 \times 10 ^ { - 6 } { KW }$ hour
0%
$8 \times 10 ^ { - 7 } { KW }$ hour

A hall is used 5 hours a day for 25 days in month. it has 6 lamps of 100 W each and 4 fans of 15 W.the total power consumed for the month is

0%
1500 kWh
0%
82.5 kWh
0%
15 kWh
0%
1.5 kWh

A train of mass

$100$ tons (

$1$ ton

$= 1000 { kg }$ ) runs on a meter gauge track (distance between the two rails is

$1{ m } .$ The coefficient of friction between the rails and the train is

$0.045 .$ The train is powered by an electric engine of

$90 \%$ efficiency. The train is moving with uniform speed of

$72$ Kmph at its highest speed limit. Horizontal and vertical component of earth's magnetic field are

$B _ { H } = 10 ^ { - 5 } T$ and

$B _ { V } = 2 \times 10 ^ { - 5 } T.$ Assume the body of the train and rails to be perfectly conducting.
The electrical power consumed by the train is -

0%
$1.11 MW$
0%
$1 MW$
0%
$0.50 MW$
0%
$0.90 MW$

In Fig . I is equal to

0%
1.5A
0%
0.4A
0%
0.9A
0%
0.7A

In an experiment of potentiometer for determining a small resistance, the balancing length for the potential difference across the large resistance is

$320\ cm$ and the balancing length for the potential difference across both resistances when connected in series, is

$360\ cm$ . The ratio of large and small resistance is

0%
$1 : 8$
0%
$8 : 1$
0%
$8 : 9$
0%
$9 : 8$

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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