Important Questions of Electricity Class 10 with Solutions
Important Questions Class 10 Science Chapter 12 – Electricity
Chapter 12 of Class 10 Science is about ‘Electricity’. One of the most fundamental elements of our society is electricity. Since the beginning of the industrial revolution, electricity has contributed to the development of our civilization by powering numerous businesses and industries. Today, life would be in complete disarray if we were to lose this energy source. Therefore, it’s crucial for students to learn ideas about how electricity functions at the molecular level and explore its applications. Chapter 12 ‘Electricity’ will teach students about the fundamentals of electricity, the movement of current, and the operation of circuits as a whole.
Extramarks has been providing great online study resources for students from grade 1 – 12. Extramarks experienced Science faculty identify important ideas and test-prep questions from every chapter to help students prepare for exams. Throughout our question bank of Chapter 12 Class 10 Science Important Questions, students are given important problems to complete so they will be familiar with the questions which are expected in their final exams.
It is not helpful to memorise the answer for Science problems since it requires an in-depth understanding of the concepts involved. For each question in our Important questions Class 10 Science Chapter 12 there is a step-by-step explanation prepared by our in-house Science subject experts. . This makes it easy for students to have complete faith and trust, to revise concepts covered in class.
The questions are compiled from the NCERT textbook, NCERT exemplar, and other reference books for providing the best and authentic source of question bank. Furthermore, there are questions from previous year’s exams also. In order to set the foundation for the chapter, it could be beneficial for students to refer to the Important Questions Class 10 Science Chapter 12.
There is a need for students to improve their understanding of the art of writing answers. They can practise representational diagrams by referring to them from NCERT textbook and from our NCERT chapter-wise solutions. Students can access our entire collection of Science Class 10 Chapter 12 Important Questions by registering on the Extramarks’ website. Besides, students can review other study materials on our Extramarks websites, such as NCERT solutions, revision notes, and the previous year’s question papers.
CBSE Class 10 Science Important Questions 2022-23
CBSE Class 10 Science Important Questions are also available for the following chapters:
CBSE Class 10 Science Important Questions | |
Sr No. | Chapters |
1 | Chemical Reactions and Equations |
2 | Acids, Bases and Salts |
3 | Metals and Non-metals |
4 | Carbon and Its Compounds |
5 | Periodic Classification of Elements |
6 | Life Processes |
7 | Control and Coordination |
8 | How do Organisms Reproduce? |
9 | Heredity and Evolution |
10 | Light Reflection and Refraction |
11 | Human Eye and Colourful World |
12 | Electricity |
13 | Magnetic Effects of Electric Current |
14 | Sources of Energy |
15 | Our Environment |
16 | Management of Natural Resources |
Important Questions Class 10 Science Chapter 12 – With Solutions
Given below is a set of questionnaires and their answers from our question bank of Class 10 Science Chapter 12 Important Questions.
For the best preparation, students are advised to go through this set of Important questions Class 10 Science Chapter 12. They can register on our website to get access to it.
Question 1: Which of the following does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
Answer 1: b) IR2
Explanation:
Electrical power is represented by the expression P = VI. (Equation 1)
According to Ohm’s law,
V = IR
Putting the value of V in ( Equation 1), we get
P = (IR) × I
P = I2R
Similarly, from Ohm’s law,
I = V/R
Putting the value of I in (Equation 1),
P = V × V/R = V2/R
It is thus clear that the equation IR2 does not represent electrical power in a circuit. Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit.
Question 2: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be _____.
(a) 75 W
(b) 100 W
(c) 50 W
(d) 25 W
Answer 2: (d) 25 W
Explanation:
This expression demonstrates how much energy the electric bulb consumes.
P = VI = V2/R
The given formula can be used to calculate the light bulb’s resistance:
R = V2/P
Putting the values, we get
R = (220)2/100 = 484 Ω
The resistance generally does not change when the voltage supply is decreased. Consequently, the amount of electricity used can be determined as follows:
P = V2/R
Putting the values, we get
P = (110)2 V/484 Ω = 25 W
As a result, the electric bulb uses 25 W of power when it is operating at 110 V.
Question 3: What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
(a) 5 Ω
(b) 10 Ω
(c) 1/5 Ω
(d) 1 Ω
Answer 3: (d) 1 Ω
Explanation: Resistance is maximum when resistors are connected in series.
R= 15+15+15+15+15
= 55
= 1Ω
Question 4: If the current ‘I’ through a resistor is increased by 100% (assuming that the temperature remains unchanged), the approximate increase in power dissipated will be
(a) 400 %
(b) 200 %
(c) 300 %
(d) 100 %
Answer 4: (c) 300 %
Explanation: The amount of heat produced by a resistor is inversely proportional to the square of the current. Therefore, the loss of heat will multiply by 2=4 when the current doubles. Accordingly, there will be a 300% increase.
Question 5: A piece of wire of resistance R is cut into five equal parts. These parts are then arranged in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is _____.
(a) 5
(b) 1/5
(c)1/25
(d) 25
Answer 5: d) 25
Explanation:
The resistance is divided into five halves, each of which has a resistance of R/5.
Since we are aware that each component is linked to the others in parallel, we can compute the equivalent resistance as follows:
1R’ = 5R + 5R + 5R + 5R +5R
= 5 + 5+ 5+ 5+ 5R = 25R
RR’ = 25
The ratio of R/R′ is 25.
Question 6: The correct representation of the series combination of cells (Figure 12.4) obtaining maximum potential is
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer 6: (a)
A cell’s positive terminal needs to be connected to the neighbouring cell’s negative terminal. The appropriate cell combination is represented by case I.
Question 7: Two pieces of conducting wire of the same material and of equal lengths and the equal diameters are first connected in series and then changed to parallel in a circuit across the same potential difference. The ratio of heat produced in both series and parallel combinations would be _____.
(a) 1:2
(b) 4:1
(c) 1:4
(d) 2:1
Answer 7: (c)
Let Rs and Rp represent the wires’ respective equivalent resistances when linked in series and parallel.
The ratio of the heat generated in the circuit is provided by
HsHp = V2Rs tV2Rpt = Rp Rs
The equivalent resistance (Rs) of resistors connected in series is R + R = 2R
The equivalent resistance (Rp) of resistors connected in parallel is 1R + 1R = R2
Hence, the estimated ratio of the heat produced in series and parallel combinations would be
HpHs = R22R = 14
Thus, the ratio of the heat produced is 1:4.
Question 8: 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
Answer 8: (b) 1/25 Ω
Explanation:
Resistance is the minimum when resistors are connected in parallel
1R= 5 + 5 + 5 +5 +5= 25 Ω
R=125Ω
Question 9: A person carries out an experiment and thus plots the V-I graph of three taken samples of nichrome wire with different resistances R1, R2 and R3, respectively (Figure.12.5). Which one of the following is true?
(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R3 > R1
Answer 9: (c)
The graph’s slope is 1R because the current (I) is plotted on the y-axis, and the potential difference (V) is plotted on the x-axis. It implies that the less resistance, the steeper the slope. R1 will therefore be the minimum and R3 the maximum.
Question 10: Two resistors of resistance 2 Ω and 4 Ω, when connected to a battery, will have
(a) the same potential difference across them when connected in series
(b) same current flowing through them when connected in series
(c) same current flowing through them when connected in parallel
(d) different p
Answer 10: (b) same current flowing through them when connected in series
Explanation:
Since the resistor gets a common current in a series arrangement, the current is not split into branches.
Question 11: What does an electric circuit mean?
Answer 11: An electric circuit is a continuous, closed path or loop composed of electronic components through which an electric current flows. Conductors, cells, Switch, and Load are the components of a simple circuit.
Question 12: An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω resistances 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 that flows through it?
Answer 12:
R1 = 100 , R2 = 50 , R3 = 500
All the devices are in parallel, so
1R = 1R1 + 1R2 + 1R3
1R = 1100 + 150 + 1500 = 5 + 10 + 1500 = 16500
R = 50016 = 1254
Current, through all the appliances
I = VR = 2201254 = 220 X 4125 = 7.04 A
Now, if only electric iron is connected to the same source such that it takes as much current as taken by all three appliances, i.e. I = 7.04 A, its resistance should be equal to 1254, i.e. 31.25 .
Question 13: How is the resistivity of alloys compared with those of pure metals from which they may have been formed?
Answer 13: An alloy often has a higher resistivity than the individual metals that make up the alloy.
Question 14: Write the relation between the resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it.
Answer 14: The relation between resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it can be represented as follows:
P = V2R
Question 15: How does the use of a fuse wire protect electrical appliances?
Answer 15: Compared to the main wiring, the fuse wire has a high resistance. Whenever there is an abrupt surge in electric current, the circuit is broken by melting fuse wire. This keeps electrical equipment from being damaged.
Question 16: Why are copper wires used as connecting wires?
Answer 16: Copper wires are used as the connecting wires because, in the case of copper, the electrical resistivity for it is low. It is ductile, inexpensive and it is an excellent electrical conductor.
Question 17: Define the SI unit of current.
Answer 17: The SI unit of current is ampere. An ampere is defined by the flow of one coulomb of Charge per second.
Question 18: 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?
Answer 18: In order to get 4 Ω, resistance 2 Ω should be connected in series with the parallel combination of 3 Ω and 6 Ω.
1RCD = 13 + 16= 2 + 16
= 36 = 12
RCD = 2 , RAB= 2
RAD = RAB + RCD
= 2 + 2 = 4
Therefore, the total resistance of the circuit is R= 4
(b) In order to get 1 , all three resistors should be connected in parallel as
1R = 12 + 13 + 16= 3 + 2+ 16 = 1
Therefore, the net equivalent resistance of the circuit is R= 1
Question 19: A rectangular block of iron has dimensions L x L x b. What will be the resistance of the block measured between the two square ends? Given p resistivity.
Answer 19: We have given that a rectangular block of iron has dimensions l x l x b. We need to find the resistance of the block measured between the two square ends.
The resistance is given by the below formula as follows :
R = LA
L is length of block
A is area of cross section
In this case,
Length of the rectangular block is l and area of block is l x b. So, resistance of the block measured between the two square ends is :
R = ll x b
R = b
So, the resistance of the block measured between the two square ends is R = b.
Question 20: Ammeter burns out when connected in parallel. Give reasons.
Answer 20: When a low resistance wire is connected in parallel, a huge quantity of current travels through it, causing it to be either burned out or short-circuited.
Question 21: Should the resistance of an ammeter be low or high? Give reason(s).
Answer 21: The resistance of an ammeter should be zero, as the ammeter should not affect the flow of current in a circuit.
Question 22: Why does the connecting rod of an electric heater not glow, but the heating element does?
Answer 22: As the resistance of the connecting rod is lower than that of the heating element, the connecting rod of an electric heater does not glow. Thus, the heating element produces more heat than the connecting cord, and it glows.
Question 23: The power of a lamp is 60 W. Find the energy in joules consumed by it in 1s.
Answer 23: Here, given the power of the lamp, P = 60 W time,
t= 1 s
So, energy consumed = power x time = (60 x 1) J = 60 J
Question 24: A wire of resistivity ‘p’ is stretched to double its length. What will be its new resistivity?
Answer 24: When a wire of resistivity p is stretched to double its length, then the new resistivity tends to remain the same because resistivity depends on the nature of the material.
Question 25: What is the resistance of any connecting wire?
Answer 25: The resistance of the connecting wire made of a good conductor is extremely low and they are assumed to have zero resistance. So, less heat is produced in them and they can be easily used in connections.
Question 26: A number of n resistors each of resistance ‘R’ are first connected in series and then in parallel connection. What is the ratio of the total effective resistance of the circuit in series combination and parallel combination?
Answer 26: Total effective resistance of the circuit in series combination Rs = nR
And for parallel combination is Rp = Rn and
RsRp = nRRn
= n2
The ratio will be n2.
Question 27: Calculate the total number of electrons constituting one coulomb of charge.
Answer 27:
We know,
The Charge of an electron = 1.6 × 10-19 C.
According to the concept of charge quantisation,
Q = nqe, where we suppose ‘n’ is the number of electrons and similarly ‘qe’ is the Charge of the electron.
Substituting these values in the said equation, the number of electrons constituting one coulomb of Charge can be calculated as follows:
1C = n X 1.6 X 10-19
n= 11.6 X 10-19 = 6.25 X 1018
Therefore, the number of electrons in one coulomb of Charge = 6. 25 × 1018.
Question 28: How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts.
Answer 28:
Here, V = 220V , R = 55
By Ohm’s law, V = IR
Therefore, 220 = 7 x 55 or I = 4A
The wattage of electric iron = Power
= V2R = (220)255 = 880 W
Question 29: A current of 1 ampere flows in a circuit of series connection containing an electric lamp and a conductor of 5 Ω and connected to a 10 V battery. Calculate the resistance of the given electric lamp.
Therefore, if the resistance of 10 Ω is connected in parallel with this series combination, what type of change (if any) in current flowing through the 5 Ω conductor and potential difference across the lamp will take place? Give reasons.
Answer 29:
Let Rlamp represent the resistance of the lamp.
Current (I) = 1 A
Resistance of conductor (Rconductor) = 5 Ω
The potential difference of battery (V) = 10 V
Given that the lamp and conductor are linked in series, the same amount of current 1 A will flow through them both.
Using Ohm’s law,
Rnet = VI
Rnet =101
Rnet = 10
We know, in series connection
Rnet = Rlamp + RConductor
10 = Rlamp + 5
Rlamp = 5
The potential difference across lamps,
Rlamp = I x Rlamp
= 1x 5 = 5 V
When a resistor of 10 Ω resistor connected parallel to the series combination of lamp and conductor
( Rnet = 5 + 5 = 10 ) then the equivalent resistance,
1Req = 110 + 110 = 210 = 15
Req = 5
Using Ohm’s law,
I’= VReq
I’= 105
I’= 2A
Equal distribution of current will occur in two parallel parts.
Thus, I’/2 = 1A current will pass through both the lamp and the resistor of 5 (because they are connected in series).
The potential difference across the lamp (Rlamp = 5 ).
V’lamp = 1×5 = 5 V
Therefore, the current flowing through the conductor of resistance 5 and the potential difference across the bulb won’t change.
Question 30: What is electrical resistivity? In a particular series electrical circuit comprising a resistor made up of a metallic wire, the ammeter generally reads 5 A. The previous reading of the ammeter decreases to half in case the length of the wire is doubled. Why?
Answer 30: Resistivity is a property of a conductor that prevents the flow of electric current. A specific material has a particular resistance. Resistance is inversely proportional to current flow and directly proportional to conductor length.
When the length is doubled, the resistance doubles and the current flow is reduced by half. This is what’s causing the ammeter value to drop.
Question 31: (i) List the three factors on which the resistance of a conductor depends.
(ii) Write the SI unit of resistivity.
Answer 31: (i) A conductor’s resistance is influenced by the following factors:
(1) Length of the conductor: The resistance (R) will increase as the conductor’s length (I) increases.
R ∝ I
(2) Area of the cross-section of the conductor: (as the cross-sectional area of the conductor increases, the resistance decreases.
R ∝ 1A
(3) Nature of conductor.
(ii) SI unit of resistivity is Ω m.
Question 32: An electric bulb which is connected to a 220 V generator and the current is 2.5 A. Calculate the power of the bulb.
Answer 32: Here, V= 220 V, I = 2.5 A
Given, Power of the bulb, P = VI = 220 × 2.5 W = 550 W
Question 33: Name a device that helps to maintain a potential difference across a conductor.
Answer 33: One of the devices that aid in maintaining a potential difference across a conductor is a battery, which can consist of one or more electric cells.
Question 34: What is the resistance of an ammeter?
Answer 34: An ammeter’s resistance generally is very minimal, and in an ideal ammeter, it is zero.
Question 35: What is the resistance of a voltmeter?
Answer 35: The resistance of a voltmeter is ideally infinite resistance.
Question 36: What is the commercial unit of electrical energy? Represent it in terms of joules.
Answer 36: The commercial unit of electrical energy is kilowatt/hr
1 kW/hr = 1 kW h
= 1000 W × 60 × 60s
= 3.6 × 106 J
Question 37: Explain two disadvantages of series arrangement for a household circuit.
Answer 37: The two drawbacks of series circuits for household wiring are:
- If one electrical appliance in a series circuit stops functioning for any reason, the entire circuit will break, and all other electrical appliances will also stop functioning.
- Because there is only one switch for every electrical device in a series circuit, they cannot be turned on or off independently.
Question 38: What is meant by the saying that the potential difference between two points is 1 V?
Answer 38: The potential difference between two points is 1V when 1 J of work is done to move a 1 C of Charge from one location to the other.
Question 39: Two equal wires of equal cross-sectional area, one of copper and the other of manganin , have the same resistance. Which one will be longer?
Answer 39: Using the equation, = RAI, where is the resistivity, R is the resistance, and A is the area.
Resistance of Copper wire = 1l1A
Resistance of Manganin wire = 2l2A
1l1 = 2l2 (As lis constant)
Since 1 <<< 2
So, l1>>>>l2
I.e. Copper wire would be longer.
Question 40: Three equal resistances are connected in series and then in parallel. What will be the ratio of their change in resistances?
Answer 40: When connected in series, Resistance Rseries = R+R+R= 3R
When connected in parallel, Resistance Rparallel = R/3
Ratio of change in resistances= RseriesRparallel = 3RR/3
Therefore, the ratio of change in resistances is 9:1
Question 41: State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.
Answer 41:
According to Ohm’s law, the potential difference (voltage) across an ideal conductor is proportional to the current flowing through it at a given temperature.
I.e. V/I = R
Verification of Ohm’s law
Make the circuit indicated in Fig., which consists of four 1.5 V cells, an ammeter, a voltmeter, and a nichrome wire of length XY, say, 0.5 m. (The metals nickel, chromium, manganese, and iron make up the alloy known as nichrome.)
Start by using a single cell as the circuit’s source. Take note of the ammeter’s reading for current (I) and the voltmeter’s reading (V) for the potential difference across the nichrome wire ‘XY’ in the circuit. Add them to the table provided.
Connect two batteries to the circuit next, and then note the ammeter and voltmeter readings for the current flowing through the nichrome wire and the potential difference across the nichrome wire values, respectively.
Use three cells in the circuit first, then four cells, and repeat the process above for each group of cells.
Question 42: If there are 3 x 1011 electrons flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron 1.6 x 1019 C.
Answer 42:
Using the equation, q = ne
= 3 x 1011 x 1.6 x 1019 C
= 4.8 x 108 C
I = q/t
=4.8 x 1082 x 60
= 4 x 107 A
The current flowing through the electric circuit is 4 x 107 A.
Question 43: Three resistors of 5 Ω, 10 Ω and 15 Ω are connected in series, and the entire combination is connected to a battery of 30 V. Ammeter and Voltmeter are connected in the circuit. Draw a circuit diagram to connect all the devices in the proper, correct order. What is the current flowing and potential difference across 10 Ω resistance?
Answer 43:
Given, Total resistance, R = R1 + R2 + R3 = 5 + 10 + 15 = 30Ω
Total potential difference, V = 30 volts
According to Ohm’s law,
V = IR ⇒ I = VR = 3030 = 1 ampere
∴ Current remains constant in this series,
∴ I1 = I2 = I3 = I;
I2 = 1amp;
R2 = 10Ω;
As V2 = I2
R2 = 1 × 10 = 10 volts
∴ The potential difference across the 10 Ω is 10 volts.
Question 44: If an electric heater rated 800 W operates 6h/day. Find the Cost of energy to operate it for 30 days at ₹3.00 per unit of consumption.
Answer 44:
Here, the Power of the heater, P = 800 W;
Time, t = 6 hour/day;
No. of days, n = 30;
Cost per unit = ₹3.00;
Thus, Consumed in 1 day = 800 × 6 = 4800 Wh
And, Energy consumed in 30 days = 4800 × 30 = 144000 Wh
1440001000 = kWh = 144 units
Now, the Cost of 1 unit = ₹3
Therefore, Cost of 144 units = 3 × 144 = ₹432
Question 45: What is the electrical resistivity of a given material? What is its unit? Discuss an experiment to study the factors on which the resistance of conducting wire depends.
Answer 45: Resistivity is an inherent property of a conductor that resists the flow of electric current. The resistivity of each material is unique.
The SI unit of resistance is Ω m.
Experiment to study the depending factors of the resistance of conducting wire.
A nichrome wire, a torch, a 10 W bulb, an ammeter (0–5 A range), a plug key, and some connecting wires are needed.
As illustrated in the figure, assemble the circuit by connecting four 1.5 V dry batteries in series with the ammeter, thereby leaving a gap XY in the circuit.
Observation:
Resistance depends on the length of the conductor, the material of the conductor, and the area of the cross-section.
Connecting the nichrome wire in the XY gap completes the circuit. Insert the key. The ammeter reading should be noted. From the plug, remove the key. [Remember: After measuring the current flowing through the circuit, always remove the key from the plug.]
Replace the nichrome wire in the circuit with the torch bulb, and then determine the current flowing through it by measuring the ammeter’s reading.
Repeat the previous process now using the 10 W bulb in the XY gap. Are there variations in the ammeter readings for the various components connected in the gap XY? What do the aforementioned observations suggest?
By leaving any material component in the gap, you are able to repeat this activity. Watch the ammeter values for each situation. Analyse the results.
Question 46: Calculate the resistance of a given metal wire of length 2m and area of cross-section 1.55 × 106 m² if the resistivity of the metal is taken to be 2.8 × 10-8 Ωm.
Answer 46: For the given metal wire,
Length, l= 2 m
Area of cross-section, A= 1.55 X 10 -6 m2
Resistivity of the metal, p = 2.8 X 10 -8 m
Since, resistance, R = lA
So, R = (2.8 X 10 -8 X 21.55 X 10 -6)
= 5.61.55 X 10 -2
= 3.6 X 10 -2
Therefore, R = 0.036
Question 47: When will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?
Answer 47:
Resistance is represented by the equation,
R = ρ l/A
where,
ρ is the resistivity of the wire material,
l is the wire
length
A is the cross-sectional area of the wire.
It is clear from the equation that the resistance is inversely proportional to the area of the wire cross-section. Therefore, the resistance increases with wire thickness and vice versa. Therefore, a thick wire conducts current more readily than a thin wire.
Question 48: What is represented by joule/coulomb?
Answer 48: The potential difference is represented by the joule/coulomb.
Question 49: A nichrome wire of resistivity 100 W m and copper wire of resistivity 1.62 ohm -m of the same length and same area of the cross-section are connected in series, and current is passed through them. Why does the nichrome wire get heated first?
Answer 49: Looking at the equation
Q = I2RT
Q= I2 (pL / A)t
Nichrome wire gets heated first because it has a higher resistance than copper wire.
Question 50: Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and thus can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.
Answer 50:
P = I2R
Where,
P = Potential difference
I = Current
R = Resistance
For resistor A,
18 = I2 x 2
I2 = 182
I2 = 9
I = 3 A
This is the maximum known current flowing through resistor A.
The maximum known current flowing through the resistors B and C, I’= 3 x 12 = 1.5 A.
Question 51: How will you infer, with the help of an experiment, that the same current flows through each and every part of the circuit containing three resistances in series connected to a battery?
Answer 51:
- You can collect three resistors, R1, R2, and R3, in series to make the circuit.
- Then use an ammeter to observe the changes in the overall current flow.
- You can remove R1 and take the readings of the potential difference of R2 and R3.
- You can remove R2 and take the reading of the potential difference between R1 and R3.
Observation:
Since the ammeter reading was the same in each case, it can be assumed that the circuit’s current is constant. One can set up an ammeter in several places and watch the current flow to double-check.
Question 52: Calculate the estimated resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance of 20 ohms.
Answer 52: 3.14 X (10 -4)2 m2
Given,
length of the wire, l = 1 m,
radius of the wire, r = 0.01 cm = 1 × 10-4 m and
given resistance, R = 20Ω
We know,
R = lA, where is the resistivity of the material of the wire.
20 = lr2 = 1 m3.14 X (10 -4)2 m2
Therefore, = 6.28 X 10 -7 m
Question 53: A charge of 2 C moves between two plates, maintained at a potential difference of IV. What is the energy acquired by the Charge?
Answer 53: The energy acquired by the Charge, W = QV
Therefore, the energy acquired is 2 J.
Question 54: Let the resistance of an electrical component remains 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?
Answer 54:
Ohm’s law is used to calculate how the current flow changes through an electrical component.
Ohm’s law states that
I = V/R, which gives the current.
The potential difference is now divided in half while maintaining the same resistance,
Let V’ = V/2 be the new voltage.
Let R’ = R be the new resistance, and the new amount of current be I’.
Ohm’s law is thus used to calculate the current change as shown below:
I’ = V’R’ = ( V2)R = 12VR = 12
As a result, the electrical component’s current is reduced by half,keeping resistance constant.
Question 55: What is the overloading of an electrical circuit? Explain two possible causes due to which overloading might occur in any household circuit. Explain one precaution, if any, that should be taken to avoid the overloading of a domestic electric circuit.
Answer 55:
Overloading: The power ratings of the appliances being utilised at a given moment determine the current flowing in household wiring. Electrical appliances with high power ratings take a tremendous amount of electricity from the circuit if too many of them are turned on at once. The overloading of the circuit takes place. . The copper wires in residential circuits get heated up due to extremely high temperatures and can immediately catch fire as a result of heavy currents running through them.
Precaution: As a result, overloading can seriously harm buildings and electrical equipment. To prevent these damages, a fuse with the appropriate rating must be used. Such a fuse wire will melt before the heated circuit wire’s temperature rises to a point where it breaks the circuit.
Question 56: What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.
Answer 56:
According to the Joule’s heating effect, the heat produced in a resistor is known to be
- (i) Directly proportional to the square of current for the given resistor.
- (ii) Directly proportional to the resistance for a given current,
- (iii) Directly proportional to the time of current flowing through the resistor.
It can be expressed as H = I2Rt
‘H’ is the heating effect, ‘I’ is the electric current, ‘R’ is resistance, and ‘t’ is time.
Experiment to demonstrate Joule’s law of heating
- Take an immersion rod for water heating and attach it to a regulator-connected socket. It’s crucial to keep in mind that a regulator regulates how much current flows through a gadget.
- Keep the regulator’s pointer at the lowest setting and time how long it takes the immersion rod to heat a specific volume of water.
- Increase the regulator’s pointer to the following level. Time the same quantity of water heating with an immersion rod.
- To measure higher amounts of the regulator, repeat the previous step.
Observation:
It has been observed that it takes less time to heat the same amount of water with an increasing electric current. This illustrates the Joule’s Law of Heating.
Application:
Electric appliances like toasters, ovens, kettles, and heaters operate using the leafing effect of current.
Question 57: Why are the coils of electric toasters and irons made of an alloy rather than any pure metal.Give reason(s).
Answer 57:
Due to its high resistivity, an alloy has a substantially higher melting point than a pure metal. Alloys are resistant to melting when temperatures are high. As a result, alloys are utilised in heating devices like electric toasters and irons.
Question 58: Name a device that helps to maintain a potential difference across a conductor in a circuit. When do we say that the potential difference across a conductor is 1 volt? Calculate the amount of work done in shifting a charge of 2 coulombs from a point A to B having potential +10V and -5V, respectively.
Answer 58: In a circuit, a battery (or cell) aids in maintaining the potential difference across a conductor.
If 1 joule of labour is expended in transporting 1 coulomb of electrical charge from one location to the other, the potential difference between the two points is said to be 1 volt.
Given, Charge, Q = 2C
Potential at A = +10 V, Potential at B = -5V
Potential difference, (V) = +10 – (-5) = 10 + 5 = 15 volts
V = wq
15 = w2
W = 15 × 2 = 30 J
Question 59: Which is a better conductor among iron and mercury?
Answer 59: Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.
Question 60: Which has more resistance, 100 W bulb or 60 W bulb?
Answer 60: As it is clearly known that R 1P, the resistance of the 60 W bulb is more.
Question 61: Find the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω, 103 Ω, and 106 Ω.
Answer 61: (a) When 1 Ω and 106 Ω when connected in parallel gives the 106equivalent resistance as follows:
1R = 11 + 1106
R= 1061+ 106 106106 = 1
Therefore, the equivalent resistance is 1 Ω. 1+ 106
(b) When 1 Ω, 103 Ω, and 106 Ω are in parallel, the equivalent resistance is given by
1R = 11 + 1103 + 1106
Solving, we get
R = 106 + 103 +1 106 = 10000001000001 = 0.999
Therefore, the equivalent resistance is 0.999 Ω.
Question 62: What are the benefits of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer 62:
There is no voltage division among the appliances when the electrical devices are connected in parallel. The supply voltage is equal to the potential difference across the devices. Devices connected in parallel lower the circuit’s effective resistance as well.
Question 63: Why does the cord of an electric heater not glow while the heating element does?
Answer 63: An electric heater’s heating element is constructed from a high-resistance alloy. The heating element glows red and gets excessively hot when the electricity passes through it. Typically, copper or aluminium, which have low resistance, is used to make the rope. Consequently, the cord doesn’t glow.
Question 64: Discuss the heat generated while transferring 96000 coulombs of Charge in one hour through a potential difference of 50 V.
Answer 64: According to Joule’s law, the heat produced can be calculated as follows:
H = VIt
where assuming,
voltage, V = 50 V
I will be current
t be the time in seconds, 1 hour = 3600 seconds
The amount of current is calculated as follows:
Amount of Current = Amount of ChargeTime flow of Charge
Substituting the value, we get
I = 960003600 = 26.66 A
Now, to find the heat generated
H = 50 X 26.66 X 3600 = 4.8 X 106 J
Therefore, the heat generated is 4.8 X 106 J
Question 65: An electric iron of resistance 20 Ω draws a current of 5 A. Calculate the heat developed in 30 s.
Answer 65: The Joule’s law of heating, which is represented by the equation, can be used to determine how the heat is produced as follows:
H = VIt
Putting the data in the above equation, we get,
H = 100 × 5 × 30 = 15,000 J
The amount of heat produced by the electric iron in 30 s is 15,000 J.
Question 66: What factors determine the rate at which energy is delivered by a current?
Answer 66: Electric power is the rate at which electric equipment uses electricity. Therefore, the power of the appliance is defined as the rate at which energy is delivered by a current.
Question 67: How is the connection of a voltmeter made in the circuit to measure the potential difference between two points?
Answer 67: The voltmeter should be connected in parallel to each of the two points in order to measure the voltage between any two points.
Question 68: Draw a circuit taking an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the resistor of 12 ohm. What would be the new reading in the ammeter and the voltmeter?
Answer 68:
Let us take the total resistance of the circuit = R
The equivalent resistance R is equal to the total resistance because all three resistors are connected in series.
R = 5 Ω + 8 Ω + 12 Ω = 25 Ω
Therefore,
V = 2V + 2V + 2V = 6V
V =IR
I = VR = 625 = 0.24 A
The reading of the voltmeter across R’ = 12 Ω is
V’ = IR’
= 0.24 X 12 = 2.88 V
Question 69: Find the following in the electric circuit given in Figure 12.9
(a) Effective resistance of the two 8 Ω resistors in the given combination
(b) Current flowing through the resistor of 4 Ω
(c) Potential difference across the resistance of 4 Ω
(d) Power dissipated in a resistor of 4 Ω (e) Difference in ammeter readings, if any
Answer 69: (i) Since two 8 resistors are in parallel, then their effective resistance Rp is given by
1Rp = 1R1+ 1R2 = 18 + 18 = 14
Rp = 4
(ii) Total resistance in the circuit
R = 4 + Rp = 4 + 4 + 8
Current, through the circuit,
I = VR = 88 = 1 A
Thus, the current through the 4 resistor is 1 A as 4 and Rp are in series and the same current flows through them.
(iii) Potential difference across 4 resistor is potential drop by the 4 resistor.
i.e. V = IR = 1 X 4 = 4 V
(iv) Power dissipated in 4 resistor
P = I2 R = 12 X 4 = 4 W
(v) There is no difference in the reading of ammeters A1 and A2 as the same current flows through all elements in a series current.
Question 70: A copper wire having a diameter of 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of the wire to make its resistance 10 Ω? How much will the resistance change if the diameter is doubled?
Answer 70: The formula provides the resistance of a copper wire with a cross-sectional area of m2 and a length in metres.
R = lA
The area of the cross-section of the wire is calculated as follows
A = (Diameter2)2
Substituting the values in the formula, we get
l = RA = 10 X 3.14 X (0.000522) 1.6 X 10 -18 = 10 X 3.14 X 254 X 1.6 = 122.72 m
The new diameter of the wire is 1mm, or 0.001m when the wire’s diameter is doubled. Therefore, the resistance can be calculated as follows:
R = lA = 1.6 X 10 -18 X 122.72 m (0.0012)2 = 250.2 X 10-2 = 2.5
The new resistance is 2.5 , and the wire’s length is 122.72 m.
Question 71: Difference features between Overloading and Short-circuiting in Domestic circuits
Answer 71:
Overloading: Overloading occurs when a circuit is used by too many electrical devices with high power ratings that are switched on simultaneously.
The copper wire used in domestic wiring heats up to an exceedingly high temperature as a result of an excessively high current running through the circuit, and a fire may subsequently ignite.
Short-circuiting: Short-circuiting is a direct result of touching bare live and neutral wires. Because the circuit’s resistance is so low in this situation, a lot of current passes through it, heating the wires to a high temperature and possibly igniting a wire. .
Question 72: When a battery of 12 V is connected across an unknown resistor, there is a current flow of 2.5 mA in the electric circuit. Find the resistance of the resistor.
Answer 72: Using Ohm’s Law, the resistor’s value can be determined as follows:
R = VI
Putting the data in the equation, we get
R = 122.5 X 10 -8 = 4.8 X 103 = 4.8
Question 73: Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance.
Answer 73:
Here, R1 = 10, R2 = 15, R3 = 5
In a parallel connection, equivalent resistance ( Req) is given by
1Req = 1R1 + 1R2 + 1R3
So, 1Req = 110 + 115 + 15
1Req = 3+ 2+630 = 1130
Therefore, Req = 3011 = 2.73
Question 74: A battery of 9 V is connected in a series system with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. What quantity of current would flow through the 12 Ω resistor?
Answer 74: There is no existing division in a series connection. An equal amount of current travels across each resistor.
We apply Ohm’s law to determine the amount of current passing through the resistors.
Let’s
first determine the equivalent resistance in the manner described below:
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
Using Ohm’s law,
I = VR = 9V13.4 = 0.671 A
The current flowing across the 12 Ω resistor is 0.671 A.
Question 75: Suppose the resistance of an electrical component remains constant, and the potential difference across the two ends of the component decreases to half of its earlier value. What type of change will occur in the current through it?
Answer 75:
Knowing that,
I = VR
If, V’ = V2
If I’ = V‘R = V2R = 12
As a result, an electrical component’s current reduces by half of what it was.
Question 76: Two same resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in both cases.
Answer 76:
Let the resistance of each resistor be R.
For series combination,
Rs = R1+ R2
So, Rs = R + R = 2R
For parallel combination,
1Rp = 1R1 + 1R2 or Rp = R1 R2R1 + R2
So, Rp = R x RR + R = R2
Required Ratio = RsRp = 2RR/2 = 4 : 1
Question 77: Explain the use of an electric fuse. What type of material is used for fuse wire and why?
Answer 77: Electric fuses guard against the very high electric current by blocking it from flowing into circuits and appliances. It is composed of a wire formed of a metal or alloy with an appropriate melting point, such as lead, copper, iron, or aluminium. The temperature of the fuse wire rises if a current more than the allowed amount runs through the circuit. The fuse wire melts, as a result, breaking the circuit.
Question 78: When an electric current flows through a conductor, it tends to become hot. Justify.List the factors on which the heat produced in a conductor depends. State Joule’s law of heating. How will the heat produced in an electric circuit be affected by this if the resistance in the circuit is doubled for the same current?
Answer 78: A conductor heats up when an electric current is carried through it. This is referred to as the current heating effect. The electrical energy is converted into heat energy to produce the heating effect of current. An electrical energy source is a cell or battery. The potential difference between the two terminals of the cell is created by the chemical reaction within, which causes the electrons to move and for current to flow through a resistor. The source must continue using up its energy. While maintaining the current, some of the source energy may be used for productive activity, while the remaining source energy may be used to generate heat.
Question 79: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer 79: Let ‘x’ be the number of resistors required.
The equivalent resistance of the resistor R in the parallel combination is given by
1R =x X 1176 = R = 176x
Now, using Ohm’s law. The number of resistors can be calculated as follows:
R = VI
Substituting the values, we get
176x = VI
x = 176 X 5220 = 4
The number of resistors required is 4.
Question 80: Two wires of the same material and same length have radii R and r. Compare their resistances.
Answer 80: Suppose R and r are resistances, then R = r as p and I are the same.
Question 81: Several electric bulbs designed and are supposed to be used on a 220 V electric supply line are rated 10 W. How many such lamps can be connected in parallel together across the two wires of a 220 V supply line if the maximum allowable current is 5 A?
Answer 81: The resistance of the bulb is calculated as follows:
P1 = V2/R1
R1 = V2/P1
Substituting the values, we get
R = (220)210 = 4840
The resistance of x number of electric bulbs is calculated as follows:
R = VI = 2205 = 44
The resistance of each electric bulb is 4840 .
The equivalent resistance of x bulbs is given by
1R = 1R1 + 1R1 + 1R1+ ………up to x times
1R= 1R1 X x
x = R1R = 484044 = 110
Hence, 110 lamps can be connected together in parallel.
Question 82: A fuse wire melts at 5 A. If it is desired that the fuse wire of the same material melt at 10 A, then in your opinion whether the new fuse wire should be of a smaller or larger radius than the earlier one? Give reasons for your answer.
Answer 82: Let R be the resistance of the wire; the heat produced in the fuse at 5A is
H = (5)2 R (H – I2 R t)
Fue melts at (5)2 R joules of heat
Let R’ be the resistance of the new wire
So, the heat produced in 1 second =(10)2 R’
To prevent it from melting
(5)2 R = (10)2R’ or R’ = R4
As R 1A
Therefore, the cross-sectional area of the new fuse wire is four times the first fuse.
Now, A = r2, so the new radius is twice as large as the old one. The new fuse wire, which is the same material and length as the old one, has a greater radius at 10 A.
Question 83: How many bulbs of 81 should be joined in parallel to draw a current of 2 A from a battery of 4V?
Answer 83:
R = V/ I
= 4/2
=2
Let n be the number of bulbs.
1/R = 1/R1 + 1/R2 +………………+ 1/Rn = n8
12 = n8
n=4
The number of bulbs is 4.
Question 84: When will current flow more easily: through a thick wire or a thin wire of the same material, when connected to the same electric source? Why?
Answer 84: A thick wire connected to the same source conducts current more easily than a short wire made of the same material. As thickness reduces, resistance increases.
Question 85: A hot plate of an electric oven connected to a 220 V supply line has two resistance coils such as A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What will be the currents in the three cases?
Answer 85:
Case (i) When coils are used separately
By using Ohm’s law, we will be able to find the current flowing through each coil as follows:
I = VR
Substituting the values, we get
I = 220 V24 = 9.166 A
When used individually, each resistor allows 9.166 A of current to pass through it.
Case (ii) When the coils are connected in series
The total resistance is 24 Ω + 24 Ω = 48 Ω in the series circuit
The current flowing through this series circuit is calculated as follows:
I = VR = 220 V48 = 4.58 A
Therefore, a current of 4.58 A will flow through the circuit in series.
Case (iii) When the coils are in parallel
connection,
the equivalent resistance is calculated as follows:
R = 24 X 2424 + 24 = 57648 = 12
By using Ohm’s law, the current flowing through the parallel circuit is given by
I = VR = 22012 = 18.33 A
The current is 18.33 A in the parallel circuit.
Question 86: What happens to the current in a circuit if its resistance is doubled?
Answer 86: As current and resistance are inversely proportional, the current is reduced to half of its previous value.
Question 87: Let the resistance of a component of electric current remain constant while the potential difference across the two ends of the component decreases to half of its earlier value. What type of change will occur in the current through it?
Answer 87: A thick wire connected to the same source conducts current more easily than a short wire made of the same material. As thickness reduces, resistance increases.
Question 88: Compare the power consumed in the 2 Ω resistor in each of the following circuit conditions: (i) a 6 V battery in series connection with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel connection with 12 Ω and 2 Ω resistors.
Answer 88: (i) Since the resistors 1 Ω and 2 Ω are connected in series, and there is a 6 V potential difference, their equivalent resistance is given by 1 Ω + 2 Ω = 3 Ω. Using Ohm’s law, the following formula is used to determine the circuit’s current:
I = VR = 63 = 2 A
2 A current will flow across all the components in the circuit because there is no division of current in a circuit of series connection.
The power in the 2 resistor is calculated as follows:
P = I2R = (2)2 X 2 = 8 W
Thus, the power consumed by the 2 Ω resistor is 8 W.
(ii) The voltage between the resistors stays constant when 12 and 2 resistors are linked in parallel. Given that a 2 Ω resistor has a 4 V voltage across it, we can use the formula below to determine how much power is used by the resistor: V2 42
P = V2R = 422 = 8 W
The power consumed by the 2 Ω resistor is 8 W.
Question 89: Two cubes, A and B, are made of the same material. The side of B is thrice that of A. Find the ratio RA/RB.
Answer 89:
The value of RA = LA and
RB = 3L9 A
RA : RB = 3: 1
Question 90: What happens to the resistance of a circuit if the current through it is doubled?
Answer 90:
Resistance is unchanged since the circuit’s resistance is independent of the current flowing through it.
Question 91: Two metallic wires, A and B, are connected. Wire A has lengths l and radius r, while B has lengths 2l and 2r. If both the wires are of the same material then find the ratio of total resistances of series combination and the resistance of wire A.?
Answer 91. Here, the Resistance of metallic wire A, R1 = lA
= lr2
Resistance of metallic wire B, R2 = 2l4r2
The total resistance in series can be expressed as R = R1 + R2
= lA + 2l4r2
= 3l2r2
The ratio of the total resistance (R) in series to the resistance of A (R1) is
RR1 = 3l2r2lr2
= 32
The ratio of the total resistance (R) in series to the resistance of wire A is 32.
Question 91: Illustrate how you would connect given three resistors, each of whose resistance is 6 Ω so that the combination has a resistance of (i) 9 Ω or (ii) 4Ω
Answer 91:
(i) When we connect R1 in series with the parallel combination of R2 and R3, as shown in Fig. (a).
The equivalent resistance is
R = R1 + R2 R3R2 + R3 = 6 + 6 X 66 + 6
= 6 + 3 = 9
(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is
R = 12 x 612 + 6 = 7218 = 4
Question 92: How does the resistance of a wire depend upon its radius?
Answer 92: As R 1A
R A
The resistance of the above-mentioned wire is directly proportional to its given radius.
Question 93: Which of the two uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes?
Answer 93:
The total energy consumed by the electrical devices is represented by the equation
H = Pt, where the power of the appliance is P and t is the time
This formula is used to determine how much energy a TV with a 250 W power rating uses:
H = 250 W × 3600 seconds
= 9 × 105 J
In the same way, the energy consumed by a toaster with a 1200 W power rating is
H = 1200 W × 600 s = 7.2 × 105 J
From the part of the above calculation, it can be said that the energy consumed by the TV is greater than the toaster.
Question 94: Two wires are of the same length and same radius, but one of them is of copper, and the other is of iron. Which will have more resistance.
Answer 94:
Since, R = 1A
But A and I have the same value. It is absolutely determined by the resistivity; hence iron has a higher resistance.
Question 95: An electric heater of 8 Ω resistance draws 15 A of current from the service mains supply for 2 hours. Calculate the rate at which heat is produced in the given heater.
Answer 95:
The rate at which the heat production takes place in the heater is thus calculated using the following formula
P = I2 R
Putting the data in the equation, we get
P = (15A) 2 × 8 Ω = 1800 watt
The electric heater produces heat at the rate of 1800 watt
Question 96: The resistance of a given wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10-8 ohm metre , find the estimated length of the wire.
Answer 96:
Here, r= 0.01 cm = 10 -4 m, p = 50 x 10 -8 m and R = 10
As, R = lA
Or, I = RA = R(r2)
So, I = 1050 X 10-8 3.14 X (10 -4)2
I = 0.628 m = 62.8 cm
Question 97: Explain the following.
- Why is tungsten used almost exclusively for the filament of electric lamps?
- Why are the conductors of electrical heating devices, like bread-toasters and electric irons, mostly made of an alloy rather than a pure metal?
- Why is the series arrangement not used in domestic circuits?
- How does the resistance of a wire vary with its area of cross-section?
- Why are copper and aluminium wires usually employed for electricity transmission?
Answer 97:
- Tungsten has a very high resistance and melting point. This characteristic prevents it from burning easily when heated. At high temperatures, electric bulbs are operated. As a result, tungsten is a popular metal choice for electric lamp filaments.
- Due to their high resistivity, alloys are used as the conductors in electric heating equipment like bread-toasters and electric irons. Because of its high resistance, it generates a lot of heat.
- Because each component in the circuit only receives a tiny voltage as a result of the voltage being divided into a series circuit, when one component fails, the circuit is broken and none of the components work. . Because of this, domestic circuits do not employ series circuits.
- The relationship between resistance and cross-sectional area is inversely proportional. This means the resistance decreases as the cross-sectional area increases and vice versa.
- Copper and aluminium are frequently used for the transmission of electricity because they are effective conductors of electricity and have low resistance.
Question 98: How will you justify that the same potential difference (voltage) will exist across three resistors connected in a parallel arrangement to a battery?
Answer 98:
You can take three resistors, R1, R2 and R3, and connect them in parallel to make a circuit, as shown in the figure.
Then use a voltmeter to take the reading of the potential difference of the three resistors in parallel combination.
Now, you can remove the resistor R1 and take the reading of the potential difference of the remaining resistors in combination.
Then, you can remove the resistor R2 and take the reading of the potential difference of the remaining resistor.
Observation:
In each case, the Voltmeter reading appears to be the same, which shows that the same potential difference tends to exist across three resistors connected in a parallel arrangement.
Benefits of Solving Important Questions Class 10 Science Chapter 12
Both teachers and students are aware of the fact that one can improve their understanding and their exam scores in Science, if they regularly engage in solving questions. For making it easy for students to find all chapter questions in one place, Extramarks team has carefully designed questions in our Important questions Class 10 Science Chapter 12 from different sources including CBSE past years’ board exam papers, CBSE mock tests, NCERT exemplar and textbooks.
Below are few benefits that a lot of students have received from solving questions from our question set of Important questions Class 10 Science Chapter 12:
- The solutions provided to each question are written in simple and are easy to understand language for the students. Also these are detailed step-by-step explanations to help students clear their doubts while solving the questions.
- Important questions Class 10 Science Chapter 12 contains a lot of important questions based on the recent syllabus of the CBSE and comes in a variety of formats that are likely to be asked in board exams. So by solving these questions students will be able to prepare for their board exams in a much better way.
- Diagrams and solutions to each of the questions from the textbook are included in the Important questions Class 10 Science Chapter 12. The understanding levels of the students are taken into consideration when responding to each and every question. To enable students to take the exams with confidence , the solutions produced fully comply with the CBSE syllabus and CBSE exam pattern. Additionally, it enhances time management abilities, which are crucial for exam preparation and getting excellent scores in the exams..
- While practising these questions students will get used to the questions with varying levels of difficulty and some of them are tweaked to test the understanding level of the students, questions are also repeated to strengthen their understanding and even provide enough practice to improve their speed along with analytical skills.
- Scoring a high percentage is not difficult. You just require the right strategies and correct understanding of the concepts to ensure 100% result in Science.
Extramarks provides students with the best learning experience and constantly strives to upgrade its products year after year to meet the changing demands of the curriculum and present its millennial generation with very simple and easy solutions for each and every student irrespective of their level.
Extramarks believes in incorporating the best learning experiences through its own repository.To enjoy the maximum benefit of these resources, students just need to register themselves at Extramarks official
website and stay ahead of the competition.Students can find various study materials which they can refer to based on their requirements.
Extramarks also provide access to additional educational resources for Class 1 to Class 12 students. Given below are few of the links for your reference:
Q.1 An object is placed in front of a convex lens at a distance of
$\text{f\xb1}\frac{\text{f}}{\text{n}},$where f is the magnitude of the focal length of the lens.
Prove that the magnification produced by the lens is n. Also find the two values of the object distances for which a convex lens of power 2.5 D will produce an image that is four times as large as the object.
Marks:5
Ans
$\begin{array}{l}\text{Given:}\\ \text{Object distance},\left|\text{u}\right|\text{}=\text{f}\frac{\text{f}}{\text{n}};\\ \text{Focal length}=\text{f;}\\ \text{Image distance, v}=\\ \text{ByLens\u2019Formula},\\ \text{}\frac{\text{1}}{\text{f}}\text{=}\frac{\text{1}}{\text{v}}\frac{\text{1}}{\text{u}}\\ \text{Object distance is always negative.}\\ \text{}\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{f}\frac{\text{f}}{\text{n}}}\\ \text{}\frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\frac{\text{1}}{\text{f}\frac{\text{f}}{\text{n}}}\\ \text{}\frac{\text{1}}{\text{v}}\text{=}\frac{\text{f}\frac{\text{f}}{\text{n}}\text{f}}{\text{f}\left(\text{f}\frac{\text{f}}{\text{n}}\right)}\\ \text{}\frac{\text{1}}{\text{v}}\text{=}\frac{\text{}\frac{\text{1}}{\text{n}}}{\left(\text{f}\frac{\text{f}}{\text{n}}\right)}\\ \text{v=}\frac{\left(\text{f}\frac{\text{f}}{\text{n}}\right)}{\text{}\frac{\text{1}}{\text{n}}}\\ \text{Now, magnification,}\left|\text{m}\right|\text{}=\left|\frac{\text{v}}{\text{u}}\right|\\ \text{|m|=n}\\ \text{Hence, proved.}\\ \text{Now, u=}\left(\text{f}\frac{\text{f}}{\text{n}}\right)\\ \text{Butf}=\frac{1}{2.5}\text{m}=0.4\text{m}\left(\u02c6\mu \text{f}=\frac{1}{\text{P}}\right)\\ \text{u}=\left(0.4\text{}\text{}\frac{0.4}{4}\right)\text{}(\u02c6\mu \text{n=4})\\ \text{u=}(0.40.1)\\ \text{u}=\text{0.5mor}0.3\text{m}\end{array}$
Q.2 An object is placed at 10cm in front of a concave mirror of focal length 15cm. Find the position, nature and size of the image.
Marks:5
Ans
$\begin{array}{l}\text{Given u}=10\text{cm};\text{f}=15\text{cm};\text{v}=;\text{m}=\\ \frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{10}}\text{=}\frac{\text{1}}{\text{15}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{15}}\text{+}\frac{\text{1}}{\text{10}}\\ \frac{\text{1}}{\text{v}}\text{=}\frac{\text{2+3}}{\text{30}}\text{=}\frac{\text{1}}{\text{30}}\\ \text{v}=30\text{cm}\\ \text{Positive sign indicates that the image is virtual and erect and form}\\ 30\text{cm behind the mirror.}\end{array}$
CBSE Class 10 Science Important Questions
FAQs (Frequently Asked Questions)
1. Why is learning about electricity important for the students in school?
In order to be thorough with the most important commodities in modern Sciences , one must understand electricity. A fundamental understanding of electricity is also necessary for many technical occupations in order to create the technologies and goods that we use in our every -day lives. For this purpose, Extramarks has curated this important questions Class 10 Science Chapter 12
2. How has electricity made our life easier in present times?
Nowadays, electricity is almost necessary for our daily work functioning. It is utilized for residential purposes such as operating fans, computer, electric stoves, air conditioning, and lighting up rooms. It is used in factories to operate heavy machines. Electricity is also used to produce various products, including food, clothing, and paper.