MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 6

A potential difference of 20V is needed to make a current of 0.05A flow through a resistor. What potential difference is needed to make a current of 300 mA flow through the same resistor?

0%
60V
0%
120V
0%
40V
0%
150V

Explanation

Answer is B.

According to Ohm's law,

$V = IR$ . That is,

$R = \dfrac VI$ .
In this case, the voltage is

$20\ V$ and the current is

$0.05\ A$ .
So, the resistance

$R = 20/0.05 = 400$ ohms.
Now, to make a current of 300 mA flow through the same resistor of 400 ohms, the potential difference to be applied is calculated as follows.
Here,

$I = 300 mA = 0.3\ A$ and

$R = 400$ ohms.
Therefore,

$V = 0.3\times 400 = 120\ V$ .
Hence, to make a current of 300 mA flow through the same resistor of 400 ohms a potential difference of 120 V is applied.

Kilowatt hour (kWh) represents the unit of :

0%
power
0%
impulse
0%
momentum
0%
energy

Explanation

1 kW h = 1 kW ×1 h = 1000 W × 3600 s = 3600000 J = 3.6

$\ast$ 10

$^{6}$ J

The energy used in households, industries and commercial establishments are usually expressed in kilowatt hour.

A cell of emf

$5$ V can supply a total energy of

$9000$ J, then the total charge that can be obtained from the cell would be _____ C

0%
$180$
0%
$18000$
0%
$1800$
0%
$18$

Explanation

EMF = 5V

Energy = 9000J

Power = Voltage

$\times$ current

P = VI (

$\varepsilon$ = energy )(t = time )(t = time)(it = q)

$\varepsilon =VIt\\ \varepsilon =Vq\\ \dfrac { 9000 }{ 5 } = 1800C\\ q=1800c$

Power of an electric heater is

$1000$ W and it is running for

$1$ hour. Calculate the energy consumed by it

0%
$3.6$ MJ
0%
$1.2$ MJ
0%
$1.8$ MJ
0%
$4.2$ MJ

Explanation

Power

$=P=1000W$

$t=1h=3600s$

Power is energy consumed per unit time.

$P=\dfrac{E}{t}$

$\implies E=Pt$

$\implies E=3600\times 1000$

$\implies E=3.6\times 10^6J$

Energy

$=E=3.6MJ$

Answer-(A)

Electric current cannot flow through

0%
Silver
0%
Wood
0%
Copper
0%
None of these

State whether given statement is True or False
Silk threads are very good conductors.

0%
True
0%
False

Explanation

Silk threads are bad conductor electricity. so it allows less amount of current at high voltage applied.

Given statement is

$False$ .

The amount of work done (or, energy spent) by a source of power

$1kW$ in

$1h$ time is:

0%
$1J$
0%
$1Wh$
0%
$1kWh$
0%
$1calorie$

Explanation

$Energy$

$spent$ =

$Power \times time$

$E=1kWh \times 1h$ =

$1kWh$

$1kWh$ is the amount of work done (or, energy spent) by a source of

$1kW$ power in

$1h$ time.

Convert 1 kWh to SI unit of energy.

0%
$3.6\times 10^6J$
0%
$3.6\times 10^8J$
0%
$1.8\times 10^6J$
0%
$1.8\times 10^8J$

Explanation

We know that,

$1\ watt = 1\ joule/sec$ and

$1\ kW=1000\ watt$

Therefore,

$1\ kWh$

$= 1000\ watt\times 1\ hour = 1000\ joule/sec \times 3600\ sec$

$= 3.6\times 10^6$

$J$

Calculate the electric current in the circuit shown.

0%
$1.5A$
0%
$0.5A$
0%
$2.5A$
0%
$2A$

Explanation

Any current that passes through the

$10 \Omega$ resistor also passes through

$5 \Omega$ , i.e.,

$10\Omega$ and

$5 \Omega$ are connected in series. Their equivalent resistance is

$R_{eq} = 10 \Omega + 5 \Omega = 15 \Omega$ .

The equivalent circuit is shown in the above figure.
The current

$\displaystyle i = \frac {V}{R_{eq}}=\frac {7.5 V}{15 \Omega}=0.5A$

Some matters do not allow the electricity to pass through it. What is called a matter which does not allow the electricity to pass though it?

0%
Conductor
0%
Insulator
0%
Semiconductor
0%
All of these
0%
None of these

Two heater coils

$A$ and

$B$ made of the same material are connected in parallel across the same mains. The length and diameter of the wire used in

$A$ are doubled as that of the wire used in

$B$ . If

$H_1$ and

$H_2$ are the quantities of heat liberated per second in

$A$ and

$B$ , respectively, then

$H_1 : H_2 =$

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

Explanation

Since the coils are in parallel across the same mains,

$V$ is the same for both

$\displaystyle \frac {H_1}{H_2} = \frac {R_2}{R_1} \left ( \because H= \frac {V^2}{R}\right)$

$\displaystyle \frac {H_1}{H_2} = \frac {2}{1}$

The substance which allows electric current to flow through it is called

0%
Conductor
0%
Insulator
0%
Semiconductor
0%
None of these

Explanation

The substance which allows electric current to pass through it are called

$conductor.$

Answer-(A)

The current in the given circuit is

0%
$0.3\;\text{amp}$
0%
$0.4\;\text{amp}$
0%
$0.1\;\text{amp}$
0%
$0.2\;\text{amp}$

Explanation

Two batteries are joined with opposite polarity so, total e.m.f.=

$5-2=3\;V$
Total resistance =

$10+20=30\Omega$
Current =

$\displaystyle\frac{3}{30}=0.1\;\text{A}$

Watt hour represents:

0%
Electric energy
0%
Current
0%
Voltage
0%
Power

Explanation

The watt-hour (Wh) is a unit of electric energy equivalent to one watt (1 W) of power expended for one hour (1 h) of time. Watt-hour is commonly used in electrical applications.

$1\ Wh = 1\times 60\times 60 = 3600\ J$

Unit used in selling electrical energy to consumer is:

0%
$Volt-Ampere$
0%
$Kilowatt-hour$
0%
$Volt/second$
0%
None of these

Explanation

The commercial unit of electrical energy is

$Kilowatt-hour\ (kWh)$ .

$1\ kWh$ is equal to

$1\ unit$ of energy.

An electric bulb of resistance

$20 \Omega$ draws a current of

$0.04 A$ . Calculate the potential difference at the ends.

0%
$0.8 V$
0%
$2 V$
0%
$0.032 V$
0%
$5 V$

Explanation

Given,

Resistance of resistor

$R = 20 \Omega$

Current draws

$I = 0.04 A$

According to ohm's law

Potential difference,

$V = IR = 0.04 \times 20 = 0.8 V$ .

Hence Option A

Number of KWh in 1 Joule.

0%
$\displaystyle 3.6\times { 10 }^{ 6 }\ KWh$
0%
$\displaystyle 2.77\times { 10 }^{ -7 }\ KWh$
0%
$\displaystyle 600\ KWh$
0%
$\displaystyle 1.6\times { 10 }^{ -19 }\ KWh$

Explanation

We know that

$KWh$ is kilo watt-hour-

In SI unit-

$K = 10^3$

$hour = 3600\ s$

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

$\therefore$

$1$

$J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$

$kWh$

Which of these units of energy is the largest?

0%
$1\ Calorie$
0%
$1\ Joule$
0%
$1\ Erg$
0%
$1\ Kilowatt-hour$

Explanation

Energy is the ability to do work.

$1\ {calorie} = 4.186\ J$

$1\ erg= 1\times 10^{-7}\ J$

$1\ kWh=3.6\times 10^{6}$

$J$

$1\ kilowatt-hour$ is the largest unit of energy.

Smallest commercial unit of energy is

0%
Kilowatt hour
0%
Watt second
0%
Watt hour
0%
Watt minutes

Explanation

Smallest commercial unit of energy is called Watt hour

$(Wh)$ . One watt hour is the energy consumed when

$1 \ watt$ of power is used for

$1 \ hour.$

Energy consumed in our house has a unit of

0%
watts
0%
horse power
0%
kilowatt-hours
0%
none

Commercial unit of energy is .......................

0%
Joule
0%
Volt
0%
Kilowatt hour
0%
Watt second

Explanation

Commercial unit of electrical energy is kilowatt hour

$\left(kWh\right)$

$1 kwh = 3.6 \times {10}^{6} J$

Calculate the monthly bill if a heater of 100 watt is used at the rate of rs. 1 per unit for 1 hour daily.

0%
Rs. 200
0%
Rs. 6000
0%
Rs. 100
0%
Rs. 3

Explanation

Given :

$P=0.1$ kW

$t = 1$ hour per day

Energy consumed per day

$E = Pt = 0.1\times 1 = 0.1$ kWh per day

Total energy consumed per month

$E_T = 30\times 0.1 = 3$ kWh per month

Cost of electricity

$=Re$

$1$ per unit i.e per kWh

$\therefore$ Monthly bill

$= Re.$

$1\times 3 =$ Rs.

$3$ per month

An electron in potentiometer with experiences a force

$2.4 \times {10}^{-19} N$ . The length of potentiometer wire is

$6 m$ . The emf of the battery connected across the wire is (electronic charge

$= 1.6 \times {10}^{-19} C$ )

0%
$6 V$
0%
$9 V$
0%
$12 V$
0%
$15 V$

Explanation

Force experienced by electron

$F = qE$ where

$E =$ Electric field intensity

$\therefore 2.4 \times {10}^{-19} = 1.6 \times {10}^{-19} E \Rightarrow E = 1.5 {N}/{m}$
Moreover

$E =$ potential gradient

$E = \dfrac{dV}{dl}$

$\therefore E \times l = V \Rightarrow V = 1.5 \times 6 = 9 V$

Copper and Carbon wires are connected in series and the combined resistor is kept at

$\displaystyle { 0 }^{ \circ }C$ . Assuming the combined resistance does not vary with temperature, the ratio of the resistances of Carbon and Copper wires at

$\displaystyle { 0 }^{ \circ }C$ is:
(Temperature coefficients of resistivity of Copper and Carbon respectively are

$\displaystyle 4\times { { 10 }^{ -3 } }/{ ^{ \circ }{ C } }$ and

$\displaystyle -0.5\times { { 10 }^{ -3 } }/{ ^{ \circ }{ C } }$ )

0%
$8$
0%
$6$
0%
$2$
0%
$4$

Explanation

It is given that combined series resistance does not change with temperature.

Thus,

$R_{Cu}+R_{C}=R_{Cu}(1+\alpha_{Cu}\Delta T)+R_C(1+\alpha_C\Delta T)$

for all values of

$\Delta T$

Thus,

$R_{Cu}\alpha_{Cu}=-R_{C}\alpha_{C}$

$\implies \dfrac{R_C}{R_{Cu}}=-\dfrac{\alpha_{Cu}}{\alpha_C}$

$=-\dfrac{4\times 10^{-3}}{-0.5\times 10^{-3}}=8$

The electric current

$i$ in the circuit shown is:

0%
$6\ A$
0%
$2\ A$
0%
$3\ A$
0%
$4\ A$

Explanation

Using Kirchhoff's junction law at A :

$3 + 2 = i_1$

$\implies i_1 = 5$ A

Using Kirchhoff's junction law at B :

$i_1 = 2 + i_2$

$\therefore$

$5 = 2 + i_2$

$\implies i_2 = 3$ A

Using Kirchhoff's junction law at C :

$i_2 + 1 = i$

$\therefore$

$3 + 1 = i$

$\implies i = 4$ A

This question relates to the DC circuit shown below.
The values for current, voltage, and resistance in the circuit are graphed from point A through point G in the graphs directly below.
Which of the graphs shows the voltage from point A to point G?

0%
A
0%
B
0%
C
0%
D
0%
E

A steady current flow in a metallic conductor of non-uniform cross-section. The quantity/quantities remaining constant along the whole length of the conductor is/are:

0%
current, electric field and drift speed
0%
drift speed only
0%
current and drift speed
0%
current only

Explanation

When a steady current flows in a metallic conductor of non-uniform cross-section then the drift speed is

$V_d=I/neA$ and the electric field

$E=I/ \rho A$

$\Rightarrow V_d\propto \dfrac{1}{A}$ and

$E\propto \dfrac{1}{A}$

$\Rightarrow$ Only current remains constant.

In a metallic conductor of non-uniform cross-section, only the current remains constant along the entire length of the conductor.

Two cells A and B of emf 2V and 1.5 V respectively, are connected as shown in figure through an external resistance

$10\Omega$ . the internal resistance of each cell is

$5\Omega$ . The potential difference

$E_A$ and

$E_B$ across the terminals of the cells A and B respectively are:

0%
$E_A = 2.0 V. E_B = 1.5 V$
0%
$E_A = 2.125 V, E_B = 1.375 V$
0%
$E_A = 1.875 V, E_B = 1.625 V$
0%
$E_A = 1.875 V, E_B = 1.375 V$

Explanation

$R_{eq} = 10 + 5 + 5 = 20 \Omega$

$V_{eq} = 2 - 1.5 = 0.5V$

$I = \dfrac{V_{eq}}{R_{eq}} \\ \\ I = \dfrac{0.5}{20} \\ I = 0.025 A$

$E_A = 2 - R_AI \\ E_A = 2 - 5 * 0.025 = 2 - 0.125 = 0.875 V$

$E_B = 1.5 + R_BI \\ E_B = 1.5 + 5 * 0.025 = 1.5 + 0.125 = 1.625V$

Hence option 'C' is correct.

A meter bridge shown in fig.is balanced.what is the value of resistance X.

0%
$20 \Omega$
0%
$10 \Omega$
0%
$5 \Omega$
0%
$2.5 \Omega$

Explanation

Since the meter bridge is balanced,

$\dfrac{10}{0.5} = \dfrac{X}{0.5}$

$X = 10 \Omega$

In shown diagram, if voltage of battery is doubled and resistance kept constant then what is the new Power dissipated at resistor(Consider

$P$ was the initial power dissipation) ?

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

Explanation

Power,

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

As R is kept constant so

$P \propto V^2$

$P'$ is new power dissipation.

So,

$\dfrac{P'}{P}=\dfrac{(2V)^2}{V^2}=\dfrac{4V^2}{V^2}$ or

$P'=4P$

Thus, option E is correct.

Potentiometer measures the potential difference more accurately than a voltmeter because :

0%
it has a wire of high resistance
0%
it has a wire of low resistance
0%
it does not draw current from external circuit
0%
it draws a heavy current from external circuit

$2V$ battery, a

$990 \Omega$ resistor and a potentiometer of

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

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

0%
$0.05Vm^{-1}$
0%
$0.5Vm^{-1}$
0%
$0.01Vm^{-1}$
0%
$0.001Vm^{-1}$

Explanation

Potential gradient

$x=\displaystyle\frac{e}{(R+R_{h}+r)}\cdot \frac{R}{L}$

$\Rightarrow \displaystyle x=\frac{2}{(990+10)}\times \frac{10}{2}=0.01Vm^{-1}$

The circuit shows a battery and a variable resistor supplying power to a light circuit. As the variable resistor is moved towards left, the bulbs dim a little. What happens if the variable resistor is moved all the way to the right?

0%
The lights brighten considerably
0%
The total circuit resistance increases
0%
The total circuit current decreases
0%
The total applied voltage increases
0%
The total power used in the circuit increases

If current in a source of e.m.f. is in the direction of e.m.f., the energy of the source

0%
Increase
0%
Decrease
0%
Remains constant
0%
Zero

The five different colored graphs above represent different possible relationship between voltage across a resistor and current through the resistor.
Which of the graphs shows the correct relationship for a resistor that obeys Ohm's law?

0%
Red
0%
Orange
0%
Blue
0%
Purple

Explanation

Ohm's law states that if physical condition are unchanged then voltage is directly proportional to current ,

$V\propto I$ ,

this relation is best depicted by blue graph , therefore blue graph obeys Ohm's law .

What do you add to distilled water for making it to conduct electricity.

0%
sulphuric acid
0%
water
0%
milk
0%
base

In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell?

0%
$\simeq 1.57 V$
0%
$\simeq 1.67 V$
0%
$\simeq 1.47 V$
0%
$\simeq 1.37 V$

Explanation

Given :

$E_1 = 1.25$ V

$l_1 = 30$ cm

$l_2 = 40$ cm

Emf of the cell

$E\propto l$

$\implies$

$\dfrac{E_2}{E_1} = \dfrac{l_2}{l_1}$

$\therefore$

$\dfrac{E_2}{1.25} = \dfrac{40}{30}$

$\implies$

$E_2 = 1.67$ V

Kirchoffs 1st and 2nd laws are based on conservation of

0%
Energy and charge respectively
0%
Charge and energy respectively
0%
Mass and charge respectively
0%
None of these above

Explanation

Kirchhoff's

$1$ st law states that the total amount of current coming at a junction must be equal to the total amount of current going away from it i.e. the total charge must be conserved. Thus Kirchhoff's

$1$ st law is based on the conservation of charge.

Kirchhoff's

$2$ nd law states that the sum of the potential drop across all the components in a loop must be zero. Thus Kirchhoff's

$2$ nd law is based on the conservation of energy.

A steady electric current is flowing through a cylindrical conductor.

0%
The electric field at the axis of the conductor is zero.
0%
The magnetic field at the axis of the conductor is zero.
0%
The electric field in the vicinity of the conductor is zero.
0%
The magnetic field in the vicinity of the conductor is zero.

Human body is a

0%
good conductor of electricity
0%
poor conductor of electricity
0%
semiconductor
0%
good insulator

Potentiometer measures potential more accurately because

0%
it measures potential in the open circuit
0%
it uses sensitive galvanometer for null deflection
0%
it uses high resistance potentiometer wire
0%
it measures potential in the closed circuit

$1$ kilowatt hr

$=$ __________ joules.

0%
$10.6 \times {10}^{6}$
0%
$3.6 \times {10}^{6}$
0%
$30.6 \times {10}^{6}$
0%
$3.6 \times {10}^{5}$

Explanation

$kWh$ is the commercial unit of energy.

$1\ kWh =1\times 1000\times 3600 = 3.6\times 10^6$ joules

For driving current of

$2A$ for

$6$ minute in a circuit.

$1000 J$ of work is to be done. The e.m.f. of the source in the circuit is

0%
$1.38V$
0%
$1.68V$
0%
$2.03V$
0%
$3.10V$

Explanation

Current passing through the circuit

$I = 2 A$

Time of operation

$t = 6$ minute

$= 360$ s

Work done

$W = 1000 J$

Using

$W = VIt$

$\therefore$

$1000 = V (2)(360)$

$\implies$

$V = 1.38$ volts

Electromotive force represents

0%
force
0%
energy
0%
energy per unit charge
0%
current

Explanation

Electromotive force is the potential difference (or voltage

$V$ ) of the cell.

Using

$V = \dfrac{W}{q}$

where

$W$ is the work done in moving a charge

$q$ from one point to another having potential difference

$V$ .

Thus electromotive force represents energy per unit charge.

Four bulbs marked

$40W, \ 250V$ are connected in series with

$250\ V$ mains, the total power consumed is

0%
$10W$
0%
$40W$
0%
$320W$
0%
$160W$

Explanation

Power of each bulb,

$P = 40 W$

Total power consumed when four bulbs are connected in series

$\dfrac{1}{P_s} = \dfrac{1}{P_1}+$

$\dfrac{1}{P_2} + \dfrac{1}{P_3}+\dfrac{1}{P_4}$

$\therefore$

$\dfrac{1}{P_s} = \dfrac{1}{P}+$

$\dfrac{1}{P} + \dfrac{1}{P}+\dfrac{1}{P}$

We get

$P_s =\dfrac{P}{4}$

$\Rightarrow P_s = \dfrac{40}{4}$

$\Rightarrow P_s =10 W$

A potentiometer has a driving cell of negligible internal resistance. The balancing length of a Daniel cell is

$5m$ . If the driving cell had internal resistance, the balancing length of the same Denial cell would have been

0%
more
0%
less
0%
same
0%
cannot be said from the data

Explanation

Let resistance of

$AB=R$ and power supply has voltage

$= V$ ,

$AB= L$ and

$AC=5m$

Current in the circuit

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

So balancing voltage

$V_{Bal} = R_{AC}\times I=(R\times \dfrac{5}{L})\times \dfrac{V}{R}=\dfrac{5V}{L}$ .... (1)

Now consider internal resistance of power supply

$= r$ and balancing length

$= l$

Current in the circuit

$I_1 = \dfrac{V}{R+r}$

Now balancing voltage

$V_{Bal}=( R\times\dfrac{l}{L})\times \dfrac{V}{R+r}$ ........(2)

By equation (1)&(2) -

$\dfrac{5V}{L} =(R\times\dfrac{l}{L})\times \dfrac{V}{R+r } \implies l = 5(1+\dfrac{r}{R}) \implies l>5m$

Which of the following devices has a source of emf inside it?

0%
Voltmeter
0%
Ammeter
0%
Ohm-meter
0%
Rectifier

A battery of e.m.f.

$E$ has an internal resistance '

$r$ '. A variable resistance

$R$ is connected to the terminals of the battery. A current

$I$ is drawn from the battery.

$V$ is the terminal P.D. lf

$R$ alone is gradually reduced to zero, which of the following best describes

$I$ and

$V$ ?

0%
$I$ approaches $E / r$ , $V$ approaches $E$
0%
$I$ approaches infinity, $V$ approaches $E$
0%
$I$ approaches zero, $V$ approaches $E$
0%
$I$ approaches $E / r$ , $V$ approaches zero

Explanation

$\textbf{Step 1: Finding Current}$

From Figure, Total resistance of the circuit

$R_{eq} = R + r$

So, Current in the circuit

$I = \dfrac{E}{R_{eq}} = \dfrac{E}{R+r}$

$....(1)$

$R\rightarrow 0$ , we get

$I \rightarrow \cfrac{E}{r}$

$\textbf{Step 2: Terminal Potential Difference(V) across the battery}$

From Figure:

$V=V_{AB} = V_{CD}=$

$I R$

$\Rightarrow V = \dfrac{ER}{R+r}$

$($ Using Equation

$(1))$

From above Equation, As

$R\rightarrow 0$ , we get

$V \rightarrow 0$ .

Thus option

$D$ is correct.

2111588_597471_ans_0b124d4c17d741e48a91dd8034c90dfb.png

$1 Wh$ (Watt hour) is equal to :

0%
$36\times { 10 }^{ 5 }J$
0%
$36\times { 10 }^{ 4 }J$
0%
$3600 J$
0%
$3600 { Js }^{ -1 }$

Explanation

Watt is a unit of power such that

$1W =1\dfrac{J}{s}$

Also we know

$1h = 3600$

$s$

$\therefore$

$1Wh = 1\dfrac{J}{s} \times 3600$

$s = 3600$

$J$

The figure shows in apart of an electric circuit, then the current I is

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

Explanation

According to the Kirchhoff's first law, amount of current entering is equal to amount of current leaving.

From the circuit,

$I+6+1=5+3+2$

$I+7=10$

$I=3A$

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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