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…

Q.29. A house hold uses the following electric appliances :- 

( i ) Refrigerator of rating 400w for two hours each day. 

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

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

Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy Rs. 300. 

Ans :- Energy consumed by refrigerator for 

                        2h  = p × t 

                              =  400w x 2h

                              = 400 / 1000 kwh × 2h

                              = 0.8 kwh

Energy consumed by electric fans for 

                      12h  =  p × t

                              =  80w × 2 × 12h

                              = 160w × 12h 

                              = 160 / 1000 kw × 12h

                              = 1.92 kwh 

Energy consumed by electric tubes for 

                       6h  =  p × t

                             =  6 × 18w × 6h 

                             =  180w × 6h

                             = 180 / 1000 kw × 6h 

                             = 0.648 kwh 

Total Energy consumed in the month of june 

                             = ( 0.8 + 1.92 + 0.648 ) kwh × 30

                             = 6.568 × 30 kwh 

                             = 197.04 kwh 

                             = 197.04 units 

Electricity bill for the months of june 

                             = 197.04 × Rs.3

                             = Rs. 591.12

                             = Rs. 591

Q.30. A torch bulb is rated 2.5v and 750mA. Calculate 

( i ) Its power.

( ii ) Its resistance. and 

( iii ) The energy consumed its this bulb is lighted for four hours.

Ans :- Here     V  =  2.5 volt

                         I   =  750 mA 

                             = 750×10⁻³A 

( i ) power    (P)    =  VI 

                             = 2.5 x 750 x 10⁻³

                             = 1875 x 10⁻³

                             = 1.875w 

( ii ) Resistance (R) = V / I

                                = 2.5 / 759 × 10⁻³ Ω

                                = 3.3 Ω

( iii ) Energy consume of 

                       E   =  p  × t 

                            = 1.875w × 4th 

                            = 1.875 J/s × 4 × 60 × 60 Sec

                            = 26000 J 

Q.31. How much current will an electric heater draw from a 220 volt line if the resistance of the heater (when hot) is 50 ohm? 

Ans :- Here      v     =     220 volt 

                         R    =     50 Ω 

                         I      =     ? 

We Have,

I = V / R

= 220 / 50 A

                                =  4.4 A 

Q.32. What is the resistance (hot) of an electric arc lamp if the lamp uses 20A when connected to a 220 volt line? 

Ans :- Hare         I    =   20A 

                           V   =   220 volt 

                           R   =   ?

We have,   

                         R     =  V / I

                               =  220 / 20 Ω

                               =  11 Ω

Q.33. A resistance of 40 ohms has a current of IA following through it. Find the potential difference across its ends.

Ans :- Here,     R    =    40 Ω

                         l     =     1A

                         v     =     ? 

We have           v     =     1R 

                                = 1 x 40 volt 

                                = 40 volt

Q.34. One billion (10⁹) electrons pass from a point P towards another point Q in 10⁻³ sec. What is the current in ampere? What is its direction? 

Ans :-Here      n     =    10⁹

                       e     =    1.6 x 10⁻¹⁹C

                        t     =    10⁻³ sec 

∴ total charge  Q   =   ne

                              =  10⁹x1.6×10⁻¹⁹C 

                              =  1.6×10⁻¹⁰C

∴ Current  I   =  Q / t

                        =  1.6×10⁻¹⁰ / 10⁻³  A 

                        =  1.6×10⁻⁷A

The direction of current is from Q to P.

Q.35. Calculate the current in a wire if a charge of 1500 coulomb flows through it in t minutes. 

Ans :- Here      Q   =  1500 c 

                          t   =  5 min 

                              =  5 x 60 sec 

                              =  300 sec 

                          I   =  ? 

We have, 

                          I  =  Q / t

                             =  1500 / 300 A 

                             =  5 A

Q.36. Calculate the charge passing through a lamp in 2 minutes if the current is 300 milliampere. 

Ans :- Here,    t   =    2 min 

                           =    2 x 60 sec 

                           =    120 sec 

                       I   =    300 mA 

                           =    300 x 10⁻³A 

                       Q  =   ?

We have,         Q  =  I × t

                             =  300 x 10⁻³x120 C 

                             =  36000 x 10⁻³ C 

                             =  36C 

Q.37. A conductor carries a current of 0.4A. Find the amount of charge that will pass through the cross-section of the conductor in 1.5 minute. How many electrons will flow in this time interval. Given charge on an electron = 1.6×10⁻¹⁹C 

Ans :- Here     I   =   0.4A 

                       Q  =   ?

                        t   =  1.5 min. 

                            =   1.5 x 60 sec 

                            =   90 sec. 

We have,        Q  =   I × t 

                            =   0.4 x 90 c 

                            =   36 c 

2nd part :-

                      n    =   ?

                      Q    =   36 c 

                       e    =   1.6×10⁻¹⁹ c 

We have,   

                         n   =  Q / e

                             =  36 / 1.6×10⁻¹⁹

                             =  22.5×10¹⁹

                             =  2.25×10²⁰ electrons

Multiple-choice questions 

Q.1. The unit of charge is 

( a ) Ampere.

( b ) Coulomb.

( c ) Farad.

( d ) Volt.

Ans :-  ( b ) Coulomb.

Q.2. The SI unit of electric current is 

( a ) Ohm.

( b ) Volt.

( c ) Ampere.

( d ) Coulomb.

Ans :- ( c ) Ampere.

Q.3. Volt is the SI unit of 

( a ) Potential difference.

( b ) Current.

( c ) Resistance.

( d ) Charge.

Ans :- ( a ) Potential difference. 

Q.4. Number of electron is IC charge is 

( a ) 1.6 × 10⁻¹⁹ 

( b ) 6.023 × 10²³

( c ) 6.25 × 10¹⁸ 

( d ) 6 × 10²⁴

Ans :- ( c ) 6.25 × 10¹⁸ 

Q.5. Filament of electric bulb is made of 

( a ) Copper.

( b ) Silver.

( c ) Tin.

( d ) Tungsten.

Ans :- ( d ) Tungsten.

Q.6. Power p in terms of potential difference v and resistance R is :- 

( a ) v²/R 

( b ) v/R 

( c ) v/R² 

( d ) v²/R 

Ans :- ( a ) v²/R 

Q.7. The resistances R₁ and R₂ are connected in parallel. The equivalent resistance of the combination is 

( a ) R₁ + R₂

( b ) R₁ – R₂

( c )  R₁ R₂/R₁ + R₂

( d )  R₁ + R₂/R₁ R₂

 Ans :- ( c )  R₁ R₂/R₁ + R₂

Q.8. The device used for measuring current is 

( a ) Galvanometer.

( b ) Ammeter.

( c ) Voltmeter.

( d ) Potentiometer.

Ans :- ( b ) Ammeter.

Q.9. Walt is the unit of 

( a ) Electric current.

( b ) Electric energy.

( c ) Electric power.

( d ) Potential difference.

Ans :- ( c ) Electric power. 

Q.10. Household electric appliances are connected in

( a ) Series.

( b ) Parallel.

( c ) Neither parallel nor series.

( d ) Both.

Ans :- ( b ) Parallel. 

Q.11. Electric power is given by 

( a ) p = V/I

( b ) p = V/I

( c ) p = I/VI

( d ) p = VI

Ans :- ( d ) p = VI

Q.12. Kilowatt hour is the unit of 

( a ) Electric power.

( b ) Electric resistance.

( c ) Electric potential.

( d ) Electric energy.

Ans :- ( d ) Electric energy. 

Q. Fill in the blanks :- 

( i ) SI unit of resistivity is —————-.

( ii ) Charge passing per unit time is called ————-.

(iii) Resistance is measured in —————.

( iv ) SI unit of conductivity is ————–. 

Ans :- ( i ) Ohm.m.

          ( ii ) Current.

          ( iii ) Ohm.

          ( iv ) Chemical, electric.

          ( v ) Ohm⁻¹ m⁻¹

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