Chapter 12 Electricity: NCERT Solutions for Class 10 Science

NCERT solutions for Class 10 of Chapter 12 Electricity provides deep insights of the concepts and formulas related to electricity backed with several experiments. Students can start preparing for the exams through these sorted solved answers. The answers are laid in a simple detailed way for better understanding and clarity among students, ensuring high marks in their scorecard.

SimplyAcad is always there to help students, scroll below to access all the answers of the entire exercise.

Chapter 12 Human eye and colourful world  : NCERT solutions for class 10 Science

1. What does an electric circuit mean?

Solution:

A continuous closed path made of electric components through which an electric current flows is known as an electric circuit. A simple circuit consists of the following components:

(a) Conductors

(b) Cell

(c) Switch

(d) Load

2. Define the unit of current.

Solution:

The unit of current is ampere. Ampere is defined by the flow of one coulomb of charge per second.

3. Calculate the number of electrons constituting one coulomb of charge. 

Solution :

We know that the magnitude of charge on one electron is
\[
1.6 \times 10^{-19} \, \text{C}
\] So, the number of electrons constituting one coulomb of charge is
\[
\frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}
\]

NCERT Solutions of Class 10 Chapter 12 – Exercise 

1. Name a device that helps to maintain a potential difference across a conductor.

Solution:

A battery consisting of one or more electric cells is one of the devices that help to maintain a potential difference across a conductor.

2. What is meant by saying that the potential difference between two points is 1 V?

Solution:

When 1 J of work is done to move a charge of 1 C from one point to another, it is said that the potential difference between two points is 1 V.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Solution:

We know that the potential difference between two points is given by the equation,

V = W/Q, where,

W is the work done in moving the charge from one point to another

Q is the charge

From the above equation, we can find the energy given to each coulomb as follows:

W = V × Q

Substituting the values in the equation, we get

W = 6V × 1C = 6 J

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V of battery.

NCERT Solutions of Class 10 Chapter 12 – Exercise 

1. On what factors does the resistance of a conductor depend?

Solution:

The resistance of the conductor depends on the following factors:

1. Temperature of the conductor
2. Cross-sectional area of the conductor
3. Length of the conductor
4. Nature of the material of the conductor

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Solution:

Resistance is given by the equation,

R = ρ l/A

where,

ρ is the resistivity of the material of the wire,

l is the length of the wire

A is the area of the cross-section of the wire.

From the equation, it is evident that the area of the cross-section of wire is inversely proportional to the resistance. Therefore, the thinner the wire, the more the resistance and vice versa. Hence, current flows more easily through a thick wire than a thin wire.

3. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? 

Solution :

Using Ohm’s law,
\[
I = \frac{V}{R} \implies I \propto V \quad (\text{since } R = \text{constant})
\] Given:
\[
V’ = \frac{V}{2}
\] \[
\therefore \frac{I’}{I} = \frac{V’}{V} = \frac{1}{2}
\] Thus, the current is halved.

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Solution:

The melting point of an alloy is much higher than a pure metal because of its high resistivity. At high temperatures, alloys do not melt readily. Therefore, alloys are used in heating appliances such as electric toasters and electric irons.

5. Use the data in the table given below and answer the following questions.

1. Which among iron and mercury is a better conductor?
2. Which material is the best conductor?

Solution:

1. Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.
2. Among all the materials listed in the table, silver is the best conductor because the resistivity of silver is lowest among all, i.e., 1.60 × 10–8.

NCERT Solutions of Class 10 Chapter 12 – Exercise 

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Solution:

A battery of three cells of 2 V each equals to battery of potential 6 V. The circuit diagram below shows three resistors of resistance 12 Ω, 8 Ω and 5 Ω connected in series along with a battery of potential 6 V.

2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Solution:

An ammeter should always be connected in series with resistors while the voltmeter should be connected in parallel to the resistor to measure the potential difference as shown in the figure below.

Using Ohm’s Law, we can obtain the reading of the ammeter and the voltmeter.

The total resistance of the circuit is 5 Ω + 8 Ω +12 Ω = 25 Ω.

We know that the potential difference of the circuit is 6 V, hence the current flowing through the circuit or the resistors can be calculated as follows:

I = V/R = 6/25 = 0.24A

Let the potential difference across the 12 Ω resistor be V1.

From the obtained current V1 can be calculated as follows:

V1 = 0.24A × 12 Ω = 2.88 V

Therefore, the ammeter reading will be 0.24 A and the voltmeter reading be 2.88 V.

In-Chapter Questions Chapter 12 Electricity

1. Judge the equivalent resistance when the following resistors are connected in parallel: (a) \(1 \, \Omega\) and \(10^6 \, \Omega\) (b) \(1 \, \Omega\), \(10^3 \, \Omega\), and \(10^6 \, \Omega\) 

Solution: (a) Two resistors: \(1 \, \Omega\) and \(10^6 \, \Omega\)} The formula for equivalent resistance \( R_{eq} \) of resistors in parallel is given by: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \] For resistors \( R_1 = 1 \, \Omega \) and \( R_2 = 10^6 \, \Omega \): \[ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{10^6} \] \[ \frac{1}{R_{eq}} = 1 + 0.000001 \] \[ \frac{1}{R_{eq}} \approx 1.000001 \] \[ R_{eq} \approx \frac{1}{1.000001} \approx 1 \, \Omega \] \(b) Three resistors: \(1 \, \Omega\), \(10^3 \, \Omega\), and \(10^6 \, \Omega\)} Using the same formula: \[ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} \] \[ \frac{1}{R_{eq}} = 1 + 0.001 + 0.000001 \] \[ \frac{1}{R_{eq}} = 1.001001 \] \[ R_{eq} = \frac{1}{1.001001} \approx 0.999 \, \Omega \] \- (a) The equivalent resistance of (1 \, \Omega\) and \(10^6 \, \Omega\) connected in parallel is approximately \(1 \, \Omega\). – \(b) The equivalent resistance of \(1 \, \Omega\), \(10^3 \, \Omega\), and \(10^6 \, \Omega\) connected in parallel is approximately \(0.999 \, \Omega\).

2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Solution:

Step 1, Given data

\begin{align*}
\text{Resistance of Lamp} &= 100 \, \Omega \\
\text{Resistance of Toaster} &= 50 \, \Omega \\
\text{Resistance of Water Filter} &= 500 \, \Omega \\
\text{Voltage} &= 220 \, \text{V}
\end{align*}

Step 2

We know, from Ohm’s law:
\[
V = IR
\]

So,
\[
I \text{ in electric lamp} = \frac{220}{100} \, \text{A}
\] \[
I \text{ in toaster} = \frac{220}{50} \, \text{A}
\] \[
I \text{ in filter} = \frac{220}{500} \, \text{A}
\]

Now, total current:
\[
\text{Total current} = \frac{220}{100} + \frac{220}{50} + \frac{220}{500}
\] \[
= \frac{220}{50} \left(\frac{1}{2} + 1 + \frac{1}{10}\right)
\] \[
= \frac{22}{5} \times \frac{8}{5}
\] \[
= \frac{176}{25} = 7 \, \text{A}
\]

Now we can write the resistance of the iron:
\[
R_{\text{iron}} = \frac{V}{I} = \frac{220}{\frac{176}{25}} = 31.25 \, \Omega
\]

Hence, the resistance of the iron is \( 31.25 \, \Omega \).

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Solution:

When the electrical devices are connected in parallel there is no division of voltage among the appliances. The potential difference across the devices is equal to supply voltage. Parallel connection of devices also reduces the effective resistance of the circuit.

4. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Solution: 

(a) To get 4 Ω equivalent resistance, 3 Ω and 6 Ω resistors are connected in parallel and 2 Ω in series.

When 3 Ω and 6 Ω are added in parallel:

\[
\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2}
\]

\[
\frac{1}{R_P} = \frac{1}{3} + \frac{1}{6}
\]

\[
\frac{1}{R_P} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2}
\]

\[
R_P = 2\,\Omega
\]

When its equivalent resistance is connected in series with a resistor of 2 Ω, then:

\[
R_S = R_1 + R_2
\]

\[
R_S = 2 + 2 = 4\,\Omega
\]

When all three resistors are connected, the equivalent resistance is 1 Ω:

\[
\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]

\[
\frac{1}{R_P} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1
\]

\[
R_P = 1\,\Omega
\]

5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Solution:

(a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest.

Their equivalent resistance connected in parallel is

Hence, the lowest total resistance is 2 Ω.

In-Chapter Questions Chapter 12 Electricity

1. Why does the cord of an electric heater not glow while the heating element does?

Solution:

The heating element of an electric heater is made of an alloy which has a high resistance. When the current flows through the heating element, the heating element becomes too hot and glows red. The cord is usually made of copper or aluminum which has low resistance. Hence the cord doesn’t glow.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Solution:

The heat generated can be computed by Joule’s law as follows:

H = VIt

where,

V is the voltage, V = 50 V

I is the current

t is the time in seconds, 1 hour = 3600 seconds

The amount of current can be calculated as follows:

2. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Solution:

The amount of heat generated can be calculated using the Joule’s law of heating, which is given by the equation

H = VIt

Substituting the values in the above equation, we get,

H = 100 × 5 × 30 = 1.5 × 104 J

The amount of heat developed by the electric iron in 30 s is 1.5 × 104 J.

In-Chapter Questions Chapter 12 Electricity

1. What determines the rate at which energy is delivered by a current?

Solution:

Electric power is the rate of consumption of electrical energy by electric appliances. Hence, the rate at which energy is delivered by a current is the power of the appliance.

2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Solution:

The power of the motor can be calculated by the equation,

P = VI

Substituting the values in the above equation, we get

P = 220 V × 5 A = 1100 W

The energy consumed by the motor can be calculated using the equation,

E = P × T

Substituting the values in the above equation, we get

P = 1100 W × 7200 = 7.92 × 106 J

The power of the motor is 1100 W and the energy consumed by the motor in 2 hours is 7.92 × 106 J.

Exercise Questions Chapter 12 Electricity

1. 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 \( R’ \), what is the ratio \( \frac{R}{R’} \)? \

Solution: Step 1:  The resistance \( R \) of the wire is given by: \[ R = \frac{\rho \ell}{A} \] where: – \( \rho \) is the resistivity, – \( \ell \) is the length of the wire, – \( A \) is the cross-sectional area. When the wire is cut into 5 equal parts, each part will have: – Length \( \frac{\ell}{5} \), – Same cross-sectional area \( A \). The resistance of each part \( R_1, R_2, R_3, R_4, R_5 \) is: \[ R_{\text{part}} = \frac{\rho \left(\frac{\ell}{5}\right)}{A} = \frac{R}{5} \] \textbf{Step 2: Equivalent Resistance} When these 5 resistances are connected in parallel, the equivalent resistance \( R’ \) can be found using: \[ \frac{1}{R’} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \] Since all \( R_i \) are equal: \[ \frac{1}{R’} = \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} \] \[ \frac{1}{R’} = 5 \times \frac{1}{\frac{R}{5}} = 5 \times \frac{5}{R} = \frac{25}{R} \] So: \[ R’ = \frac{R}{25} \] \textbf{Ratio \( \frac{R}{R’} \)}: \[ \frac{R}{R’} = \frac{R}{\frac{R}{25}} = 25 \] \The ratio \( \frac{R}{R’} \) is 25.

2. Which of the following does not represent electrical power in a circuit?

(a) I2R

(b) IR2

(c) VI

(d) V2/R

Solution:

Answer: b) IR2

Explanation:

Electrical power is given by the expression P = VI. (1)

According to Ohm’s law,

V = IR

Substituting the value of V in (1), we get

P = (IR) × I

P = I2R

Similarly, from Ohm’s law,

I = V/R

Substituting the value of I in (1), we get

P = V × V/R = V2/R

From this, it is clear that the equation IR2 does not represent electrical power in a circuit.

3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be _____.

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Solution:

Answer: 25 W

Explanation:

The energy consumed by the appliance is given by the expression

P = VI = V2/R

The resistance of the light bulb can be calculated as follows:

R = V2/P

Substituting the values, we get

R = (220)2/100 = 484 Ω

Even if the supply voltage is reduced, the resistance remains the same. Hence, the power consumed can be calculated as follows:

P = V2/R

Substituting the value, we get

P = (110)2 V/484 Ω = 25 W

Therefore, the power consumed when the electric bulb operates at 110 V is 25 W.

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be _____.

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Solution:

Let Rs and Rp be the equivalent resistance of the wires when connected in series and parallel respectively.

For the same potential difference V, the ratio of the heat produced in the circuit is given by

Hence, the ratio of the heat produced is 1:4.

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Solution:

To measure the voltage between any two points, the voltmeter should be connected in parallel between the two points.

6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Solution:

The resistance of the copper wire of length in meters and area of cross-section m2 is given by the formula

The length of the wire is 122.72 m and the new resistance is 2.5 Ω.

7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

Plot a graph between V and I and calculate the resistance of that resistor.

Solution:

The plot between voltage and current is known as IV characteristic. The current is plotted in the y-axis while the voltage is plotted in the x-axis. The different values of current for different values of voltage are given in the table. The I V characteristics for the given resistor is shown below.

The slope of the line gives the value of resistance.

The slope can be calculated as follows:

Slope = 1/R = BC/AC = 2/6.8

To calculate R,

R = 6.8/2 = 3.4 Ω

The resistance of the resistor is 3.4 Ω.

8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. \textbf

Solution: Ohm’s Law states that: \[ V = IR \] where: – \( V \) is the potential difference (voltage), – \( I \) is the current, – \( R \) is the resistance. Given: – \( V = 12 \, \text{V} \) – \( I = 2.5 \, \text{mA} = 2.5 \times 10^{-3} \, \text{A} \) To find the resistance \( R \): \[ R = \frac{V}{I} \] \[ R = \frac{12 \, \text{V}}{2.5 \times 10^{-3} \, \text{A}} \] \[ R = \frac{12}{2.5 \times 10^{-3}} \] \[ R = \frac{12}{0.0025} \] \[ R = 4800 \, \Omega \] \ The resistance of the resistor is \( 4800 \, \Omega \).

9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Solution:

In series connection, there is no division of current. The current flowing across all the resistors is the same.

To calculate the amount of current flowing across the resistors, we use Ohm’s law.

But first, let us find out the equivalent resistance as follows:

R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Now, using Ohm’s law,

The current flowing across the 12 Ω is 0.671 A.

10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Solution:

Let us consider the number of resistors required as ‘x.’

The equivalent resistance of the parallel combination of resistor R is given by

The number of resistors required is 4.

11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Solution:

If we connect all the three resistors in series, their equivalent resistor would 6 Ω + 6 Ω + 6 Ω =18 Ω, which is not the desired value. Similarly, if we connect all the three resistors in parallel, their equivalent resistor would be

which is again not the desired value.

We can obtain the desired value by connecting any two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

The third resistor is in series, hence the equivalent resistance is calculated as follows:

R = 6 Ω + 3 Ω = 9 Ω

Case (ii)

When two resistors are connected in series, their equivalent resistance is given by

R = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is calculated as follows:

12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Solution:

The resistance of the bulb can be calculated using the expression

P1 = V2/R1

R1 = V2/P1

Substituting the values, we get

Hence, 110 lamps can be connected in parallel.

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each with a resistance of \( 24 \, \Omega \). The coils can be used separately, in series, or in parallel. What are the currents in the three cases?

Case 1: Coils used separately.
\[
V = IR
\] \[
220 = 24I
\] \[
I = 9.167 \, \text{A}
\]

Case 2: Coils used in series
\[
R_{\text{eq}} = R_1 + R_2 = 24 + 24 = 48 \, \Omega
\] \[
V = IR_{\text{eq}}
\] \[
220 = 48I
\] \[
I = 4.583 \, \text{A}
\]

Case 3: Coils used in parallel
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}
\] \[
R_{\text{eq}} = 12 \, \Omega
\] \[
V = IR_{\text{eq}}
\] \[
220 = 12I
\] \[
I = 18.334 \, \text{A}
\]

14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Solution:

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series, hence their equivalent resistance is given by 1 Ω + 2 Ω = 3 Ω. The current in the circuit can be calculated using the Ohm’s law as follows:

Therefore, the power consumed by the 2 Ω is 8 W.

(ii) When 12 Ω and 2 Ω resistors are connected in parallel, the voltage across the resistors remains the same. Knowing that the voltage across 2 Ω resistor is 4 V, we can calculate the power consumed by the resistor as follows:

The power consumed by the 2 Ω resistor is 8 W.

15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Solution:

Since both the bulbs are connected in parallel, the voltage across each of them will be the same.

Current drawn by the bulb of rating 100 W can be calculated as follows:

P = V × I

I = P/V

Substituting the values in the equation, we get

I = 100 W/220 V = 100/220 A

Similarly, the current drawn by the bulb of rating 60 W can be calculated as follows:

I = 60 W/220 V = 60/220 A

Therefore, the current drawn from the line is

16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Solution:

The energy consumed by electrical appliances is given by the equation

H = Pt, where P is the power of the appliance and t is the time

Using this formula, the energy consumed by a TV of power ration 250 W, can be calculated as follows:

H = 250 W × 3600 seconds = 9 × 105 J

Similarly, the energy consumed by a toaster of power rating 1200 W is

H = 1200 W × 600 s = 7.2 × 105 J

From the calculations, it can be said that the energy consumed by the TV is greater than the toaster.

17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Solution:

The rate at which the heat develops in the heater can be calculated using the following formula

P = I2 R

Substituting the values in the equation, we get

P = (15A) 2 × 8 Ω = 1800 watt

The electric heater produces heat at the rate of 1800 watt

18. Explain the following.

• Why is tungsten used almost exclusively for filament of electric lamps?
• Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
• Why is the series arrangement not used for domestic circuits?
• How does the resistance of a wire vary with its area of cross-section?
• Why copper and aluminium wires are usually employed for electricity transmission?

Solution:

1. The resistivity and melting point of tungsten is very high. Due to this property, it doesn’t burn readily when heated. Electric lamps operate at high temperature. Hence, tungsten is a choice of metal for the filament of electric lamps.
2. The conductors of electric heating devices are alloys because of their high resistivity. Alloys have higher resistivity than pure metals. Due to its high resistivity, a large amount of heat is produced when current passes through it.
3. The series arrangement is not used for domestic circuits due to the following reasons:

• The overall voltage gets distributed in a series circuit. As a result, electric appliances may not get the rated power for their operation.
• All the connected appliances cannot be operated independently. If one device is defective, then the entire circuit will not function.
• The total resistance becomes large, and as a result, the current is reduced.

4. Resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa.

5. Copper and aluminium are good conductors of electricity and have low resistivity, because of which they are usually employed for electricity transmission. Due to low resistivity, the power losses in the form of heat are also significantly less when electricity is transmitted through them.

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