Class 10 Science Chapter 12 Electricity

Class 10 Science Chapter 12 Electricity The answer to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 10 Science Chapter 12 Electricity and select need one.

Class 10 Science Chapter 12 Electricity

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given Assam Board Class 10 Science Chapter 12 Electricity Solutions for All Subjects, You can practice these here…

Additional Questions and Answers

Q.1. What is frictional electricity? 

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Ans :- The electricity developed by rubbing or friction is called frictional electricity.

Q.2. What is the fundamental law of electrostatics? 

Ans :- Like charges repel and unlike charges attract each other. 

Q.3. Define law of conservation of charge.

Ans :- Electric charges can neither be created nor destroyed they can only be transferred from one body to another. 

Q.4. What do you mean by static and current electricities? 

Ans :- Static electricity deals with the electric charges at rest while other current electricity deals with the electric charges in motion. 

Q.5. Define conductor. Give examples. 

Ans :- A substance which allows passage of electric charges through it easily is called a conductor. A conductor offers very low resistance to the flow of current. For example copper, silver, aluminium etc. 

Q.6. What is insulator? Give examples. 

Ans :- A substance that has infinitely high resistance does not allow electric current to flow through it. It is called an insulator.

For example – rubber, glass, plastic etc.

Q.7. Define electric current. 

Ans :- The electric current is defined as the rate of flow of electric charge through any section of a conductor. 

Electric current  =  Charge / Time

                   i.e. I  = Q / t

Electric current is a scalar quantity.

Q.8. What is electric field? 

Ans :- Electric field is the region around a charged body within which its influence can be experienced. 

Q.9. Define ohm’s law. 

Ans :- This law states that the current passing through a conductor is directly proportional to the potential difference across its ends, provided the physical conditions like temperature, density etc. remain unchanged.

∴   i.e. v ∞ I 

    Or   v = IR 

Where R is called resistance of the conductor.

Q.10. What do you mean by heating effect of current?

Ans :- When an electric current is passed through a conductor, heat is produced in it. This is known as heating effect of current. 

Q.11. What is meant by resistance? What is SI unit of nonresistance? 

Ans :- Resistance is the opposition to the flow of current in the conductor.

The SI unit of resistance is ohm ( Ω )

Q.12. What is the potential difference between two points in the electric field? Name and define its SI unit. 

Ans :- The potential difference between two points in the electric field is defined as the amount of work done in moving a unit positive charge from one point to the other against electrostatic force due to electric field. 

SI unit of potential difference is volt. 

Potential difference between two points is said to be 1 volt if 15 of the work is done in moving a charge of IC from one point to the other. 

Q.13.Two resistors when connected in parallel, give resultant value of 2 Ω. When connected in series, the value becomes 9 Ω. Calculate the value of each resistance.

Q.14. How is voltmeter connected in circuit ?

Ans :- Voltmeter is put in parallel to resistance across which difference is to be measured. 

Q.15. What is material of wires used in household? Give reason.

Ans :- It is copper. Its resistance Is very small. Hence energy lost in transmission is very small. 

Q.16. Name a low resistance device. 

Ans :- An ammeter. 

Q.17. What is electric energy? What is its practical unit? 

Ans :- Electric energy is the total amount of energy consumed in an electric circuit in a given time. 

The total energy consumed not only depends upon the power of the appliance but also upon the time for which the power is maintained. 

If a power P is maintained for t second, the work done or the energy consumed 

                               w = p x t 

                             W  =  p ( J / s ) × t ( s ) 

                                   =  Pt joule 

                       But  P  =  vI 

                      ∴     w  =  vI t joul 

Practical unit of Electric energy :- Practical unit of electric energy called kwh is usually used. 

This unit kwh is equal to the work done or energy consumed when a power of IKW is consumed for I hour. 

        Energy (in kwh) = kw × 1 hour 

                                        = watt × hour / 1000

or Energy (in kwh)  = volt × ampere × hour / 1000

Q.18. A 100w and 500w bulb are joined in parallel to the mains, which bulb will glow brighter? 

Ans :- In parallel same voltage v is applied to both the bulbs. But 500 bulb has a smaller resistance. So it will produce more heat per second and will glow brighter than 100w bulb. 

Q.19. An electric geyser has the ratings 2000w, 220v marked on it. What should be the minimum rating, in whole number of a fuse wire, that may be required for safe use with this geyser? 

Ans :- Here,   P   =  2000w 

                      V   =  220v 

                       I    =  ? 

          we have, 

                        P  =  vI

              ⇒       I    =  P / V

                              =  2000 / 220 A 

                              =  9.1 A 

Hence, the minimum rating of fuse wire should be 10A for safe use with the geyser. 

Q.20. The element of heater is very hot while the wires carrying current are cold. Why?

Ans :- Both heater element and conducting wire carry the same current. But the heater element becomes hot due to its high resistance and the conducting wires remain cold due to their low resistance. 

Q.21. Out of 60w and 40w lamps, which one has a higher electrical resistance when in use? 

Ans :- Resistance 

                                R  =  v² / p

                                     =  i.e. R ∞ P 

So, 40w lump has move resistance than a 60w lamp. 

Q.22. Tap water conducts electricity whereas distilled water does not. Why? 

Ans :- Tap water contains dissolved salts and minerals which ionise in water. Tap water conducts electricity due to the presence of these ions. Distilled water is a covalent compound containing very few ions and almost does not conduct electricity. 

Q.23. Name a metal which offers higher resistance to the passage of electricity rather than copper. 

Ans :- Aluminium. 

Q.24. What is meant by the statement, “The potential difference between two points in an electric field is ivolt”? 

Ans :- The potential difference between two points in an electric field is said to one volt if one joule of work has to be done in brining a positive charge of one coulomb from one point to another. 

Q.25. An electric fan is connected to 220v supply and draws a current of 0.5A. What is the power rating of the fan? 

Ans :- Here    V   =  220 volt

                       I    =  0.5 A 

                       P   =  ? 

      We have,  P   =  VI  

                             =  220×0.5 w 

                             =  110w 

Q.26. Calculate the energy transferred by 5A current flowing through a resistor of 2 Ω for 30 minutes. 

Ans :- Here     I   =  5A 

                       R  =  2 Ω

                       t   =  30 minutes 

                           =  60×30 sec. 

                           =  1800 seconds 

∴ Required energy transferred  

=  I²Rt

=  5²x2x1800 

=  25x2x1800 

=  90,000 J 

Q.27. A radio set of 60w runs for 50 hours. Find the electrical energy consumed in terms of units of electricity. 

Ans :- Here       P    =   60w 

                                =  6 / 1000 kw

                                =  6 / 100 kw 

                        t       =   50 hours 

∴  Energy consume =  p × t

= 6 / 100 × 50 kwh

                                 = 3 kwh 

                                  = 3 units 

Q.28. In an office, two 60w bulbs remain lighted for 5 hours and three 100w bulbs for 4 hours everyday. Find the electrical energy consumed in the month of April.

Ans :- For two 60w bulbs

                                    P  =  2 × 60w

= 120W = 120 / 1000 km

                                    t    = 5h 

∴ Energy consumed   = p × t

for one day = 120 / 1000 × 5 kWh

= 60 / 100 kwh

= 6 / 10 kwh

∴ Energy consumed for 30 day = 6 / 100 × 30 kwh

                                                  = 18kwh 

     for 100w bulbs   p   =    3×100w

                                     =    300w

= 300 / 1000 kw

= 3 / 10 kw

                                     = 300 / 1000 kw

                                      = 3 / 10 kw

                                 t    =  4h 

∴ Energy consumed for 1 day  =  p × t 

                                                 = 3 / 10 × 4 kwh

                                                  = 12 / 10 kwh

∴ Energy consumed for 30 day = 12 / 10 × 30 kwh

                                                      10

                                                   = 36 kwh 

∴ Total Energy consumed for 30 day 

                                                   = ( 18 + 36 ) kwh 

                                                   = 54 kwh 

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