MCQ Questions for Class 10 Physics Electricity Quiz 7

What is the minimum resistance that one can obtain by connecting all the five resistances each of

$0.5\Omega$ ?

0%
$\frac{1}{10} \Omega$
0%
$\frac{1}{5} \Omega$
0%
$\frac{1}{50} \Omega$
0%
$\frac{1}{25} \Omega$

Explanation

Given

$R=0.5=\dfrac{1}{2}\Omega$

For obtaining minimum resistance, all five resistances should be connected in parallel.

And the equivalent resistance =

$\dfrac{R}{5}$

$=\dfrac{0.5}{5}=\dfrac{1}{10}\Omega$

Answer-(A)

An electric bulb is rated 220 volt and 100 watt, Power consumed by it when operated on 110 volt is :

0%
50 W
0%
85 W
0%
90 W
0%
45 W

What is the effective resistance between points P and Q?

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

Explanation

$\textbf{Hint: Here all the resistors are connected in parallel}$

Step1: Find the type of connection for each resistance

Here, there are 3 resistance of 12 Ω, 15 Ω and 20 Ω. From point P to Q, there are 3 paths in which all resistance are different. (i.e. current can pass from any resistance). So, all the resistance are in parallel connection with each other.

Step2: Find the net resistance of the circuit

Assume that net resistance of the circuit is R_{net}. So, from the formula of parallel resistance connection,

$\dfrac{1}{R_{net}} = \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{20}$

LCM of 12, 15, 20 is 60. So,

$\dfrac{1}{R_{net}} = \dfrac{5\ +\ 4\ +\ 3}{60} = \dfrac{12}{60} = \dfrac{1}{5}$

$\Rightarrow R_{net} = 5\ Ω$

$Answer:$

Hence, option A is the correct answer.

If the specific resistance of a wire of length

$l$ and radius

$r$ is

$k$ then resistance is

0%
${k\pi {r}^{2}}/{l}$
0%
${\pi {r}^{2}}/{lk}$
0%
${kl}/{\pi {r}^{2}}$
0%
${k}/{l{r}^{2}}$

Explanation

Given : Specific resistance

$\rho = k$

Cross-section area of the wire

$A = \pi r^2$

Resistance of wire is given by

$R = \rho \dfrac{l}{A} = \dfrac{kl}{\pi r^2}$

n identical resistors each of resistance r when connected in parallel, have a total resistance R. When these resistors are connected in series, then effective resistance in terms of R is:

0%
R/n $^2$
0%
n $^2$ R
0%
nR
0%
R/n

Explanation

$\displaystyle \frac{1}{R} = \frac{1}{r} + \frac{1}{r} + ........ n$ terms

$= \displaystyle \frac{n}{r}$

$\therefore \displaystyle R = \frac{r}{n} \Rightarrow r = Rn$
When they are connected in series

$R_{eq} = r + r + ...... n$ terms

$= nr = n \times Rn = n^2 R$

The number of joules contained in

$1 kWh$ is

0%
$36 \times {10}^{2}$
0%
$36 \times {10}^{3}$
0%
$36 \times {10}^{4}$
0%
$3.6 \times {10}^{6}$

Explanation

1 KWh = 1000

$\ast$ joule/sec

$\ast$ 60

$\ast$ 60 sec

1KWh = 1000

$\ast$ 60

$\ast$ 60joue/sec

$\ast$ sec

1KWh = 1000

$\ast$ 60

$\ast$ 60 joule

1KWh = 36,00,000 joule = 3.6

$\ast$ 10

$^{6}$

A man has five resistors each of value

$\frac {1}{5}\Omega$ . What is the minimum resistance he can obtain by connecting them ?

0%
$\dfrac {1}{10}\Omega$
0%
$\dfrac {1}{5}\Omega$
0%
$\dfrac {1}{50}\Omega$
0%
$\dfrac {1}{25}\Omega$

Explanation

Minimum resistance is obtained when all the resistance are connected in parallel.

Effective resistance in parallel connection

$1\dfrac{1}{R_{min}} = \dfrac{1}{R}+\dfrac{1}{R}+....$ 5 terms

$\implies$ Minium resistance

$R_{min} = \dfrac{R}{5} =\dfrac{1/5}{5} =\dfrac{1}{25}\Omega$

The smallest resistance which can be obtained with ten 0.1 ohm resistors is

0%
1 ohm
0%
0.1 ohm
0%
0.01 ohm
0%
0.001 ohm

Explanation

When the resistances are connected in parallel, the effective resistance becomes less than the smallest individual resistance. Thus smallest resistance is obtained when ten

$0.1\Omega$ resistors are connected in parallel.

$\therefore$ Effective resistance

$R_p = \dfrac{r}{10} =\dfrac{0.1}{10} = 0.01\Omega$

A 4

$\Omega$ resistance is bent through 180

$^o$ at its midpoint and the two halves are twisted together. Then the resistance is

0%
1 $\Omega$
0%
2 $\Omega$
0%
5 $\Omega$
0%
8 $\Omega$

Explanation

As the

$4\Omega$ resistance is folded at its midpoint, thus the resistance of each segment which are now connected in parallel is

$2\Omega$ .

Equivalent resistance between A and B

$\dfrac{1}{R_{AB}} = \dfrac{1}{2}+\dfrac{1}{2}$

$\implies$

$R_{AB} = 1\Omega$

We have n resistors each of resistance R. The ratio of the combination for maximum and minimum values is

0%
n
0%
$n^2$
0%
$n^3$
0%
$\displaystyle \frac{1}{n}$

Explanation

The maximum resistance is obtained when all the

$n$ resistors are connected in series.

Thus maximum resistance

$R_{max} = nR$

Minimum resistance is obtained when all the

$n$ resistors are connected parallel to each other.

$\therefore$ Minimum resistance

$R_{min} = \dfrac{R}{n}$

$\implies$

$\dfrac{R_{max}}{R_{min}} = \dfrac{nR}{\frac{R}{n}} = n^2$

A wire had a resistance of

$12\ \Omega$ . It is bent in the form of a circle. The effective resistance between two points on any diameter is:

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

Explanation

Resistance of the wire,

$R=12\ \Omega$

As the length of each segment is half of the complete wire, thus the resistance of each segment,

$R' =\dfrac{12}{2} =6\Omega$

Effective resistance between P and Q (

$R'$ in parallel),

$R_{eq}$

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

$\dfrac{1}{R_{eq}}=\dfrac{1}{R'}+\dfrac{1}{R'}$

$\dfrac{1}{R_{eq}}=\dfrac{2}{R'}$

$R_{eq}= \dfrac62$

$R_{eq} =3\ \Omega$

You are given

$5 m$ length of heating wire, it has resistance of

$24 \Omega$ . It is cut into two and connected to

$110$ volt line individually. The total power for the two half lengths is:

0%
500 watts
0%
1000 watts
0%
2000 watts
0%
2500 watts

Explanation

Resistance of the full wire

$= 24 ohm$

Since resistance is directly proportional to length.

$\therefore$ Resistance of the half wire

$= \displaystyle\frac{1}{2} \times 24 = 12 ohm$
Power in one half

$= V. I.$

and

$V=IR$

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

$V=110volts$

$R=12ohms$
Power in one half=

$= 110 \times \displaystyle\frac{110}{12}$

$= 55 \times \displaystyle\frac{55}{3} watts$
Total power of two half lengths

$= 2 \times 55 \times \displaystyle\frac{55}{3} = \displaystyle\frac{6050}{3}$

$= 2000 watts$ (approximately)

$100 watt$ bulb is connected to

$200 volts$ main. The resistance of the filament of the bulb is

0%
$400 ohm$
0%
$100 ohm$
0%
$200 ohm$
0%
$50 ohm$

Explanation

Power in resistor is given by

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

$P=100Watt$

$V=200Volts$

$\therefore R = \displaystyle\frac{{V}^{2}}{P} = \displaystyle\frac{200 \times 200}{100} = 400 ohm$ .

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 lamp is marked 60 W, 220 V. If it operates at 200 V, the rate of consumption of energy will _____

0%
decrease
0%
increase
0%
remain unchanged
0%
first increase then decrease

Explanation

Given power of bulb =60W

Voltage=220V

Power

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

$60=\dfrac{220\times220}{R}$

$R=(\dfrac{220\times220}{60})\Omega$

$R=(\dfrac{220\times22}{6})\Omega$ (1)

When voltage = 200V, resistance will be the same

$P=\dfrac{200\times200}{R}$

From 1,

$=\dfrac { 200\times 200\times 6 }{ 220\times 22 }$

$\dfrac { 200\times 20\times 6 }{ 22\times 22 }$

$=49.5 W$

We see that power gets decreased.

Power means rate of consumption of energy, so rate of consumption of energy decreases.

Option A is correct.

Calculate the effect the resistance when two resistors of resistances 10

$\Omega$ and 20

$\Omega$ are connected in parallel.

0%
0.15 $\Omega$
0%
6.66 $\Omega$
0%
3.33 $\Omega$
0%
0.25 $\Omega$

Explanation

We know that in parallel connection of resistors

$R_1,R_2,R_3...R_n$ the equivalent resistance is given by-

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...+\dfrac{1}{R_n}$

For two given resistances-

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\implies R=\dfrac{R_1R_2}{R_1+R_2}$

$\implies R=\dfrac{10\times 20}{10+20}=\dfrac{200}{30}$

$\implies R=6.66\Omega$

Answer-(B)

An electric heater has a rating of 2 kW, 220 V. The cost of running the heater for 10 hours at the rate of Rs 350 per unit is Rs ____

0%
216
0%
165
0%
209
0%
105

Explanation

Power is energy consumed per unit time.

Here, energy consumed

$=P=2kW=2000W$ and time

$=t=10hr=36000s$

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

$\implies E=Pt=2kW\times 10h=20kWh=20units$

Since

$1unit=1kWh$

Cost

$=20\times 350=Rs 7000$

What is the effective resistance between points P and Q?

0%
$\dfrac {r}{2}$
0%
$\dfrac {r}{3}$
0%
$\dfrac {2r}{3}$
0%
$\dfrac {4r}{3}$

Explanation

In the given combination, point

$B$ is equivalent to point

$P$ as they are connected by a conducting wire. Similarly, point

$A$ is equivalent to point

$Q$ for the same reason.

So, the first resistance (

$2r$ ) is connected between

$P$ and

$A$ , i.e.

$P$ and

$Q$ .

The second resistance (

$2r$ ) is connected between

$A$ and

$B$ , i.e.

$P$ and

$Q$ .

the third resistance (

$r$ ) is connected between

$B$ and

$Q$ , i.e.

$P$ and

$Q$ .

Therefore,

$\text{the three resistors are connected in parallel to each other.}$

$\therefore$ Equivalent resistance between P and Q

$\dfrac{1}{R_{eq}} = \dfrac{1}{2r}+\dfrac{1}{2r}+\dfrac{1}{r}$

$\implies$

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

Find the effective resistance, when

$1\Omega$ ,

$10\Omega$ and

$4\Omega$ resistances are connected in series.

0%
$25\Omega$
0%
$15\Omega$
0%
$30\Omega$
0%
$\frac {20}{27}\Omega$

Explanation

In a series combination of resistors, the equivalent resistance is given by-

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

Here-

$R_{eq}=1+10+4$

$\implies R_{eq}=15\Omega$

Answer-(B)

Calculate the power of an electric bulb which consumes

$2400\ J$ in a minute

0%
$80\ W$
0%
$120\ W$
0%
$60\ W$
0%
$40\ W$

Explanation

Power is the energy consumed per unit of time.

Here, energy consumed,

$E=2400J$ ,

and time

$t=1\ min=60s$

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

$\Rightarrow P=\dfrac{2400}{60}$

$\Rightarrow P=40W$

Two wires of resistances 10

$\Omega$ and 5

$\Omega$ are connected in series. The effective resistances is _______

$\Omega$

0%
15
0%
20
0%
30
0%
40

Explanation

we know that in the series combination of two resistors, the equivalent resistance is given by-

$R_{eq}=R_1+R_2$

Here-

$R_{eq}=10+5$

$\Rightarrow R_{eq}=15\Omega$

A dynamo develops

$0.5 A$ at

$6 V$ , the power delivered is ______.

0%
$30 W$
0%
$3 W$
0%
$1.5 W$
0%
$2.5 W$

Explanation

Given,

$i=0.5\ A\\V=6\ volt$

Power is given by

$P=VI$

$\implies P=6\times 0.5$

$\implies P=3W$

Answer-(B)

If n number of identical resistors of resistance R are connected in parallel combination, then the effective resistance of the combination is ______

0%
nR
0%
n/R
0%
R/n
0%
$nR-\frac {n}{R}$

Explanation

For parallel combination we know that for equivalent resistance-

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...$

Here-

$\dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+... n \hspace{2mm}times$

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

$\implies R_{eq}=\dfrac{R}{n}$

Answer-(C)

Calculate the electrical energy consumed in units when a 60 W bulb is used for 10 hours

0%
1
0%
0.6
0%
0.3
0%
2

Explanation

Power is energy consumed per unit time.

Here, energy consumed

$=P=60W$ and time

$=t=10hr=36000s$

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

$\implies E=Pt=60\times 36000J=3.6\times 10^5\times 6J$

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

$\implies E=\dfrac{3.6\times 6\times 10^5}{3.6\times 10^6}$

$\implies E=0.6units$

Answer-(B)

The equivalent resistance of the combination of resistors shown in the figure between the points

$A$ and

$B$ is:

0%
$2R$
0%
$\displaystyle \frac {R}{4}$
0%
$\displaystyle \frac {R}{2}$
0%
$4R$

Explanation

$\displaystyle R_{AB} = \frac {R}{4}$

Effective resistance between

$A$ and

$C$ is:

0%
$2 \Omega$
0%
$6 \Omega$
0%
$9 \Omega$
0%
None

Explanation

$R_{eqv} = \dfrac{6}{3} = 2 \Omega$

$1$ ampere is same as ______.

0%
$1 Cs^{-1}$
0%
$1 Cs$
0%
$1 JC^{-1}$
0%
$1 VC^{-1}$

Explanation

$1$ ampere

$(A)$ is defined as the current when

$1$ coulomb

$(C)$ of charge passes a given cross section in

$1$ second

$(s)$ .

$1 \ A=\dfrac{1 \ C}{1 \ s}=1Cs^{-1}$

Answer: (A)

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

In effective resistance between

$C$ and

$B$ is .

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

The electricity tariff in a town is

$Rs.3.00$ per unit. Calculate the cost of running an

$80 W$ fan for

$10 h$ a day for the month of June.

0%
$Rs. 12.00.$
0%
$Rs. 52.00.$
0%
$Rs. 72.00.$
0%
$Rs. 92.00.$

Explanation

Total time for which the fan is used

$= (10 hours) \times (30 days) = 300$ hours.
Energy consumed

$= (80 W) \times (300$ hours

$)$

$= 24,000 Wh = 24 kWh.$
But

$1 kWh =1$ unit.
Therefore,

$24 kWh = 24$ units
The cost of this power

$= 24$ units

$\times Rs. 3= Rs. 72.00.$

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.

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.

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,

$1\ kWh$

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

$= 3.6\times 10^6$

$J$

Energy consumed in our house has a unit of

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

What will be the current drawn by an electric fan of 805 W, when it is connected to a source of 230 V ?

0%
$3$ A
0%
$4$ A
0%
$3.5$ A
0%
$4.5$ A

Explanation

Given:

Power of the electric fan

$P=805\ W$

Voltage of the source

$V=230\ V$

We know that, Power

$P=VI$

$\therefore Current\ I=\dfrac{P}{V}=\dfrac{805}{230}=3.5\ A$

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)

An experiment is being performed to test the conduction of electricity through a sample liquid as shown in the image, What happens when the current passing through the circuit is too weak?

0%
The filament does not get heated sufficiently and the bulb does not glow.
0%
The filament gets heated but the bulb does not glow.
0%
The filament does not get heated but the bulb glows.
0%
The filament gets heated and the bulb glows.

Equivalent resistance of the above combination will be:

0%
$\displaystyle 5\Omega$
0%
$\displaystyle 5/4\Omega$
0%
$\displaystyle 10\Omega$
0%
$\displaystyle 20\Omega$

Explanation

Equivalent resistance of resistors connected in series

$R_{eq} = R_1+R_2+R_3+R_4$

$\therefore$

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

A wire has resistance of

$\displaystyle 12\Omega$ . It is cut into two parts and both halve are connected in parallel. The new resistance is

0%
$\displaystyle 3\Omega$
0%
$\displaystyle 1.5\Omega$
0%
$\displaystyle 12\Omega$
0%
$\displaystyle 6\Omega$

Explanation

As the wire is bent at its mid point, thus resistance of each segment

$R = \dfrac{12}{2} = 6\Omega$

Both are in parallel

$\therefore$ Effective resistance between A and B

$R' = \dfrac{R_1\times R_2}{R_1+R_2} = \dfrac{6\times 6}{6+6} = 3\Omega$

Two metallic wires of

$\displaystyle 6\Omega$ and

$\displaystyle 3\Omega$ are in connection. What will be the mode of connection so that to get effective resistance of

$\displaystyle 2\Omega$ ?

0%
Parallel
0%
Series
0%
Both
0%
None

Explanation

$R_1$ =

$6\Omega$ ,

$R_2$ =

$3\Omega$

Case (1): In Series connection :

$R_{eq}$ =

$R_1$ +

$R_2$

$R_{eq}$ =

$6$ +

$3$

$R_{eq}$ =

$9\Omega$

So series connection not possible.

Case (2): In Parallel connection :

$\frac{1}{R_{eq}}$ =

$\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}$ =

$\frac{1}{6}+\frac{1}{3}$

$R_{eq}$ =

$\frac{6\times3}{6+3}$ =

$\frac{18}{9}$ =

$2\Omega$

Hence the two resistors are connected in parallel.

Three resistors each having the same resistance are connected in parallel. Their equivalent resistance is

$\displaystyle 5\ \Omega$ . If they are connected in series, their equivalent resistance is.

0%
$\displaystyle 3\ \Omega$
0%
$\displaystyle 45\ \Omega$
0%
$\displaystyle 5\ \Omega$
0%
$\displaystyle 8\ \Omega$

Two resistors are connected a) in series b) in parallel.
The equivalent resistance in two cases are

$9\Omega$ and

$2\Omega$ respectively. Then the resistance of the component resistors are.

0%
$2\Omega$ and $7\Omega$
0%
$3\Omega$ and $6\Omega$
0%
$3\Omega$ and $9\Omega$
0%
$5\Omega$ and $4\Omega$

Explanation

$\textbf{Step 1: Equivalent resistance for series connection}$

Let the resistances be

$R_1$ and

$R_2$ .

$R_{s} = R_1+R_2$

$\Rightarrow$

$9 = R_1+R_2$

$\Rightarrow R_1 = 9-R_2$

$......(1)$

$\textbf{Step 2: Equivalent resistance for parallel connection}$

$R_{p} = \dfrac{R_1R_2}{R_1+R_2}$

$\Rightarrow$

$2 = \dfrac{R_1 R_2}{R_1 +R_2}$

$......(2)$

$\textbf{Step 3: Solving Equations}$

Put the value of

$R_1$ from

$(1)$ in equation

$(2)$

$2 = \dfrac{(9-R_2) R_2}{9}$

$\Rightarrow R_2^2 - 9R_2 +18 = 0$

$\Rightarrow (R_2-6)(R_2 -3)= 0$

$\Rightarrow R_2= 6\Omega$ or

$R_2 =3\Omega$

From equation (1)

$\Rightarrow R_1= 3\Omega$ or

$R_1 =6\Omega$

Hence, option(B) is correct.

A lamp rated

$20W$ and an electric iron rated

$50W$ are used for

$2$ hour everyday. Calculate the total energy consumed in

$20$ days.

0%
$14kwh$
0%
$2.8kwh$
0%
$40kwh$
0%
None of these

Explanation

Energy consumed by lamp in 2 hour

$E_l = 0.02\times 2 = 0.04$ kWh per day

Energy consumed by iron in 2 hour

$E_i = 0.05\times 2 = 0.1$ kWh per day

$\therefore$ Total energy consumed by both appliance in

$20$ days,

$E_T = (E_l+E_i)\times 20 = (0.04+0.1)\times 20 = 2.8$ kWh

$\displaystyle { R }_{ 1 }$ and

$\displaystyle { R }_{ 2 }$ are respectively the filament resistance of a

$200 \ W$ bulb and

$100 \ W$ bulb designed to operate on the same voltage then

0%
$\displaystyle { R }_{ 1 }$ is two times $\displaystyle { R }_{ 2 }$
0%
$\displaystyle { R }_{ 2 }$ is two times $\displaystyle { R }_{ 1 }$
0%
$\displaystyle { R }_{ 2 }$ is four times $\displaystyle { R }_{ 1 }$
0%
$\displaystyle { R }_{ 1 }$ is four times $\displaystyle { R }_{ 2 }$

Explanation

Given :

$P_1 = 200$ W

$P_2 = 100$ W

Power, voltage and resistance are related as

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

For constant voltage,

$\implies$

$R\propto \dfrac{1}{P}$

$\therefore$

$\dfrac{R_1}{R_2} = \dfrac{P_2}{P_1} = \dfrac{100}{200}$

$\implies R_2 = 2R_1$

Find the effective resistance when

$\displaystyle 1\Omega ,2\Omega$ and

$\displaystyle 3\Omega$ resistances are connected in series

0%
$\displaystyle 2\Omega$
0%
$\displaystyle 1\Omega$
0%
$\displaystyle 6\Omega$
0%
$\displaystyle 5\Omega$

Explanation

Effective resistance of resistors connected in series

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

$\implies$

$R_s = 1+2+3 = 6\Omega$

For three resistors connected in series.

0%
$\displaystyle R_{eff}={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$
0%
$\displaystyle V={ V }_{ 1 }={V }_{ 2 }={ V}_{ 3 }$
0%
$\displaystyle \dfrac {1}{R_{eff}}=\dfrac {1}{{ R }_{ 1 }}+\dfrac {1}{ {R }_{ 2 }}+\dfrac {1}{ {R }_{ 3 }}$
0%
None of these

Explanation

Same current flows through the resistors connected in series but potential difference is different across individual resistors and sum of individual potential difference is equal to the total potential difference.

$\therefore$

$V = V_1+V_2+V_3$

$IR_{eff} = IR_1+IR_2+IR_3$

$\implies$

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

A wire of resistance

$12\ ohm$ per meter is bent to form a complete circle of radius

$10\ cm$ . The resistance b/w its two diametrically opposite points

$A$ and

$B$ as shown in the figure is

0%
$\displaystyle 6\pi \Omega$
0%
$\displaystyle 3\Omega$
0%
$\displaystyle 0.6\pi \Omega$
0%
$\displaystyle 6\Omega$

Explanation

Length of semi-circle

$L = \pi r = \pi (0.1)$

$\therefore$ Resistance of each semi-circle

$R = 12\times 0.1\pi = 1.2\pi \Omega$

Equivalent resistance between point A and B

$R_{eq} = 1.2\pi\parallel 1.2\pi = \dfrac{1.2\pi}{2} = 0.6\pi \Omega$

CBSE Questions for Class 10 Physics Electricity Quiz 7 - MCQExams.com

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CBSE Questions for Class 10 Physics Electricity Quiz 7 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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