MCQ Questions for Class 10 Physics Electricity Quiz 12

There are

$n$ similar conductor each of resistance

$R$ . The resultant resistance comes out to be

$x$ when connected in parallel. If they are connected in series, the resistance comes out to be

0%
$x/n^{2}$
0%
$n^{2}x$
0%
$x/n$
0%
$nx$

Explanation

In parallel, combination of n resistance of value R each , Equivalent resistance

$x=\dfrac{R}{n}\implies R=nx$

In series,

$R_{eq}= R+R+R.... n$ times

$=nR=n(nx)=n^2x$

If three resistors of resistance

$2\Omega, 4\Omega$ and

$5 \Omega$ are connected in parallel then the total resistance of the combination will be

0%
$\dfrac{20}{19}\Omega$
0%
$\dfrac{19}{20}\Omega$
0%
$\dfrac{19}{10}\Omega$
0%
$\dfrac{10}{19}\Omega$

Explanation

For parallel resistance connection we know that,

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{19}{20}\Rightarrow R_{eq}=\dfrac{20}{19}\Omega$

Two bulbs, one of 50 watt and another of 25 watt are connected in series to the mains. The ratio of the currents through them is

0%
$2 : 1$
0%
$1 : 2$
0%
$1 : 1$
0%
Without voltage, cannot be calculated

Which of the following statement is false

0%
Heat produced in a conductor is proportional to its resistance
0%
Heat produced in a conductor is proportional to the square of the current
0%
Heat produced in a conductor is proportional to charge
0%
Heat produced in a conductor is proportional to the time for which current is passed

Explanation

According to Joule's heating law -

$H = i^{2} Rt$

We also know that current

$i = \dfrac{q}{t}$ .

Hence

$H = \dfrac{q^{2}R}{t}$

$\therefore H \propto q^{2}$

An electric heater kept in vacuum is heated continuously by passing electric current. Its temperature

0%
Will go on rising with time
0%
Will stop after sometime as it will lose heat to the surroundings by conduction
0%
Will rise after sometime and there after will start falling
0%
Will become constant after sometime because of loss of heat due to radiation

$10$ wires (same length, same area, same material) are connected in parallel and each has

$1\Omega$ resistance, then the equivalent resistance will be

0%
$10\Omega$
0%
$1\Omega$
0%
$0.1\Omega$
0%
$0.001\Omega$

Explanation

When

$n$ no. of resistances are connected parallelly then

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

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

$\Rightarrow$

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

$\Rightarrow$

$R_{eq}=\dfrac {R}{n}=\dfrac {1}{10}=0.1\Omega$

Three equal resistors connected in series across a source of emf together dissipate

$10$ watt. If the same resistors are connected in parallel across the same emf, then the power dissipated will be:

0%
$10\ watt$
0%
$30\ watt$
0%
$\dfrac{10}{3}\ watt$
0%
$90\ watt$

A nichrome wire

$50\ cm$ long and one square millimeter cross-section carries a current of

$4A$ when connected to a

$2V$ battery. The resistivity of nichrome wire in ohm meter is

0%
$1\times 10^{-6}$
0%
$4\times 10^{-7}$
0%
$3\times 10^{-7}$
0%
$2\times 10^{-7}$

Explanation

Given,

$V=2\ volt\\l=50\ cm=50\times 10^{-2}\ cm\\A=1\ mm^2\ =10^{-6}\ m^2\\i=4\ A\\ \rho=?$

According to Ohm's law,

$R=\dfrac Vi$ ............(1)

Resistance or any resistor can also be written as

$R=\rho \dfrac{l}{A}$ ............(2)

From equation 1 and 2,

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

$\Rightarrow \dfrac {2}{4}=\rho \dfrac {50\times 10^{-2}}{( 10^{-6})}$

$\Rightarrow \rho =1\times 10^{-6}\Omega m$

If you are provided three resistances

$2 \Omega, 3 \Omega$ and

$6 \Omega$ . How will you connect them so as to obtain the equivalent resistance of

$4\Omega$

0%
0%
0%
0%
None of these

The resistors of resistances

$2\ \Omega, 4\ \Omega$ and

$8\ \Omega$ are connected in parallel, then the equivalent resistance of the combination will be:

0%
$\dfrac{8}{7}\ \Omega$
0%
$\dfrac{7}{8}\ \Omega$
0%
$\dfrac{7}{4}\ \Omega$
0%
$\dfrac{4}{9}\ \Omega$

Explanation

Equivalent of the parallelly connected resistance is given as,

$\dfrac {1}{R_{eq}}=\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}=\dfrac {4+2+1}{8}$

$\Rightarrow R_{eq}=\dfrac {8}{7}\Omega$ .

Two bulbs of

$100 W$ and

$200 W$ working at

$220$ volt are joined in series with 220 volt supply. Total power consumed wll be approximately.

0%
$65$ watt
0%
$33$ watt
0%
$300$ watt
0%
$100$ watt

Explanation

The two bulbs of 100W and 200W are working at 220V are joined in series with 220V supply

Thus ,

$P_{s}=\dfrac{P_{1}P_{2}}{P_{1}+P_{2}}=\dfrac{100\times 200}{100+200}=\dfrac{200}{3}\approx 65watt$

If a high power heater is connected to electric mains then the bulbs in the house become dim, becuase there is a

0%
Current drop
0%
Potential drop
0%
No current drop
0%
No Potential drop

Explanation

Brightness depend on power of bulb and

power

$\propto i^2$

If a high power heater is connected to electric mains then the bulbs in the house bcome dim, becuase there is a change in voltage across bulb, which result in current drop.

An electric iron draws 5 amp, a TV set draws 3 amp and refrigerator drawn 2 amp from a 220 volt main line. The three appliances are connected in parallel. If all the three are operating at the same time, The fuse may be of

0%
$20$ amp
0%
$5$ amp
0%
$15$ amp
0%
$10$ amp

Explanation

Given,

Current in iron

$i_1=5\ A$

Current in TV

$i_2=3\ A$

Current in refrigerator

$i_3=2\ A$

Total load current

$i=i_1+i_2+i_3= 5+3+2 = 10$

The current capacity of a fuse wire should be slightly greater than the total rated load current.
1770079_1817457_ans_283d375a2eb448f8a2df773e086b08c7.png

1770079_1817457_ans_283d375a2eb448f8a2df773e086b08c7.png

The electric passing through a metallic wire produces heat because of

0%
Collisions of conduction electrons with each other
0%
Collisions of the atoms of the metal with each other
0%
The energy released in the ionization of the atoms of the metal
0%
Collisions of the conduction electrons with the atoms of the metallic wires

$60 \ Watt$ bulb operates on

$220 \ V$ supply. The current flowing through the bulb is

0%
$\dfrac{11}{3} \ amp$
0%
$\dfrac{3}{11} \ amp$
0%
$3 \ amp$
0%
$6 \ amp$

Explanation

The power rated on the bulb

$P = 60 \ W$ ,it operates on

$V = 220 \ V$ supply .

Power is given by

$P=Vi$

Hence,

$\Rightarrow i = \dfrac{P}{V} = \dfrac{60}{220} = \dfrac{3}{11} \ amp$

1723369_1816967_ans_62d819f3b5cd490d94ec0283db286960.png

Electric radiator which operates at 225 volts has resistance of 50 ohms. Power of the radiator is approximately

0%
$1000$ W
0%
$450$ W
0%
$750$ W
0%
$1500$ W

Explanation

It is known that power =

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

$P=\dfrac{225^2}{50}=1012.5\approx 1000\,W$

so Power of the radiator is approximately 1000 W

A steel wire has a resistance twice that of an aluminium wire. Both of them are connected with a constant voltage supply. More heat will be dissipated in:

0%
Steel wire when both are connected in series
0%
Steel wire when both are connected in parallel
0%
Aluminium wire when both are connected in series
0%
Aluminium wire when both are connected in parallel

Explanation

Let the resistance of steel wire

$=R_{steel}$

Resistance of aluminium wire

$=R_{aluminium}$

Given,

$R_{steel} = 2R_{aluminium}$

We know,

Heat dissipated,

$H=I^2R$

Also,

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

In series, the current

$I$ will be the same in both the wires.

$H=I^2R$

$H\propto R$ \

$\therefore$ More heat will dissipate in the steel wire when both are connected in series.

In parallel, the voltage

$V$ will be the same across both the wires.

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

$H\propto \dfrac 1R$

$\therefore$ More heat will dissipate in the aluminium wire when both are connected in parallel.

An electric bulb marked 40 W and 200 V, is used in a circuit of supply voltage 100 v, Now its power is

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

Explanation

The power genrated is given by

$P=\dfrac{V^{2}}{R}\Rightarrow \dfrac{P_{2}}{P_{1}}=\dfrac{V^{2}_{1}}{V^{2}_{1}}(\because R is constant )\Rightarrow \dfrac{P_{2}}{P_{1}}=\left ( \dfrac{100}{200} \right )^{2}=\dfrac{1}{4}\Rightarrow P_{2}=\frac{P_{1}}{4}=\dfrac{40}{4}=10W$

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated

$220 \ V$ and

$100 \ W$ is connected to

$(220\times 0.8) \ V$ source, then the power would be :

0%
$100\times 0.8watt$
0%
$100\times (0.8)^{2}watt$
0%
Between $100\times 0.8watt$ and 100 watt
0%
Between $100\times (0.8)^{2}watt$ and $100\times 0.8watt$ watt

Explanation

Since Power

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

For the first case, when

$V=220 \ V$

$P_{1}=\dfrac{(220)^2}{R_1}$

For the second case, when

$V=220 \times 0.8 \ V$

$P_{2}=\dfrac{(220\times 0.8)^{2}}{R_{2}}$

Now,

$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )^{2}}{(220)^{2}}\times \dfrac{R_{1}}{R_{2}}$

$\Rightarrow \dfrac{P_{2}}{P_{1}}=(0.8)^{2}\times \dfrac{R_{1}}{R_{2}}$

Since

$V_2<V_1$ , voltage has been decreased and is directly proportional to the resistance, from Ohm's law.

Hence,

$R_{2}< R_{1}$

$\Rightarrow \dfrac{R_1}{R_2} > 1$

$\Rightarrow \dfrac{P_2}{P_1} > (0.8)^2$

$\Rightarrow P_2 > 100 \times (0.8)^2 \ W$

Also, Since Power

$P = Vi$

Hence

$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )i_2}{220i_1},$

Since

$V_2<V_1$ , voltage has been decreased and is directly proportional to the current, from Ohm's law.

Hence,

$i_{2}< i_{1}$

$\Rightarrow \dfrac{i_2}{i_1} <1$

$\dfrac{P_{2}}{P_{1}}< 0.8\Rightarrow P_{2}< (100\times 0.8) \ W$

Hence the actual power would be between

$100 \times (0.8)^{2} \ W$ and

$100\times (0.8) \ W$

A wire of resistance

$12\Omega$ per metre is bent to form a complete circle of radius

$10\ cm$ . The resistance between its two diametrically opposite points

$A$ and

$B$ as shown in the figure is

0%
$6\Omega$
0%
$0.6\pi \Omega$
0%
$3\Omega$
0%
$4\Omega$

An electric lamp is marked

$60 \ W$ ,

$230 \ V$ . The cost of

$1 \ kilowatt-hour$ of power is

$Rs \ 1.25$ . The cost of using this lamp for

$8 \ hours$ is

0%
$Rs. \ 1. 20$
0%
$Rs. \ 4. 00$
0%
$Rs. \ 0. 25$
0%
$Rs. \ 0. 60$

Explanation

Total energy consumed is given as

$E=P\times t$

where

$P$ is power and

$t$ is time for which it is used.

Given,

$P=60\,W$ and

$t=8\,h$

$E = 60 \times 8 \ Wh=\dfrac{60\times 8}{1000} = 0.48 \ kWh$

Given that, cost of

$1 \ kWh$ is

$Rs. \ 1.25$

Hence, total cost is

$\text{Cost}=0.48\times 1.25 = Rs. \ 0.6$

The current flowing through a lamp marked as

$50 \ W$ and

$250 \ V$ is

0%
$5 \ amp$
0%
$2.5 \ amp$
0%
$2 \ amp$
0%
$0. 2 \ amp$

Explanation

The power dissipated is given by

$P=Vi$ ,

where

$V$ =voltage across lamp, and

$i$ =current flowing through lamp

Hence,

$i = \dfrac{P}{V} = \dfrac{50}{250} = 0.2 \ amp$

If a

$5 \,V$ battery is provoding

$2 \,A$ current in a conductor then what is the resistance of conductor ?

0%
$3$ Ohm
0%
$2.5$ Ohm
0%
$10$ Ohm
0%
$2$ Ohm

Explanation

Given that:

Battery supplies a voltage of

$V=5 \ V$

Current provided by the battery

$I=2 \ A$

Resistance= ?

By Ohm's law,

$V=IR$

$\Rightarrow R=\dfrac{V}{I}=\dfrac{5}{2}=2.5 \Omega$

A heater coil is cut into two equal parts and only one part is now use in the heater. The heat generated will now be

0%
One fourth
0%
Halved
0%
Doubled
0%
Four times

Explanation

The heat generation is given by

$Heat=Power\times time$

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

$H=\dfrac{V^2 \times t}{R}$

After cutting equally, length becomes half.

The resistance is directly proportional to

$l$

It means, resistance becomes half.

$H=\dfrac{V^2 \times t}{R/2}$

$H=2\ \dfrac{V^2 \times t}{R}$

Therefore the heat generated will be doubled.

The conductors of resistance

$1 \Omega, 2 \Omega$ and

$3\Omega$ are in series in a circuit. What is the equivalent resistance in the circuit ?

0%
Less than $1\Omega$
0%
Less than $3\Omega$
0%
More than $1 \Omega$
0%
More than $3 \Omega$

Explanation

Given:

$R_1=1 \Omega$

$R_2=2 \Omega$

$R_3=3 \Omega$

All are in series

$R_{eq}=R_1+R_2+R_3=1+2+3=6 \Omega$

Kilowatt hour is unit of

0%
Energy
0%
Power
0%
Impulse
0%
Force

What happens to the current in each resistor when resistors of different values in parallel combination are connected to a source of electricity?

0%
Values of current and potential difference are different
0%
Values of current and potential difference are same
0%
Values of current are different but value of potential difference is same
0%
Values of current are the same but values of potential difference are different

Two wires A and B of same metal are connected in parallel. Wire A has length

$l$ and radius

$r$ and wire B has length

$2l$ and radius

$2r$ . then the ratio of total resistance of parallel combination and the resistance of wire A is

0%
$1 : 2$
0%
$1 : 3$
0%
$1 : 4$
0%
$1 : 5$

If the five equal pieces of a resistance wire having

$5$ resistance each is connected in parallel, then their equivalent resistance will be

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

Explanation

Number of resistance given

$n=5$

Resistance of each resistor

$R=5\Omega$

All resistors are connected in parallel

$\dfrac 1{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_5}+\dfrac{1}{R_5}=\dfrac{5}{R}$

$\dfrac 1{R_{eq}}=\dfrac{5}{5}=1\ \Omega$

Five resistances of R

$\Omega$ were taken. First three resistances are connected in parallel combination and rest two are connected in series combination, then the equivalent resistance is :

0%
$\frac{3}{7} R \ \Omega$
0%
$\frac{7}{3} R \ \Omega$
0%
$\frac{7}{8} R \ \Omega$
0%
$\frac{8}{7} R \ \Omega$

Explanation

$R_{equi} = \frac{R}{3} + R + R = \frac{7R}{3} \ \Omega$

The resistance of one conducting wire is

$10$ . How much electric current will flow by connecting it with a battery of

$1.5 V$ ?

0%
$0.15 mA$
0%
$1.5 mA$
0%
$15 mA$
0%
$150 mA$

Explanation

From Ohm's law,

$V = IR$

$\Rightarrow I = \dfrac{V}{R}= \dfrac{1.5}{10}= 0.15A = 150 mA$

Materials which allow current to pass through them easily are called?

0%
Insulator
0%
Conductor
0%
Switch
0%
Battery

Explanation

Materials which allow current to pass through them easily are called as conductor. The most common conductors are metals: copper, silver, gold, platinum and aluminum etc.
So, the correct option is

$(B)$

The amperage of the fuse wire used in a circuit that works on

$230 \ V$ is

$2.2 \ A$ . If so. the power of the device is

0%
less than $300 \ W$
0%
$300 \ W$ to $500 \ W$
0%
between $500 \ W$ and $510 \ W$
0%
more than $510 \ W$

Explanation

The amperage of the fuse wire used in a circuit that works on

$V = 230 \ V$ is

$i = 2.2 \ A$ .

Power,

$P = Vi = 230 \times 2.2 = 506 \ W$

Hence, the power of the device is between

$500 \ W$ and

$510 \ W$ .

The correct option is C.

Several resistors are connected in series. Which of the following statements is correct? Choose all that are correct.

0%
The equivalent resistance is greater than any of the resistances in the group.
0%
The equivalent resistance is less than any of the resistances in the group.
0%
The equivalent resistance depends on the voltage applied across the group.
0%
The equivalent resistance is equal to the sum of the resistances in the group.
0%
None of those statements is correct.

Explanation

According to the relationship for resistors in series,

$R_{oq}=R_1+R_2+...$
the sum

$R_{oq}$ is always larger than any of the resistances

$R_1, R_2,$ etc.

Answer the question:
Two

$5.0\Omega$ resistors are connected as shown in the diagram.
What is the total resistance of the combination ?

0%
less than $5.0\Omega$
0%
$5.0\Omega$
0%
more than $5.0\Omega$ but less than $10.0\Omega$
0%
$10.0\Omega$

Explanation

As we can see that both resistance are connected in parallel.

For parallel combination:

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

$R_{eq}=\dfrac{5\times 5}{5+5}=2.5\Omega$

Hence option A is correct, equivalent resistance is less than 5 ohm

$500\Omega$ of resistance is made by adding five

$100\Omega$ resistance of tolerance

$4\%$ , then the tolerance of the combination is .

0%
$5\%$
0%
$4\%$
0%
$20\%$
0%
$10\%$

Explanation

Individual resisistors are having resistance:

$R=100\pm 4$
On combining 5 similar resistors we get:

$5R=500\pm 20$
So, tolerance of the combination is also

$4\%$

In which direction conventional current flow in an electric circuit:

0%
Negative to positive
0%
Positive to negative
0%
Both A & B
0%
None of these

The equivalent resistance between points

$A$ and

$B$ is

0%
$\dfrac {91}{2}\Omega$
0%
$\dfrac {65}{2}\Omega$
0%
$\dfrac {45}{2}\Omega$
0%
$\dfrac {5}{2}\Omega$

The equivalent resistance of eight equal resistances in series is

$48$ ohms. What would be the equivalent resistance if they are connected in parallel?

0%
$\displaystyle \frac {3}{4} \Omega$
0%
$\displaystyle \frac {4}{3} \Omega$
0%
$4 \Omega$
0%
$3 \Omega$

Explanation

Let the resistance of each resistor is

$R$ .

In series we know, equivalent resistance is

$R_s=nR=8R=48\Omega$

$\implies R=6\Omega$

In paralle we know, equivalent resistance is

$R_p=\dfrac{R}{n}$

$\implies R_p=\dfrac{6}{8}=\dfrac{3}{4}\Omega$

Answer-(A)

On a bulb it is written

$220\ V$ and

$60\ W$ . Find out the resistance of the bulb and the value of the current flowing through it:

0%
$806.67\ \Omega \ \text{and} \ 0.27\ A$
0%
$500\ \Omega \ \text{and} \ 2\ A$
0%
$200\ \Omega \ \text{and} \ 4\ A$
0%
$100\ \Omega \ \text{and} \ 1\ A$

Explanation

$\textbf{Step1:Resistance of bulb}$

We know,
Power, $P=\dfrac {V^2}{R}$

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

$R=\dfrac {(220)^2}{60}$

$R=806.67\ \Omega$

$\textbf{Step2:Apply Ohm's law}$
According to Ohm's law,
$V=IR$

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

$I=\dfrac {220}{806.67}$

$I=0.27\ A$

1 kWh is equal to

0%
$3.6 \times 10^6 MJ$
0%
$3.6 \times 10^5 MJ$
0%
$3.6 \times 10^2 MJ$
0%
$3.6 MJ$

Explanation

1 kilowatt hour is the energy produced by 1 kilowatt power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

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

$\implies 1kWh=3.6MJ$

Answer-(D)

An electric lamp whose resistance of 10 ohm, and a conductor of 2 ohm resistance are connected in series with a 6 V battery. The total current through the circuit and the potential difference across the electric lamp are :

0%
3.6 A, 6 V
0%
0.5 A, 5 V
0%
2.0 A, 0.2 V
0%
0.3 A, 3 V

If the filament of bulb D breaks then does bulb C give more light, less light, the same light or no light?

0%
more light
0%
less light
0%
same light
0%
no light

Explanation

It is assumed that every bulb is identical and has same resistance(

$R$ ).

So C and D are in parallel.And A , B and combination of (C and D) all these three are in series.

Initially resistance here was

$R/2$ (C and D both have

$R$ in parallel).So resistances in series are

$R$ ,

$R$ and

$R/2$ . Since resistance across third element is less than first two elements so less portion of cell voltage will be shared here.

Now bulb D breaks.So combination of C and D will reduced to only C.Now only C is there so net resistance here is

$R$ .So now A (

$R$ resistance) , B (

$R$ resistance) and C (

$R$ resistance) are in series and have equal resistances so voltage across C will be

$1/3$ of cell voltage.

In other words after breaking of D voltage across C has increased from before hence it will glow more brightly than before.

Do the arrows in the diagram indicate the direction of electron flow or conventional flow of electricity?

0%
Conventional flow of current
0%
Flow of Electron
0%
Flow of positive charge
0%
None of the above

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

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

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

For a circuit shown in the figure, the total current in the circuit is:

0%
$0.1\ A$
0%
$0.2\ A$
0%
$18\ A$
0%
$1\ A$

Explanation

Given,

Resistance of each resistors,

$R = 1\ \Omega$

$R_1 = R_2 = R_3 = R$

Voltage across the battery,

$V = 6\ V$

Equivalent resistances of 3 resistors connected in parallel

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

$\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R} =3$

$R_{eq} = \dfrac{1}{3}\ \Omega$

Current,

$I = \dfrac{V}{R} = \dfrac{6}{1}\times 3$

$I = 18\ A$

Identify the cases where the bulbs are connected totally in series.

Arrange the order of power dissipated in the given circuits, if the same current enters at point

$A$ in all the circuits and resistance of each resistor is

$R$ .

0%
$(ii) > (iii) > (iv) > (i)$
0%
$(iii) > (ii) > (iv) > (i)$
0%
$(iv) > (iii) > (ii) > (i)$
0%
$(i) > (ii) > (iii) > (iv)$

When current is passed through a conductor, the heat produced in it depends upon
I. The material of the conductor
II. The current flowing through the conductor.
III. The time for which the current flows.

0%
I and II only
0%
I and III only
0%
II and III only
0%
All I, II and III

Two heater coils separately take

$10$ minute and

$5$ minute to boil certain amount of water. If both the coils are connected in series, the time taken will be

0%
$15$ min
0%
$7.5$ min
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CBSE Questions for Class 10 Physics Electricity Quiz 12 - MCQExams.com

After cutting equally, length becomes half.

It means, resistance becomes half.

$\textbf{Step1:Resistance of bulb}$

We know,
Power, $P=\dfrac {V^2}{R}$

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

$R=\dfrac {(220)^2}{60}$

$R=806.67\ \Omega$

$\textbf{Step2:Apply Ohm's law}$
According to Ohm's law,
$V=IR$

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

$I=\dfrac {220}{806.67}$

$I=0.27\ A$

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 12 - MCQExams.com

After cutting equally, length becomes half.

It means, resistance becomes half.

Step1:Resistance of bulb\textbf{Step1:Resistance of bulb}

We know,Power, P=V2RP=\dfrac {V^2}{R}⟹R=V2P\implies R=\dfrac {V^2}{P}R=(220)260R=\dfrac {(220)^2}{60}R=806.67 ΩR=806.67\ \OmegaStep2:Apply Ohm's law\textbf{Step2:Apply Ohm's law}According to Ohm's law,V=IRV=IR⟹I=VR\implies I=\dfrac {V}{R}I=220806.67I=\dfrac {220}{806.67}I=0.27 AI=0.27\ A​

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

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$\textbf{Step1:Resistance of bulb}$

We know,
Power, $P=\dfrac {V^2}{R}$

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

$R=\dfrac {(220)^2}{60}$

$R=806.67\ \Omega$

$\textbf{Step2:Apply Ohm's law}$
According to Ohm's law,
$V=IR$

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

$I=\dfrac {220}{806.67}$

$I=0.27\ A$