NCERT Exemplar Solutions Class 10 Science Electricity

Read NCERT Exemplar Solutions Class 10 Science Electricity below, students should study NCERT Exemplar class 10 Science available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 10 Science Exemplar problems have been prepared by teacher of Grade 10. These NCERT Exemplar class 10 Science solutions have been designed as per the latest NCERT syllabus for class 10 and if practiced thoroughly can help you to score good marks in standard 10 Science class tests and examinations

 

Multiple Choice Questions.......................


Question 1:  A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of given figure. The current recorded in the ammeter will be

NCERT Exemplar Solutions Class 10 Science Electricity

(a) maximum in (i)

(b) maximum in (ii)

(c) maximum in (iii)

(d) the same in all the cases 

Solution 1:  (d) the same in all the cases 

Since none of the circuits have changed, the current would be the same in both of them.

 

Question 2: In the following circuits figure given below, the heat produced in the resistor or combination of resistors connected to a 12 V battery will be

NCERT Exemplar Solutions Class 10 Science Electricity

(a) same in all the cases

(b) minimum in case (i)

(c) maximum in case(ii)

(d) maximum in case(iii) 

Solution 2:

NCERT Exemplar Solutions Class 10 Science Electricity

Two transistors are connected in series here. Since the individual resistances are linked parallel in figure (iii), the total resistance would be less than the individual resistances. Option c) is the correct response since higher resistance produces more heat.

 

Question 3: Electrical resistivity of a given metallic wire depends upon

(a) its length

(b) its thickness

(c) its shape

(d) nature of the material 

Solution 3:  (d) nature of the material 

The nature of the material determines the electrical resistivity of a given metallic wire.

 

Question 4:  A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

(a) 1020

(b) 1016

(c) 1018

(d) 1023 

Solution 4:  (a) 1020 

I = Q/t 
Q = It 
Q = 1 x 16 
Q = 16 q 
Q = ne 
n = Q/e 
n = 16/1.6  x 10-19 
n = 10 x 1019 
n = 1020 electrons  
The total number of electrons flowing is 1020
 

Question 5:  Identify the circuit in which the electrical components have been properly connected.

(a) (i)

(b) (ii)

(c) (iii)

(d) (iv) 

Solution 5:   (b) (ii)

NCERT Exemplar Solutions Class 10 Science Electricity


Question 6:  What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

(a) 1/5 Ω

(b) 10 Ω

(c) 5 Ω

(d) 1 Ω 

Solution 6: (d) 1 Ω

When resistors are connected in series, the maximum resistance is obtained.

R = 15 + 15 + 15 + 15 + 15

    = 5/5

    = 1Ω

 

Question 7:  What is the minimum resistance which can be made using five resistors each of 1/5 Ω? 

(a) 1/5 Ω 

(b)1/25 Ω 

(c) 1/10 Ω 

(d) 25 Ω  

Solution 7: (b) Ω 

When resistors are connected in parallel, the lowest resistance is obtained. 

1/R = 5 + 5 + 5 +5 + 5 

    = 25 Ω 

R = 1/25 Ω

 

Question 8:  The proper representation of the series combination of cells obtaining maximum potential is

NCERT Exemplar Solutions Class 10 Science Electricity

(a) (i)

(b) (ii)

(c) (iii)

(d) (iv) 

Solution 8:  (a) (i) 

NCERT Exemplar Solutions Class 10 Science Electricity

The next cell's positive terminal is right next to the previous cell's negative terminal.

 

Question 9:  Which of the following represents voltage?

(a)WorkdoneCurrent × Time

(b) Work done × Charge

(c)Workdone × TimeCurrent

(d) Work done × Charge × Time 

Solution 9:  (a)WorkdoneCurrent × Time 

Voltage is represented by WorkdoneCurrent × Time.


Question 10: A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has an area of cross-section
(a)  A/2
(b) 3A/2
(c) 2A
(d) 3A
 
Solution 10:  (c) 2A
 
P = RAl
When Length doubles
P = RAl 
RAl = P = RA2l
A = 2A
  
Question 11: A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?
NCERT Exemplar Solutions Class 10 Science Electricity

(a) R1 = R2 = R3

(b) R1 > R2 > R3

(c) R3 > R2 > R1

(d) R2 > R3 > R1 

Solution 11: (c) R3 > R2 > R1 

Resistance is inversely proportional to current flow. The highest resistance would result in the least amount of current flow, so the response is (c).

 

Question 12:  If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

(a) 100 %

(b) 200 %

(c) 300 %

(d) 400 % 

Solution 12:  (c) 300 % 

The amount of heat generated by a resistor is proportional to the square of the current. As a result, when the current doubles, the heat dissipation multiplies by 2 = 4. This means there would be a 300 percent boost.

 

Question 13:  The resistivity does not change if

(a) the material is changed

(b) the temperature is changed

(c) the shape of the resistor is changed

(d) both material and temperature are changed 

Solution 13: (c) the shape of the resistor is changed 

If the shape of the resistor is changed, the resistivity doesn’t change.

 

Question 14:  In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) The brightness of all the bulbs will be the same

(b) The brightness of bulb A will be the maximum

(c) The brightness of bulb B will be more than that of A

(d) The brightness of bulb C will be less than that of B 

Solution 14: (c) Brightness of bulb B will be more than that of A 

Since the bulbs are bound in parallel, the combined resistance is less than the arithmetic number of all the bulbs' resistance. As a result, there will be no negative impact on current flow. As a result, bulbs will glow in a wattage-dependent manner.

 

Question 15: In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

(a) 5 J

(b) 10 J

(c) 20 J

(d) 30 J 

Solution 15: (c) 20 J 

Equivalent resistance of the circuit is R = 4 + 2 = 6Ω

current, I = V/R = 6/6 = 1A

The heat dissipated by 4-ohm resistor is, H = IRt = 20J

 

Question 16:  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?

(a) 1 A

(b) 2 A

(c) 4 A

(d) 5 A 

Solution 16:  (d) 5 A 

P = V x I 
1000 w = 220 v x I 
I = 1000w/220v = 4.54 A 
= 5 A 
 
Question 17: Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have 
(a) same current flowing through them when connected in parallel 
(b) same current flowing through them when connected in series 
(c) the same potential difference across them when connected in series 
(d) different p
 
Solution 17:  (b) same current flowing through them when connected in series 
Since the resistor receives a common current, the current in a series combination is not divided into branches. 

 

Question 18: Unit of electric power may also be expressed as 
(a) volt-ampere 
(b) kilowatt-hour 
(c) watt-second 
(d) joule second 

Solution 18: (b) kilowatt-hour 
The apparent power in an electrical circuit is measured in volt-amperes (VA). A watt-second (also watt-second, symbol W s or W. s) is a derived energy unit equal to a joule. Planck's constant is measured in joules-seconds. 

 

Short Answer Questions.............................

 

Question 19:  A child has drawn the electric circuit to study Ohm’s law as shown in below figure. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

 NCERT Exemplar Solutions Class 10 Science Electricity

Solution 19:

NCERT Exemplar Solutions Class 10 Science Electricity

 

Question 20:  Three 2 Ω resistors, A, B and C, are connected as shown in below figure. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

NCERT Exemplar Solutions Class 10 Science Electricity

Solution 20:  

Current P = I2R

          18W = I2 x 2Ω

                I= 18W/ 2Ω

                   = 9A

                 I = 3A

This is the highest current that the three resistors can handle.

 

Question 21:  Should the resistance of an ammeter be low or high? Give reason. 

Solution 21:   Since the ammeter should not influence the flow of current, its resistance should be zero.

 

Question 22:  Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason. 

Solution 22:   

NCERT Exemplar Solutions Class 10 Science Electricity

Total resistance for parallel combination of 40 resistors can be calculated as follows:
1/R = 1/4  + 1/4  = 1/2
R = 2 Ω

As a result, the resistance of a parallel combination is equal to the resistance of a set of resistors. As a result, the potential difference between 20 resistors would be the same as the potential difference across the other two parallel resistors.

 

Question 23: How does use of a fuse wire protect electrical appliances? 

Solution 23:  The resistance of fuse wire is even higher than that of the main wiring. When the electric current is significantly increase. The circuit is broken when the fuse wire melts. This prevents the electrical appliance from being damaged.

 

Question 24: What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why? 

Solution 24:  Resistivity is the property of a conductor that prevents the movement of electric current. A material's resistance is one-of-a-kind. The resistance of a conductor is proportional to its length and inversely proportional to its current flow.

When the length is doubled, the resistance doubles and the current flow is cut in half. The drop in ammeter reading is due to this.

 

Question 25: What is the commercial unit of electrical energy? Represent it in terms of joules. 

Solution 25:   The kilowatt-hour is a commercial unit of electrical energy.

1 kw/hr  = 1 kW h
= 1000 W × 60 × 60s
= 3.6 × 106 J
 
  
Question 26:  A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductors and potential difference across the lamp will take place? Give reason.
 
Solution 26:

1) Let R denote the electric lamp's resistance. Complete resistance in sequence = 5 + R
           I = v/r
           1 = 10/5 + R
           R = 5 ohm
 
2) V across Lamp + conductor = 10 V
     V across Lamp = I × R = 1 × 5 = 5 Volt


Question 27:  Why is parallel arrangement used in domestic wiring? 

Solution 27:    Since it offers the same possible difference for all electrical appliances, parallel wiring is used in domestic wiring.

 

Question 28:  B1, B2 and B3 are three identical bulbs connected as shown in given figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

  1. What happens to the glow of the other two bulbs when the bulb B1 gets fused?
  2. What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?
  3. How much power is dissipated in the circuit when all the three bulbs glow together?

 Solution 28:

1. In a parallel circuit, the potential difference is not separated. As a result, when bulb one is fused the glowing of other bulbs will not be affected. 

2. The reading on ammeter A is 3A. This indicates that Al. A2, and A3 both have IA readings.

3. R = V/I = 4.5V/3A = 1.5Ω

         P = I2R

            = (3A)2  x 1.5 Ω

            = 13.5 W

 

Long Answer Questions......................


Question 29:  Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason. 

Solution 29:  

(a) The resistance of a series of bulbs would be three times that of a single bulb. As a result, when opposed to the current in each bulb in a parallel combination, the current in the series combination would be one-third of the current in the parallel combination. The bulbs in the parallel combination would have a brighter light. 

(b) When the circuit is broken and the current is zero, the bulbs in a series combination will stop glowing. The bulbs in a parallel mix, on the other hand, would continue to glow at the same brightness.

 

Question 30:  State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment. 

Solution 30:   At constant temperature, the potential difference (voltage) through an ideal conductor is proportional to the current flowing through it, according to Ohm's law.

V/I = R

NCERT Exemplar Solutions Class 10 Science Electricity

Set up a circuit with a nichrome wire XY of 0.5 m length, an ammeter, a voltmeter, and four 1.5 V cells, as shown in Figure (Nichrome is a metal alloy that contains nickel, chromium, manganese, and iron).

To begin, use only one cell as the circuit's source. Note the current reading on the ammeter I and the potential difference around the nichrome wire XY in the circuit on the voltmeter V.

Connect two cells in the circuit and record the ammeter and voltmeter readings for the current through the nichrome wire and the potential difference across the nichrome wire, respectively.

Repeat the steps above with three and then four cells in the circuit, in that order.

NCERT Exemplar Solutions Class 10 Science Electricity

 

Question 31:  What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends. 

Solution 31:

A conductor's intrinsic property of resistivity is its ability to resist the movement of electric current. Each material's resistance is different. The SI unit of resistance is the metre.

Experiment to learn about the factors that influence the resistance of conducting wire.

Take a nichrome cable, a torch bulb, a 10 W bulb, a plug key, and some connecting wires, as well as an ammeter (0 – 5 A range).

To make the circuit, connect four 1.5 V dry cells in series with the ammeter, leaving an XY gap in the circuit.

Observation:

Resistance is observed to be dependent on conductor content.

Resistance is determined by the length of the conductor.

Resistance is proportional to the cross-sectional area.

NCERT Exemplar Solutions Class 10 Science Electricity

Attach the nichrome wire to the gap XY to complete the circuit. Insert the key. Make a note of the ammeter's reading. Remove the key from the socket. [Note: After measuring the current through the circuit, always remove the key from the plug.]

Replace the nichrome wire with the torch bulb in the circuit and use the ammeter to determine the current flowing through it.

Repeat the previous move with the 10 W bulb in the XY gap. Are the ammeter readings for the various components linked in the XY gap different? What do the preceding statements imply?

Keep any material part in the gap to replicate this Operation. In each case, take note of the ammeter readings. Analyze the findings.

 

Question 32:  How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery? 

Solution 32:

• To make the circuit, connect three resistors R1, R2, and R3 in series.

• Use an ammeter to check for changes in the current flow.

• Remove R1 and measure the possible difference between R2 and R3.

• Remove R2 and measure the possible difference between R1 and R3.

NCERT Exemplar Solutions Class 10 Science Electricity

Observation:

Since the ammeter readings were consistent in each case, it can be assumed that the circuit current remains constant. To double-check, position an ammeter in various locations and observe the current flow.

 

Question 33: How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery? 

Solution 33:

To make a circuit, connect three resistors RI, R2, and R3 in parallel as shown in the diagram.

Using a voltmeter, measure the potential difference between three parallel resistors.

Now, take the reacting of the potential difference of the remaining resistors' combination after removing the resistor R1.

Then, take a reading of the potential difference of the remaining resistor after removing the resistor R.

NCERT Exemplar Solutions Class 10 Science Electricity

Observation:

The Voltmeter readings were identical in each case, indicating that the same potential difference occurs between three parallel resistors.

 

Question 34:  What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life. 

Solution 34:   The heating effect of heat in a resistor, according to Joules, is

  1. For the given resistor, directly proportional to square of current.
  2. For a given current, directly proportional to resistance.
  3. The time it takes for current to pass through the resistor is directly proportional to the time it takes for current to flow through the resistor.

This can be depicted as

H = I2Rt

H stands for heating effect, I stands for electric current, R stands for resistance, and t stands for time.

Experiment with the Joules law of heating to illustrate it.

  • Attach a water heating immersion rod to a socket that is connected to the regulator. It's important to remember that the amount of current flowing through a system is regulated by a regulator.
  • Set the regulator's pointer to the lowest setting and time how long it takes the immersion rod to heat a certain amount of water.
  • Transfer the regulator's pointer to the next environment. Calculate how long it takes the immersion rod to heat the same volume of water.
  • Count the time by repeating the above steps for higher levels on the regulator. 

Observation:

When the amount of electric current is raised, it takes less time to heat the same amount of water. This is a demonstration of Joule's Law of Heating. 

Application:

The leafing effect of current is used in electric toasters, ovens, electric kettles, and electric heaters, among other things.

 

Question 35:  Find out the following in the electric circuit given in the below figure:
 
(a) Effective resistance of two 8 Ω resistors in the combination
 
(b) Current flowing through 4 Ω resistor
 
(c) Potential difference across 4 Ω resistance
 
(d) Power dissipated in 4 Ω resistor
 
(e) Difference in ammeter readings, if any 
 
Solution 35: 

(a) As, two 8 Ω resistors placed parallel, the effective resistance would be   

1/( Rp) = 1/R1 + 1/R2 

 
                     1/( Rp) = 1/8 + 1/8 
 
                     1/( Rp) = 1/4 
 
                      Rp = 4 Ω
 
(b) Total resistance in the circuit
                   R = 4 Ω + Rp 
                                   = 4 Ω + 4 Ω 
                      = 8 Ω
        Current through the circuit,
             I = V/R = 8/8 = 1A
 
(c) Potential difference across 4 Ω resistance
            V = IR = 1 × 4 = 4V
 
(d) Power dissipated in 4 Ω resistor
             P = I2
                = 12 × 4 
                = 4 W
 
(e) The readings of ammeters A1 and A2 are identical since the same current flows through all elements in a series current.
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