CBSE Class 10

HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

Question 1.
One end of a bar magnet is brought near the south pole of a magnetic compass needle. It was noticed that the needle of the magnetic compass deflected away from the end of the bar magnet. Name the pole of the bar magnet at its end pointing towards the south pole of the magnetic compass. Justify your answer.
Answer:
The pole of the bar magnet at the end pointing towards the south pole of the magnetic compass is south pole. This is because, like magnetic poles repel each other.

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Magnetic field lines around a bar magnet are shown in figure. A student makes a statement that magnetic field at point A is stronger than at point B. State, whether the statement is correct or incorrect. Explain.
(CBSE 2012)

Answer:
Statement is correct. This is because field lines are crowded in a region of strong magnetic field and field lines diverge in a region of weak magnetic field.

Question 3.
Magnetic lines of force of two pairs of magnets are shown in figure A and B. Out of these two figures, which one represents the correct pattern of field lines. Name the poles of magnets facing each other. (CBSE 2012)

Answer:
Figure B represents the correct pattern of magnetic field lines of a pair of magnets. Figure A does not represent the correct pattern of field lines because magnetic field lines never cross each other.

Poles of magnets facing each other are north poles as the magnetic field lines emerge from a magnet at N-pole.

Question 4.
A student performs an experiment to study the magnetic effect of current around a current carrying straight conductor. He reports that

1. the direction of deflection of the north pole of a compass needle kept at a given point near the conductor remains unaffected even when the terminals of the battery sending current in the wire are interchanged.
2. for a given battery, the degree of deflection of a N-pole decreases when the compass is kept at a point farther away from the conductor. Which of the above observations of the student is incorrect and why ?
   (CBSE Sample Paper Question)

Answer:

1. The first statement is incorrect because the direction of the magnetic field around the conductor is reversed when the direction of the current flowing in the conductor changes. Hence, the direction of deflection of N-pole of the compass needle cannot remain unaffected when the terminals of the battery sending current in the wire or conductor are interchanged.
2. The second statement is correct because the magnetic field decreases with the increase in the distance from the conductor.

Question 5.
A current carrying conductor is placed perpendicular to the magnetic field of horse-shoe magnet. The
conductor is displaced upward. What will happen to the displacement of the conductor if

1. current in the conductor is increased,
2. a horse-shoe magnet is replaced by another stronger horse-shoe magnet and
3. the length of the conductor is increased. (Similar Text Book Question)

Answer:

1. The force acting on a current carrying conductor placed perpendicular to a magnetic field increases with the increase in the current flowing through a conductor. Thus, the displacement of the conductor will increase if the current in the conductor is increased.
2. When a horse-shoe magnet is replaced by a stronger magnet, then magnetic field increases. Since, force acting a conductor increases with the increase in the magnetic field, therefore, the displacement of the conductor will increase.
3. Since, the force acting on the conductor increases with the increases in the length of the conductor, therefore, the displacement of the conductor will increase.

Hope given HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity

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

(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases.
Answer:
(d).
Explanation : Same current flows through every part of the circuit having resistance connected in series to a cell or battery.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
In the following circuits shown in figure, power produced in the resistor or combination of resistors connected to a 12V battery will be

(a) same in all cases
(b) minimum in case (i)
(c) maximum in case (ii)
(d) maximum in case (iii)
Answer:
(d).

Question 3.
Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of material.
Answer:
(d).
Explanation : Resistivity of a given metallic wire does not depend on its dimensions and shape but depends on the nature of the material of the wire.

Question 4.
A current of 1A 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
Answer:
(a).

Question 5.
Identify the circuit (Figure) in which electrical components have been properly connected.

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
(CBSE 2010)
Answer:
(b).
Explanation : Voltmeter is always connected in parallel and ammeter is connected in series in an electric circuit. Moreover, positive terminals of voltmeter and ammeter are connected with the +ve terminal of a cell or battery and negative terminals of voltmeter and ammeter are connected with -ve terminal of a cell or battery.

Question 6.
What is the maximum resistance which can be made using five resistors each 1/5 Ω ?
(a) 10 Ω
(b) 10 Ω
(c) 5 Ω
(d) I Ω (CBSE 2010)
Answer:
(d).

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

Answer:
(d).

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

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
(CBSE 2011,2012)
Answer:
(d).
Explanation : Cells are connected in series if -ve terminal of one cell is connected to the +ve terminal of other cell.

Question 9.
Which of the following represents voltage ?

Answer:
(a).

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 area of cross-section.
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Answer:
(c).

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 as shown in figure. Which of the following is true ?
(a) R₁ = R₂ = R₃
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ > R₁
(d) R₂ > R₃ > R₁

Answer:
(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%
Answer:
(c).

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.
Answer:
(c).
Explanation : Resistivity of material does not depend on its di¬mensions and shape. However, resistivity depends upon the tem¬perature and the nature of material.

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) Brightness of bulb A will be maximum
(b) Brightness of bulb B will be more than that of A
(c) Brightness of bulb C will be less than that of B
(d) Brightness of all the bulbs will be same
Answer:
(b).
Explanation : Brightness of bulb depends upon the heat produced per second. Since resistance of bulb is inversely proportional to the power, so resistance of bulb A is more than that of B and resistance of bulb B is more than that of C. Heat produced per second = V²/R
where V is same for all bubls.
Therefore, brightness of bulb B is more than that of A.

Question 15.
In an electrical circuit two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by 4Ω resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
Answer:
(c).

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
Answer:
(d).

Question 17.
Two resistors of resistance 2Ω and 4Ω 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) same potential difference across them when connected in series
(d) different potential difference across them when connected in parallel
Answer:
(b)
Explanation : Resistors are said to be connected in series if same current flows through them as series combination has a single path for the flow of current. Resistors are said to be connected in parallel if potential difference across them is

Question 18.
Unit of electric power may also be expressed as
(a) Volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second
Answer:
(a).
Explanation : P = VI.

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

Answer:
Ammeter is connected in series in an electric circuit and volt¬meter is connected in parallel to the resistor R. Positive terminals of ammeter and voltmeter are connected with the positive terminal of battery and negative terminals of ammeter and voltmeter are connected with the negative terminal of the battery. The correct circuit is shown below.

Question 20.
Three 20 resistors, A, B and C are connected as shown in figure. Each of them dissipates energy that can withstand a maximum power of 18W without melting. Find the maximum current that can flow through three resistors.
(CBSE, 2010)

Answer:

Since resistors B and C are connected in parallel, so potential difference across B and C is same. Let I₁, be the current flowing through resistor.B and I₂ be the current flowing through resistor C

Question 21.
Should the resistance of an ammeter be low or high ? Give reason.
Answer:
Ammeter is connected in series in an electric circuit to measure electric current. If its resistance is high, then the net resistance of the electric circuit will increase and hence current in the electric circuit will decrease. Hence, ammeter will not read the actual value of the current in the circuit. If resistance of the ammeter is low, then the net resistance of the circuit will not be effected much. Hence, the current in the circuit is not affected. Ideal ammeter has zero resistance.

Question 22.
Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a reistor of 2Ω in series with a combination of two resistors (4Ω each) in parallel and a voltmeter across the energy. 1 kWh = 1000 Watt x 3600 s parallel combination. Will the potentional difference across the 2Ω resistor be the same as that across the parallel combination of 4Ω resistors ? Give reason.
Answer:

Therefore, potential difference across 2Ω resistor will be same ; as that of across the parallel combination of 4Ω resistors. V = IR.
As R and I in both the cases is same so V = same.

Question 23.
How does use of a fuse wire protect electrical appliances ?
(CBSE 2012)
Answer:
When large current flows in an electric circuit, fuse wire melts due to the large heat produced in it. Therefore, current stops flowing in the circuit and electrical appliances connected in the circuit are protected from burning due to large current in j the circuit.

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

Question 25.
What is the commercial unit of electrical energy ? Represent it in terms of joule.
Answer:
kilowatt hour (kWh) is the commercial unit of electrical energy.

Question 26.
A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω when connected to a 10V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω condcutor and potentional difference across the lamp will take place ? Give reason. Draw circuit diagram.
(CBSE 2010)
Answer:

Net resistance of lamp and conductor connected in series = 5 + 5 = 10Ω
Net resistance of the combination oflamp and conductor and resistance 10Ω connected in parallel,

It means 1A current will flow through 10Ω resistance and 1A current will flow through the lamp and conductor of 5Ω resistance. Hence, there is no change in the current through 5Ω conductor.
Now potential difference across the lamp in case (i)
V = IR = 1 x 5 = 5V
Potential difference across the lamp in case (ii)
V = IR = 1 x 5 = 5V
Hence, there is no change in the potential difference across the lamp.

Question 27.
Why is parallel arrangement used in domestic wiring ?
(CBSE 2012)
Answer:

1. If any one of the electric devices in parallel fuses, then the working of other devices will not be affected.
2. When different devices are connected in parallel, they draw the current as per their requirement and hence they work properly.

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

1. What happenes to the glow of other 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 ?
   

Answer:

1. The glow of bulb depends upon the energy disspated per second i.e. P =V²/R.
   Since V and R of both the bulbs B₂ and B₃ remain the same even if bulb B, gets fused so glow of B₂ and B₃ remain the same.
2. Since bulbs are identical, so their resistance is equal (i.e. resistance of each bulb = R ohm).
   When all bulbs glow, net resistnace of the circuit is given by
   
   When B₂ gets fused, only two bulbs B₁ and B₂ in parallel are in the circuit.
   .’. Net resistance of the circuit is given by
   
   Thus, reading of ammeter A = 2A
   Since B₁ and B₃ are in parallel and have same resistance, so 2A current will be equally distributed between B₁ and B₃. Therefore, reading of ammeter A₁ = 1A Reading of ammeter A₃ = 1A Circuit containing B₂ is broken, so no current flows through this circuit. Hence reading of ammeter A₂ = zero.
3. Power dissipated in the circuit,
   P =V x I
   = 4.5 x 3 = 13.5 W

Question 29.
Three incandescent bulbs of 100W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same voltage 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.
(CBSE 2012)
Answer:
(a) Power dissipated in a circuit = V²/R. Since resistance of the circuit containing bulbs connected in series is more than the resistnace of the circuit containing bulbs in parallel, therefore, power dissipated in parallel combination is more than that in series combination. Hence, bulbs connected in parallel will glow more brightly.
(b) In series combination, there is only one path for the flow of current. So when one bulb gets fused, circuit is broken and hence the bulbs stop glowing.
In parallel combination, each bulb has its own path for the flow of current. So when one bulb gets fused, other bulbs will continue to glow as the current is flowing in the circuits of these bulbs.

Question 30.
State Ohms law. How can it be verified experimentally ? Does it hold good under all conditions ? Comment.
(CBSE 2010)
Answer:
For Ohm’s law: Ohm’s law states that the electric current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor, provided the temperature and . other physical conditions of the conductor remain the same.
For experimental verification: Verify Ohm’s law
Apparatus : A conductor of resistance R, an ammeter, a voltmeter, a battery, a variable resistance (or rheostat used to change the current in the circuit), connecting wires, a key and sand paper.
Procedure:

1. 1. Connect the various components as shown in figure 12.
      
   2. Close the key, so that current begins to flow in the circuit.
   3. Note down the potential difference (V) across the conductor PQ of resistance R shown by the voltmeter and the corresponding current (I) shown by the ammeter.
   4. Now move the knob of rheostat so that the current in the circuit increases.
   5. Again note down the potential difference (V) across the conductor PQ of resistance R in the voltmeter and current in the circuit shown by ammeter.
   6. Repeat the experiment at least five times by increasing the current in the circuit by moving the knob of the rheostat in steps.

Observations:

Plot a graph between V and I by taking V along X-axis and I along Y-axis. We get a straight line passing through origin as shown in figure 11.

Conclusion: From the graph between V and I, we conclude that I x V, which is Ohm’s law. Hence Ohm’s law is verified experimentally.
Precautions: While verifying Ohm’s law experimentally, the following precautions should be taken :

1. Current should not be allowed to pass through the circuit continuously for a long time, which may cause the increase in temperature of the conductor. Therefore, the plug of the key must be taken out every time after noting the readings of ammeter and voltmeter.
2. Connections should be tight.
3. The conductor used in the experiment should be such that its resistance is not changed with increase in temperature of the conductor.

Ohm’s law holds good if the temperature of the conductor remains the same.

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. (CBSE 2012)
Answer:
For Electrical resistivity: Electrical resistivity of a material is defined as the resistance of an object (made of the material) of unit length and unit area of cross-section.
Unit of electrical resistivity is ohm-metre.
For experiment:
Connect the various electrical components as shown in figure 15.

1. Dependence of length of a conductor: 
   Take a copper wire of length l and connect it between the terminals A and B. Note the reading of ammeter. Now take another copper wire of same area of cross-section but of length 2l. Connect it between the terminals A and B by disconnecting the previous wire. Again, note the reading of ammeter. It will be found that the reading of ammeter (i.e., electric current) in the second case is half of the reading of ammeter in the first case. Since R = V/I, so resistance of second wire is double than the resistance of the first wire. Thus, resistance of a conductor is directly proportional to the length of the conductor,
   i.e      resistance ∝ length of the conductor
   Thus, more is the length of a conductor, more is its resistance.
   Thus, the resistance of’a conductor is ‘inversely proportional to the area of cross -section of the conductor.
2. Dependence on area of cross-section of a conductor:
   Now take two copper wires of same length but of different area of cross-sections. Let area of cross-section of first wire is more than the area of cross-section of the second wire. Connect first wire between the terminals A and B in the circuit shown in figure 15. Note the reading of ammeter. Now disconnect the first wire and connect the second wire between the terminals A and B. Again note the reading of the ammeter. It will be found that the reading of ammeter (i.e. electric current) is more when first wire (i.e. thick wire) is connected between A and B than the reading of the ammeter when second wire (i.e. thin wire) is connected between the terminals A and B.
   
   Thus, the resistance of a thin wire is more than the resistance of a thick wire.
3. Effect of the Nature of material:
   Take two identical wires, one of copper and other of aluminium. Connect the copper wire between the terminals A and B. Note the reading of ammeter. Now, connect the aluminium wire between the terminals A and B. Again note the reading of ammeter. It is found that the reading of ammeter when copper wire is connected in the circuit is more than the reading of the ammeter when aluminium wire is connected in the circuit.
   Therefore, resistance of copper wire is less than the resistance of aluminium wire. Hence, resistance of a wire or a conductor depends upon the nature of the material of the conductor.
4. Effect of temperature of conductor:
   If the temperature of a metallic conductor connected in the circuit increases, its resistance increases.
   Thus, factors on which resistance of a conductor depends are :
   1. its length,
   2. its area of cross-section,
   3. the nature of its material and
   4. its temperature.

The various factors affecting resistance of a conductor are given in table 1.
Table 1. Factors affecting resistance of a conductor:

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 ?
Answer:
Perform an activity to show that in series combination of resistors, same current flows through each resistor.

1. Connect three resistors of resistances R₁ = 1Ω, R₂ = 2Ω and R₃ = 3Ω in series. (C.B.S.E. 2014)
2. Connect the series combination of resistors with a battery of 6 V, a plug key K and an ammeter A as shown in figure 17.
3. Note the reading of ammeter after plugging the key. Let it be I.
   
4. Now disconnect ammeter and connect it in between the resistors R₁ and R₂ as shown in figure 18.
5. Again plug the key and note the reading of ammeter. It is found that again it is I.
6. Now disconnect the ammeter and connect it in between the resistors R₂ and R₃.
7. Plug the key and note the reading of ammeter. Again it is found to be I.

Conclusion : Same amount of current flows through each resistor or element connected in series combination.

Question 33.
How will you conclude that the same potential difference (Voltage) exists across three resistors connected in parallel arrangement to a battery ?
Answer:
Perform an activity to investigate the relation between potential difference across parallel combination of resistors and the potential difference across each individual resistors,

1. Connect three resistors of resistances R₁, R₂ and R₃ in parallel. One end of each resistor is joined at a common point ‘a’ and the other end of each resistor is connected at another common point ‘b’.
2. Connect the parallel combination of resistors with a battery, a plug key K and an ammeter A as shown in figure 22(A).
   
3. Now connect a voltmeter across the parallel combination of resistors between a and b points.
4. Note the reading of voltmeter. Let it be V. This is the potential difference across the parallel combination of resistors.
5. Now, disconnect the voltmeter and connect it across R₁ as shown in figure 22(B).
   
6. Note the reading of voltmeter. It is found to be V.
7. Disconnect the voltmeter and connect it across R₂. Note the reading of voltmeter. It is found to be V.
8. Again disconnect the voltmeter and connect it across R₃. Note the reading of voltmeter. It is found to be V.

Conclusion : When resistors are connected in parallel to each other, potential difference across each resistor is equal to the potential difference across the parallel combination of resistors.

Question 34.
What is Joule’s heating effect ? How can it be demonstrated experimentally ? List its four applications in daily life.
(CBSE 2012)
Answer:
Joule’s law can be stated as : The amount of heat produced in a conductor is
(i) directly proportional to the square of the electric current flowing through it.
This is                  H ∝ I²                                                           –
(ii) directly proportional to the resistance of the conductor or resistor.
That is,                H ∝ R                                          ‘
(iii) directly proportional to the time for which the electric current flows through the conductor or resistor.
That is,               H ∝ t
Combining (i), (ii) and (iii), we get H ∝ I² Rt.
or         H = KI²Rt, where K is constant of proportionality
If          K = 1, then H = I²Rt joule
This is known as Joule’s law of heating.
Four Applications:

1. When electric appliances like electric heater, electric iron and water heater etc. are connected to the main supply of electricity, these appliances become hot but the connecting wires remain cold.
   We know, heat produced is directly proportional to the resistance of the material through which current flows. The element of electric heater is made of nichrome. Since, resistance of nichrome is high, so a large amount of heat is produced in the element of the electric heater. Thus, filament of electric heater becomes red hot. However, heat produced in connecting wires made of copper or aluminium is very small and hence they are not heated up.
   
2. Filament of an electric bulb is made of a thin wire of tungsten. The melting point of filament is high i.e., about 3380 °C. The filament of the bulb is enclosed in a glass envelope fixed over an insulated support as shown in figure 28. The glass envelope of electric bulb is filled with inactive gases like nitrogen and argon to increase the life of the tungsten filament.
   Since resistance of thin filament is very high, so a large heat is produced as the electric current flows through the filament. Due to this large amount of heat produced, filament of the bulb becomes white hot. Hence, the filament of the bulb emits light and heat.
   
3. Electric fuse is a safety device connected in series with the electric circuit. Electric fuse is a wire made of a material whose melting point is very low. Examples of the materials of making fuse wire are copper or tin-lead alloy. When large current flows through a circuit and hence through a fuse wire, a large amount of heat is produced. Due to this large amount of heat, the fuse wire melts and the circuit is broken so that current stops flowing in the circuit. This saves the electric circuit from burning due to the passage of large current through it.
   

Question 35.
Find out the following in the circuit given in the figure.

(a) Effective resistance of two 8Ω resistors in the combination.
(b) Current flowing through 40 resistor
(c) Potentional difference across 40 resistor
(d) Power dissipated in 4Ω resistor
(e) Difference in ammeter readings, if any.
(CBSE 2010,2012)
Answer:
(a) Two 80 resistors are in parallel, so their effective resistance is given by

(e) Since same current flows through every part in a series circuit and both the ammeters are connected in series, so there will be no difference in ammeter readings.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

Value Based Questions in Science for Class 10 Chapter 12 Electricity

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 12 Electricity

Question 1.
In the house of Ram, there are 20 incandescent bulbs each of 100 W, three geysers each of 2000 W and 20 tubes each of 40 W. All these appliances work for 5 hours in a day. Every month, he pays heavy amount as electricity bill. His son Sham studying in X standard asked his father to replace all incandescent bulbs with CFL bulb each of 40 W to save electricity.

1. How much units of electricity are saved per month (30 days) by replacing incandescent bulbs with CFL ?
2. What values are shown by Sham ?

Answer:

1. Electric energy consumed by 20 incandescent bulbs in 30 days = P x t = 20 x 100 x 5 x 30
   = 300000 Wh = 300 kWh
   = 300 units
   Electric energy consumed by 20 CFL bulbs = 23 x 40 x 5 x 30 = 120000 Wh
   = 120 kWh =120 units
   Units of electricity saved = 300 – 120 = 180 units.
2. Sham helped his father to pay less electricity bill. He also believes that saving energy contributes for the development of our nation.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Electricity plays an important role in the development of a country. Ram, a student of class X was studying in the library after school hours. When he left the library, he found that electric fans of all rooms were ‘ON’ although there was no one in the class rooms. He immediately switched ‘OFF’ all the fans and reported the matter to Principal of the school.

1. Comment on the attitude of Ram.
2. Why, Ram reported the matter to the Principal ?

Answer:

1. Ram knows the importance of electricity. He believes that loss of electricity is the loss of school as well as the loss of nation. He is against the misuse of national resources.
2. He reported the matter to the Principal so that the Principal may ask the peon to ensure that such incident should not occur in future.

Question 3.
A welder was asked by Mr. Sumit to weld an iron grill in his house. He started using electricity by connecting the wires of welding set directly with the transmission wires and not through the energy meter. Sumit’s neighbour objected the action of welder but Sumit sided with the welder. However, the son of Sumit appreciated the neighbour.

1. Why was the action of welder objected by Sumit’s neighbour ?
2. Why, Sumit supported the action of the welder ?
3. Why, Sumit s son appreciated the action of his neighbour ?
4. Write the commercial unit of electrical energy.

Answer:

1. It is a crime to steal electrical energy and no honest person can support it.
2. Sumit thought the welder was doing a favour to him.
3. Sumit’s son knows that welder is doing wrong.
4. Kilowatt hour (kWh).

Hope given Value Based Questions in Science for Class 10 Chapter 12 Electricity are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

HOTS Questions for Class 10 Science Chapter 12 Electricity

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 12 Electricity

Question 1.
Following table gives the resistivity of three samples in (Ωm)

Which of them is a good conductor ? And which of them is an insulator ? Explain why ? (CBSE 2012)
Answer:
A material having low resistivity is a good conductor. Since, resistivity of sample A is the least among all other materials, so sample A is a good conductor. A material having high value of resistivity is an insulator. Therefore, sample C is an insulator.

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
The electrical resistivity of few materials is given below in ohm-meter. Which of these materials can be used for making element of a heating device ?

(CBSE Sample Paper, CBSE 2012, Similar CBSE 2014)
Answer:
For making element of a heating device, we use alloy instead of pure metals.The resistivity of material D lies in the range of resistivities of alloys. Therefore, material D can be used for making element of a heating device.

Question 3.
Electrical resistivities of some substances at 20°C are given below :


Answer the following questions in relation to them :

1. Among silver and copper, which one is a better conductor ? Why ?
2. Which material would you advise to be used in electrical heating devices ? Why ? (CBSE 2012)

Answer:

1. A material whose electrical resistivity is low is a good conductor of electricity. Since the electrical resistivity of silver is less than that of the copper, so silver is a better conductor than the copper.
2. For making the elements of heating devices, alloy is used instead of a pure metal. This is because the resistivity of an alloy is more than that of a metal and alloy does not burn (or oxidise) even at higher temperature. Out of the given substances, nichrome is an alloy, so nichrome is used in electrical heating devices.

Question 4.
The following table gives the resistivity of three samples :

Which of them is suitable for heating elements of electrical appliances and why ? (CBSE, 2010, 2012)
Answer:
For making the heating elements of electrical appliances, alloy is used instead of a pure metal. This is because alloy does not burn even at higher temperature. The resistivity of sample C is of the order of an alloy, so sample C is suitable for heating elements of electrical appliances.

Question 5.
Two metallic wires A and B are connected in series. Wire A has length l and radius r, while wire B has length 2l and radius 2r. Find the ratio of the total resistance of series combination and the resistance of wire A, if both the wires are of same material
Answer:

Question 6.
Two metallic wires A and B of same material are connected in parallel. Wire A has length l and radius r and wire B has length 2l and radius 2r. Compute the ratio of the total resistance of parallel combination and the resistance of wire A.
(CBSE Sample Paper)
Answer:

Question 7.
Two students perform the experiments on series and parallel combinations of two given resistors R₁ and R₂ and plot the following V-I graphs.

Which of the graphs is (are) correctly labelled in terms of the words ‘series’ and ‘parallel’. Justify your answer.
(CBSE Sample Paper, 2012)
Answer:
In first graph, slope of I-V graph = Resistance.
Since in series combination, resistance is more than the resistance in parallel combination, therefore slope of I-V graph for series combination is more than the slope of resistance.

Hence, graph is correctly labelled. In second graph, slope of resistance

Question 8.
Two electric circuits I and II are shown in figure.

(i) Which of the two circuits has more resistance ?
(ii) Through which circuit, more current passes ?
(iii) In which circuit, the potential difference across each resistor is equal.
Answer:
(i) Equivalent resistance of series combination of resistors is more than the equivalent resistance of the parallel combination of resistors. So, the resistance of circuit I is more than the resistance of circuit II.

∴ So current in circuit II is more than the current in circuit I.
(iii) Potential difference across each resistor is equal in circuit II.

Question 9.
An electrician puts a fuse of rating 5 A in that part of domestic electrical circuit in which an electrical heater of rating 1.5 kW, 220 V is operating. What is likely to happen in this case and why ? What change, if any needs to be made ?
(CBSE Sample Paper)
Answer:
The fuse will melt and the circuit breaks if electric current more than the rating of fuse (i.e., 5 A) flows in the circuit. Electric current flowing in the circuit,

Since, current flowing in the circuit (6.82 A) is more than the rating of fuse (5 A), therefore, the fuse will melt and the electrical heater does not work. To operate the heater, fuse of rating 10 A is to be put in the circuit.

Question 10.
The electric power consumed by a device may be calculated by using either of the two expressions: P = I²R or P = V²/R . The first expression indicates that the power is directly proportional to R,whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained ?
(CBSE Sample Paper)
Answer:
P = I²R is used when current flowing in every component of the circuit is constant. This is the case of series combination of the devices in the circuit.
P = V²/R is used when potential difference (V) across every component of the circuit is constant. This expression is used in case of parallel combination in the circuit. In series combination, R is greater than the value of R in parallel combination.

Question 11.
Three V-I graphs are drawn individually for two resistors and their series combination. Out of A, B, C which one represents the graph for series combination of the other two. Give reason for y or ansiver. (CBSE 2011)
Answer:
Slope of V-I graph = resistance of a resistor.
When two resistors are connected in series,Volts the resistance of this combination (R = R₁ + R₂) is more than the resistance of both the resistors.

Since, slope of C is greater than the slopes of A and B. Therefore, C represents the graph for series combination of the other two.

Question 12.
V-I graphs for the two wires A and B are shown in the figure. If we connect both the wires one by one to the same battery, which of the two will produce more heat per unit time ? Give justification for your answer (CBSE 2014, 2015)
Answer:
Heat produced per unit time = V²/R
Now slope of V-I graph = R (resistance of wire).

Since slope of V-I graph for wire A is greater than the slope of V-I graph for wire B, therefore, resistance of wire A is greater than the resistance of wire B. Hence, more heat will be produced per unit time in wire B than in wire A.

Question 13.
The resistivities of some substances are given below :

Answer the following questions in relation to them giving justification for each :
(i) Which material is best for making connecting cords ?
(ii) Which material do you suggest to be used in heater elements ?
(iii) You have two wires of same length and same thickness. One is made of material A and another of material D. If the resistance of wire made of A is 2 Ω, what is the resistance of the other wire ?
Answer:
(i) Material A is best for making connecting cords as its resistivity is the lowest one.
(ii) For heater elements, material of high resistivity is used. Therefore, the material E is to be used in heater elements.

Hope given HOTS Questions for Class 10 Science Chapter 12 Electricity are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Short Answer Questions

Question 1.
Prepare a list of items that you use daily in the school. Identify from the list five such items that can be recycled.
Answer:
Items. Rexin bag, steel lunch box, steel spoon, steel compass, steel dividers, paper, plastic box, pen, pencil, blade, eraser, handkerchief.
Recycleable Items. Steel lunch box, steel spoon, steel compass, steel dividers, blade, paper, plastic box.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
List the advantages associated with water harvesting at the community level. (CCE 2012)
Answer:
Water harvesting at the community level is capturing, collection and storage of rain water and surface run off for filling either small water bodies or recharging ground water. This is carried out through water shed management, check dams, earthen dams, roof top harvesting and filter wells in flood drains.
Benefits:

1. It ensures water availability in non-rainy season,
2. It reduces the chances of flooding during rainy season,
3. Ground water level does not fall as it is regularly recharged,
4. Ground water recharge is the best form of water harvesting as the water is filtered and free from contaminations. It also does not evaporate,
5. Water becomes available for drinking as well as irrigation.

Question 3.
In a village in Karnataka people started cultivating crops all around a lake which was always filled with water. They added fertilizers to their field in order to enhance the yield. Soon they discovered that the water body was completely covered with green floating plants and fishes started dying in large numbers.
Analyse the situation and give reasons for excessive growth of plants and death of the fish in the lake.
Answer:
Fertilizer rich run off from fields must have passed into the lake. It caused nutrient enrichment of lake water. The result is excessive growth of algae and other aquatic plants which float on the water surface and produce water bloom. Old dead plants produce a lot of organic matter. The submerged plants are also killed due to shading. BOD of water increases. As more and more oxygen is consumed by decomposers little is left for respiration of aquatic animals. Therefore, fish begin to die. The phenomenon of nutrient enrichment of water body that causes formation of water bloom and subsequent killing of aquatic life is called eutrophication.

Question 4.
What measures would you take to conserve electricity in your house ? (CCE 2012)
Answer:

1. Judicious use of electricity by switching off lights and electrical appliances not required,
2. Replacement of incandescent bulbs with fluorescent, compact fluorescent ones and LED bulbs.
3. Replacement of electricity or gas operated geysers with solar water heaters,
4. Replacement of electricity generating sets with solar light,
5. Having more natural light and ventilation with design supporting warming during winters and cooling during summer.

Question 5.
Although coal and petroleum are produced by degradation of biomass, yet we need to conserve them. Why ? (CCE 2012)
Answer:
Coal and petroleum have been produced from large amounts of biomass entrapped inside the earth under high temperature, pressure and anaerobic conditions. Such a situation develops only rarely like big upheavals on earth. At present no more coal or petroleum is being formed. All that is available has been formed millions of years ago. Being rich source of energy, coal and petroleum are being consumed in ever increasing amount in industry, transport, kitchens, etc. If the trend continues, soon they will be exhausted. Therefore, they must be conserved by developing more efficient machines, hybrid engines and using hydrogen as a fuel.

Question 6.
Suggest a few measures for controlling carbon dioxide levels in the atmosphere.
Answer:

1. Increasing Vegetation Cover. It will increase utilisation of atmospheric CO₂ in photosynthesis.
2. Seeding of Oceans With Phytoplankton. Increased photosynthetic activity of oceans will result in decreasing CO₂ concentration.
3. Carbonation. CO₂ released during combustion should not be allowed to pass into atmosphere. Instead, it can be changed into carbonates.
4. Alternate Sources of Energy. Instead of fossil fuels, hydrogen fuel and solar energy should be used.
5. Burning of Litter. Litter and crop residue should not be burnt but instead converted into manure.

Question 7.
(a) Locate and name the water reservoirs in figures (i) and (ii).
(b) Which has advantage over the other and why ?

Answer:
(a) Water reservoir in figure (i) is pond while it is underground water body (ground water) in figure (ii).
(b) Ground water is more advantageous than pond water.
For Benefits: 

1. Prevents flooding,
2. Checks soil erosion.
3. Retains water underground and prevents drought,
4. Increases life of downstream reservoirs and dams,
5. Higher biomass production and income of water shed community,
6. Maintenance of ecological balance.

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Long Answer Questions

Question 8.
In the context of conservation of natural resources, explain the terms reduce, recycle and reuse. From among the materials that we use in daily life, identify two materials for each category.
Answer:
Three Rs — reduce, recycle and reuse.

1. Reduce: It is to reduce consumption by preventing wastage.
   1. Switching off unnecessary lights, fans and other electrical appliances,
   2. Repair of leaky taps.
   3. Reducing food wastage,
   4. Walking down to nearby market instead of using vehicle.
2. Recycle: Separation of recyclable wastes from non-recyclable wastes. The former are taken by rag pickers for sending them to industries involved in recycling, e.g., paper, plastic, metal, glass.
3. Reuse: Carry bags, packing material, plastic containers and other reusable articles should not be thrown away if the same are uncontaminated. For example, plastic bottles and jars containing various food items brought from market can be washed and used for storing things in the kitchen.

Question 9.
Prepare a list of five activities that you perform daily in which natural resources can be conserved or energy utilisation can be minimised.
Answer:

1. Judicious use of electricity by switching off lights and electrical appliances not required,
2. Replacement of incandescent bulbs with fluorescent, compact fluorescent ones and LED bulbs.
3. Replacement of electricity or gas operated geysers with solar water heaters,
4. Replacement of electricity generating sets with solar light,
5. Having more natural light and ventilation with design supporting warming during winters and cooling during summer,
6. Reducing wastage of water, food and other articles.
7. Separation of recyclable waste from non-cyclable waste prior to disposal.
8. Increasing reuse of containers,
9. Using cloth bags instead of polythene, plastic or paper bags.

Question 10.
Is water conservation necessary ? Give reasons.
Answer:

1. Distribution of fresh water is highly uneven. Large tracts are deficient in rain as well as ground water,
2. At most places more water is withdrawn from reservoir and underground source than their recharging
3. Requirement in urban and industrial areas is nearly always higher than the availability,
4. Further demand for water is rising by 4 – 8% annually in all fields, whether agriculture, industry or domestic use.

Therefore, water conservation is necessary. Wastage of the resource should be prevented. Waste water should be recycled. Water harvesting involving recharging of ground water should be practised.

Question 11.
Suggest a few useful ways of utilising waste water.
Answer:
Waste or used water can also become a resource.

1. Treated municipal water can be poured in irrigation channels for supply to crop fields,
2. Treated waste water can be used in urban areas for watering gardens, lawns and washing vehicles,
3. Industries can treat their waste water and recycle the same,
4. Waste water passed into ponds recharges the ground water,
5. Sewage sludge, separated from waste water is a source of manure, compost and biogas.

Question 12.
What is the importance of forests as a resource ?
Answer:
Economic Reasons:

1. Food: Tribals obtain most of their food requirements from the forests, e.g., fruits, tubers, fleshy roots, leaves.
2. Nuts: Pine Nut (Chilgoza), Almond, Walnut and Cashewnut are obtained from forests trees.
3. Spices: Cardamom, Cinnamon, Nutmeg and Cloves are spices obtained from forest plants.
4. Commercial Products: A number of forest products are of commercial importance, g., rubber, resin, tannins, tendu, lac, cork, camphor, essential oils, soap pod and drugs.
5. Fuel Wood: Nearly two billion persons depend upon forests for fuel wood.
6. Timber: Wood for the manufacture of furniture, household fitments and several other articles mostly comes from forests. Bamboo is called poorman’s timber as it is used in thatching huts, preparing baskets and a number of other articles including furniture.
7. Paper: It is prepared from cellulose rich plants like bamboos, Boswellia, Eucalyptus, grasses and several

Protective Functions:                                                                                  _i

1. Forests provide shelter to wild animals. Over 40 million tribals and villagers live in forests.
2. Plant roots hold the soil firmly. Vegetation protects the soil from action of wind and water. Forests, therefore, protect the soil from erosion and landslides.
3. Pollution. Forests reduce atmospheric pollution by absorbing gases, collecting suspended particles and reducing noise.

Regulative Functions:

1. Absorption and Retention of Water. Forests reduce run off, hold water like a sponge and allow slow percolation to form perennial springs and rivulets.
2. Forests increase atmospheric humidity, increase frequency of rainfall and moderate temperature.
3. Atmospheric Gases. Forests absorb large quantity^ of C0₂ from the atmosphere, reducing the threat of global warming. They also release a lot of oxygen.

Question 13.
Why are Arabari forests of Bengal known to be good example of conserved forests.
Answer:
Regeneration of Sal Forests — An Example of People’s Participation in the Management of Forests Despite best efforts, the West Bengal Forest Department could not revive the degraded Sal forests of Southwestern districts of the state. Excessive surveillance and policing of the degraded forests not only alienated the people but also resulted in frequent clashes between villagers and forest officials. This also fueled the militant peasant movement led by Naxalites. Realising the failure, the forest department revised its strategy in 1972. It allowed forest officer A.K. Banerjee of Arabari forest range of Midnapore to involve villagers in regeneration of 1272 hectares of badly degraded Sal forest. Banerjee provided employment to villagers in silviculture (cultivation of trees) and harvesting, 25% of final harvest and allowed collection of fuel wood as well as fodder at nominal fee. By 1983, the Arabari forest had been revived and was then valued at 12-5 crores.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.