MP Board Class 10th Science Chapter 12 Electricity Solutions

MP Board Class 10th Science Chapter 12 Electricity Solution with Pdf file . These solutions are solved by subjects experts.

Board Madhya Pradesh
Class10th
SubjectScience
Chapter 10 Electricity
MP Board 10th Science Chapter 10 Solutions

MP Board Class 10th Science Chapter 12 NCERT Textbook Exercises

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 RR‘ is –
(a) 125
(b) 15
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2R
Answer:
(b) IR2

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W

Question 4.
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 combinations would be-
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Given, Diameter = 0.5 mm = 0.0005 m

So, the resistance is one-fourth if the diameter is doubled.

Question 18.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why arc the conductors of electric heating devices, such as bread-toasters and electric irons, majle of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(a) Tungsten used almost exclusively for filament of electric lamps because its melting point is very high and resistivity of tungsten is low, it has special property of glowing on heating, it does not oxidized.

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is higher than metal so alloy produces large amount of heat. They do not get oxidized. Melting point is also high.

(c) As in series, circuit voltage is divided. Components of a series circuit receives only small voltage so the amount of current decreases and the device turn hot and do not work properly. Hence, series arrangement is avoided being used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A) i.e., when area of cross section decreases the resistance increases or vice versa.

(e) Copper and aluminium are usually used for electricity transmission because they are good conductors of electricity and also have low resistivity and do not get heated.

MP Board Solutions

MP Board Class 10th Science Chapter 12 Additional Important Questions

Multiple Choice Questions

Question 1.
SI unit of an electric current is equal to
(a) 6 × 108 C
(b) 9.8 × 1010 C
(c) 1.6 × 10-19 C
(d) None of these
Answer:
(d) None of these

Question 2.
Which materials have changing resistivity?
(a) Conductors
(b) Semiconductors
(c) Non-metals
(d) Super metals
Answer:
(b) Semiconductors

Question 3.
If wire in a circuit is doubled its resistivity will be?
(a) Two times
(b) Ten times
(c) Zero
(d) half
Answer:
(d) half

Question 4.
A wire used in circuit must be
(a) Highly resistant
(b) Non resistance
(c) Have high melting point
(d) None of these
Answer:
(b) Non resistance

Question 5.
A heating w ire must have
(a) High melting point
(b) High conductivity to heat
(c) Both
(d) None
Answer:
(b) High conductivity to heat

Question 6.
SI unit of resistance is:
(a) Coulomb
(b) Ampere
(c) Ohm
(d) Ohm – meter
Answer:
(c) Ohm

Question 7.
Symbol for use of voltmeter in circuit is:

Answer: (d)

Question 8.
Symbol for use of battery in circuit is:

Answer : (a)

Question 9.
Relation of potential difference and current is represented by:
(a) Joule’s law
(b) Ohm’s law
(c) Dalton’s law
(d) None
Answer:
(b) Ohm’s law

Question 10.
Formula which represents the relation among potential difference and current following in a circuit is:
(a) V = IR
(b) V = I/R
(c) l = Q/t
(d) R = p lA
Answer:
(a) V = IR

Question 11.
Kilo watt is a unit for:
(a) Power
(b) Electrical energy commercial
(c) Resistance
(d) Current flow in circuit
Answer:
(a) Power

Question 12.
Which one has less resistance?
(a) 10 watt bulb
(b) 15 watt LED
(c) 60 watt bulb
(d) 60 watt LED.
Answer:
(a) 10 watt bulb

Question 13.
In which type of combinations resistance is more?
(a) Series Combination
(b) Parallel combination
(c) both
(d) Single resistance unit.
Answer:
(a) Series Combination

Question 14.
Joule’s law formulation is
(a) H = IRT
(b) H = I2RT
(c) P = VI
(d) RT = PV
Answer:
(d) RT = PV

MP Board Solutions

Question 15.
Hardest metal which can resist maximum heat is:
(a) Lead
(b) Tungsten
(c) Silver
(d) Carbon
Answer:
(b) Tungsten

Question 16.
If current passing a circuit is 10 A and wire is connected to 100 volt line resistant in the wire will be:
(a) 10 A
(b) 10 Ω
(c) 100 Ω
(d) O Ω
Answer:
(b) 10 Ω

Question 17.
An insulator has a resistivity equals to:
(a) 10-8 Ω m
(b) 10 Ω m
(c) 5 Ω m
(d) 10-18 Ω
Answer:
(a) 10-8 Ω m

Question 18.
Calculate current-(I) in circuit given below:
(a) 0.6 A
(b) 20 A
(c) 50 A
(d) 10 A

Answer: (a) 0.6 A

Question 19.
If resistors are arranged in parallel with value 1 Ω, 2 Ω and 3 Ω Total resistant will be:
(a) more than 3 Ω
(b) less than 3 Ω
(c) less than 1 Ω
(d) equal to 1 Ω
Answer:
(b) less than 3 Ω

Question 20.
The changing 1 kWh to joule will be equal to
(a) 3.6 × 106 J
(b) 3.6 × 10-6 J
(c) 1.6 × 106 J
(d) 1 × 106 J
Answer:
(a) 3.6 × 106 J

MP Board Solutions

MP Board Solutions

Very Short Answer Type Questions

Question 1.
What kind of circuit is required for flow of current ?
Answer:
Closed.

Question 2.
Draw a schematic diagram of a simple electric with battery, bulb, ammeter and key.
Answer:

electricity

Question 3.
Express in formula form:

  1. net charge of a circuit
  2. time taken for production of a unit amount of current.

Answer:

  1. Q = It.
  2. t = Q/I.

Question 4.
What is the value of an electric charge for an electron?
Answer:
-1.6 × 10-19 C.

Question 5.
Formulate relationship between potential difference and current.
Answer:
V = IR.

Question 6.
Which material has highest conductivity and least resistance?
Answer:
Super conductors.

MP board Class 10th Science Electricity

Question 7.
Which quality of electrical energy is inversely proportional to electrical power?
Answer:
Resistance.

Question 8.
If a circuit has open key will current flow from it?
Answer:
No.

Question 9.
If ammeter is not placed in circuit will current flow?
Answer:
Yes.

Question 10.
Which gas is filled in electric bulb?
Answer:
Argon.

Question 11.
Write the formula representing relationship between heat produced and current flowing in a circuit.
Answer:
H = I2RT.

Question 12.
Write the formula to calculate power generated by a circuit a with current M.
Answer:
Since, P = VI
Putting value of I, P = VM (I = m, Given)

Question 13.
Convert 5 kWh in joule.
Answer:
1 kWh = 3.6 × 106 Joule
5 kWh = 5 × 3.6 × 106 J.
= 16 × 106 J
or 1.6 × 107 J

Question 14.
One volt ampere is equal to how many watt power?
Answer:
1 Watt Power.

MP Board Solutions

MP Board Solutions

Short Answer Type Questions

Question 1.
What is electric current?
Answer:
Total charge which pass through a particular area in unit time is called electric current of that particular conductor.

Question 2.
What is potential difference?
Answer:
Movement of electron need a pressure difference among conductors. Hence, battery or current flow suppliers are added to circuit. This difference in potential is called potential difference.

MP Board Solutions

Question 3.
Define Ohm’s law.
Answer:
Ohm’s law states that current flowing in a circuit is directly proportional to its potential difference. It is represented by formula,
V α I
or V = RI
or V = IR
Here, R is a constant known as resistance.

Question 4.
Why resistance is applied to a circuit?
Answer:
To control the current flow in desired way resistance is applied. Example- Regulator of fan give desired speed of wind flow by controlling flow of current which is being converted to physical energy by fan.

Question 5.
Write down the factors at which the resistance of the conductor depends.
Answer:

  1. Length of wire: Resistance is directly proportional to length R ∝ L.
  2. Area cross section: Resistance is inversely proportional to area cross section of a wire R ∝ 1/A.
  3. Nature of material.
  4. Temperature of the conductor.

Question 6.
Arrange the resistivity in decreasing order of following material,
Iron, Silver, Tungsten, Manganin and Glass
Answer:
Glass > Manganin > Iron > Tungsten > Silver.

Question 7.
Give reason for the following:

  1. Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon.
  2. Copper and aluminium wires are usually employed for electricity transmission.
  3. Fuse wire is placed in series with the device.

Answer:

  1. Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon because these gases do not react with the hot tungsten filament and hence, prolong the life of filament of the electric bulb.
  2. Copper and aluminium wires are usually employed for electricity transmission because copper and aluminium have low resistivity and thus, they are very good conductors of electricity.
  3. Fuse wire is placed in series with the device because when large current passes through the circuit the fuse wire gets heated up and melts and whole circuit breaks and the device is protected from the damage.

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