MCQ Questions for Class 10 Physics Electricity Quiz 4

The smallest resistance that can be obtained by combining 10 resistors each of resistance

$10 \Omega$ is

0%
$10\Omega$
0%
$0.5\Omega$
0%
$2\Omega$
0%
$20\Omega$
0%
$1\Omega$

Explanation

When the resistors are connected in parallel combination, the equivalent resistance will decrease. Thus, if the 10 resistors are connected in parallel, we will get minimum resistance.

Thus,

$\dfrac{1}{R_{eq}}=\dfrac{n}{R}$ where

$n=10$ number of resistors which have same resistance and

$R=10 \Omega$ resistance of each resistor.

So,

$\dfrac{1}{R_{eq}}=\dfrac{10}{10}$

$R_{eq}=1$

In the series combination of two or more than two resistances.

0%
The current through each resistance is same
0%
The voltage through each resistance is same
0%
Neither current nor voltage through each resistance is same
0%
Both current and voltage through each resistance are same

In a house, individual powers of two elements are $$100\ W\ 100\ W$$. The effective power of their series combination will be :

0%
$120\ W$
0%
$75\ W$
0%
$50\ W$
0%
$60\ W$

Explanation

$\begin{array}{l}I = \dfrac{V}{{{R_1}}}\\{I^2}R = 100\\I' = \dfrac{V}{{{R_2}}}\\I{'^2}{R_2} = 150\\{R_1} = \dfrac{{100}}{{{I^2}}}\\{R_1} = \dfrac{{100}}{{{{\left( {V/{R_1}} \right)}^2}}}\\{R_1} = \dfrac{{{V^2}}}{{100}}\\{R_2} = \dfrac{{{V^2}}}{{150}}\\{R_1} + {R_2} = {V^2}\left( {\dfrac{1}{{100}} + \dfrac{1}{{150}}} \right)\\{R_3} = {V^2}\dfrac{5}{{100}} = \dfrac{{{V^2}}}{{60}}\\effective\,power = \dfrac{{{V^2}}}{{{R_3}}} = \dfrac{{{V^2}}}{{\left( {{V^2}/60} \right)}} = 60W\end{array}$

A light bulb is rated at 100 W for a 220 V ac supply. The resistance of the bulb is then

0%
$284 \Omega$
0%
$384 \Omega$
0%
$484 \Omega$
0%
$584 \Omega$

Explanation

Here, P = 100 W ,

$V \, = \, 220 \, V$
Resistance of the bulb is

$R \, = \, \dfrac{V^{2}}{P} \, = \, \dfrac{(220)^2}{100} \, = \, 484 \, \Omega$

If five equal pieces of

$25\Omega$ are connected in parallel, then their equivalent resistance will be............

0%
$25\Omega$
0%
$1\Omega$
0%
$5\Omega$
0%
$\dfrac{1}{5}\Omega$

Explanation

Resistance of each piece :

$R_1 = R_2=R_3=R_4 = R_5 = 25 \ \Omega$

Equivalent resistance in parallel

$\dfrac{1}{R_p} = \dfrac{1}{R_1}+\dfrac{1}{R_2} +\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}$

$\therefore$

$\dfrac{1}{R_p} = \dfrac{1}{25}+\dfrac{1}{25} +\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}$

$\implies \ \dfrac{1}{R_p} = \dfrac{5}{25}$

We get,

$R_p = 5\Omega$

Combine three resistors

$5\Omega, 4.5\Omega$ and

$3\Omega$ in such a way that the total resistance of this combination is maximum:

0%
$12.5\Omega$
0%
$13.5\Omega$
0%
$14.5\Omega$
0%
$16.5\Omega$

Explanation

Given resistance,

$R_1=5\Omega\\R_2=4.5\Omega\\R_1=3\Omega$

In series combination effective resistance will be maximum.

ie,

$R_{eq}=R_1+R_2+R_3$

$R_{eq}=5 \Omega + 4.5 \Omega + 3 \Omega = 12.5 \Omega$

Option A

$1$ micro-ampere

$=$ ______ ampere

0%
$10^{-9}$
0%
$10^{-6}$
0%
$10^{6}$
0%
$10^{3}$

Explanation

$1\ micro- ampere$ =

$10^{-6} A$

Which of the following is a conductor ?

0%
Sugar
0%
Graphite
0%
Ceramic
0%
Wood

A (75 - j40 )

$\Omega$ load is the connected to a coaxial line of

$z _ { 0 } = 75 \Omega$ The load matching on the line can be accomplished by connecting

0%
A short-circuit stub at the load
0%
An inductance at the load
0%
A short circuited stub at a specific distance from the load
0%
A capacitance at a specific distance from the load

A cell of emf

$6\ V$ and resistance

$0.5\ ohm$ is short circuited. The current in the cell is

0%
$3\ amp$
0%
$12\ amp$
0%
$24\ amp$
0%
$6\ amp$

Explanation

$i=\dfrac Er=\dfrac{6}{0.5}=12\ amp$

Miniature circuit breaker (MCB) works on the heating effect of current.

0%
True
0%
False

The current in a car headlamp is

$3.0$ A when connected to a

$12$ V battery. What is the resistance of the lamp when it is lit?

0%
$0.25\Omega$
0%
$4.0\Omega$
0%
$15\Omega$
0%
$36\Omega$

Explanation

Given that,

Current in a car headlamp ,

$I=3.0\ A$

Voltage applied across it ,

$V= 12\ V$

Resistance of the lamp ,

$R=\dfrac VI= \dfrac{12}{3}= 4.0\ \Omega$

Write true or false for the following statements :
Electric power is also called BOTU.

0%
True
0%
False

If four resistances are connected as shown in the fig. between A and B, the effective resistance is:

0%
$3\Omega$
0%
$1\Omega$
0%
$2.4\Omega$
0%
$2\Omega$

Explanation

Simplified, or equivalent circuit diagram is shown in the figures.

From figure, effective resistance:

$R_{eq} = \dfrac{4 \times4}{2 \times 4}$

$= 2 \Omega$

The resistance between A and B in the given figure will be(in ohm):

0%
20
0%
30
0%
90
0%
more than 10 but less than 20

Explanation

In the given circuit, suppose,

$r=10\Omega$

$r_1=10\Omega$

$r_2=20\Omega$
and,

$r_3=50\Omega$
(Please refer the diagram.)
Now,

$r_1, r_2, r_3$ are connected in parallel. Suppose, resultant of these three resistances is

$R_e$ .
Therefore,

$\dfrac{1}{r_e}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} = \dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{50} = \dfrac{17}{100}$
Hence,

$r_e = \dfrac{100}{17} \Omega$
Now, resistances

$r$ and

$r_e$ are connected in series.
Therefore, equivalent resistance in the given circuit,

$R=r+r_e$

$R=10+\dfrac{100}{17} = \dfrac{270}{17} \Omega$ , which is more than

$10\Omega$ , but less than

$20 \Omega$ .

The resistivity of a conductor with a current density of

$2.5 Am$

$^{-2}$ when an electric field

$5\times 10^{-8}Vm^{-1}$ is applied to it, is

0%
$1.0\times 10^{-8}\Omega m$
0%
$2.0\times 10^{-8}\Omega m$
0%
$0.5\times 10^{-8}\Omega m$
0%
$12.5\times 10^{-8}\Omega m$

Explanation

$\textbf{Step 1:}$ $\textbf{Relation between Electric field(E), Resistivity and Current density(J)}$

Suppose length of conductor

$= l$

We know that, Potential difference,

$V=E.l$

$\textbf{&}$ Current Density

$J=I/A$
So,

$I = J A$

and

$R=\dfrac{\rho l}{A}$

From

$V = IR$

$\Rightarrow E.l = J A \times \dfrac{\rho l}{A}$

$\Rightarrow E = J(\rho)$

$\therefore \rho = \dfrac{E}{J}$

$\textbf{Step 2: Putting the values in above equation}$

$\Rightarrow \rho =\dfrac{E}{J}= \dfrac{5 \times 10^{-8}Vm^{-1}}{2.5Am^2}$

$= 2 \times 10^{-8} \Omega m$

Hence, Option B is correct

A certain piece of copper is to be shaped into a conductor of minimum resistance. Which of the following options will give the minimum resistance?

0%
$L$ and $A$
0%
$2L$ and $\dfrac{A}{2}$
0%
$\dfrac{L}{2}$ and $2A$
0%
$3L$ and $\dfrac{A}{3}$

Explanation

Resistance depends on length and area as follows (resistivity

$\rho$ is same in all the cases)

$R \propto \dfrac{L}{A}$

Case A)

$R_A$

$= \rho \dfrac{L}{A}$

Case B)

$R_B$

$= \rho \dfrac{2L}{A}\times2 = \dfrac{\rho L}{A} \times4$

Case C)

$R_C$

$= \rho \dfrac{L/2}{2A}=\rho \dfrac{L}{A} \times \dfrac{1}{4}$

Case D)

$R_D$

$=\rho \dfrac{3L}{A/3} = \rho \dfrac{L}{A} \times 9$

$\Rightarrow$ case

$C$ gives minimum resistance.

$\text{Assertion(A)} :$ However long a fuse wire may be,the safe current that can be allowed is the same.

$\text{Reason (R)} :$ The safe current that can be allowed to pass through a fuse wire depends on the radius of the wire.

0%
Both (A) and (R) are true and (R) is the correct explanation of A.
0%
Both (A) and (R) are true but (R) is not the correct explanation of A.
0%
(A) is true but (R) is false.
0%
(A) is false but (R) is true.

The least resistance that one can have from six resistors each of 0.1 ohm resistance is:

0%
0.167 $\ \Omega$
0%
0.00167 $\Omega$
0%
1.67 $\Omega$
0%
0.0167 $\Omega$

Explanation

Least resistance is achieved when all the resistors are connected in parallel.

$\therefore R_{eq} = \dfrac{R}{6} = \dfrac{0.1}{6} = 0.0167\ \Omega$

Three resistances each of

$3\Omega$ are connected as shown in figure. The resultant resistance between A and F is:

0%
$9\Omega$
0%
$2\Omega$
0%
$4\Omega$
0%
$1\Omega$

Explanation

To draw the circuit bring point B to D and Point C to point E. From the diagram, inner two resistances are in parallel, and its equivalent is also parallel to the third resistance.

$\therefore R_{AF} = \dfrac{\dfrac{R}{2} \times R}{\dfrac{3R}{2}}$

$= \dfrac{R}{3}$

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

A wire 50 cm long and

$1 \ mm^2$ in cross-section carries a current of 4 A when connected to a 2 V battery. The resistivity of the wire is

0%
$2\times 10^{-7}\Omega m$
0%
$5\times 10^{-7}\Omega m$
0%
$4\times 10^{-6}\Omega m$
0%
$1\times 10^{-6}\Omega m$

Explanation

By Ohm's law, resistance can be found from current and voltage as below.

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

$= \dfrac{2}{4} = \dfrac{1}{2} \Omega$

The resistance R is dependent on resistivity, length and area of cross-section as below,

$R = \rho \dfrac{l}{A}$

$\Rightarrow \dfrac{1}{2} = \dfrac{\rho \times 0.50\ m}{{10^{-6}m^2}}$

This implies the resistivity of the wire is:

$\Rightarrow 10^{-6} \Omega m$

A wire of resistance

$20\Omega$ is bent in the form of a square. The resistance between the ends of diagonal is:

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

Explanation

Since the wire is bend in form of the square and we know that sides of square are equal so each side will have same resistance i.e 5 ohm. Now if we make diagonal two faces in each side of the diagonal will have two sides and resistance will be 10 ohm.

$R=20 \Omega$
So,

$\dfrac{R}{4}=5 \Omega$

$\Rightarrow$

$R_{1} = 10 \Omega$ ;

$R_{2} = 10\Omega$
So, resistance between the ends of a diagonal =

$R_{D} = \dfrac{10 \times10}{2 \times10}\Omega = 5\Omega$

The equivalent conductance of two wires of resistance 10 ohm and 5 ohm joined together in parallel is:

0%
$\dfrac{50}{15}$ mho
0%
$\dfrac{35}{50}$ mho
0%
$0.3$ mho
0%
$\dfrac{2}{3}$ mho

Explanation

Effective resistance of the parallel combination:

$\dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}$
So, conductance

$= \dfrac{3}{10}=0.3$ mho

The resistance between A and B is:

0%
$9\Omega$
0%
$2\Omega$
0%
$12\Omega$
0%
$8\Omega$

Explanation

$Req \ of \ AC = \dfrac{6\times6}{12} = 3$

$Req \ of \ AD = \dfrac{6\times6}{12} = 3$

$Req \ of \ AE = \dfrac{6\times6}{12} = 3$

$R_{AB} = \dfrac{6\times3}{9} = 2\Omega$

There are

$5$ tube-lights each of

$40\ W$ in a house. These are used on an average for

$5\ hrs/day$ . In addition, there is an immersion heater of

$1500\ W$ used on an average for

$1\ hr/day$ . The number of units of electricity that are consumed in a month of

$30\ days$ is:

0%
$25\ units$
0%
$50\ units$
0%
$75\ units$
0%
$100\ units$

Explanation

$E=n_1{ P }_{ 1 }{ t }_{ 1 }+n_2{ P }_{ 2 }{ t }_{ 2 }$

$E= 5 \times 40\times 5\times 30+1 \times 1500\times 30$

$=200\times 150+45000$

$E=75\ kWh$

$\Rightarrow E = 75\ units$

If an electric iron of

$1200 \ W$ is used for

$30 \ minutes$ everyday, find electric energy consumed in the month of April.

0%
$12\ kWh$
0%
$18 \ kWh$
0%
$36\ kWh$
0%
$6\ kWh$

Explanation

$Power \ \ P= 1200 W = \dfrac{1200}{1000}=1.2 kW$

$time \ \ t = 30 \ min =\dfrac{30}{60}=0.5 h$

Number of days in the month of april =

$30$

Hence, Total time for which iron is used is given by (in hrs) =

$0.5 \times 30$

$Energy \ \ E =Power\times Total \ time$

$= 1.2\times 0.5\times 30$

$= 18 kW h$

The resistance of a conductor is 1.08

$\ \Omega$ . To reduce it to 1

$\ \Omega$ , the resistance that must be connected is:

0%
$0.08$ $\Omega$ in series
0%
$13.5$ $\Omega$ is parallel
0%
$0.08$ $\Omega$ in parallel
0%
$13.5$ $\Omega$ in series.

Explanation

To reduce the resistance value we should connect another resistance in parallel.
So,

$R = \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}$
or,

$1 = \dfrac{1.08 \times R_{2}}{1.08 + R_{2}}$
or,

$1.08 + R_{2} = 1.08 R_{2}$
or,

$1.08 = 0.08 R_{2}$

$\therefore$

$R_{2} = \dfrac{108}{8} = 13.5\ \Omega$

The resultant resistance of two resistors when connected in series is 48 ohm. The ratio of their resistances is 3 :The value of each resistance is

0%
20 $\Omega$ ,28 $\Omega$
0%
32 $\Omega$ ,16 $\Omega$
0%
36 $\Omega$ , 12 $\Omega$
0%
24 $\Omega$ , 24 $\Omega$

Explanation

$R_{1} + R_{2} = 48$ ohm

$\frac{R_{1}}{R_{2}} = \frac{3}{1}$

$\Rightarrow R_{1} = 3 R_{2}$

$\Rightarrow 4R_{2} = 48 \Rightarrow R_{2} = 12$ ohm

$R_{1} = 36$ ohm

Three unequal resistors in parallel are equivalent to a resistance

$1\ \Omega$ . If two of them are in the ratio

$1 : 2$ and if no resistance value is fractional, the largest of the three resistances in

$\Omega$ is:

0%
$4$
0%
$6$
0%
$8$
0%
$12$

Explanation

Let

$R_1, R_2,$ and

$R_3$ be the three resistances connected in parallel.

$R_{p}$ is equivalent resistance in parallel combination.
Hence,

$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {1}{R_{2}}+\dfrac {1}{R_{3}}$
Let,

$R_{2} = 2R_{3}$
Therefore,

$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {1}{2R_{3}}+\dfrac {1}{R_{3}}$

$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {3}{2R_{3}}$

$R_p = 1\ \Omega$ ; given

$\dfrac {1}{R_{1}} = 1- \dfrac {3}{2R_{3}}$

$\therefore R_{1} = \dfrac {2R_{3}}{(2R_{3}-3)}$

$R_{3} = 3\Omega$ ,

$R_{1} = 2\Omega$ and

$R_{2} = 6\Omega$

$\therefore$ Largest resistance

$= 6\Omega$

The electrical energy in kilowatt hours consumed in operating ten

$50\ W$ bulbs for

$10\ hrs/day$ in a month of

$30\ days$ is

0%
$1500$
0%
$15000$
0%
$15$
0%
$150$

Explanation

Let

$n$ be the number of bulbs,

$n= 10$

$P$ be the power of each bulb,

$P= 50\ W$
total time to burn the bulb,

$t =10 \times 30 = 300\ hr$

Now the energy

$E$ , consumed in operating the bulbs is:
Energy consume by each built in 30 days

$E=Pt=50\times 300$

Energy consume by 10 bulbs in 30 days

$E=nPt=10 \times 50 \times 300\ Wh$

$E=150\ kWh$ (Option D)

The current that passes through 20 ohm resistance when it is connected in parallel with a 30 ohm resistance and this set is connected to a battery of 2V is

0%
0.2A
0%
0.3A
0%
0.1A
0%
0.016A

Explanation

Given,

$R_1=20\Omega\\R_2=30\Omega\\E=2\ V$

According to question, both resistance are in parallel and it is connected with

$2\ V$ battery.

Potential difference across each resistance

$2\Omega$

Current in

$20\Omega$ resistance

$I_{20\Omega}=\dfrac{\Delta V}{R}=\dfrac{2}{20}=0.1\ A$

A house is fitted with ten lamps of each

$60\ W$ . Each lamp burns for

$5\ hrs$ a day on an average. The cost of consumption in a month of

$30\ days$ at

$80$ rupees per unit is:

0%
$Rs. 152$
0%
$Rs. 252$
0%
$Rs. 352$
0%
$Rs. 452$

Explanation

Total energy spent is given by:

$E = nE'$

where

$n:$ number of bulbs

$E':$ Energy spent by each bulb

$E' = Pt$

$E' = 60 \times 5 \times 30\ Wh$

$E = 60 \times 5 \times 30 \times 10$

$= 9000 \times 10$

$= 90 KWh$

Cost

$= 90 \times 2.8$

$= 28 \times 9$

$= 252 Rs$

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

0%
1A
0%
2A
0%
4A
0%
5A

Explanation

Hint : A fuse wire should be able pass that amount of current which the appliance requires

Step 1 : Find the maximum current that the appliance requires

Given, Power

$= 1KW= 1000W$ & Voltage

$= 220\ V$

$P= V \times I$

So,

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

$I=\dfrac{1000}{220}= 4.545 A$

The fuse wire should be able to pass at least

$4.545\ A$ current.

So, the rating should be

$5A$ .

Observe the figure given below.
Find the current passing through

$6\Omega$ resistor.

0%
0.72 A
0%
0.80 A
0%
0.48 A
0%
Cannot be said

Explanation

Let the equivalent resistance between

$E$ and

$F$ consisting of

$R_1=6 \Omega$ and

$R_2=4 \Omega$ in parallel be

$R_{eq}.$

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}=\dfrac{R_1+R_2}{R_1R_2}$

$\Rightarrow R_{eq}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{(6)(4)}{6+4}=2.4 \Omega$

Let the voltage across

$E$ and

$F$ be

$V_{EF}.$

Applying Ohm's law for

$R_{eq}$

$V_{EF}=IR_{eq}$

$V_{EF}=(1.2 \ A) \times (2.4 \ \Omega)=2.88 \ V$

The voltage

$V_{EF}=2.88 \ V$ also appears across

$R_1$ and

$R_2$ .

$\therefore V_{EF}=V_{BC}=V_{AD}=2.88 \ V$

Let the current through

$6\Omega$ resistance be

$I_{BC}.$

Applying Ohm's Law for resistance

$R_1=6 \Omega$

$V_{BC}=I_{BC}R_1$

$\Rightarrow I_{BC}=\dfrac{V_{BC}}{R_1}=\dfrac{2.88}{6}=0.48 \ A$

Find power dissipated in

$4 \ \Omega$ resistor :

0%
12 W
0%
16 W
0%
4 W
0%
6 W

Explanation

Power dissipated

$P = I^2 R = 1^2 \times 4 = 4 W$

Name the effect of current responsible for the glow of the bulb in an electric circuit.

0%
Chemical
0%
Heating
0%
Magnetic
0%
Heating and Magnetic

You have a large supply of light bulbs and a battery. You start with one light bulb connected to the battery and notice its brightness. You then add one light bulb at a time, each new bulb being added in series to the previous bulbs then

0%
Brightness of the bulbs will increase
0%
Current through the bulbs will increase
0%
Power transferred from the battery will decrease
0%
Lifetime of the battery will decrease

In the electric circuit shown in the figure, the effective resistance of two 8

$\Omega$ resistors in the combination is :

0%
$4 \Omega$
0%
$16 \Omega$
0%
$2 \Omega$
0%
$12 \Omega$

Explanation

In this case, the 8 ohms resistances are connected in parallel.
So the effective resistance (

$R_{eff}$ ) is given as:

$\dfrac{1}{R_{eff}}=\dfrac { 1 }{ 8 } +\dfrac { 1 }{ 8 } =\dfrac { 1 }{ 4 }$ .
Hence, the effective resistance of two 8 ohms resistors in the combination is given as

$4$ ohm.

Answer question numbers 72 and 73 based on the following information.
Solar surface radiates energy uniformly at a rate of

$4 \times 10^{26}W$ . This energy spreads or distributes uniformly and normally outwards.

Considering Earth and the Sun to be spherical object, the amount of radiant energy received by the Earth per second is nearly

0%
$1360 \:W.$
0%
$1.75 \times 10^{17}\:W$
0%
$3.5 \times 10^{17}\:W$
0%
$7 \times 10^{17}\:W$

Two students set up their circuits for finding the equivalent resistance of two resisto connected in series in two different ways as shown.
The circuit (s) likely to be labelled as incorrect :

0%
are neither of the two circuits
0%
is only circuit I
0%
is only circuit II
0%
are both one circuits

Which effect of current is responsible for a compass placed in an electric field to be deflected?

0%
Heating effects of current.
0%
Magnetic effects of current.
0%
Conducting effects of heat.
0%
Chemical effects of current

When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be:

0%
R/8
0%
4R
0%
R/2
0%
R/4

Explanation

$\textbf{Step 1: Draw resistance structure}$

$\textbf{[Ref. Fig.]}$

Resistance of both part will be

$R/2$ because wire is uniform.

$\textbf{Step 2: Equivalent resistance between A and B}$ ,

Each resistor has both terminals connected to both terminals of the others so both are in parallel-

$R_{eq \ parallel} = \dfrac{R_1R_2}{R_1+R_2}$

$\Rightarrow R_{AB} = \dfrac{\dfrac{R}{2}\times \dfrac{R}{2}}{\dfrac{R}{2} + \dfrac{R}{2}} = \dfrac{R}{4}$

Hence equivalent resistance between any two of diametrically opposite points will be

$\dfrac{R}{4}$

Option

$(D)$ correct.

Combined resistance is the sum of separate resistances provided that the various conductors are connected in______ .

0%
Series
0%
Parallel
0%
Both series and parallel
0%
Data insufficient

A wire of resistance R is cut into ten equal parts which are then joined in parallel. The new resistance is:

0%
0.01 R
0%
0. 1 R
0%
10 R
0%
100 R

Explanation

We know,
1. When a resistance is divided into n identical resistances, each of resistance (say R') become

$\dfrac{R}{n}$ .
2. When n identical resistors each of resistance R', are connected in parallel, the equivalent resistance is

$\dfrac{R'}{n}$
The resistor R, when divided into ten parts becomes resistors of resistance

$\dfrac{R}{10}$ each.
The recombination of

$10$ resistors, each of resistance

$\dfrac{R}{10}$ , gives a resistor of equivalent resistance

$R'$ of value

$\dfrac{\frac{R}{10}}{10} = \dfrac{R}{100}$
Hence,

$\dfrac{R'}{R} = \dfrac{1}{100}$
Or,

$R' = 0.01 R$

Two wires of resistance

$R_1$ and

$R_2$ are joined in parallel. The equivalent resistance of the combination is:

0%
$R_1R_2/(R_1+R_2)$
0%
$R_1+R_2$
0%
$R_1 \times R_2$
0%
$R_1R_2$

Explanation

The equivalent resistance of N resistors connected in parallel connection is given by,

$\dfrac{1}{R^{}_{Equivalent}} = \dfrac{1}{R^{}_{1}} + \dfrac{1}{R^{}_{2}} + \dfrac{1}{R^{}_{3}} + . . . + \dfrac{1}{R^{}_{N}}$
Hence,

$\dfrac{1}{R^{}_{12}} = \dfrac{1}{R^{}_{1}} + \dfrac{1}{R^{}_{2}}$
or,

$R^{}_{12} = \dfrac{R^{}_{1}R^{}_{2}}{R^{}_{1} + R^{}_{2} }$

Three resistors of

$4.0\ \Omega$ ,

$6.0\ \Omega$ and

$10.0\ \Omega$ are connected in series. What is their equivalent resistance?

0%
$20\ \Omega$
0%
$7.3\ \Omega$
0%
$6.0\ \Omega$
0%
$4.0\ \Omega$

Explanation

The equivalent resistance of N resistors connected in series is given by,
$R^{}_{Equivalent} = R^{}_{1} + R^{}_{2} + R^{}_{3} + . . . +R^{}_{N}$
Hence,

$R^{}_{Equivalent} = 4.0 + 6.0 + 10.0 = 20.0 \Omega$

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is:

0%
1/25
0%
1/5
0%
5
0%
25

Explanation

We know,
1.When a resistance is divided into n identical resistances, each of resistance (say R') become

$\dfrac{R}{n}$ .
2.When n identical resistors each of resistance R', are connected in parallel, the equivalent resistance is

$\dfrac{R'}{n}$
The resistor R ,when divided into five parts become resistors of resistance

$\dfrac{R}{5}$ each.
The recombination of five resistors, each of resistance

$\dfrac{R}{5}$ , gives a resistor of equivalent resistance R' of value

$\dfrac{\dfrac{R}{5}}{5} = \dfrac{R}{25}$
Hence,

$\dfrac{R}{R'} = \dfrac{25}{1}$

Two wires of same metal have the same length but their cross-sections area in the ratio 3 :They are joined in series. The resistance of the thicker wire is

$10\Omega$ .The total resistance of the combination will be:

0%
$40 \Omega$
0%
$40/3 \Omega$
0%
$5/2 \Omega$
0%
$100 \Omega$

Explanation

(Let R' denote thicker and R denote thinner resistor)
We know,

$R = \dfrac{\rho \times L}{A} \implies \dfrac{R'}{R} = \dfrac{A}{A'}$

$\implies \dfrac{10}{R} = \dfrac{1}{3} \implies R = 30 \Omega$

Since they are connected in series,the equivalent resistance is given by,

$R^{}_{eq} = R + R' \implies R^{}_{eq} = 30 + 10 = 40 \Omega$

Three resistances of 2, 3 and 5

$\Omega$ are connected in parallel to a lay battery 10V with negligible internal resistance. The potential difference across the

$3 \Omega$ resistance will be

0%
2 V
0%
3 V
0%
5 V
0%
10 V

Explanation

Emf of cell

$=E=10V$

In parallel combination of resistors, the potential difference across each resistor is same and is equal to the potential difference between the point between which resistors are attached.

Here potential difference across

$3\Omega$ resistor is equal to potential difference of cell

$=10V$

Answer-(D)

Kilowatt-hour is the unit of :

0%
potential difference.
0%
electric power.
0%
electrical energy.
0%
charge.

Explanation

Kilowatt-hour $(KWh)$ is unit of Electrical Energy.

We know that,

$\text{Power} \times \text{time} = \text{Energy}$

$\Rightarrow\ \ \text{Kilo-Watt} \times \text{Hour} = \text{Energy}$

$1 KWh$ is electrical energy consumed when an electrical equipment of power $1\ KiloWatt$ is operated for $1\ Hour$ .
It is also used as commertial unit of energy in households/Buildings/Factories etc where 1 Unit of electricity is equivalent of 1KW-h of energy.

Hence, Option C is correct.

CBSE Questions for Class 10 Physics Electricity Quiz 4 - MCQExams.com

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 4 - MCQExams.com

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.