case study questions from electricity class 10

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Case Study Questions Class 10 Science Electricity

Case study questions class 10 science chapter 12 electricity.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

Case study: 1

Thus, V= IR

Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity.

The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

1) What is SI unit of resistivity?

2) What is variable resistance?

3) Why tungsten is used in electric bulbs?

4) 1M ohm = ?

1) The SI unit of resistivity is ohm meter.

2) The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance.

3) Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high.

4) 1M ohm = 10 6 ohm

Case study: 2

Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3

Again, IR = IR1 + IR2 + IR3

Thus, R = R1 + R2 + R3

Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit.

Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively.

Thus, total current through the circuit is the sum of current flowing through each resistor.

I = I1 + I2 + I3

Again, V/R= V/R1 + V/R2 + V/R3

Thus, 1/R = 1/R1 + 1/R2 + 1/R3

Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Questions:

1) In which case the equivalent resistance is more and why?

2) In our home, which type of combination of electric devices is preferred? Why?

3) If n resistors of resistance R are connected in parallel then what is the equivalent resistance?

1) In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel.

Since, 1/R = 1/R1 + 1/R2 + 1/R3 +….

2) At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices.

3) If n resistors of resistance R are connected in parallel then equivalent resistance is given by,

1/Re = 1/R + 1/R + 1/R +….n times 1/R

Thus, 1/Re = n/R

Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study:3

This effect was discovered by Joule, hence it is called as Joule’s law of heating.

Also, we can write, H = I 2 Rt

Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point.

In case of electric circuit, this heating effect is used to protect the electric circuit from damage.

The rate of doing work or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI

The SI unit of electric power is watt.

1) What is the SI unit of electric energy?

2) How heating effect works to protect electric circuit?

3) 1KW h = ?

4) If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb?

1) The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h.

2) In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

3) 1kW h = 3.6*10 6 joule

4) Given that,

V = 200V, I = 1A

Then, P = VI = 200*1 = 200 J/s = 200 W

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Case Study Questions Class 10 Science Chapter 12 Electricity

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Case study Questions Class 10 Science Chapter 12 are very important to solve for your exam. Class 10 Science Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Science Chapter 12 Electricity

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In CBSE Class 10 Science Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electricity Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser, etc. all are based on the heating effect of current.

(i) What are the properties of heating elements? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point.

Answer: (b) Low resistance, high melting point

(ii) What are the properties of an electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

Answer: (c) 750J

Question 2:

The relationship between potential difference and the current was first established by George Simon Ohm. This relationship is known as Ohm’s law. According to this law, the current passed through a conductor is proportional to the potential difference applied between its ends provided the temperature remains constant i.e. I ∝ V or V = IR where R is the constant for the conductor and it is known as the resistance of the conductor. Although Ohm’s law has been found valid over a large class of materials, there are some materials that do not hold Ohm’s law.

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

Case Study 3

3.1) The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one fourth

Answer(c) four times

3.2) The heat produced in a wire of resistance ‘a’ when a current ‘b’ flows through it in time ‘c’ is given by : (a) a 2 bc (b) abc 2 (c) ab 2 c (d) abc

Answer(c) ab2c

3.3) What are the properties of heating element ? (a) high resistance, high melting point (b) low resistance, high melting point (c) low resistance, high melting point (d) low resistance, low melting point

Answer (a) high resistance, high melting point

3.4) Calculate the heat produced when 96,000 coulombs of charge is transferred in one hour through a potential difference of 50 volts. (a) 4788 J (b) 4788 kJ (c) 478 kJ (d) 478 J

Answer (b) 4788 kJ

3.5) Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) low melting point (c) thick and short (d) high resistance

Answer (c) thick and short

Case Study 4

Substance through which charges cannot pass is called insulators. Glass, pure water, and all gases are insulators. Insulators are also called dielectrics. In insulators, the electrons are strongly bound to their atoms and cannot get themselves freed. Thus, free electrons are absent in insulators. Insulators can easily be charged by friction. This is due to the reason that when an electric charge is given to an insulator, it is unable to move freely and remains localized. But this does not mean that conductors cannot be charged. A metal rod can be charged by rubbing it with silk if it is held in a handle of glass or amber

4.1) Calculate the current in a wire if a 1500 C charge is passed through it in 5 minutes. (a) 2 A (b) 5 A (c) 3 A (d) 4 A

Answer (b) 5 A

4.2) Electrons and conventional current flows in : (a) The same direction (b) The opposite direction (c) Any direction (d) Can’t say

Answer (b) The opposite direction

4.3) If the current passing through a lamp is 5 A, what charge passes in 10 second ? (a) 0.5 C (b) 3 C (c) 5 C (d) 50 C

Answer (d) 50 C

4.4) One-coulomb charge is equivalent to the charge contained in : (a) 6.2 × 10 19 electrons (b) 2.6 × 10 18 electrons (c) 2.65 × 10 19 electrons (d) 6.25 × 10 18 electrons

Answer (d) 6.25 × 1018 electrons

4.5) When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is : (a) 0.5 W (b) 6 W (c) 12 W (d) 24 W

Answer (b) 6 W

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Case Study Questions for Class 10 Science Chapter 12 Electricity

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Question 1:

Read the following and answer the questions from (i) to (v) given below: The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule (as shown in figure).

case study questions from electricity class 10

Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved.

(i) The SI unit of electric energy per unit time is (a) joule (b) joule-second (c) watt (d) watt-second

(ii) Kilowatt-hour is equal to (a) 3.6 ×10 4 J (b) 3.6 ×10 6 J (c) 36 ×10 6 J (d) 36 ×10 4 J

(iii) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is (a) doubled (b) half (c) remains same (d) four times

(iv) The power of a lamp is 60 W. The energy consumed in 1 minute is (a) 360 J (b) 36 J (c) 3600 J (d) 3.6 J

(v) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 Ω for 30 minutes. (a) 40 kJ (b) 60 kJ (c) 10 kJ (d) 90 kJ

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Electricity Case Study Based Questions Class 10

Students who are studying in CBSE class 10 board, need to get the knowledge about the Electricity Case Study Based Questions. Case based questions are generally based on the seen passages from the chapter Electricity. Through solving the case based questions, students can understand each and every concept.

With the help of Electricity Case Study Based Questions, students don’t need to memorise each answer. As answers for these case studies are already available in the given passage. Questions are asked through MCQs so student’s won’t take time to mark the answers. These multiple choice questions can help students to score the weightage of Electricity.

Electricity Case Study Based Questions with Solutions

Selfstudys provides case studies for the Class 10 Science chapter Electricity with solutions. The Solutions can be helpful for students to refer to if there is a doubt in any of the case studies problems. The solutions from the Selfstudys website are easily accessible and free of cost to download. This accessibility can help students to download case studies from anywhere with the help of the Internet.

Electricity Case Study Based Questions with solutions are in the form of PDF. Portable Document Format (PDF) can be downloaded through any of the devices: smart phone, laptop. Through this accessibility, students don't need to carry those case based questions everywhere.

Features of Electricity Case Study Based Questions

Before solving questions, students should understand the basic details of Electricity. Here are the features of case based questions on Electricity are:

These case based questions start with short or long passages. In these passages some concepts included in the chapter can be explained.
After reading the passage, students need to answer the given questions. These questions are asked in the Multiple Choice Questions (MCQ).
These case based questions are a type of open book test. These case based questions can help students to score well in the particular subject.
These Electricity Case Study Based Questions can also be asked in the form of CBSE Assertion and Reason .

Benefits of Solving Electricity Case Study Based Questions

According to the CBSE board, some part of the questions are asked in the board exam question papers according to the case studies. As some benefits of solving Electricity Case Study Based Questions can be obtained by the students. Those benefits are:

Through solving case studies students will be able to understand every concept included in the chapter Electricity
Passages included in the case study are seen passages, so students don’t need to struggle for getting answers. As these questions and answers can be discussed by their concerned teacher.
Through these students can develop their observation skills. This skill can help students to study further concepts clearly.
Case studies covers all the concepts which are included in the Electricity

How to Download Electricity Case Based Questions?

Students studying in CBSE class 10 board, need to solve questions based on case study. It is necessary for students to know the basic idea of Electricity Case Study Based Questions. Students can obtain the basic idea of case based questions through Selfstudys website. Easy steps to download it are:

Open Selfstudys website.

Electricity Case Study, Electricity Case Based Questions

Bring the arrow towards CBSE which is visible in the navigation bar.

A pop-up menu will appear, Select case study from the list.

New page will appear, select 10 from the list of classes.

Select Science from the subject list.

And in the new page, you can access the Electricity Case Study Based Questions.

Tips to solve Electricity Case Study Questions-

Students should follow some basic tips to solve Electricity Case Study Based Questions. These tips can help students to score good marks in CBSE Class 10 Science.

Generally, the case based questions are in the form of Multiple Choice Questions (MCQs).
Students should start solving the case based questions through reading the given passage.
Identify the questions and give the answers according to the case given.
Read the passage again, so that you can easily answer the complex questions.
Answer according to the options given below the questions provided in the Electricity Case Study Based Questions.

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Class 10 Physics (India)

Course: class 10 physics (india) > unit 3.

Electricity class 10 numerical: CBSE board practice

Electricity class 10: CBSE previous question paper problems

Electric current and circuit.

I = 10 A ‍ ,
t = 2 mins = 120 s ‍ .
net charge, Q = 1200 C ‍ ,
charge of electron, e = 1.6 × 10 − 19 C ‍ ,
number of electrons, n = ? ‍
Understand the concept better with this video : Unit of charge (Coulombs)

Electric potential and potential difference

W = 100 J ‍ ,
Q = 20 C ‍ .
Get more practice with this exercise : Voltage and work
Understand the concept better with these videos : ( i ‍ ) Intro to potential difference (& voltage) , ( ii ‍ ) Solved example: Potential difference & work done

Circuit diagram

Solution A4. [ 1 ‍ mark]
Solution A5. ( i ‍ ) Electric cell: [ 0.5 ‍ marks] ( ii ‍ ) Open plug key: [ 0.5 ‍ marks] ( iii ‍ ) Wires crossing without connection: [ 0.5 ‍ marks] ( iv ‍ ) Variable resistor: [ 0.5 ‍ marks] ( v ‍ ) Battery: [ 0.5 ‍ marks] ( vi ‍ ) Electric bulb: [ 0.5 ‍ marks] ( vii ‍ ) Resistance: [ 0.5 ‍ marks]

Ohm's law

Ohm's law: The electric current flowing through a conducting wire, I ‍ , is directly proportional to the potential difference across its ends, V ‍ , provided its temperature stays unchanged. Mathematically, V ∝ I ‍ .
The graph obtained by plotting the potential difference against the current flowing in a conductor is a straight line .
From Ohm's law, V ∝ I ‍ , or V = I R ‍ , where R ‍ is the resistance. So, R = V I ‍ , which represents the slope of the V − I ‍ graph.
Understand the concept better with these videos : ( i ‍ ) Introduction to circuits and Ohm's law , ( ii ‍ ) Solved example: Ohms law , ( iii ‍ ) Ohm's law graph (verifying Ohm's law) , ( iv ‍ ) Solved example: (Ohm's law graph) .

Factor on which the resistance of a conductor depends

Solution A7. ( a ‍ ) The resistance of a conductor: ( i ‍ ) is directly proportional to the length of the conductor, l ‍ . R ∝ l ‍ ( ii ‍ ) is inversely proportional to the cross-sectional area of the conductor, A ‍ . R ∝ 1 A ‍ ( iii ‍ ) depends on the material of the conductor. [ 1 ‍ mark] ( b ‍ ) Metals have a much lower resistivity than glass. This makes metals better conductors of electricity. [ 1 ‍ mark] ( c ‍ ) Alloys have a high resistivity and do not oxidise or burn readily at high temperatures , making them perfect for use in heating devices. [ 1 ‍ mark]
radius of wire, r = 0.01 cm ‍ = 10 − 4 m ‍ ,
resistivity, ρ = 50 × 10 − 8 Ω m ‍ ,
resistance, R = 10 Ω ‍ .
length of wire, l = ? ‍
length of wire, l = 50 cm ‍ = 50 × 10 − 2 m ‍ ,
cross-sectional area, A = 0.01 mm 2 ‍ = 0.01 × 10 − 6 m 2 ‍ ,
resistivity, ρ = 5 × 10 − 8 Ω m ‍ .
Understand the concept better with this video : Resistivity and conductivity

Resistance of system of resistors

Resistance, R 1 = 5 Ω ‍ ,
Resistance, R 2 = 10 Ω ‍ ,
e.m.f., V = 6 V ‍ .
R 1 ‍ = R 2 ‍ = R 3 ‍ = 2 Ω ‍ ,
potential difference, V = 5 V ‍ ,
current, I = ? ‍
Solution A12. Derivation of equivalent parallel resistance: The following diagram shows a parallel combination of the three resistors R 1 ‍ , R 2 ‍ , and R 3 ‍ . Currents I 1 ‍ , I 2 ‍ , and I 3 ‍ flow through the resistors R 1 ‍ , R 2 ‍ , and R 3 ‍ respectively. The total current, I ‍ , is the sum of the currents through the three branches: I = I 1 + I 2 + I 3 ‍ If the equivalent resistance is R P ‍ , we have using Ohm's law, I = V R P ‍ The potential difference across each individual resistor is also V ‍ . Applying Ohm's law for each resistor, we find, I 1 = V R 1 ‍ , I 2 = V R 2 ‍ , and I 3 = V R 3 ‍ [ 1 ‍ mark] Tying it all up, I = I 1 + I 2 + I 3 V R P = V R 1 + V R 2 + V R 3 V × ( 1 R P ) = V × ( 1 R 1 + 1 R 2 + 1 R 3 ) 1 R P = 1 R 1 + 1 R 2 + 1 R 3 ‍ ∴ ‍ The equivalent resistance, R P ‍ , is related to the individual resistors as: 1 R P = 1 R 1 + 1 R 2 + 1 R 3 ‍ [ 1 ‍ mark]
Get more practice with these exercises : ( i ‍ ) Finding equivalent resistance , ( ii ‍ ) Identifying types of resistor combinations , ( iii ‍ ) Simplifying resistor networks , ( iv ‍ ) Finding currents and voltages (pure circuits) .
Understand the concept better with these videos : ( i ‍ ) Series resistors , ( ii ‍ ) Parallel resistors (part 1) , ( iii ‍ ) Parallel resistors (part 2) , ( iv ‍ ) Parallel resistors (part 3) , ( v ‍ ) Example: Analyzing a more complex resistor circuit , ( vi ‍ ) Solved example: Finding current & voltage in a circuit .

Heating effect of electric current

Solution 1 (in words) A13 (in words). The heat produced every second in a fuse wire of resistance, R ‍ , and with a current, I ‍ , in it, is given as, H = I 2 R ‍ . [ 1 ‍ mark] The heat produced for I = 5 A ‍ melts the wire. It follows from the relation that for the same heat to be produced for I = 10 A ‍ , the resistance must be smaller . [ 1 ‍ mark] Since, R ∝ 1 cross-sectional area , ‍ a smaller resistance has a larger cross-section . A larger cross-section implies a larger radius . So, the new fuse wire has a larger radius. [ 1 ‍ mark]
Solution 2 (calculation) A13 (calculation). For current I 1 = 5 A ‍ passing through the older fuse of resistance R 1 ‍ , the heat produced every second is, H = I 1 2 R 1 ‍ The same heat is desired for I 2 = 10 A ‍ . If the new fuse wire has a resistance, R 2 ‍ , we have, H = I 1 2 R 1 = I 2 2 R 2 ⇒ R 2 = I 1 2 I 2 2 R 1 = 25 1 A 100 4 A R 1 = 1 4 R 1 ‍ [ 1 ‍ mark] ∵ R ∝ 1 A ⇒ 1 A 2 = 1 4 × 1 A 1 ⇒ A 2 = 4 A 1 ‍ where A ‍ is the cross-sectional area. [ 1 ‍ mark] Since, A = π r 2 ‍ , we get from the result above, π r 2 2 = 4 π r 1 2 ⇒ r 2 = 2 r 1 ‍ The radius of the new fuse wire is twice that of the old one. [ 1 ‍ mark]
Solution A14. Derivation: For work done, W ‍ , on moving a net charge, Q ‍ , the potential difference is defined as, V = W Q ⇒ W = V Q ‍ Let t ‍ be the time it takes to move the net charge Q ‍ . Multiplying and dividing the R.H.S. by t ‍ , we get, W = V × Q t × t = V I t ‍ where Q t ‍ is the current, I ‍ , by definition. [ 1 ‍ mark] Since this work is converted into heat energy, we can write, W = H = V I t = ( I R ) × I t H = I 2 R t ‍ where R ‍ is the resistance in the circuit and Ohm's law ( V = I R ‍ ) is applied in the second step. [ 1 ‍ mark] Second part: We obtained, W = H = V I t ‍ . Writing I = V R ‍ from Ohm's law, we get, H = V × ( V R ) × t = V 2 R t H ∝ 1 R ‍ [ 1 ‍ mark] If two equal resistances, R ‍ each, are connected in series , the equivalent resistance, R S = R + R = 2 R ‍ . When connected in parallel , the equivalent resistance, R P ‍ , is found as, 1 R P = 1 R + 1 R R P = R 2 ‍ [ 1 ‍ mark] Let the heat produced in the series combination be H S ‍ , and the parallel combination be H P ‍ . We have, H S H P = 1 / R S 1 / R P = R P R S = R 2 × 1 2 R H S = 1 4 H P H P = 4 H S ‍ Therefore, the heat produced in the parallel combination is four times that of the series combination. [ 1 ‍ mark]
Get more practice with this exercise : Calculating heat dissipated in circuits .
Understand the concept better with these videos : ( i ‍ ) Heating effect of current , ( ii ‍ ) Solved example - Calculating power & heat dissipated .

Electric power

Solution A15. Since the lamps are connected in parallel to the mains supply, the voltage across each lamp is 220 V ‍ . The current drawn by a lamp = power rating voltage applied ‍ . Current drawn by the 100 W ‍ lamp, I 1 = 100 W 220 V = 100 220 A ‍ Current drawn by the 60 W ‍ lamp, I 2 = 60 220 A ‍ [ 1 ‍ mark] Net current drawn from the mains supply, I ‍ , is, I = I 1 + I 2 = 100 220 A + 60 220 A = 160 220 A ≈ 0.73 A ‍ [ 1 ‍ mark]
Solution A16. ( a ‍ ) Power, P = V I ‍ , where V ‍ is the voltage and I ‍ is the current. Power consumed is minimum when the current passing is minimum. So the resistors should be connected in series . [ 1 ‍ mark] ( b ‍ ) Bulbs: Power of each bulb = 100 W ‍ . Total power of three bulbs, P B = 300 W ‍ . Energy consumed by three bulbs in a day, E B = 300 W × 5 h = 1500 Wh = 1.5 kWh ‍ [ 1 ‍ mark] Fans: Power of each fan = 50 W ‍ . Total power of two bulbs, P F = 100 W ‍ . Energy consumed by two fans in a day, E F = 100 W × 10 h = 1000 Wh = 1 kWh ‍ [ 1 ‍ mark] Heater: Energy consumed by heater in a day, E H = 1 kW × 1 2 h = 0.5 kWh ‍ [ 1 ‍ mark] Total energy consumed, E = E B + E F + E H = ( 1.5 + 1 + 0.5 ) kWh = 3 kWh ‍ [ 0.5 ‍ marks] Energy consumed in a month of 31 ‍ days = E × 31 = 93 kWh ‍ . Therefore, the cost of energy consumed ₹ ₹ = ₹ 3.60 / kWh × 93 kWh = ₹ 334.80 ‍ [ 0.5 ‍ marks]
Get more practice with this exercise : Bulbs connected in series or parallel .
Understand the concept better with these videos : ( i ‍ ) Electric power & energy , ( ii ‍ ) Solved example: Power dissipated in bulbs , ( iii ‍ ) Solved example - Cost of operation of electrical device .

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Class 10 Science Chapter 11 Case Based Questions - Electricity

Case study - 1.

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt This effect was discovered by Joule, hence it is called as Joule’s law of heating. Also, we can write, H = I 2 Rt Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point. In case of electric circuit, this heating effect is used to protect the electric circuit from damage. The rate of doing work or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI The SI unit of electric power is watt.

Q1: What is the SI unit of electric energy? Ans: The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h. Q2: How heating effect works to protect electric circuit? Ans: In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

Q3: 1KW h = ? Ans: 1kW h = 3.6*10 6 joule Q4: If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb? Ans: Given that, V = 200V, I = 1A Then, P = VI = 200*1 = 200 J/s = 200 W

Case Study - 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively. Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3 Again, IR = IR1 + IR2 + IR3 Thus, R = R1 + R2 + R3 Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit. Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively. Thus, total current through the circuit is the sum of current flowing through each resistor. I = I1 + I2 + I3 Again, V/R= V/R1 + V/R2 + V/R3 Thus, 1/R = 1/R1 + 1/R2 + 1/R3 Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Q1: In which case the equivalent resistance is more and why? Ans: In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel. Since, 1/R = 1/R1 + 1/R2 + 1/R3 +…. Q2: In our home, which type of combination of electric devices is preferred? Why? Ans: At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices. Q3: If n resistors of resistance R are connected in parallel then what is the equivalent resistance? Ans: If n resistors of resistance R are connected in parallel then equivalent resistance is given by, 1/Re = 1/R + 1/R + 1/R +….n times 1/R Thus, 1/Re = n/R Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study - 3

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature. Thus, V= IR Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity. The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

Q1: What is SI unit of resistivity? Ans: The SI unit of resistivity is ohm meter. Q2: What is variable resistance? Ans: The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance. Q3: Why tungsten is used in electric bulbs? Ans: Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high. Q4: 1M ohm = ? Ans: 1M ohm = 10 6 ohm

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Class 10 Science: Case Study Chapter 12 Electricity PDF Download

Here we are providing you with Class 10 Science Chapter 12 Electricity Case Study Questions, by practicing these Case Study and Passage Based Questions will help you in your Class 10th Board Exam.

Case Study Chapter 12 Electricity

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

Answer: (b) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

Answer: (c) 750J

Question 2:

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

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Important Questions for CBSE Class 10 Science Chapter 12 - Electricity 2024-25

CBSE Class 10 Science Chapter-12 Electricity Important Questions with Answers - Free PDF Download

Class 10 is an essential stage in every student's career. Their performance in Class 10 acts as a base for their future studies. So it's important to score well in Class 10. The most important and challenging subject in Class 10 is Science and many students seem to get confused here and lose marks. If students want a successful career in future, then they can't afford to lose marks in this stage of their life. It's important to understand every chapter in science thoroughly to score good marks. Chapter 12 of Class 10 Science which is about electricity is one of the difficult chapters. A student who is incapable of understanding this chapter must practice Important Questions for Class 10 Science Chapter 12 . These important questions of Electricity Class 10 can make the students through on the concepts of this chapter. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions, they can download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Download CBSE Class 10 Science Important Questions 2024-25 PDF

Also, check CBSE Class 10 Science Important Questions for other chapters:

Related Chapters

Study Important Questions for Class 10 Science Chapter 12 – Electricity

Very Short Answer Questions (1 Mark)

1. Which two circuit components are connected in parallel in the following circuit diagram?

(a) \[{{\text{R}}_{\text{1}}}\] and ${{\text{R}}_{\text{2}}}$ only

(b) ${{\text{R}}_{\text{1}}}{\text{,}}{{\text{R}}_{\text{2}}}$ only

(d) ${{\text{R}}_{\text{1}}}$ and ${\text{V}}$only

Ans: (a) The two circuit components that are connected parallel in the circuit diagram is and ${R_2}$ only.

2. A metallic conductor has loosely bound electrons called free electrons. The metallic conductor is

(a) negatively charged

(b) positively charged

(d) Either positively charged or negatively charged

Ans: (c) The metallic conductor is neutral.

3. Which of the following expressions does not represent the electric power in the circuit?

(a) ${\text{VI}}$.

(b) ${{\text{I}}^{\text{2}}}{\text{/R}}$

(d) ${{\text{I}}^{\text{2}}}{\text{R}}$

Ans: (b) The expression which does not represent the electric power in the circuit is ${I^2}/R$ .

4. Resistivity of a metallic wise depends on

(a) its length

(b) its shape

(d) nature of material

Ans: (d) Resistivity of a metallic wire depends on the nature of the material.

5. If the current I through a resistor is increased by \[{\mathbf{100}}\% \] the increased in power dissipation will be (assume temperature remain unchanged)

(a) \[{\mathbf{100}}\% \]

(b) \[{\mathbf{200}}\% \]

(d) \[{\mathbf{400}}\% \]

Ans: (c) The increase in power dissipation will be \[300\% \] .

6. For the circuit arrangement shown below, a student would observe.

(a) Some reading in both ammeter and voltmeter.

(b) No reading in either the ammeter or the voltmeter.

(d) Some reading in the voltmeter but no reading in the ammeter.

Ans: (c) A student will observe some reading in the ammeter but no reading in the voltmeter.

7. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected in parallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is

(a) $\dfrac{{\text{1}}}{{\text{5}}}$

(b) ${\text{5}}$

(d) ${\text{25}}$

Ans: (d)The ratio $\dfrac{R}{{R'}}$ is $25$ .

8. The resistance of the conductor is $R$ . If its length is doubled, then its new resistance will be

Ans: (c) The new resistance is $4R$ .

9. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances ${R_1},{R_2},{R_3}$ respectively as shown in the figure. Which of the following is live?

(a) ${R_3} > {R_2} > {R_1}$

(b) ${R_2} > {R_3} > {R_1}$

(d) ${R_1} = {R_2} = {R_3}$

Ans: (a)According to the graph , ${R_3} > {R_2} > {R_1}$

10. The nature of the graph between potential difference and the electric current flowing through a conductor is

(a)parabolic

(b) circle

(d) hyperbolic

Ans: (c)The nature of the graph between potential difference and the electric current flowing through a conductor is a straight line.

11. An electric heater is salted at $1500$ w. How much heat is produced per hour?

(i) $5400$ J

(ii) $54000$ J

(iii) $5.4 \times {10^5}$

(iv) $5.4 \times {10^6}$.

Ans: (iv) The electric heater produces $5.4 \times {10^6}$ J per hour.

12. A student says that the resistance of two wires of the same length and same area of cross section is the same. This statement is correct if

(a) Both wires are of different materials

(b) Both wires are made of the same material and are at different temperatures.

(d) Both wires are made of different materials and are at the same temperature.

Ans: This statement is correct if (c)The resistance of two wires of the same length and same area of cross section is the same if both wires are made of the same material and are at the same temperature.

13. In an experiment ohm, s law a student obtained a graph as shown in the diagram. The value of resistance of the resistor is

(a) $0.1\Omega $

(b) $1.0\Omega $

(d) $100\Omega $

Ans: (d) The value of resistance of the resistor is $100\Omega $ .

14. Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is

(a) $1$ Joule

(b) $10$ Joule

(d) $100$ Joule

Ans: (c) Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is zero.

15. Three resistors are shown in the figure. The resistance of the combination is

(a)$3\Omega $

(b) $6\Omega $

(d) $7\Omega $

Ans: (c) The resistance of the combination is $9\Omega $ .

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

Ans: A device that helps to maintain a potential difference between conductors is the battery.

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

Ans: The rate at which energy is delivered by a current is determined by electric power.

18. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is

(a) $\dfrac{1}{5}$

Ans: (d)A wire of resistance ${\text{R}}$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are ${\text{R'}}$. In this cases, the ratio$\dfrac{R}{{R'}}$ is $25$ .

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

(a) ${I^2}R$

(b) $I{R^2}$

(d) ${V^2}/R$

Ans: (b) The term that does not represent electrical power in a circuit is $I{R^2}$ .

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

(a) $220$ W

(b) $75$ W

(d) $25$ W

Ans: (d)The power consumed will be $25$ W.

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

Ans: (c) The ratio of heat produced in series and parallel combination will be $1:4$ .

22. A wire of resistance $R$ is bent in the form of a closed circle, what is the resistance across a diameter of the circle?

Ans: The resistance across a diameter of the circle

23. A charge of $6$ C is moved between two points $P$ and $Q$ having, potential $10$ V and $5$ V respectively. Find the amount of work done.

Ans: The amount of work done, $W = q({V_2} - {V_1})\;$

$\; = {\text{ }}6\left( {10 - 5} \right)$

\[ = {\text{ }}30\] joule

24. Name the physical quantity whose SI unit is JC

Ans: The physical quantity whose SI unit is JC is Potential.

25. Why are copper wires used as connecting wires?

Ans: Copper wires are used as connecting wires because in case of copper the electrical resistivity for it is low.

26. A wire of resistivity $p$ is stretched to double its length. What is its new resistivity?

Ans: When a wire of resistivity ${\text{p}}$ is stretched to double its length, then the new resistivity remains the same because resistivity depends on the nature of material.

27. What is the resistance of the connecting wire?

Ans: The resistance of a connecting wire made of a good conductor is extremely low.

28. What is the resistance of an ammeter?

Ans: An ammeter's resistance is very minimal, and in an ideal ammeter, it is zero.

29. What is the resistance of a Voltmeter?

Ans: An ideal voltmeter's internal resistance is infinite.

Short Answer Questions (2 Marks)

1. How does use of fuse wire protect electrical appliances?

Ans: When a large quantity of current passes through the circuit, the temperature of the wire rises and the fuse wire melts. This prevents current from flowing into the house's other circuits, saving electrical appliances.

2. Calculate the resistance of an electric bulb which allows a $10$A current when connected to a $220$ V power source?

Ans: It is given from the question that an electric bulb allows a $10$A current when connected to a $220$ V power source.

\[I = 10\] A ,

\[V = 220\]V

\[R = \dfrac{V}{I}\]

\[ = \dfrac{{220}}{{10}}\]

\[ = 22\] ohm

3. (i) Identify the $V - I$ graphs for ohmic and non-ohmic materials.

Ans: The $V - I$ graphs for ohmic and non-ohmic materials respectively can be represented as shown below:

(ii) Give one example of each.

Ans: (ii) Some examples of ohmic material are Copper, Nichrome and some examples of Non-ohmic material are Diode, Transistor.

4. What do the following symbols represent in a circuit? Write the name and one function of each?

Ans: (i) It symbolises a battery that maintains a potential difference across the circuit element to allow current to flow.

Ans: It's an ammeter that measures how much current is flowing across a circuit.

5. Define the term “volt”?

Ans: If $1$joule of energy is transferred between two points A and $B$, the potential difference between them is one volt. In an electric circuit, work is done to move one coulomb of charge from one point to another field.

6. Why does the connecting rod of an electric heater not glow while the heating element does?

Ans: As its resistance is lower than that of the heating element, the connecting cord of an electric heater does not glow. As a result, the heating element produces more heat than the connecting cord, and it glows

7. A number of \[n\] resistors each of resistance \[R\] are first connected in series and then in parallel. What is the ratio of the total effective resistance of the circuit is series combination and parallel combination?

Ans: Total effective resistance of the circuit when in series combination \[{R_s} = nR\]

And for parallel combination is \[{R_p} = \dfrac{R}{n}\] and

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

The ratio will be ${n^2}$ .

8. Draw a schematic diagram of a circuit consisting of \[3\] V battery, \[5\] ohm, \[3\Omega \] and \[1\Omega \] resistor, an ammeter and a plug key, all connected in series.

Ans: The circuit diagram of a circuit consisting of \[3\] V battery, \[5\] ohm, \[3\Omega \] and \[1\Omega \] resistor, an ammeter and a plug key, all connected in series can be represented as show below,

9. A copper wire has diameter \[0.5\] mm and Resistivity of \[1.6 \times {10^{ - 8}}\Omega m\] What is the length of this wire to make its resistance ? How much does the resistance change if diameter is doubled?

Ans: Diameter of the copper wire, $D = 0.5 \times {10^{ - 3}}$ m

$P = 1.6 \times {10^{ - 8}}$

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

$ = \dfrac{{\rho l}}{{\pi {r^2}}}$

$ = \dfrac{{4\rho l}}{{\pi {D^2}}}$

$ \Rightarrow l = \dfrac{{\pi R{D^2}}}{{4\rho }}$

$l = \dfrac{{3.14 \times 10 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{4 \times 1.62 \times {{10}^{ - 8}}}}$

$ = 121.14$

Length of the wire, $l = 121.14$ m

New $R' = \dfrac{{4\rho l}}{{\pi {{(D')}^2}}}$

$ = \dfrac{1}{4}\dfrac{{4\rho l}}{{\pi {D^2}}}$

$ = \dfrac{1}{4}R$

Length of the wire to make its resistance $10\Omega $ is $121.14$m and when the diameter is doubled the new resistance will be one fourth that of the old one.

10. Alloys are used in electrical heating devices rather than pure metals. Give a reason.

Ans: Alloys are utilised in electricity heating devices rather than pure metals because alloys have a higher resistivity and hence produce more heat. Furthermore, alloy is non-combustible (or oxidises easily at higher temperature).

11. On what factor does the resistance of a conductor depend?

Ans: The factors on which Resistance depends are:-

(a) Length of the conductor

(b) Area of cross - section

(d) Nature of material

12. Calculate the number of electrons consisting of one coulomb of charge?

Ans: Let $x = $ no. of electrons

Charge on $1$ electron $ = 1.6 \times {10^{ - 19}}$ C , that is

$x = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$x = 6.25 \times {10^{18}}$

The number of electrons consisting of one coulomb of charge is $6.25 \times {10^{18}}$ .

13. What does an electric circuit mean?

Ans: An electric circuit is a current route that is both continuous and closed. Current can flow through an electric circuit if it is complete.

14. Define the unit of current.

Ans: The ampere is the SI unit for electric current. If $1$ coulomb charge flows per second across a conductor cross-section, the current is said to be $1$ ampere.

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

Ans: The charge on one electron $ = 1.6 \times {10^{ - 19}}$ coulomb.

Number of electrons in one coulomb of charge$ = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$ = 6.25 \times {10^{18}}$

16. What is meant by saying that the potential difference between two points is \[1\] v?

Ans: If $1$ joule of labour is required to move a charge of $1$ coulomb from one location to another, the potential difference between the two points is said to be $1$ volt.

17. Ammeter burns out when connected in parallel. Give reasons.

Ans: When a low-resistance wire is connected in series, a huge quantity of current travels through it, causing it to be burned, or short-circuited.

18. Judge the equivalent resistance when the following are connected in parallel:

(a) Equivalent resistance of \[1\Omega \] and \[{10^6}\Omega \]

Ans: When the resistances are connected in a parallel arrangement, the resultant resistance is given by:

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \cdots $

$\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^6}}}$

$ = 1 + {10^{ - 6}}$

$R = 1\Omega $

The equivalent resistance is $1\Omega $ .

(b) Equivalent resistance of \[1\Omega \] , \[{10^3}\Omega \] and \[{10^6}\Omega \]

Ans: $\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^6}}}$

$ = 1 + {10^{ - 3}} + {10^{ - 6}}$

19. An electric iron of resistance \[20\] takes a current of \[5\] A. Calculate the heat developed in \[30\] s.

Ans: Resistance of electric iron, $R = 20\Omega $ , current, $I = 5$ A and time $ = 30$ s.

Heat generated $H = {I^2}Rt$

$ = {5^2} \times 20 \times 30$

$ = 15000$ j

The heat developed in $30$ second is $15000$ j .

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

Ans: The given information is as shown below,

Charge transferred, $Q = 96000$

Potential Difference, $V = 50$ V.

Heat generated, $H = VQ$

$ = 50 \times 96000$

$ = 4800000$ j

$ = 4.8 \times {10^6}$ j

The heat generated while transferring $96000$ coulomb of charge in one hour

through a potential difference of $50$ V is $4.8 \times {10^6}$ j

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

Ans: Given that current drawn by electric motor $I = 5$ A.

The line voltage $V = 220$ V

Time, $t = 2$ h

Power of motor , $P = VI$

$ = 220 \times 5$

$ = 1100$ W and

the energy consumed $E = Pt$

$ = 2 \times 1100$

$ = 2.2$ KWh

The power of the motor and energy consumed in $2$ h are $1100$ W and $2.2$ kWh respectively.

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

Ans: A voltmeter is connected in parallel to the resistance across the place where the potential difference is to be determined.

23. When a \[12\] v battery is connected across an unknown resistor, there is a current of \[2.5\] mA in the circuit. Find the value of the resistance of the resistor.

Ans: Given that Voltage of battery, $V = 12$ V,

Current, $I = 2.5$ mA

$ = 2.5 \times {10^{ - 3}}$ A

Resistance, $R = V/I$

$ = \dfrac{{12}}{{2.5 \times {{10}^{ - 3}}}}$

$ = 4800\Omega $

The value of the resistance of the resistor is $4800\Omega $ .

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

Ans: The given information is as shown below

Each bulb is rated as $10$ W, $220$ V,

It draws current, $I = P/V$

$ = \dfrac{{10}}{{220}}$ V

$ = 1/22$ A.

The maximum allowable current is $5$ A and all lamps are connected in parallel. Therefore the maximum number of bulbs joined in parallel with each other $ = 5 \times 22$whichis$110$.

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

Ans: Current drawn by ${1^{st}}$lamp rated $100$ W at $220$ , $V = P/V$

\[ = 100/220\]

\[ = 5/11\] A.

Current drawn by ${2^{nd}}$lamp rated $60$ W at \[220\] , \[V = 60/220\]\[ = 3/11\;\]A.

In parallel arrangement the total current \[ = {\text{ }}3/11 + {\text{ }}5/11\]

\[ = {\text{ }}8/11\]

\[ = {\text{ }}0.73\] A.

Current drawn from the line if the supply voltage is $220$ V is $0.73$ A .

26. Which uses more energy, a \[250\] W TV set in \[1\] hour, or a \[1200\] W toaster in \[10\] minutes?

Ans: Energy used by a TV set of power $250$ W in $1$ hour $ = Pt$

Energy used by toaster of power \[1200\] W in \[10\] minute \[\left( {10/60h} \right) = {\text{ }}200\] Wh.

A $250$ W TV set in $1$ hour uses more energy.

27. An electric heater of resistance \[8\] draws \[15\] A from the service mains for \[2\] hours. Calculate the rate at which heat is developed in the heater.

Ans: Resistance of electric heater, \[R{\text{ }} = {\text{ }}8\Omega \],

current, \[I{\text{ }} = {\text{ }}15\] A.

Rate at which heat developed in the heater $ = \dfrac{{{I^2}Rt}}{t}$

$ = 1800$ W.

The rate at which heat is developed in the heater is $1800$ W.

28. In the given figure what is the ratio of current in \[A\]

Ans: Observe that it I clearly known to us that $V = IR$

$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{R}{{2R}}$

$ = \dfrac{1}{2}$

The ratio of current is$1/2$.

29. Two wires of equal cross sectional area, one of copper and other of managing have the same resistance. Which one will be longer?

Ans: Using the equation $\rho = \dfrac{{RA}}{l}$ , where $\rho $ is the resistivity, $R$ is the resistance and $A$ the area

We can see that copper wire has a lower resistance than manganin, hence copper will last longer.

30. A Rectangular block of iron has dimensions $L \times L \times b$ . What is the resistance of the block measured between the two square ends? Given $p$ resistivity.

Ans: $R = \dfrac{{pb}}{{{L^2}}}$is the resistance of the block measured between the two square ends

31. Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances?

Ans: ${R_{series}} = 3R$

${R_{parallel}} = R/3$

The ratio of Resistances is $9$ .

32. Justify for any pair of resistance the equivalent resistance equivalent resistance in parallel.

Ans: As $R = V/I$

From the graph for any pair of resistance the equivalent resistance in series is greater than equivalent resistance in parallel.

$A = $ Series, $B = $ Parallel

33. How many bulbs of \[81\] should be joined in parallel to draw a current of \[2\] A from a battery of \[4\] V?

Ans: $R = V/I$

$ = 2\Omega $

let $n$ be the number of bulbs.

$1/R = 1/{R_1} + 1/{R_2} + \cdots + 1/{R_n} = \dfrac{n}{8}$

$ \Rightarrow \dfrac{1}{2} = \dfrac{n}{8}$

$ \Rightarrow n = 4$

Number of bulbs are $4$ .

34. Two cubes \[A\] and \[B\] are of the same material. The side of \[B\] is thrice as that of \[A\] . Find the ratio \[{R_A}/{R_B}\] .

Ans: The value of ${R_A} = \dfrac{{pL}}{A}$ and

${R_B} = \dfrac{{p3L}}{{9A}}$

${R_A}:{R_B} = 3:1$

35. If there are $3 \times {10^{11}}$ electrons are flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron $1.6 \times {10^{19}}$

Ans: Observe as shwn below,

Using the equation

$ = 3 \times {10^{11}} \times 1.6 \times {10^{19}}$C

$ = 4.8 \times {10^8}$C

$ = \dfrac{{4.8 \times {{10}^8}}}{{2 \times 60}}$

$ = 4 \times {10^7}$ A

The current flowing through the circuit is $4 \times {10^7}$ A.

36. A nichrome wire of resistivity $100$ W m and copper wire of resistivity $1.62$ M ohm-m of the same length and same area of cross section are connected in series , current is passed through them, why does the nichrome wire get heated first?

Ans: Looking at the equation

$Q = {I^2}RT$

$Q = {I^2}(pL/A)t$

Henceforth, because nichrome wire has a higher resistance than copper wire, it must be heated first.

37. What is represented by joule/coulomb?

Ans: The potential difference is represented by the joule/coulomb.

38. A charge of $2$ C moves between two plates, maintained at a p.d of $IV$ . What is the energy acquired by the charge?

Ans: The energy acquired by the charge, $W = QV$

The energy acquired is $2$ J

39. Which has more resistance: $100$ W bulb or $60$ W bulb?

Ans: As, it is clearly known that$R \propto \dfrac{1}{P}$, thus the resistance of $60$ W bulb is more.

40. What happens to the current in a circuit if its resistance is doubled?

Ans: As current and resistance are inversely proportional, the current is reduced to half of its previous value.

41. What happens to the resistance of a circuit if the current through it is doubled?

Ans: Resistance is unaffected since the circuit's resistance is independent of the current flowing through it.

42. How does the resistance of a wire depend upon its radius?

Ans: As $R \propto \dfrac{1}{A}$

$ \Rightarrow R \propto V$

Resistance of a wire is directly proportional to its radius.

43. Two wires are of the same length, same radius, but one of them is of copper and the other is of iron. Which will have more resistance.

Ans: Since \[R{\text{ }} = {\text{ }}p1/A\] ,

but A and I are the same. It is solely determined by resistivity, hence iron has a higher resistance.

44. Two wires of same material and same length have radii $R$ and $r$ Compare their resistances.

Ans: Suppose $R$ and $r$ are resistances, then $R = r$ as $p$ and $I$ are the same.

Short Answer Questions (3 Marks)

1. Two metallic wires A and B are connected, wire A has length I and radius $r$ , while wire B has length $2l$ and radius $2r$ . Find the ratio of total resistance of series combination and the resistance of wire $A$ , if both the wires are of the same material?

Ans: Observe as shown below,

Resistance of metallic wire $A$ , ${R_1} = \dfrac{{\rho l}}{A}$

Resistance of metallic wire $B$ , ${R_2} = \dfrac{{\rho 2l}}{{4\pi {r^2}}}$

Total resistance in series is $R = {R_1} + {R_2}$

$ = \dfrac{{\rho l}}{{\pi {r^2}}} + \dfrac{{2\rho l}}{{4\pi {r^2}}}$

$ = \dfrac{{3\rho l}}{{2\pi {r^2}}}$

The ratio of the total resistance is series to the resistance of $A$ is

$\dfrac{R}{{{R_1}}} = \dfrac{{\dfrac{{\rho l}}{{\pi {r^2}}}}}{{\dfrac{{3\rho l}}{{2\pi {r^2}}}}}$

The ratio of the total resistance is series to the resistance of $A$ is $2/3$ .

2. Should the heating element of an electric iron be made of iron, silver or nichrome wire? Justify giving three reasons?

Ans: The following reasons can be found, why the heating element of an electric iron is composed of nichrome wire.

(1) Due to the high resistance, the passage of current generates additional heat.

(2) High melting point.

(3) At high temperatures, it does not easily oxidised (or burn).

3. (a) Define electric resistance of a conductor?

Ans: A conductor's electric resistance is defined as the resistance it provides to the flow of current.

That is $R = V/I$ and its S.I. unit is ohm , $\Omega $ .

(b) A wire of length $L$ and resistance $R$ is stretched so that its length is doubled and the area of the cross section is halved. How will its

(i) resistance change

Ans: It is clearly known that resistance, $R = \dfrac{{\rho l}}{A}$

New length $L' = 2L$ and

$A' = \dfrac{A}{2}$

Therefore ${R^1} = \dfrac{{\rho L'}}{{A'}}$

Therefore, the resistance of a wire becomes 4 times its original resistance.

(ii) resistivity change?

Ans: The size of a wire has no bearing on its resistance. As a result, resistance does not vary.

4. Two resistors of resistance $R$ and $2R$ are connected in parallel in an electric circuit. Calculate the ratio of the electric power consumed by $R$ and $2R$ ?

Ans: Power consumed by $R$ , ${\rho _1} = \dfrac{{{V^2}}}{R}$

Power consumed by$2R$ , ${\rho _2} = \dfrac{{{V^2}}}{{2R}}$

Ratio $\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{{\dfrac{{{V^2}}}{R}}}{{\dfrac{{{V^2}}}{{2R}}}}$

The ratio of the electric power consumed by $R$ and $2R$ is $2:1$ .

5. The length of different metallic wires but of same area of cross section and made of the same material are given below

(i) Out of these two wires which wire has higher resistance.

Ans: As $R \propto l$ (length of the conductor) and since length of wire $C$ is more than $A$ and $B$ , wire $C$ has higher resistance.

(ii) Which wire has higher electrical resistance? Justify your answer.

Ans: The electrical resistivity of a wire is determined by the nature of the material, not by its dimensions. As a result, the resistivity of all wires is the same as the substance of all wires.

6. Two resistors of resistances $R$ and $2R$ are connected in series with an electrical circuit? Calculate the ratio of the electric power consumed by $R$ and $2R$ ?

Ans: It is clearly known thatElectric power consumed by $R$ , ${P_1} = {I^2}R$

Also, electric power consumed by $2R$ , ${P_2} = {I^2}2R$

$\dfrac{{{P_1}}}{{{P_2}}} = 1/2$

The ratio of the electric power consumed by $R$ and $2R$ is$1:2$.

7. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. the ratio of heat produced in series and parallel combinations would be

Ans: (c)Let resistance of each wire is $R$

In series, resistance is $ = 2R$

Heat produced, ${H_1} = \dfrac{{{V^2}}}{{2R}}t$

In parallel total resistance $ = R/2$

Heat produced, ${H_2} = \left( {\dfrac{{{V^2}}}{{\dfrac{R}{2}}}} \right)t$

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

${H_2} = 4{H_1}$

$\dfrac{{{H_1}}}{{{H_2}}} = 1/4$

The ratio of heat produced in series and parallel combinations is $1:4$ .

8. Calculate the following of a circuit shown in the figure.

(i) effective resistance

Ans: Effective resistance, $R = {R_1} + {R_2}$

$ = 5 + 10$

The effective resistance is $15\Omega $ .

(ii) current

Ans: Current, $I = V/R$

The current is $0.133$ A.

(iii) Potential difference across $10\Omega $ resistor

Ans: Potential difference across $10\Omega $

$ = \dfrac{2}{{15}} \times 10$

The potential difference across $10\Omega $ is $1.33$ volt.

9. A Piece of wire of resistance $20\Omega $ is drawn out so that its length is increased to twice its original length to calculate the resistance of the wire is the new situation?

Ans: Resistance of wire$ = 20\Omega $

As $R = \dfrac{{\rho l}}{A}$

And as length of a wire is increased, its area of cross- section decreases, and volume of the wire remains constant.

$R' = \dfrac{{\rho l'}}{{A'}}$

$ = \dfrac{{4\rho l}}{{A'}}$

$\dfrac{{R'}}{R} = \dfrac{4}{1}$

$ \Rightarrow R' = 80$

The resistance of the wire is the new situation is $80\Omega $ .

10. A battery made of $5$ cells each of $2$ V and have internal resistance $0.1\Omega ,0.2\Omega ,$$0.3\Omega ,0.4\Omega $ and \[0.5\Omega \] is connected across \[10\Omega \] resistance. Draw a circuit diagram and calculate the current flowing through$10\Omega $resistance?

Ans: The internal resistance , \[0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5\Omega \]

Total resistance \[ = 1.5 + 10\]

\[ = 11.5\]

\[I = V/R\]

\[ = 10/11.5\]

\[I = {\text{ }}0.869\] A

Current flowing through $10\Omega $ resistance is $0.869$ A.

11. In the circuit diagram given here Calculate-

(a) The total effective resistance

Ans: As resistances are in parallel

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$

$ = \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{10}}$

$ = \dfrac{8}{{10}}$

$R = \dfrac{{10}}{8}$

Total effective resistance is $\dfrac{{10}}{8}\Omega $

(b) The total current

Ans: Total current, $I = V/R$

$ = \dfrac{6}{{10/8}}$

The total current is $4.8$ A.

Ans: From the circuit diagram, ${I_1},{I_2}$ and ${I_3}$ be the current through $2\Omega ,5\Omega $ and $10\Omega $ respectively

Therefore, ${I_1} = \dfrac{V}{{{R_1}}}$

${I_2} = \dfrac{V}{{{R_2}}}$

${I_3} = \dfrac{V}{{{R_3}}}$

Current through each$2\Omega ,5\Omega $ and $10\Omega $ is $3$ A ,$1.2$ A, and $0.6$ A respectively.

12. You have two circuits Compare the power used in \[2\Omega \] resistor in each case.

(i) a \[6V\] battery is series with \[1\Omega \] and \[2\Omega \] resistor

Ans: Potential difference, $V = 6V$

${R_1} = 1\Omega $

${R_2} = 2\Omega $

Total Resistor $ = {R_1} + {R_2}$

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

$ = \dfrac{6}{3}$

${P_1} = \dfrac{{{V^2}}}{R}$

The power used in $2\Omega $resistor is $8$ W.

(ii) a \[4V\] battery in parallel with \[12\Omega \] and \[2\Omega \] resistor

Ans: Potential difference, $V = 4V$

${R_1} = 12\Omega $

${P_2} = \dfrac{{{V^2}}}{R}$

The ratio of both power is $1:1$ .

13. How much energy is given to each coulomb of charge passing through a \[6\] volt battery?

Charge, $Q = 1C$

Energy = total work done

\[ = {\text{ }}Q{\text{ }}x{\text{ }}V\]

\[ = {\text{ }}1x6\]

\[ = {\text{ }}6\] joule.

Energy given to each coulomb of charge passing through a $6$ volt battery is $6$ joules.

14. On what factor does the resistance of a conductor depend?

Ans: A conductor's resistance is determined by the following factors:

(i) length of conductor

(ii) Area of cross-section

(iii) Temperature

(iv) Conductors are made from a variety of materials.

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

Ans: When linked to the same source, current flows more freely through a thick wire than via a small wire of the same material. It's because resistance rises as thickness decreases.

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

Ans: T he electrical component's resistance R remains unchanged, but the potential difference across its ends falls to half of its original value. As a result of Ohm's law, new current is reduced to half of its initial value.

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

Ans: The following reasons are why the coils of electric toasters and electric irons are built of an alloy rather than a pure metal:

(i) An alloy's resistivity is higher than that of pure metal.

(ii) An alloy does not rust quickly at high temperatures.

18. Draw a schematic diagram of a circuit consisting of a battery of three cells of \[2V\] , each, a \[5\Omega \] resistor, \[8\Omega \] resistors and a \[12\Omega \] and a plug key, all connected in series.

Ans: The diagram of circuit is as follows

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

Ans: It can be found from the question, voltage, $V = 220V$

${R_1} = 100\Omega $

${R_2} = 50\Omega $

${R_3} = 500\Omega $

\[1/R = 1/100 + 1/50 + 1/500\]

\[R{\text{ }} = {\text{ }}500/16\]

\[ = {\text{ }}31.25\Omega \]

The resistance of electric iron, which draws as much current as all three appliances take together is $31.25\Omega $ .

Current passing through electric iron, \[I = V/R\]

\[ = 220/31.25\]

\[ = 7.04\] A.

That is the current passing is $7.04$ A.

20. What is (a) the highest total resistance that can be secured by a combination of four resistance of \[4\Omega ,8\Omega ,12\Omega \] and \[24\Omega \] ?

Ans: When all four resistances must be connected in series, the highest resistance is achieved. In that instance, the outcome

$R = {R_1} + {R_2} + {R_3} + {R_4}$

\[ = {\text{ }}4 + 8 + 12 + 24\]

\[ = {\text{ }}48\Omega \]

The highest resistance is \[48\Omega \] .

(b) The lowest total resistance that can be secured by a combination of four resistance of \[4\Omega ,8\Omega ,12\Omega \] and \[24\Omega \] ?

Ans: All four resistances must be connected in parallel to produce the lowest resistance.

$1/R = 1/{R_1} + 1/{R_2} + 1/{R_3} + 1/{R_4}$

\[ = {\text{ }}1/4{\text{ }} + 1/8{\text{ }} + 1/12{\text{ }} + {\text{ }}1/24\]

\[ = {\text{ }}12/24\]

The lowest resistance is $2\Omega $ .

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

Ans: When connected to the voltage source, the cord of a heater and the cord of an electric heater are connected in series and carry the same current.

Because the resistance of the cord is so low in comparison to the resistance of the heater element.

As a result, the amount of heat created in the cord is extremely low, but significantly higher in the heater element. As a result, the heating element glows, but the cord does not.

22. A copper wire has diameter \[0.5\] mm and resistivity of \[1.6 \times {10^{ - 8}}\] m. What will be the length of this wire to make its resistance \[10\] ? How much does the resistance change if the diameter is doubled?

Ans: The diameter of wire, d = 0.5 mm,

Resistivity, $\rho = 1.6 \times {10^{ - 8}}$

resistance $R = {\text{ }}10{\text{ }}\Omega $

\[R = \rho L/A\]

$L = \dfrac{{\pi {D^2}R}}{{4\rho }}$

\[ = \dfrac{{22 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{7 \times 4 \times 1.6 \times {{10}^{ - 8}}}}\]

\[ = {\text{ }}122.5\] m

As resistance is inversely proportional to the cross-section area of wire, when the diameter is doubled for a given length of material, the resistance reduces.

23. A battery of \[9V\] is connected in series with resistance of \[0.2\Omega ,0.3\Omega ,0.4\Omega ,0.5\Omega \] and \[12\Omega \]respectively. How much current would flow through the \[12\] resistor?

Ans: Potential difference $V = 9V$ .

Total resistance \[ = {\text{ }}0.2{\text{ }} + 0.3{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}12{\text{ }}\]

\[ = {\text{ }}13.4{\text{ }}\Omega \]

Current in the circuit \[I{\text{ }} = {\text{ }}V/R{\text{ }}\]

\[ = {\text{ }}9{\text{ }}V{\text{ }}/{\text{ }}13.{\text{ }}4{\text{ }}\] A.

\[ = {\text{ }}0.67\]

In a series circuit the same current flows through all the resistance, hence current of \[0.67\] A will flow through \[12{\text{ }}\Omega \] resistor.

24. How many \[176\Omega \] resistors (in parallel) are required to carry \[5\] A on a \[220\] V line?

Ans: Let the resistors of \[176\Omega \] be joined in parallel.

Their combined resistance,

\[1/R = 1/176 + 1/176 \ldots \ldots \] times

\[ = n/176\;\] or

$ \Rightarrow R = 176/n\Omega $

Given that \[V = 220V\] and

\[I = 5\] A

\[R = V/I\]

\[ = 176/n\]

\[ = 220/5\]

\[ = 44\Omega \;\;\;\;\;\;\]

\[n{\text{ }} = {\text{ }}176/44{\text{ }}\]

\[ = {\text{ }}4\] ,

A number of $4$ resistors should be joined in parallel.

25. Show how you would connect three resistors, each of resistance \[6\Omega \] so that the combination has resistance of

(i) \[9\Omega \]

Ans: From the question, ${R_1} = {R_2}$

Join three resistors as below to get net resistance of $9\Omega $ :

(ii) \[4\Omega \]

Ans: Join three resistors as below to get \[4\Omega \] net resistance :

26. A hot plate of an electric oven connected to a \[220V\] line has two resistance coils \[A\] and \[B\] . Each of \[24\Omega \] resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?

Ans: From the question, potential difference\[V = {\text{ }}220{\text{ }}V\] .

Resistance of coil $A = $ Resistance of coil \[B\]

\[ = 24{\text{ }}\Omega \]

When coil is used separately, the circuit \[I = V/R\]

\[ = 220V/24\Omega \]

\[ = {\text{ }}9.2{\text{ }}A\]

The current is\[9.2{\text{ }}A\].

When coils are used in series total resistance $R = {R_1} + {R_2}$

\[ = 24 + 24\]

\[ = 48\Omega \]

The current flowing, \[I = V/R\]

\[ = 220V/48\Omega \]

\[ = 4.6\] A

The current is $4.6$ A.

Two coils are joined in parallel.

Total resistance \[R = {\text{ }}1/24{\text{ }} + {\text{ }}1/24\]

\[ = {\text{ }}2/24\]

\[R = 12{\text{ }}\Omega \] .

Current \[I = V/R\]

\[ = 220V/12\Omega \]

\[ = 18.3\] A.

The current is $18.3$ A.

27. Compare the power used in the \[2\Omega \] resistor in each of the following circuits:

(i) a \[6\] volt battery in series with \[1\Omega \] and \[2\Omega \] resistors and,

Ans: Suppose a \[2\Omega \] resistor is joined to a $6$ V battery in series with \[1\Omega \] and \[2\Omega \] resistors.

Total resistance \[R = 2 + 1 + 2\]

\[ = 5\Omega \]

Current \[I = 6V/5\Omega \]

\[ = 1.2\] A

Power used in $2A$ resistor \[ = {I^2}R\]

\[ = 2.88\] W

Power used is $2.88$ W.

(ii) a \[4V\] battery in parallel with \[12\Omega \] and \[2\Omega \] resistors.

Ans: Suppose \[2\Omega \] resistor is joined to a $4V$ battery in parallel with \[12\Omega \] resistor and \[2\Omega \] resistors,

the current flowing in \[2\Omega = 4V/2\Omega \]

Power used in \[2\Omega \] resistor \[ = {I^2}R\]

\[ = 8\] W

Ratio \[ = 2.88/8\]

\[ = 0.36:1\;\]

The ratio of power used is \[0.36:1\] .

28. In the given figure what is ratio of ammeter reading when \[J\] is connected to \[A\] and then to \[B\]

Ans: Connect $J$ to $A$ , then

\[ = 0.6A\]

When $J$ is connected to $B$\[V = 1 + 2 + 3 + 4\]

Ratio of ammeter reading when J is connected to $A$ and then to $B$ is $3:10$ .

29. Given a resistor each of resistors \[R\] . How will you combine them to get the

(i) maximum effective resistance?

Ans: For maximum resistance $R = nr$ , this is the same as combining a series of numbers.

(ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

Ans: For minimum resistance $R' = r/n$ , this is the same as combining a series of numbers.The ratio of the maximum to minimum resistance is $R/R' = {n^2}$.

30. A wire of length \[L\] and resistance \[R\] is stretched so that its length is doubled. How will its

(a) Resistance change

Ans: The resistance of a wire is determined by its length, cross-sectional area, and resistivity as\[{\text{R = }}\dfrac{{{{\rho l}}}}{{\text{A}}}\]

Hence, if the length is doubled and area is halved, then we have

$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho {l_2}}}{{{A_2}}}}}{{\dfrac{{\rho {l_1}}}{{{A_1}}}}}$

$ = \dfrac{{{l_2}{A_1}}}{{{l_1}{A_2}}}$

Therefore, ${R_2} = 4{R_1}$

Hence, resistance of the wire becomes four times the original value.

(b) Resistively change?

Ans: The substance from which wire is formed has a property called wire resistivity. As a result, changing the wire's size has no effect on its resistivity.

31. Two students perform the experiments on series and parallel combinations of two given resistors \[{R_1}\] and \[{R_2}\] and plot the following \[V - I\] graphs.

Ans: Both students are correct because

\[AV/A1 = \;\] resistance $R$ and

\[A1/AV = l/R\;\]

The term "series" refers to high resistance, while "parallel" refers to low resistance.

32. A household uses the following electric appliances. Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. \[3.00\]

(i) Refrigerator of rating \[4\] for ten hours each day.

Ans: Month of June has \[30\] days.

Refrigerator of $400$ W is running $2$ hours each day.

Total hours it is run in $30$ days \[ = 2 \times 30\]

\[ = 60\] h

Energy consumed in kWh is \[ = 400{\text{ }} \times {\text{ }}60/1000{\text{ }}\]

\[ = {\text{ }}24\] kWh

(ii) Two electric fans of rating \[80\]each for twelve hours each day.

Ans: Two electric fans of \[80\] W are run \[12\] hours each day.

Total hours they are run in \[30\] days \[ = 12 \times 30\]

Energy consumed in kWh is \[ = 2 \times 80 \times 360/1000\]

$ = 57.6$ kWh

(iii) Six electric tubes of rating \[18\] W each for 6 hours each day.

Ans: Six electric tubes each of \[18\] W are run $6$ hours daily.

Total hours it is run in $30$ days \[ = 6 \times 30\]

\[ = 180\] h

Energy consumed in kWh is \[ = 6 \times 18 \times 180/1000\]

\[ = 19.44\] kWh

Net energy consumed in the month of June is \[ = 24 + 57.6 + 19.44\]

\[ = 101.04\] kWh

Thus, the electric bill is \[ = 3 \times 101.04\]

\[ = Rs303.12\]

Long Answer Questions (5 Marks)

1. Two wires \[A\] and \[B\] are of equal length, different cross sectional areas and made of the same metal.

(a) (i) Name the property which is same for both the wires,

Ans: Resistivity - As resistivity is a property of a substance, it is constant for both wires.

(ii) Name the property which is different for both the wires.

Ans: Resistances - As the cross sectional areas of each wires are different, they are treated as separate objects.

(b) If the resistance of wire \[A\] is four times the resistance of wire \[B\] , calculate

(i) the ratio of the cross sectional areas of the wires and

Ans: Since \[R = \dfrac{{\rho l}}{A}\]

For wire $A$ , ${R_1} = \dfrac{{\rho l}}{{{A_1}}}$

For wire $B$ , ${R_2} = \dfrac{{\rho l}}{{{A_2}}}$

\[ \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{A_1}}}{{{A_2}}}\]

Since ${R_1} = 4{R_2}$

$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = 1:4$

\[\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\pi {r_1}^2}}{{\pi {r_2}^2}}\]

\[ = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\]

Ratio is \[{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\] .

(ii) The ratio of the radii of the wire.

Ans: \[{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} = 1/4\]

Ratio is $1:2$ .

2. (a) State ohm’s law?

Ans: If the temperature and other physical conditions of the conductor stay constant, the electric current flowing through the conductor is precisely proportional to the potential difference across the conductor's end.

(b) The value of \[(I)\] current flowing through a conductor for the corresponding values of \[(V)\] potential difference are given below

Plot a graph between \[V\] and \[I\] and also calculate resistance.

Ans: Along $x$ -axis \[IV = 1\] cm

The resistance is $1.67\Omega $ .

3. (a) Define electrical energy with S.I. unit?

Ans: The effort done by a source of electricity to sustain current in a circuit is known as electrical energy. The joule is its SI unit.

(b) A household uses the following electric appliance; Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. \[3.00\] .

(i) Refrigerator of rating \[400\] w for ten hour each day.

Ans: Electricity consumed by refrigerator in one day $ = $ power $ \times $ time

\[ = 400 \times \;10\]

\[ = {\text{ }}4000\] Wh

\[ = 4\] kwh

Therefore the electricity consumed is $4$ KWh.

(ii) Two electric fans of rating \[80\] w each for twelve hours each day.

Ans: Electricity consumed by $2$ electric fans in $1$ day, power $ \times $ time

\[ = 2 \times 80 \times 12\]

\[ = {\text{ }}1.92\] kwh

Therefore the electricity consumed is $1.92$ KWh.

(iii) Six electric tubes of rating \[18\] w each for 6hours each day.

Ans : Electricity consumed by $6$ electric tubes in $1$ day \[ = 6 \times 18 \times \;6\]

\[ = {\text{ }}0.648\] kwh

Therefore the electricity consumed is $0.648$ KWh.

Total energy consumed in one day \[ = 4 + 1.92 + 0.648\]

\[ = 6.548\] kwh

Total energy consumed in one month \[ = 6.568 \times 30\]

\[ = 197.04\] kwh

Cost of $1$ unit (kwh) $ = $ Rs \[3.00\]

Cost of \[197.04\] kwh \[ = 197.04 \times 3\]

Electricity bill \[ = Rs591.12\]

The electricity bill of the household for the month of June is Rs. $591.12$.

4. Redraw the circuit of question \[1\] , putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the \[12\Omega \] resistors. What would be the reading in the ammeter and voltmeter?

Ammeter $A$ has been joined in series with circuit and voltmeter $V$ is joined in parallel to $12$ ohms resistor.

Total voltage of battery \[V = 3x2\]

\[ = 6\] V.

Total resistance \[R = {R_1} + {R_2} + {R_3}\]

\[ = 5\Omega + 8\Omega + 12\Omega \]

\[ = 25\Omega \]

Ammeter reading \[ = I\]

\[ = 6/25\]

\[ = 0.24\] A.

Voltmeter reading \[ = {\text{ }}IR\]

\[ = 0.24{\text{ }}x{\text{ }}12\]

\[ = 2.88\] V.

The reading in the ammeter and voltmeter is $0.24$ A and $2.88$ V respectively.

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

Ans: The following are the benefits of connecting electrical equipment in parallel with the battery rather than in series:

(i) Each connecting electrical device will have the same voltage, and the device will take current according to its resistance.

(ii) It is possible to use separate on/off switches.

(iii) As the total resistance in the parallel circuit falls, a large current can be pulled from the cell.

(iv) Even if one electrical gadget is broken, other devices continue to function normally.

6. How can three resistors of resistance \[2\Omega ,3\Omega \]and \[6\Omega \] be connected to give a total resistance of

(a) \[4\Omega \]

Ans: If we connect resistance of $3\Omega $ and $6\Omega $ in parallel and resistance of $2\Omega $ is connected in series with the combination, then total resistance of combination is $4\Omega $ .

(b) \[9\Omega \]

Ans: If all the three resistance are joined in parallel the resulting resistance will be $3\Omega $ .

7. The value of current \[I\] flowing in a given resistor for the corresponding values of potential difference \[V\] across the resistor are given below:

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

Ans: From the given data the \[I - V\] graph is a straight line as shown below:

Resistance of resistor \[R{\text{ }} = {\text{ }}{V_A} - {V_B}/{1_A} - {1_B}\]

\[ = {\text{ }}12{\text{ }}V{\text{ }}--{\text{ }}6{\text{ }}V/{\text{ }}3.6{\text{ }}A{\text{ }}--{\text{ }}1.8{\text{ }}A\]

\[ = {\text{ }}6V/{\text{ }}1.8{\text{ }}A{\text{ }}\]

\[ = {\text{ }}3.3{\text{ }}\Omega \]

8. Explain the following:

(a) Why is tungsten used almost exclusively for filament of electric lamps?

Ans: For the filament of electric lamps, we need a robust metal with a high melting point. Because of its high melting point, tungsten is utilised only for electric lamp filament.

(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Ans: Electric heating device conductors are composed of an alloy because it has a higher resistance than pure metal and a higher melting point, which prevents it from oxidising at high temperatures.

Ans: As the current to all appliances remains constant despite varying resistance, each appliance cannot be turned on or off independently.

(d) How does the resistance of wire vary with its area of cross-section?

Ans: As resistance of a wire is inversely proportional to its cross-section area, the resistance will decrease when the area of cross section increases.

(e) Why are copper and aluminum wires usually employed for electric transmission?

Ans: As copper and aluminium wires are good conductors with low resistance, they are commonly utilised for electrical transmission. They can also be drawn into thin wires since they are ductil

Important Questions of Chapter 12 Class 10 Science - Free PDF Download

Students who are weak in Science and do not have a strong core knowledge might find the chapter electricity quite confusing. This chapter is full of theories and diagrams which confuse students and act as a barrier in their way of achieving good marks. Students must plan to avoid this situation so that they can score the highest possible marks in the final exams. The best way to overcome this problem is a continuous practice. Students can solve some of Class 10 Science Chapter 12 Important Questions regularly. They must induct these practice hours in their preparation schedule. So that every day some time is given to revise and practice. This will make the students efficient and thorough.

Students who are unable to solve the questions must take the help of Important Questions of Electricity Class 10 with Solutions which is available on Vedantu for free. Students can download these questions in PDF format. The downloaded CBSE Class 10 Science Chapter 12 Important Questions act as a guide in preparation for the final exams.

Class 10 Science Ch 12 Important Questions

Students will learn about the concepts and theories of electricity when they will study Chapter 12 of Class 10. They will gain a detailed knowledge of this chapter after practising Important Questions for Class 10 Science Chapter 12. Some of the knowledge that the students will learn are as follows:

Electricity

Electricity is considered as a set of physical phenomena which are associated with the presence and motion of electric charge. It is also believed that electricity is somehow related to magnetism which is why both are part of a phenomenon which is called the phenomenon of electromagnetism. This was described in Maxwell's equations. There are many other phenomena which are related to electricity such as lightning, static electricity, electric heating, electric discharges and many others.

An electric field gets produced, when there is some kind of presence of positive or negative charge. The positive and negative charges are considered as electric charges. When the electric charge moves it is known as Electric current, and it produces a magnetic field nearby.

Attributes of Electricity

There are two primary attributes of electricity: voltage and current. They both have different properties and are quite different from each other, but only in electronic circuits, they are interrelated. Absence of any of these attributes can harm the circuit operation and ruin it. Let's discuss both these attributes:

Voltage is considered as a force which is responsible for making the current flow in a circuit. Voltage gets measured in volts. In order to get a better understanding of voltage, try to think of a water faucet, in which voltage can get compared to the pressure at which the water is coming out of the faucet. A slow-flowing stream from a faucet resembles a low voltage circuit. In contrast, a high flowing stream resembles a high voltage circuit. A voltage is a mandatory requirement for a current to flow.

The movement of electrical charges is termed as current. When electrons flow through an electronic circuit, it generates a current. Current is measured in terms of amperes (amp). Here also we can take the example of a water faucet to understand the concept of current. It is assumed that the current is at a high level when more water flows in an hour through the faucet, whereas when the flowing water level is low, the current is also low. After the rain, you must have seen that river flows faster than the usual speed because the current remains high at that point of time as more amount of water passes during rains.

Conductors and Insulators

Another thing that students are going to learn while practising the Electricity Class 10 Important Questions is about the significant classification of elements which is done on the basis of their conductivity of electric charge, i.e. conductors and insulators.

Conductors:

In simple terms, conductors is the name given to those materials which allow electricity to flow through them easily without any difficulties. The conductors' property that makes them capable of allowing electricity to flow through them is termed as conductivity. When electrons start flowing in the conductors, they start to produce an electric current. The force which is required for the current to move through the conductor is known as the voltage. Some examples of conductors are copper, gold and iron. Let's discuss some properties of a Conductor.

Materials which are considered as conductors have a minimal resistance because electricity flows through them.

The inductance of the electric conductor occurs when there is a high voltage drop in the conductor.

Inside a perfect conductor, the electric field is considered to be zero. It is zero because it helps to keep the electrons calm so that they don’t accelerate.

There is no electric charge inside the material which is considered as a conductor.

Insulators are considered to be materials within which free flow of electrons from one particle of the element to the other particle is interrupted. If a certain amount of charge is transferred to such an element at a given point of time, then the charge does not get distributed in the surface and remains at the same position. The most common process to charge these elements is by rubbing or charging it through induction. Some examples of insulators are wood, plastic and glass. Let's discuss some properties of insulators:

There are no free electrons in such material because all the electrons are tightly held with each other.

The ability of these materials to stop the electric current from passing through them is known as resistance.

Dielectric length of insulators is vast. Dielectric strength is considered as the maximum electric field that an insulator can handle without suffering an electrical breakdown.

High air permeability is a feature of good insulators as they allow air to pass through their pores.

Important Questions for Electricity Class 10

To give the students an overview of the Important Questions of Chapter 12 Class 10 Science , we are listing here some of the questions which are most likely to come in the exams:

What is Electricity?

What are the attributes of electricity?

Write the SI unit of resistivity.

What do you mean by electric current? Mention and define the SI unit of electric current.

What do you understand by electrical resistivity of a material? Describe an experiment which will show the factors on which the resistance of a conducting wire depends.

What do you understand by a conductor? State its properties.

What do you understand by an insulator? State its properties.

Differentiate between indicator and conductor.

State the difference between electric energy and power.

Mention the commercial unit of electric energy.

Convert the unit of electric energy into joules.

Benefits of Important Questions for Class 10 Science Chapter 12

Class 10 Science Chapter 12 Important Questions offer significant advantages for board exam preparation.

Vedantu's team conducts thorough research to compile a list with a high probability of exam inclusion.

Expert reviews from seasoned professionals with extensive experience ensure accuracy.

Questions adhere to the CBSE board format, providing students insight into exam patterns.

Detailed and explanatory solutions accompany these important questions for comprehensive understanding.

Important Related Links for CBSE Class 10 Science

These important questions serve as a valuable reference, elucidating key concepts and aiding comprehensive exam preparation. Whether unravelling Ohm's Law or comprehending circuit configurations, these important questions offer understanding. For students grappling with scientific complexities, relying on "Electricity Class 10 Important Questions" proves beneficial, promising improved comprehension and heightened confidence. This resource is not just a tool for academic success but a key to unlocking one's potential in understanding and excelling in the captivating subject of electricity.

FAQs on Important Questions for CBSE Class 10 Science Chapter 12 - Electricity 2024-25

1. How will Vedantu’s important questions for Class 10 Science Chapter 12 will help you to score well?

Ans: Vedantu’s important questions for Class 10 Science Chapter 12 helps the students by providing an excellent exam strategy to prepare for the board examinations. Science is an essential subject for students who are aspiring for various competitive exams like JEE, NEET, etc in future. Every student aspiring to excel in the board examinations requires deep knowledge and thorough understanding of this subject. Therefore, the important questions for CBSE Class 10 Science help the students in preparing for their exams. All the questions present in these materials are framed under the latest CBSE curriculum and guidelines.

2. Can I download the important questions for Class 10 Science Chapter 12 from Vedantu website for free?

Ans: Yes, the download option is available on Vedantu website and mobile application for Class 10 Science Chapter 12. Any student can download these study materials from Vedantu website in PDF format at absolutely free of cost. These are high-quality study materials created by our in-house subject matter experts as per the latest CBSE curriculum and guidelines.

3. Define electricity.

Ans: Electricity is one of the most important aspects of our society. Electricity is a set of physical phenomena that has been shaping up our civilization ever since the wake of the industrial revolution all over the entire industries and businesses. Today, life without electricity would be unimaginable and would lead to total chaos if we somehow lose this important source of energy.

4. What are the topics and subtopics covered under this chapter of Class 10 Science?

Resistivity and Resistance

Factors that affect the Resistance of a Conductor

Parallel and Series Combination of Resistors and their applications

Heating Effect of Electric Current and its Applications

Electric Power

The interrelation between P, V, I and R

5. Define the following terms-

Potential difference

Ans: One volt- One volt is defined as the potential difference between two points in a conductor that carries current. Here, one joule of work is done to locomote a charge from one place to the other. It is the SI unit of potential difference.

Potential Difference- Potential difference is defined as the work done to move a charge from one point to another in a current-carrying conductor. The formula of potential difference is 1 volt- 1 joule/ 1coulomb. A voltmeter is used to measure the potential difference. A voltmeter is connected parallelly in a circuit.

6. Define resistance.

Ans: Resistance is defined as the internal property of a substance that offers obstruction to the current flowing through it. The electric resistance of a substance is inversely proportional to its area of cross-section (A) and directly proportional to its length (l). This means when the area will increase the resistance will decrease and vice-versa. And if the length increases, the resistance will also increase.

The formula is: R=⍴*L/A

Here R is the resistance, L is the length, A is the Area and ⍴ is a constant. The SI unit of resistance is the ohm.

7. Define resistivity.

Ans: The resistivity can be defined as the resistance offered by a current-carrying conductor that has an area of cross-section equal to one centimetre square with a length equal to one centimetre. This means it is the resistance of a one-centimetre cube of the current-carrying conductor. The formula for resistivity is ⍴=R*L/A

Here R is the resistance, L is the length and A is the area. The SI unit of resistivity is ohm*metre. To study more about resistivity, students can download the Important Questions free of cost from the Vedantu website or mobile app.

8. Define ohm’s law.

Ans: Ohm’s law states that the potential difference (V) across a current-carrying conductor is directly proportional to the current (I) flowing through it. Mathematically, V ∝ I or V= IR.

Here, V is the potential difference, I is the current and R is a constant called resistance of the conductor. Ohm’s law is followed by both electrolytic conductors as well as metallic objects.

9. Define Joule’s law of heating.

Ans: Joule’s law of heating states that the heat generated in a current-carrying resistor is directly proportional to the square of the current, resisting capability of the resistor and time till which the current flows through the resistor. The formula of Joule’s Law of Heating is given as

H= I square RT. Here, H is the heat produced in the resistor, I is current, R is the resistance and T is the time during which the current flows in the resistor. The SI unit of heat is Joule.

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CBSE Class 10 Science Chapter Wise Important Case Study Questions

Chapter wise important case study questions cbse class 10 science: cbse class 10 science board exam 2024 is just around the corner and students are working hard to score maximum marks. check these case study questions from class 10 science to ace your examination this year also download the solutions from the pdf attached towards the end. .

CBSE Class 10 Science Chapter Wise Important Case Study Questions: While the CBSE Board exam for Class 10 students are ongoing, the CBSE Class 10 Science board exam 2024 is to be held on March 2, 2024. With the exams just a few days away, CBSE Class 10th Board exam candidates are rushing to prepare the remaining syllabus, practising their weak portions, trying to revise the important questions from the past year papers, practise questions, etc.

Why are CBSE Class 10 Science Case Study Questions Important?

Section A : 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B : 6 Very Short Answer type questions carrying 2 marks each. Answers to these questions should be in the range of 30 to 50 words.
Section C : 7 Short Answer type questions carrying 3 marks each. Answers to these questions should be in the range of 50 to 80 words.
Section D : 3 Long Answer type questions carrying 5 marks each. Answers to these questions should be in the range of 80 to 120 words.
Section E : 3 Case Based/ Source Based units of assessment (4 marks each) with sub-parts.

How to solve case study questions in CBSE Class 10 Science?

Read the case given and the associated questions carefully.
Read the questions attentively and analyse what they are asking.
Apply your subject knowledge and theories in the given case to decide what the correct answers should be.

1.A chemical reaction is a representation of chemical change in terms of symbols and formulae of reactants and products. There are various types of chemical reactions like combination, decomposition, displacement, double displacement, oxidation and reduction reactions. Reactions in which heat is released along with the formation of products are called exothermic chemical reactions. All combustion reactions are exothermic reactions.

(i) The massive force that pushes the rocket forward through space is generated due to the

(a) combination reaction

(b) decomposition reaction

(d) double displacement reaction

(ii) A white salt on heating decomposes to give brown fumes and yellow residue is left behind. The yellow residue left is of

(a) lead nitrate

(b) nitrogen oxide

(d) oxygen gas

(iii) Which of the following reactions represents a combination reaction?

(a) CaO (s) + H2O (l) → Ca (OH)2 (aq)

(b) CaCO3 (s) → CaO (s) + CO2(g)

(d) 2FeSO4(s) → Fe2O3 (s) +SO2(g) + SO3(g)

(iv) Complete the following statements by choosing correct type of reaction for X and Y.

Statement 1: The heating of lead nitrate is an example of ‘X’ reaction.

Statement 2: The burning of magnesium is an example of ‘Y’ reaction.

(a)X-Combination,Y-Decomposition

(b)X-Decomposition,Y-Combination

(c)X-Combination,Y-Displacement

(d) X- Displacement, Y-Decomposition

2.The earlier concept of oxidation and reduction is based on the addition or removal of oxygen or hydrogen elements so, in terms of oxygen and hydrogen, oxidation is addition of oxygen to a substance and removal of hydrogen from a substance. On the other hand, reduction is addition of hydrogen to a substance and removal of oxygen from a substance. The substance which gives oxygen to another substance or removes hydrogen from another substance in an oxidation reaction is known as oxidising agent, while the substance which gives hydrogen to another substance or removes oxygen from another substance in a reduction reaction is known as reducing agent. For example,

(i) A redox reaction is one in which

(a) both the substances are reduced

(b) both the substances are oxidised

(d) one substance is oxidised while the other is reduced.

(ii) In the reaction, H2S+Cl2⟶S+2HCl

(a) H2S is the reducing agent.

(b) HCl is the oxidising agent.

(d) Cl2 is the reducing agent.

(iii) Which of the following processes does not involve either oxidation or reduction?

(a) Formation of slaked lime from quicklime.

(b) Heating mercuric oxide.

(d) Formation of zinc from zinc blende.

(iv) Mg+CuO⟶MgO+Cu

Which of the following is wrong relating to the above reaction?

(a) CuO gets reduced

(b) Mg gets oxidised.

(d) It is a redox reaction.

3.A copper vessel gets tarnished due to formation of an oxide layer on its surface. On rubbing lemon on the vessel, the surface is cleaned, and the vessel begins to shine again. This is due to the fact that which reacts with the acid present in lemon to form a salt which is washed away with water. As a result, the layer of copper oxide is removed from the surface of the vessel and the shining surface is exposed.

1.Which of the following acids is present in lemon?

(a) Formic acid

(b) Acetic acid

(d) Hydrochloric acid

2.The nature of copper oxide is

d) amphoteric

3.Name the salt formed in the above reaction

a) copper carbonate

b) copper chloride

c)copper citrate

d) copper citrate

4.The phenomenon of copper getting tarnished is

a) corrosion

b) rancidity

c) displacement

d)none of these

4.Metals as we know, are very useful in all fields, industries in particular. Non-metals are no less in any way. Oxygen present in air is essential for breathing as well as for combustion. Non-metals form a large number of compounds which are extremely useful, e.g., ammonia, nitric acid, sulphuric acid, etc. Non-metals are found to exist in three states of matter. Only solid non-metals are expected to be hard however, they have low density and are brittle. They usually have low melting and boiling points and are poor conductors of electricity.

i.____________ is a non-metal but is lustrous

A.Phosphorus

ii.Which of the following is known as 'King of chemicals'?

C. Sulphuric acid

D. Nitric acid

iii.Which of the following non-metals is a liquid?

iv.Hydrogen is used

A.for the synthesis of ammonia

B. for the synthesis of methyl alcohol

C.nitrogenous fertilizers

D. all of these

5.Nisha observed that the bottoms of cooking utensils were turning black in colour while the flame of her stove was yellow in colour. Her daughter suggested cleaning the air holes of the stove to get a clean, blue flame. She also told her mother that this would prevent the fuel from getting wasted.

a) Identify the reasons behind the sooty flame arising from the stove.

b) Can you distinguish between saturated and unsaturated compounds by burning them? Justify your answer.

c) Why do you think the colour of the flame turns blue once the air holes of the stove are cleaned?

6.Blood transport food, Oxygen and waste materials in our bodies. It consists of plasma as a fluid medium. A pumping organ [heart] is required to push the blood around the body. The blood flows through the chambers of the heart in a specific manner and direction. While flowing throughout the body, blood exerts a pressure against the wall or a vessel.

Pulmonary artery
Pulmonary vein
Very narrow and have high resistance
Much wide and have low resistance
Very narrow and have low resistance
Much wide and have high resistance
It is a hollow muscular organ
It is four chambered having three auricles and one ventricle.
It has different chambers to prevent O2 rich blood from mixing with the blood containing CO2
Both A & C
Blood = Plasma + RBC + WBC + Platelets
Plasma = Blood – RBC
Lymph = Plasma + RBC
Serum = Plasma + RBC + WBC

7.A brain is displayed at the Allen Institute for Brain Science. The human brain is a 3-pound (1.4-kilogram) mass of jelly-like fats and tissues—yet it's the most complex of all known living structures The human brain is more complex than any other known structure in the universe. Weighing in at three pounds, on average, this spongy mass of fat and protein is made up of two overarching types of cells—called glia and neurons— and it contains many billions of each. Neurons are notable for their branch-like projections called axons and dendrites, which gather and transmit electrochemical signals. Different types of glial cells provide physical protection to neurons and help keep them, and the brain, healthy. Together, this complex network of cells gives rise to every aspect of our shared humanity. We could not breathe, play, love, or remember without the brain.

1)Animals such as elephants, dolphins, and whales actually have larger brains, but humans have the most developed cerebrum. It's packed to capacity inside our skulls and is highly folded. Why our brain is highly folded?

b) Learning

3)Which among these protects our brain?

a)Neurotransmitter

b) Cerebrospinal fluid

d) Grey matter

4.Ram was studying in his room. Suddenly he smells something burning and sees smoke in the room. He rushes out of the room immediately. Was Ram’s action voluntary or involuntary? Why?

8.Preeti is very fond of gardening. She has different flowering plants in her garden. One day a few naughty children entered her garden and plucked many leaves of Bryophyllum plant and threw them here and there in the garden. After few days, Preeti observed that new Bryophyllum plants were coming out from the leaves which fell on the ground.

1.What does the incident sited in the paragraph indicate?

(a). Bryophyllum leaves have special buds that germinate to give rise to new plant.

(b). Bryophyllum can propagate vegetatively through leaves.

(c). Bryophyllum is a flowering plant that reproduces only asexually

(d). Both (a) and (b).

2.Which of the following plants can propagate vegetatively through leaves like Bryophyllum?

3.Do you think any other vegetative part of Bryophyllum can help in propagation? If yes, then which part?

4.Which of the following plant is artificially propagated (vegetatively) by stem cuttings in horticultural practices?

(b)Snakeplant

(d)Water hyacinth

9.The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of the population.

1) What are common signs of sexual maturation in boys?

a) Broadening of shoulders

b) Development of mammary glands

c) Broadening of waist

d) High pitch of voice

2) Common sign of sexual maturation in girls is

a) Low pitch voice

b) Appearance of moustache and beard

c) Development of mammary glands

d) Broadening of shoulders

3) Which contraceptive method changes the hormonal balance of the body?

b) Diaphragms

c) Oral pills

d) Both a) and b)

4) What should be maintained for healthy society?

a) Rate of birth and death rate

b) Male and female sex ratio

c) Child sex ratio

d) None of these

10.Pea plants can have smooth seeds or wrinkled seeds. One of the phenotypes is completely dominant over the other. A farmer decides to pollinate one flower of a plant with smooth seeds using pollen from a plant with wrinkled seeds. The resulting pea pod has all smooth seeds.

i) Which of the following conclusions can be drawn?

(1) The allele for smooth seeds is dominated over that of wrinkled seeds.

(2) The plant with smooth seeds is heterozygous.

(3) The plant with wrinkled seeds is homozygous.

(b) 1 and 2 only

(d) 1, 2 and 3

ii) Which of the following crosses will give smooth and wrinkled seeds in same proportion?

(a) RR X rr

(b) Rr X rr

(d) rr X rr

iii) Which of the following cross can be used to determine the genotype of a plant with dominant phenotype?

(a) RR X RR

(b) Rr X Rr

(d) RR X rr

iv) On crossing of two heterozygous smooth seeded plants (Rr), a total of 1000 plants were obtained in F1 generation. What will be the respective number of smooth and wrinkled seeds obtained in F1 generation?

(a) 750, 250

(b) 500, 500

(d) 950, 50

11.Food chains are very important for the survival of most species.When only one element is removed from the food chain it can result in extinction of a species in some cases.The foundation of the food chain consists of primary producers.Primary producers or autotrophs,can use either solar energy or chemical energy to create complex organic compounds,whereas species at higher trophic levels cannot and so must consume producers or other life that itself consumes producers. Because the sun’s light is necessary for photosynthesis,most life could not exist if the sun disappeared.Even so,it has recently been discovered that there are some forms of life,chemotrophs,that appear to gain all their metabolic energy from chemosynthesis driven by hydrothermal vents,thus showing that some life may not require solar energy to thrive.

1.If 10,000 J solar energy falls on green plants in a terrestrial ecosystem,what percentage of solar energy will be converted into food energy?

(d)It will depend on the type of the terrestrial plant

2.Matter and energy are two fundamental inputs of an ecosystem. Movement of

(a)Energy is by directional and matter is repeatedly circulating

(b)Energy is repeatedly circulating and matter is unidirectional

(c)Energy is unidirectional and matter is repeatedly circulating

(d)Energy is multidirectional and matter is bidirectional

3.Raj is eating curd/yoghurt. For this food intake in a food chain he should be considered as occupying

(a)First trophic level

(b)Second trophic level

(c)Third trophic level

(d)Fourth trophic level

4.Which of the following, limits the number of trophic levels in a food chain

(a)Decrease in energy at higher trophic levels

(b)Less availability of food

(c)Polluted air

5.The decomposers are not included in the food chain. The correct reason for the same is because decomposers

(a) Act at every trophic level at the food chain

(b) Do not breakdown organic compounds

(d) Release enzymes outside their body to convert organic material to inorganic forms

12.Shyam participated in a group discussion in his inter school competition on the practical application of light and was very happy to win an award for his school. That very evening his father gave treat to celebrate Shyam’s win. Shyam while sitting saw an image of a person sitting at his backside in his curved plate and could see that person’s mobile drop in the flower bed. Person was not aware until Shyam went and informed him. He thanked Shyam for his clever move.

a)From which side of his plate Shyam observed the incident –

i)outward curved

ii)inward curved

iii)plane surface

b)Part of plate from which Shyam observed the incident acted like a-

i)concave mirror

ii)convex mirror

iii)plane mirror

c)The nature of the size of the image formed in above situation is –

i)real, inverted and magnified

ii)same size , laterally inverted

iii)virtual, erect and diminished

iv)real , inverted and diminished

d)Magnification of the image formed by convex mirror is –

more than 1

iii)equal to 1

iv)less than 1

The location of image formed by a convex lens when the object is placed at infinity is

(a) at focus

When the object is placed at the focus of concave lens, the image formed is

(a)real and smaller

(b) virtual and smaller

The size of image formed by a convex lens when the object is placed at the focus ofconvex lens is

(a) highly magnified

(b) point in size

When the object is placed at 2F in front of convex lens, the location of image is

(b) between F and optical center

(d) none of the above

14.One of the wires in domestic circuits supply, usually with a red insulation cover, is called live wire. with black insulation is called neutral wire. The earth wire, which has insulation of green colour, is usually connected to a metal plate deep in the earth near the house appliances that has a metallic body. Overloading contact, in such a situation the current in the circuit abruptly increases. circuit prevents damage to the appliances and the circuit due to overloading.

1 When do we say that an electrical appliance

2 Mention the function of earth wire in electrical line

3 How is an electric fuse connected in a domestic circuit?

4 When overloading and short circuiting are said to occur?

5 What is a live wire?

15.Light of all the colours travel at the same speed in vacuum for all wavelengths. But in any transparent medium(glass or water), the light of different colours travels at different speeds for different wavelengths, which means that the refractive index of a particular medium is different for different wavelengths. As there is a difference in their speeds, the light of different colours bend through different angles. The speed of violet colour is maximum and the speed of red colour is minimum in glass so, the red light deviates least and violet colour deviates most. Hence, higher the wavelength of a colour of light, smaller the refractive index and less is the bending of light.

(i)Which of the following statements is correct regarding the propagation of Light of different colours of white light in air?

(a) Red light moves fastest.

(b) Blue light moves faster than green light.

(d) Yellow light moves with the mean speed as that of the red and the violet light.

(ii)Which of the following is the correct order of wavelength?

(a) Red> Green> Yellow

(b) Red> Violet> Green

(d) Red> Yellow> Orange

(iii)Which of the following is the correct order of speed of light in glass?

(a) Red> Green> Blue

(b) Blue> Green> Red

(d) Green> Red> Blue

(iv)Which colour has maximum frequency?

16.The region around a magnet where magnetism acts is represented by the magnetic field.The force of magnetism is due to moving charge or some magnetic material. Like stationary charges produce an electric field proportional to the magnitude of charge, moving charges produce magnetic fields proportional to the current. In other words, a current carrying conductor produces a magnetic field around it. The subatomic particles in the conductor, like the electrons moving in atomic orbitals, are responsible for the production of magnetic fields. The magnetic field lines around a straight conductor (straight wire) carrying current are concentric circles whose centres lie on the wire.

1)The magnetic field associated with a current carrying straight conductor is in anti- clockwise direction. If the conductor was held horizontally along east west direction,what is the direction of current through it?

2)Name and state the rule applied to determine the direction of magnetic field in a straight current carrying conductor.

3)Ramus performs an experiment to study the magnetic effect of current around a current carrying straight conductor with the help of a magnetic compass. He reports that

a)The degree of deflection of magnetic compass increases when the compass is moved away from the conductor.

b)The degree of deflection of the magnetic compass increases when the current through the conductor is increased.

Which of the above observations of the student appears to be wrong and why?

Case Study Questions Class 10 Science CBSE Chapter Wise PDF

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Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023

Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10 Science Subject - Electricity, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Electricity case study questions with answer key.

Final Semester - June 2015

The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current. The electric potential is defined as the amount of work done in bringing a unit positive test charge from infinity to a point in the electric field. The amount of work done in bringing a unit positive test charge from one point to another point in an electric field is defined as potential difference. $\begin{equation} V_{A B}=V_{B}-V_{A}=\frac{W_{B A}}{q} \end{equation}$ The SI unit of potential and potential difference is volt. (i) The 2 C of charge is flowing through a conductor in 100 rns, the current in the circuit is

(ii) Which of the following is true? (a) Current flows from positive terminal ofthe cell to the negative terminal of the cell outside the cell. (b) The negative charge moves from lower potential to higher potential. (c) The direction of flow of current in same as the direction of flow of positive charge. (d) All of these (iii) The potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to other is

(iv) The number of electrons flowing per second in a conductor if 1A current is passing through it

(v) The voltage can be written as

The relationship between potential difference and current was first established by George Simon Ohm called Ohm's law. According to this law, the current through a metallic conductor is proportional to the potential difference applied between its ends, provided the temperature remain constant i.e. I $\begin{equation} \propto \end{equation}$ V or V = IR; where R is constant for the conductor and it is called resistance of the conductor. Although Ohm's law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold. (i) If both the potential difference and the resistance in a circuit are doubled, then

(ii) For a conductor, the graph between V and I is there. Which one is the correct?

(iii) The slope of V - I graph (V on x-axis and I on y-axis) gives

(iv) When battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is

(v) By increasing the voltage across a conductor, the

The obstruction offered by a conductor in the path of flow of current is called resistance. The SI unit of resistance is ohm ( $\begin{equation} \Omega \end{equation}$ ). It has been found that the resistance of a conductor depends on the temperature of the conductor. As the temperature increases the resistance also increases. But the resistance of alloys like mangnin, constantan and nichrome is almost unaffected by temperature. The resistance of a conductor also depends on the length of conductor and the area of cross-section of the conductor. More be the length, more will be the resistance, more be the area of cross-section, lesser will be the resistance. (i) Which of the following is not will desired in material being used for making electrical wires?

(iii) Two wires of same material one of length L and area of cross-section A, other is of length 2L and area A/2 . Which of the following is correct?

(iv) For the same conducting wire (a) resistance is higher in summer (b) resistance is higher in winter (c) resistance is same is summer or in winter (d) none of these (v) A wire of resistance 20 $\begin{equation} \Omega \end{equation}$ is cut into 5 equal pieces. The resistance of each part is

(ii) When the three resistors each of resistance R ohm, connected in series, the equivalent resistance is

(iii) There is a wire oflength 20 cm and having resistance 20 $\begin{equation} \Omega \end{equation}$ cut into 4 equal pieces and then joined in series. The equivalent resistance is

Several resistors may be combined to form a network. The combination should have two end points to connect it with a battery or other circuit elements. When the resistances are connected in series, the current in each resistance is same but the potential difference is different in each resistor. When the resistances are connected in parallel, the voltage drop across each resistance is same but the current is different in each resistor. (i) The household circuits are connected in

(v) Two resistances 10 $\begin{equation} \Omega \end{equation}$ and 3 $\begin{equation} \Omega \end{equation}$ are connected in parallel across a battery. If there is a current of 0.2 A in 10 .Q resistor, the voltage supplied by battery is

The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current. (i) What are the properties of heating element? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point. (ii) What are the properties of electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point (iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 $\begin{equation} \Omega \end{equation}$ , the amount of heat produced is

The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule. Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved. So for commercial purposes we use a bigger unit of electrical energy which is called kilowatt hour. 1 kilowatt-hour is equal to 3.6 x 106 joules of electrical energy. (i) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is

(ii) The power of a lamp is 60 W The energy consumed in 1 minute is

(iii) The electrical refrigerator rated 400 W operates 8 hours a day. The cost of electrical energy is $\begin{equation} ₹ \end{equation}$ 5 per kWh. Find the cost of running the refrigerator for one day?

(iv) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 $\begin{equation} \Omega \end{equation}$ for 30 minutes?

(v) Which of the following is correct? (a) 1 watt hour = 3600 J (b) lkWh = 36x10 6 J (c) Energy (in kWh) = power (in W) x time (in hr) (d) $\begin{equation} \text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000} \end{equation}$

(i) Total resistance of parallel combination is : (a) 2.4 Ω (b) 3 Ω (c) 6 Ω (d) 2 Ω (ii) Equivalent resistance of total circuit is : (a) 5 Ω (b) 9 Ω (c) 11 Ω (d) 13 Ω (iii) Total current in the circuit is : (a) 2 A (b) 4.5 A (c) 0.5 A (d) 10 A (iv) Current in 6 ohm resistance is (a) 0.3 A (b) 0.2 A (c) 4 A (d) 6 A (v) Potential across 3.6 ohm resistance will be : (a) 1.8 V (b) 2.6 V (c) 9 V (d) 4.5 V

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Important Questions For Class 10 Science Chapter 12 Electricity

Important Questions for Class 10 Science Chapter 12 -Electricity

According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11 in NCERT Class 10 Science Textbook.

Chapter 12 of CBSE Class 10 is on Electricity. This chapter has fewer theory questions and more numerical problems. Most of the students find this chapter difficult to understand as it contains questions on ohms law, resistors, circuits, etc. So, to help students prepare effectively for the exam, we have provided Important Questions for CBSE Class 10 Science Chapter 12 Electricity. Practising these questions before the board exam will help in revision as well as boost the confidence level of students.

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CBSE MCQ for Class 10 Science Chapter 12 Electricity Free PDF

The CBSE MCQ for Class 10 Science Chapter 12 Electricity are provided below, in detailed and free to download PDF format. The solutions are latest , comprehensive , confidence inspiring , with easy to understand explanation . To download NCERT Class 10 MCQ PDF for Free, just click ‘ Download pdf ’ button.

CBSE MCQ for Class 10 Science Chapter 12 Electricity PDF

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Case Study Questions Class 10 Science
At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks. Case study: 1. We can see that, as the applied voltage is increased the current through the wire also increases.
Case Study Questions Class 10 Science Chapter 12 Electricity
Case Study/Passage Based Questions. Question 1: The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire.
CBSE 10th Science Electricity Case Study Questions With Solution 2021
CBSE 10th Standard Science Subject Electricity Case Study Questions With Solution 2021. The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current. The electric potential is defined as the amount of ...
Case Study Questions for Class 10 Science Chapter 12 Electricity
Case Study Questions for Class 10 Science Chapter 12 Electricity. Question 1: The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule (as shown in figure). Actually, Joule represents a very small quantity of energy and ...
Case Study Chapter 12 Electricity
Case Study Questions Chapter 12 Electricity. Case/Passage - 1. Two tungston lamps with resistances R1 and R2 respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: R1 > R2.
Electricity Case Study Based Questions Class 10
Students should follow some basic tips to solve Electricity Case Study Based Questions. These tips can help students to score good marks in CBSE Class 10 Science. Generally, the case based questions are in the form of Multiple Choice Questions (MCQs). Students should start solving the case based questions through reading the given passage.
Electricity class 10: CBSE previous question paper problems
The switch if OFF. Q2. A current of 10 A flows through a conductor for two minutes. ( i ) Calculate the amount of charge passed through any area of cross section of the conductor. ( ii ) If the charge of an electron is 1.6 × 10 − 19 C , then calculate the total number of electrons flowing.
Class 10 Science Chapter 11 Case Based Questions
The Case Based Questions: Electricity is an invaluable resource that delves deep into the core of the Class 10 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
Class 10 Science: Case Study Chapter 12 Electricity PDF Download
Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity. Case Study/Passage Based Questions. Question 1: The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way ...
Top 15 Case Based Questions
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NCERT Exemplar Class 10 Science Solutions Chapter 12
NCERT Exemplar Solutions Class 10 Science Chapter 12 - Free PDF Download. NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity are the study materials necessary for you to understand the questions that can be asked from the Class 10 Science Electricity chapter. It is crucial for students to get acquainted with this chapter in order to score excellent marks in their CBSE Class ...
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NCERT Solutions for Class 10 Science Chapter 12 Electricity
NCERT Solutions for Class 10 Science Chapter 12 Textbook Chapter End Questions. Question 1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is : (a) 125.
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Electricity Case Study Questions (CSQ's) Practice Tests. Timed Tests. Select the number of questions for the test: Select the number of questions for the test: TopperLearning provides a complete collection of case studies for CBSE Class 10 Physics Electricity chapter. Improve your understanding of biological concepts and develop problem ...
Important Questions for CBSE Class 10 Science Chapter 12
H= I square RT. Here, H is the heat produced in the resistor, I is current, R is the resistance and T is the time during which the current flows in the resistor. The SI unit of heat is Joule. Get chapter-wise important questions for CBSE Class 10 Science Chapter 12 Electricity with answers on Vedantu.
NCERT Solutions for Class 10 Science Chapter 12 Electricity
NCERT Solutions for Class 10 Science Electricity - CBSE Free PDF Download *According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11. NCERT Solutions for Class 10 Science Chapter 12 Electricity provides answers and explanations to all the exercise questions provided in the textbook. These NCERT Solutions has questions related to electric cells, electric bulbs ...
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The Chapter wise Important case study based questions with their solved answers in CBSE Class 10 Science can be accessed from the table below: CBSE Class 10 Science Chapter 1 Chemical Reactions ...
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10th Standard CBSE Science free Online practice tests. Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023 - Complete list of 10th Standard CBSE question papers, syllabus, exam tips, study material, previous year exam question papers, centum tips, formula, answer keys, solutions etc..
Important Questions For Class 10 Science Chapter 12 Electricity
So, to help students prepare effectively for the exam, we have provided Important Questions for CBSE Class 10 Science Chapter 12 Electricity. Practising these questions before the board exam will help in revision as well as boost the confidence level of students. Important Questions for Class 10 Science Chapter 12 Electricity- Download Free PDF
Electricity Class 10 Important Questions with Answers Science Chapter
Electricity Class 10 Important Questions Very Short Answer Type. Question 1. The power (P) in watts is found by multiplying the potential difference (V) in volts by the current (I) in amperes. Question 2. A charge of 150 coulomb flows through a wire in one minute.
Electricity Class 10 Important Questions with Answers Science Chapter
Important Questions of Electricity Class 10 Science Chapter 12. Question 1. A current of 10 A flows through a conductor for two minutes. (i) Calculate the amount of charge passed through any area of cross section of the conductor. (ii) If the charge of an electron is 1.6 × 10 -19 C, then calculate the total number of electrons flowing.
CBSE MCQ for Class 10 Science Chapter 12 Electricity Free PDF
The CBSE MCQ for Class 10 Science Chapter 12 Electricity are provided below, in detailed and free to download PDF format. The solutions are latest, comprehensive, confidenceinspiring, with easy to understandexplanation. To download NCERT Class 10 MCQ PDF for Free, just click ' Download pdf ' button.

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