CBSE Class 10

These NCERT Solutions for Class 10 Science Chapter 14 Sources of Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Sources of Energy NCERT Solutions for Class 10 Science Chapter 14

Class 10 Science Chapter 14 Sources of Energy InText Questions and Answers

In-text Questions (Page 243)

Question 1.
What is a good source of energy ?
Answer:
Sun.

Question 2.
What is a good fuel ?
Answer:
A fuel which burns in air/oxygen with a moderate rate and produces a large amount of energy is called a good fuel, e.g. L.P.G.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
L.P.G. because it has high calorific value.

In-text Questions (Page 248)

Question 1.
What are the disadvantages of fossil fuels ?
Answer:
Disadvantages of fossil fuels :
(i) It is a non-renewable source of energy. If we were to continue consuming these courses as such fast rates, we would soon face energy crisis.

(ii) Fossil fuels cause air pollution. The oxides of carbon, nitrogen and sulphur that are released on burning fossil fuels are acidic oxides. These lead to acid rain which affects our water and soil resources.

Question 2.
Why are we looking at alternate sources of energy ?
Answer:
In ancient time, wood was the most common source of heat energy. The exploitation of coal and petroleum of energy made the industrialisation possible. The growing demand for energy was largely met by the fossil fuels, coal and petroleum. Our technologies were also developed for using these energy sources. But these fuels were formed over millions of years ago and there are only limited resources. If we were to continue consuming these sources at such alarming rates, we would soon face energy crises, In order to avoid this, we are looking at alternate source of energy.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience ?
Answer:
The kinetic energy of the wind can be used to do work. This energy was harnessed, by windmills in the past to do mechanical work such as in a water lifting pump. The rotatory motion of windmill is utilised to lift water form a well. But nowadays, wind energy is used to generate electricity. To generate electricity, the rotatory motion of the windmill is used to turn the turbine of the electric generator.

The kinetic energy of flowing water or potential energy of water at a height can be converted into electricity. The hydropower plants are associated with dams. The water from the high level in the dam is carried through pipes, to the turbine and thus electricity is generated.

In-text Questions (Page 253)

Question 1.
What kind of mirror concave, convex or plain-would Be best suited for use in a solar cooker ? Why?
Answer:
Plane mirror, because it is a good reflector and a temperature inside the cooker goes upto 100° to 140°c. This heat is sufficient to cook the food easily. If other mirrors are used, the temperature is very low or high. At low temp, food is not cooked well and at very high temperature food is damaged or looses its taste.

Question 2.
What are the limitations of the energy that can be obtained form the oceans ?
Answer:

1. The energy potential from the oceans is quite large, but efficient commerical exploitation is difficult.
2. The setting of a ocean power plant is very difficult.

Question 3.
What is geothermal energy ?
Answer:
The energy which is obtained due to geological changes is called geothermal energy.

Question 4.
What are the advantages of nuclear energy ?
Answer:

• It does not create air pollution.
• A small amount of fuel is able to produce a tremendous amount of energy.
• The nuclear energy produced can be converted into electricity easily.
• Once the nuclear fuel is loaded into a nuclear power plant, than it will go on releasing energy for two or three years.

In-text Questions (Page 253)

Question 1.
Can any source of energy be pollution fee ? Why or why not ?
Answer:
C. N. G is considered as a pollution free fuel because it does not produce any poisonous or polluting gas on burning and it also does not produce particulate matter on burning. Solar energy is also in the same category. A source of energy be considered polluted if it produce poisonous gaseous and particulates matter on burning eg. coal.

Question 2.
Hydrogen has been used as a rocket fuel. Would consider it a cleaner fuel than CNG? Why or why not ?
Answer:
Hydrogen has been used as a rocket fuel. The chemical reaction involved is :
H₂(g) + O₂ (g) H₂O (l) + 286 kJ

It is considered a cleaner fuel than C. N. G, because its combustion product is water. So hydrogen as a fuel provides pollution free atmosphere.

In-text Questions (Page 254)

Question 1.
Name Two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:
Renewable sources :

1. Wind energy.
2. Water energy Because if these sources are consumed continously, they will not get exhausted.

Question 2.
Give the name of two energy sources that you would consider to be exhaustible. Give reasons for your choice.
Answer:
Exhaustible energy sources :
(i) Coal, (ii) Petroleum.
Because if these sources are consumed, they will get exhausted.

Class 10 Science Chapter 14 Sources of Energy Textbook Questions and Answers

Page no. 254-255

Question 1.
A solar water heater cannot be used to get hot water on:
(a) a sunny day
(b) a cloudy day
(c) a hot day
(d) a windy day
Answer:
(b) A cloudy day.

Question 2.
Which of the following is not an example of a biomass energy source :
(a) wood
(b) gobar-gas
(c) atomic energy
(d) coal.
Answer:
(c) Atomic energy

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the sun’s energy:
(a) geothermal energy
(b) wind energy
(c) fossil fuels
(d) bio-mass.
Answer:
(a) Geothermal energy.

Question 4.
Compare and constrast fossil fuels and the sun as direct sources of energy.
Answer:
Fossils fuel :

1. It is a non-renewable sources.
2. Fossil fuel are formed in nature long ago under special conditions of temperature and pressure.
3. It is exhaustable source of energy.
4. Coal and petroleum are the main fossils fuel.
5. It causes pollution.

Solar energy or sun as direct source of energy:

1. It is a renewable source of energy energy.
2. The sun is the ultimate source of all forms of energy available on earth.
3. It is in exhaustible source of energy.
4. Solar energy, wind energy Ocean energy and hydro energy are the main forms of sun energy.
5. It does not causes pollution.

Question 5.
Compare and constrast bio-mass and hydroelectricity as sources of energy
Answer:
Bio-mass: Wood has been used as a fuel for a long time. If we can ensure that enough trees are planted, a continuous supply of fire-wood can be assured. We are also familiar with the use of cow-dung cakes as a fuel. Given the large live-stock population in India, this can also assure us a steady source of fuel. Since these fuels are plant and animal products, the source of these fuels is said to be bio-mass. These fuels, however do not produce much heat on burning and a hot of smoke is given out when they are burnt. So, technological inputs to improve the efficiency of these fuels are necessary. When wood is burnt in a limited supply of oxygen, charcoal is formed. Charcoal bums without flame, is comparatively smokeless and has a higher heat generation efficiency.

Similarly cow dung, various plant materials like the residue after harvesting the crops, vegetable waste and sewage are decomposed in the absence of oxygen to give biogas. Biogas is an excellent fuel as it contains up to 75% methane. It has high heat capacity, it burns without smoke.

Hydro-electricity: The flowing water has both kinetic as well as potential energy The energy of the flowing water is called hydro-energy. It is an important renewable source of energy. The hydro energy has been traditionally used in past for grinding of grains just by turning the water wheel. Presently, the hydro energy is used to generate electricity by turning the blades of a turbine. The electricity, thus generated by flowing water is called hydro electricity For generating hydro-electricity, a dam is constructed across a river to block the flow of water in the reservoir behind the dam. The huge turbine blades are turned by the water as it flows down from the dam. The hydro-electricity generated by flowing water is transmitted to far away from the power station for supply to factories and homes through conducting wires.

Question 6.
What are the limitations of extracting energy from: (a) the wind ? (b) waves ? (c) tides ?
Answer:
(a) Limitations of wind energy:

• The appliances or machines operating with wind energy stop working as soon as wind stops.
• The minimum speed of wind to operate generator to produce electricity is about 15 km/h. As soon as the speed of the wind becomes less than 15 km/h, the generator stops working.
• The use of wind energy is limited to certain places where wind is in plenty and blows most of the time.
• Wind energy is not sufficient to operate very heavy machines

(b) Limitations of waves energy:

• Wave energy would be available proposition only where waves are very strong.
• Strong winds blowing across the sea only can general wave.

(c) Limitations of tidal energy :
To operate tidal power plant, the difference between the water levels of high tide and low tide should be very large. This much level of tide is not available at all coastal places, Thus, tidal power plants cannot be installed everywhere.

Question 7.
On what basis would you classify energy sources as :
(a) renewable and non-renewable ?
(b) exhaustible and inexhaustible? Are the options given in (a) and (b) the same ?
Answer:
(a) Renewable and non-renewable sources : The sources of energy which keep on renewing themselves regularly are called renewable sources, e.g. solar energy, wind energy. The sources of energy that cannot be renewed quickly are known as non-renewable sources eg. coal, petroleum etc.

(b) Exhaustible and non-exhaustible : If the sources of energy consumed continously, they will get exhausted, is called exhaustible sources of energy e.g. coal, petroleum. If the source of energy are consumed continuously, they will not get exhausted is called non exhaustible sources of energy, e.g. water energy, wind energy, solar energy etc.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
Qualities of an ideal source of energy.

• It should supply enough amount of useful energy.
• It should be easily stored.
• It should be easily transported.
• It should supply useful energy in a controlled manner.
• It should occupy less space for storage.
• It should be easily available.
• It should be cheap.
• It should be pollution free.

Question 9.
What are the advantages and disadvantages of using a solar cooker ? Are there places where solar cookers would have limited utility ?
Answer:
Advantages of solar cooker:

• Economical. The cost of cooking food in the solar cooker is very small as money is only spent to purchase the solar cooker.
• It saves the costly fuel like wood, gas, kerosene etc.
• It can cook two or three dishes at a time.
• Nutrition value of food is preserved as the food is cooked at low temperature.
• Pollution. No pollution is caused as there is no burning of fuel.

Disadvantages :

• Food cannot be cooked at night.
• Food cannot be cooked on a cloudy day.
• Food cannot be cooked quickly as solar cookers takes 4 to 5 hours to cook it.
• Large quantity of food cannot be cooked with the solar cooker.
• Chapatti cannot be made with this cooker.
• Food cannot be fried.
• The position of the reflecting plane mirror has to be changed time and again so that it always faces the sun.

Solar cooker have limited utility in the hill area and the cloudy places.

Question 10.
What are the environmental consequences of increasing demand for energy? What steps would you suggest to reduce energy consumption.
Answer:
We have studied a large number of sources of energy. Exploiting any source of energy disturbs the environment in some way or the other. In any given situation, the source we would choose depends on factors such as the ease of extracting energy from that source, the economics of extracting energy from the source, the efficiency of the technology available and the environmental damage that will be caused by using that source. We have already seen that burning fossil fuels causes air pollution Research continues in these areas to produce longer lasting device that will cause less damage throughout their life.

We can contribute a lot to save energy by adopting the following strategies in our day-to-day activities.

• We must switch off the lights, fans and other home appliances running by electricity when they are not required.
• We should always use efficient appliances.
• We should close the water taps immediately after using it.
• We should always cover the vessels while cooking our food.
• Solar cookers, solar water heaters and solar dryers should be preferred wherever possible.
• Pulses, rice, etc., should be soaked in water for at least 1520 minutes before cooking.
• We must use pressure cooker for cooking rice, dal, meat, vegetables etc.
• We should prefer to use public transport for travelling as they consume less fuel energy per head as compared to the individual vehicles.

Class 10 Science Chapter 14 Sources of Energy Textbook Activities

Activity 14.1 (Page 242)

• List four forms of energy that you use from morning, when you wake up, till you reach the school.
• From where do we get these different forms of energy ?
• Can we call these ‘sources’ of energy? Why or why not?

Observation : The muscular energy for carrying out physical work, electrical energy for running various appliances, chemical energy for cooking food or running a vehicle all come from some source. We need to know how do we select the source needed for obtaining the energy in its usable form.

Activity 14.2 (Page 243)

• Consider the various options we have when we choose a fuel for cooking our food.
• What are the criteria you would consider when trying to categorise something as a good fuel?
• Would your choice be different if you lived
  (a) in a forest?
  (b) in a remote mountain village or small island?
  (c) in New Delhi?
  (d) lived five centuries ago?
• How are the factors different in each case?
  Criteria for a good fuel:
• A good fuel would do a large amount of work per unit volume or mass.
• be easily accessible.
• be easy to store and transport and
• perhaps most importantly be economial.
  (a) wood
  (b) cow dung cake or biomass fuel
  (c) L.P.G.
  (d) fats obtain from animals.
• The availability of the fuel is tht main factor in each case.

Activity 14.3 (Page 244)

• Take a table tennis ball and make three slits into it.
• Put semicircular ( ) fins cut out of a metal sheet into these slits.
• Pivot the tennis ball on an axle through its centre with a straight metal wire fixed to a rigid support. Ensure that the tennis ball rotates freely about the axle.
• Now connect a cycle dynamo to this.
• Connect a bulb in series.
• Direct a jet of water or steam produced in a pressure cooker at the fins (Fig.). What do you observe?

Observation :
This is our turbine for generating electricity. The simplest turbines have one moving part, a rotor-blade assembly. The moving fluid acts on the blades to spin them and impart energy to the rotor. Thus, we see that basically we need to move the fan, the rotor blade, with speed which would turn the shaft of the dynamo and convert the mechanical energy into electrical energy – the form of energy which has become a necessity in today’s scenario. The various ways in which this can be done depends upon availability of the resources.

Activity 14.4 (Page 248)

Find out from your grand-parents or other elders –

Question 1.
How did they go to school?
Answer:
Our grand-parents or other eiders went to school on foot.

Question 2.
How did they get water for their daily needs when they were young?
Answer:
They got water from ponds and well for their daily needs when they were young.

Question 3.
What means of entertainment did they use?
Answer:
Games such as kabaddi, kho-kho etc. were the main source of entertainment.

Question 4.
Compare the above answers with how you do these tasks now.
Answer:
Now a days we go to school by bus. We get water through water pipelines or through tube-well. T.V.V.CR movies etc. are the means of entertainment.

Question 5.
Is there a difference? If yes, in which case more energy from external sources is consumed?
Answer:
Yes there is a difference. Now a days (modern age) more energy is consumed from external sources.

Activity 14.5 (Page 249)

• Take two conical flasks and paint one white and the other black. Fill both with water.
• Place the conical flasks in direct sunlight for half an hour to one hour.
• Touch the conical flasks. Which one is hotter? You could also measure the temperature of the water in the two conical flasks with a thermometer.
• Can you think of ways in which this finding could be used in your daily life?

A black surface absorbs more heat as compared to a white or a reflecting surface under identical conditions. Solar cookers and solar water heaters use this property in their working. Some solar cookers achieve a higher temperature by using mirrors to focus the rays of the Sun. Solar cookers are covered with a glass plate.

Activity 14.6 (Page 249)

• Study the structure and working of a solar cooker and/or a solar water-heater, particularly with regard to how it is insulated and maximum heat absorption is ensured.
• Design and build a solar cooker or water- heater using low-cost material available and check what temperatures are achieved in your system.

Question 1.
Discuss what would be the advantages and limitations of using the solar cooker or water-heater.
Answer:
It is easy to see that these devices are useful only at certain times during the day. This limitation of using solar energy is overcome by using solar cells that convert solar energy into electricity. A typical cell develops a voltage of 0.5-1 V and can produce about 0.7 W of electricity when exposed to the Sun A large number of solar cells are, combined in an arrangement called solar cell panel (Fig.) that can deliver enough electricity for practical use.

The principal advantages associated with solar cells are that they have no moving parts, require little maintenance and work quite satisfactorily without the use of any focusing device. Another advantage is that they can be set up in remote and inaccessible hamlets or very sparsely inhabited areas in which laying of a power transmission line may be expensive and not commercially viable.

Activity 14.7 (Page 252)

Question 1.
Discuss in class the question of what is the ultimate source of energy for bio-mass, wind and ocean thermal energy.
Answer:
Sun is the ultimate source of energy for bio-mass, wind and ocean thermal energy.

Question 2.
Is geothermal energy and nuclear energy different in this respect? Why?
Answer:
Geothermal energy and nuclear energy is different because geothermal energy is produced due to geological changes and nuclear energy is produced by splitting of heavier nuclei into smaller or combination of lighter nuclei into heavier.

Question 3.
Where would you place hydroelectricity and wave energy?
Answer:
They are in the category of wind energy and biomass energy.

Activity 14.8 (Page 253)

• Gather information about various energy sources and how each one affects the environment.
• Debate the merits and demerits of each source and select the best source of energy on this basis.

We have studied various sources of energy like coal, petroleum, CNG, LPG, wind energy, ocean energy etc. Exploiting any source of energy disturbs the environment in some way or the other. In any given situation, the source we would choose depends on factors such as the ease of extracting energy from that source, the economics of extracting energy from the source, the efficiency of the technology available and the environmental damage that will be caused by using that source. But the fuels like CNG is said to be clean fuel, it would be more exact to say that a particular source is cleaner than the other. We have already seen that burning fossil fuels causes air pollution. In some cases, the actual operation of a device like the solar cell may be pollution free, but the assembly of the device would have caused some environmental damage. Research continues in these areas to produce longer lasting devices that will cause less damage throughout their life. C. N. G. and solar cell is the best source of energy.

Activity 14.9 (Page 254)

Question 1.
Debate the following two issues in class.
(a) The estimated coal reserves are said to be enough to last us for another two hundred years. Do you think we need to worry about coal getting depleted in this case? Why or why not?
(b) It is estimated that the Sun will last for another five billion years. Do we have to worry about solar energy getting exhausted? Why or why not?
Answer:
(a) The growing demand for energy is largely met by the fossil fuels like coal. Our technologies were also developed for using these energy sources. But coal was formed over millions of years ago and there are only limited reserves. The fossil fuels are non-renewal source of energy, so we need to conserve them. If we were to antique consuming these source at such alarming rates, we would soon run out of energy. In order to avoid this, alternate source of energy were explored. So we need to worry about coal getting depleted. But actually a large number of alternative source of energy also exist like wind energy, ocean energy etc. there is no need to worry about coal getting depleted.

(b) Sun is the ultimate source of energy on earth. Animals and plants can survive only in presence of sunlight. If solar energy gets exhausted then all the living things will also die. So we have to worry about solar energy getting exhausted.

But actually renewable energy is available in our natural environment, in the form of some continuing or repetitive currents of energy, or is stored in such large underground reservoirs that rate of depletion of the reservoir because of extraction of usable energy is practically negligible. So no need to worry about solar energy getting exhausted.

Question 2.
On the basis of the debate, decide which energy sources can be considered
(i) exhaustible,
(ii) inexhaustible,
(iii) renewable and
(iv) non-renewable.
Give your reasons for each choice.
Answer:
(i) Exhaustible : Coal, petroleum
(ii) Inexhaustible : Wind energy, ocean energy
(iii) Renewable : Wind energy, solar energy.
(iv) Non-renewable : Coal, petroleum.

Class 10 Science Chapter 14 Sources of Energy Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Give two forms of energy which is mainly utilised at our house ?
Answer:
Heat and electricity.

Question 2.
Name any two renewable sources of energy ?
Answer:
Wind energy, ocean energy.

Question 3.
Name the radiation of sunlight that mainly carries heat with it.
Answer:
Infrared radiation.

Question 4.
Name any two radiations of solar energy that are not visible to us.
Answer:

1. Ultraviolet,
2. Intra-red.

Question 5.
Name the type of radiation emitted by a hot electric iron.
Answer:
Intra-red radiation

Short Answer Type Questions

Question 1.
Mention any four areas where solar cells are being used a source of energy.
Answer:

1. In operating radio and TV sets in remote/rural areas.
2. In calculators and in wrist watches.
3. In light houses and offshore drilling operations.
4. In research centres located in remote areas.

Question 2.
Why wind energy farms on be established only at specific locations ? Give reasons to support your answer.
Answer:
Wind energy farms should be built at specific locations because :

• Wind should blow for greater part of the year.
• Wind should be steady.
• Wind should be strong blowing atleast the speed of 15 km/hr.
• Moreover, the establishment of wind energy farms require large areas of land besides a high cost of construction.

Question 3.
It is difficult to use hydrogen as a source of energy, although its calorific value is quite high. Explain.
Answer:
It is difficult to use hydrogen as a source of energy although its calorific value is quite high due to following reasons :

• It is explosive in nature because it catches fire easily with oxygen.
• It is an expensive fuel because a lot of money is required for its production.
• Its storage and transportation is not very easy.

Long Answer Type Question

Question 1.
All heat produced by the burning of cow dung cake of mass 20 gm was used to heat water of 50 gm. If the temperature of water is raised to 2°C, calculate the calorific value of the cow dung cake. Assume specific heat capacity of water = 4.2 J/g°C,
Answer:
Mass of cow dung cake, m = 200 m
Mass of water m = 50 gm
Specific heat capacity of water, C = 4.2 J/gm°C
Rise of temperature of water, ΔQ = 2°C
Heat absorbed by water = Q
= m, CΔQ = 50gm × 4.2J/gm 2°C × 420 J.
This heat absorbed = Heat produced by burning of cow dung cake.
Caloric value of cow dung = \(\frac{\mathrm{Q}}{\mathrm{m}}=\frac{420 \mathrm{~J}}{200 \mathrm{gm}}\) = 2.1 Jgm^-1

Multiple Choice Questions

Question 1.
A renewable source of energy :
(a) LPG
(b) Coal
(c) Petroleum
(d) Wind energy
Answer:
(d) Wind energy

Question 2.
Non-renewable source of energy :
(a) Natural gas
(b) Ocean energy
(c) Geothermal energy
(d) Wind energy
Answer:
(a) Natural gas

Question 3.
Solar cooker is based on :
(a) Solar energy
(b) Wind energy
(c) Hydro energy
(d) Chemical energy
Answer:
(a) Solar energy

Question 4.
Solar cooker is not used for the following purpose :
(d) Preparation of rice
(b) Preparation of pulses
(c) Preparation of grams
(d) Frying of foods.
Answer:
(d) Frying of foods.

Question 5.
The estimated energy of sun reaches per second per square meter is :
(a) 1.4 kJ
(b) 1.5 kJ
(c) 1.6 kJ
(d) 1.3 kJ
Answer:
(a) 1.4 kJ

These NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.1

Question 1.
How many tangents can a circle have?
Solution:
There can be infinitely many tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………… point(s).
(ii) A line intersecting a circle in two points is called a ………… .
(iii) A circle can have parallel tangents at the most ………… .
(iv) The common point of a tangent to a circle and the circle is called ……….. .
Solution:
(i) One
(ii) Secant
(iii) Two
(iv) Point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) \( \sqrt{199} \) cm
Solution:
We have given that PQ is a tangent of a circle whose centre is O are radius 5 cm. Length of OQ = 12 cm.

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
OQ = 12 cm
∴ ∠OPQ = 90°
[The tangent to a circle is perpendicular to the radius through the point of contact]
PQ² = OQ² – OP² [By Pythagoras theorem]
PQ² = 12² – 5² = 144 – 25 = 199
∴ PQ = \( \sqrt{199} \) cm.
Hence correct option is (d).

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:

PQ is a tangent and LM a secant.

These NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).

Solution:
In this fig. AB is height of the pole and AC is length of the rope. The inclination of rope with the ground is 30°.
Now, In ∆ABC
sin 30° = \(\frac { AB }{ AC }\)
or, \(\frac { 1 }{ 2 }\) = \(\frac { AB }{ 20 }\)
AB = \(\frac { 20 }{ 2 }\) = 10m
Therefore, the height of the pole is AB = 10m.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Let AB be the tree broken at a point C such that the broken part CB takes the position CO and strikes the ground at O. It is given that OA = 8 metres and ∠AOC = 30°.

Height of the true = (x + y)m

Hence height of the tree is 8\(\sqrt{3}\) m

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
I – case (The children below 1 the age of 5 years)

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Solution:
Let A and B be the top and foot of the tower AB, Let BC be the horizontal ground. It is given that BC = 30m, ∠ACB = 30° and ∠B = 90°
Let x be the height of the tower AB, i.e., AB = x m
We have cos 30° = \(\frac { BC }{ AC }\)

Hence, height of the tower is 10\(\sqrt{3}\) m.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
In this fig. AB is the height of kite and AB is length of string which is tied at point C on the i ground. At point C the inclination of the string with the ground is 60°.
Now, In ∆ABC,

Therefore, the length of string = 40\(\sqrt{3}\) m.

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
In this fig, AB is height of the building and CD is the height of the boy. Angle of elevation from boy’s eyes to the top of the building is 30° and after y m walk’s towards building the angle of elevation becomes 60°.

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
In this fig. AB is height of tower which is kept on the to a 20m high tower BC Angle of elevation of the top and bottom of tower from a 1 point D on ground is 60° and 45° repectivelv.

Hence, the height of the tower is 20(\(\sqrt{3}\) – 1) m.

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
In this fig. AB is the height of the statue and BC is the height of pedestal. At the poind D on the ground the angle of elevation of the top of the statue and top of pedestal is 60° and 45° respectively.

Therefore, the height of pedestal is BC = 2.19 m (approx)

Question 9.
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
In this fig, ABC is the height of tower and CD is the height of building. Angle of elevation of the top of the building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of a building is 60°.

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
Solution:
In this fig, AB and CD are poles of equal height on either side of 80 m wide road. There is a point P on the road in which the angle of elevation of the poles are 60° and 30° respectively.
Let the distance between the point P and the first pole is x m
∴ The distance between the point P and the second pole is 80 – x m

Therefore, height of each poles is 20\(\sqrt{3}\) m and distance of the point from first pole is x = 20m and (80 – x) = (80 – 20) = 60m

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.

Solution:
In this fig, AB is the height of the cable tower and CD is the height of the building. The angle of elevation of the top of a cable tower from the top of the building is 60° and angle of depression of the foot of a cable tower from the top of the building is 45°
In ∆ABC,

So, height of the three = AB = 10\(\sqrt{3}\) m and width of the river = BC = 10m

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
In this fig. AB is the height of the cable towar and CD is the height of the building. The angle of elevation of the top of a cable towar from the top of the building is 60 and the angle of depression of the foot of a cable tower from the top of tire building is 45°.

Therefore, height of the tower AB = AE + EB = 7\(\sqrt{3}\) + 7 = 7(\(\sqrt{3}\) + 1)

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
In this fig, AB is the height of the lighthouse and at point C and D there are two ships just behind each other. Angle of depression from the top of the lighthouse to the ships are 30° and 45° respectively.

Therefore, the distance between the two ships are
CD = BD – BC

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
Solution:
Initial height = 88.2 – height of the girl
= 88.2 – 1.2 = 87 m

Therefore, the distance travelled by the ballon during the interval CQ \(\frac{296 \sqrt{3}}{5}\) m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let AB be a tower and a man be at A.

A man observes a car at D at an angle of depression of 30°
i.e., ∠EAD = 30°
⇒ ∠EAD = 30° [Both are alternate angels; AE || BD and AD cutsthen at A and D]
The car is approaching towards B with a uniform speed. After travelling for 6 seconds let the car be at C.
From A, the angle of depression of the car at C is 60°.
i.e., ∠EAC= 60°
⇒ ∠EAC = ∠ACB
[Both are alternate angles; AE || BD and AC cuts them at A and C]
In right triangle ABC, we have,

From equations (i) and (ii), we get

Let V be the velocity of the car
Now in 6 sec distance covered = CD

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Angles are complementary.
∠DAC = 9
and ∠DBC = 90° – 0
In ∆DBC,

Hence, height of tower is 6m Hence Proved.

These NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Magnetic Effects of Electric Current NCERT Solutions for Class 10 Science Chapter 13

Class 10 Science Chapter 13 Magnetic Effects of Electric Current InText Questions and Answers

In-text Questions (Page 224)

Question 1.
Why does a compass needle get deflected when brought near a bar magnet ?
Answer:
A compass needle is a small bar magnet, so it is deflected when brought near a bar magnet.

In-text Questions (Page 228)

Question 1.
Draw magnetic field lines around a bar magnet.
Answer:

Question 2.
List the properties of magnetic field lines.
Answer:

• The magnetic line of force emerge from north pole and merge at the south pole.
• Inside the magnet, the direction of field lines is from its south pole to its north pole.
• The closeness of the lines shows the degree relative strength of the magnetic field.
• No two field lines are closed to each other.

Question 3.
Why don’t two magnetic field lines intersect each other ?
Answer:
If two magntic lines are found to cross each other, it would mean that at the point cf inter-section, the compass needle would point towards two directions, which is not possible.

In-text Questions (Page 229)

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
The direction of the magnetic field will be downward and perpendicular to the table inside the loop. And outside the loop magnetic field lies in the upward direction.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
Uniform magnetic field in a region is represented by parallel straight line as follows :

Question 3.
Choose the correct option.
The magnetic field inside a long straight solenoid-carrying current
(a) is zero.
(b) decreases as we move towards its end.
(c) increases as we move towards its end.
(d) is the same at all points.
Answer:
(d) is the same at all points.

In-text Questions (Page 231)

Question 1.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) Velocity
(d) Momentum

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if (i) current in rod AB is increased; (ii) a stronger horse-shoe magnet is used; and (iii) length of the rod AB is increased ?
Answer:
(i) Displacement of the rod towards left increases.
(ii) Displacement of the rod again increases.
(iii) Displacement increases.

Because as current, length and magnetic field increase, the force of attraction or repulsion between the wire and magnet increases.

Question 3.
A positively-charged particle (alpha- particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(c) Downward.

In-text Questions (Page 233)

Question 1.
State Fleming’s left-hand rule.
Answer:
According to this, stretch the thumb, forefinger and middle finger of you left hand such that they are mutually perpendicular. If the first finger points in the direction of magnetic field and the second finger in the direction of current then the thumb will point in the direction of motion or the force acting on the conductor.

Question 2.
What is the principle of an electric motor?
Answer:
It is based on the principle that when a rectangular coil is placed in a magnetic field and a current is passed through it, a torque acts on the coil which rotates it continuously.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
In electric motor, the split ring acts as a commutator.

In-text Questions (Page 236)

Question 1.
Explain different ways to induce current in a coil.
Answer:
The ways to induce current in as coil are:

1. By changing the magnetic field : Current can be induced in a coil by changing the magnitude of the magnetic field.
2. By moving a coil in the magnetic field : Current can be induced in a coil by moving coil in the magnetic field.

In-text Questions (Page 237)

Question 1.
State the principle of an electric generator.
Answer:
It is based on the phenomenon of electromagnetic induction i.e., when a coil is moved in a magnetic field, a induced current is produced in the coil.

Question 2.
Name some sources of direct current.
Answer:
D.C. generator, Dry Cell, Battery.

Question 3.
Which sources produce alternating current?
Answer:
A.C. generator, Dynamo.

Question 4.
Choose the correct option.
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution
Answer:
(c) Half revolution

In-text Questions (Page 238)

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Earth wire, Electric fuse.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
The resistance of electric oven, say R
∴ P = I² × R
2 × 100 W = (5 A)² × R
R = \(\frac{2 \times 100 \mathrm{~W}}{25 \mathrm{~A}^{2}}\) = 80 ohm
Now, this electric oven is connected with 220 V so the current drawn by the appliances.
V = 1
⇒ 1 = \(\frac{V}{R}=\frac{220 \mathrm{~V}}{80 \mathrm{ohm}}\) = 2.73 A

Question 3.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:
A fuse in a circuit is used to avoid overloading. Overloading can occur when the live wire and neutral wire come into direct contact. In such a situation, the current in the circuit abruptly increases. The use of an electric fuse prevents the electric circuit and the appliance from a possible damage by stopping the flow of unduly high electric current.

Class 10 Science Chapter 13 Magnetic Effects of Electric Current Textbook Questions and Answers

Page no. 240

Question 1.
Which of the following correctly describes the magnetic Field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the Wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centered on the wire.
Answer:
(d) The field consists of concentric circles centered on the wire.

Question 2.
The phenomenon of electromagnetic induction is wire.
(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to a relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) Producing induced current in a coil due to relative motion between 1 magnet and the coil.

Question 3.
The device used for producing electric current is called a
(a) galvanometer
(b) generator
(c) motor
(d) ammeter
Answer:
(a) Generator.

Question 4.
The essential difference between AC generator and a DC generator is that
(a) AC generator has an electromagnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) AC generator has slip rings while the DC generator has a commutator.

Question 5.
At the time of short circuit, the current in the circuit.
(a) reduces substantially
(b) does not change
(c) increases heavily
(d) vary continuously
Answer:
(c) Increase heavily

Question 6.
State whether the following statements are true or false:
(a) An electric motor converts mechanical energy into electric energy
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the Centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire.
Answer:
(a) False,
(b) True,
(c) True,
(d) False.

Question 7.
List three sources of magnetic fields.
Answer:
Solenoid, Bar magnet, Magnetite.

Question 8.
How does a solenoid behave like a magnet ? Can you determine the north and south poles of a current carrying solenoid with a help of bar magnet ? Explain.
Answer:
A solenoid is a long, helically round coil of insulated Wire.

When an electric current flows through a solenoid a magnetic field is set up which is similar to the magnetic field of a bar magnet. One end of the solenoid acts as south pole and the other end acts as north pole. If the current flows in a clockwise direction when the coil is seen end-on, then that end of the solenoid acts as a South pole. On the other hand if the current flows in anticlockwise direction when the coil is seen-end- on, then that end of the solenoid acts as a North pole.

When the North pole of a bar magnet is placed near the one end of the solenoid and if it is repelled, then end of the solenoid will be North pole and if it is attracted than end of the solenoid will be South pole.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field is largest ?
Answer:
The force will be largest when the direction of current is at right angles to the direction of the magnetic field.

Question 10.
Imagine you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. Hat is the direction of magnetic field?
Answer:
Magnetic field is horizontally towards front wall.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in An electric motor ?
Answer:
Principle: Electric motor is based on the principle that “when a current carrying conductor is placed in a magnetic field Then it experiences a mechanical force tending to rotate the Conductor” and the direction of the force experienced is given by Fleming’s Left Hand Rule.

Working : Suppose D.C. current (from a battery) is passed via Brushes, and commutator through armature, when the plane of The coil ABCD is parallel to the magnetic field as shown in Fig. (a) According to Fleming’s left hand rule, the limb AB of the coil Experiences an upward force. While the limb CD experiences a Downward force: Now these two equal, but opposite forces acting At AB and CD constitutes a couple, which rotates the coil in Clockwise direction. It may be noted that the moment of the couple


is maximum to start with, but it goes on decreasing and becomes Zero, when the coil becomes perpendicular to magnetic lines of Force. Consequently, the coil should stop at this position. However Due to inertia of the moving coil, it crosses this position and the Coil again becomes parallel to the magnetic lines of force Contact of the brushes (B₁ and B₂) with the segments (S₁ and S₂) Gets reversed and the direction of current through the coil also gets Reversed, thereby the current starts flowing along DCBA as shown In Fig. (b): Consequently, the limb DC of the coil starts moving Upwards; while the limb AB os the coil starts moving downwards In accordance with Fleming’s left hand rule. In this way, the Motion of the armature is always same (clockwise, in this case) And becomes continuous.

Question 12.
Name some devices in which electric motors are used.
Answer:
Fans, Cooler, Air conditioner, CD Player etc.

Question 13.
A coil of insulated copper wire is connected to a Galvanometer. What will happen if a bar magnet is (i) pushed into the coil (ii) with drawn from inside the coil (iii) held stationary Inside the coil ?
Answer:
(i) We will see a deflection in the galvanometer.
(ii) The needle of galvanometer again shows deflection, but in The opposite direction.
(iii) No deflection in the galvanometer.

Question 14.
Two circular coils A and B are placed closed to each Other. If the current in the coil “A” is changed, will some current Be induced in the coil ‘B’ ? Give reason.
Answer:
Yes, some current will be induced in the coil B as Current is changed in coil A due to electromagnetic induction. In electromagnetic induction by changing magnetic field in a conductor induces a current in another conductor because as the Current changes in coil ‘A the magnetic field associated with it. Also changes. Thus the magnetic field lines around the coil B Also change. Hence the chance magnetic field lines associated With the coil ‘B’ is the cause of the induced current in it.

Question 15.
State the rule to determine the direction of a (i) Magnetic field produced around a straight conductor Current (ii) force experienced by a Current-carrying straight Conductor placed in a magnetic field which is perpendicular it and (iii) current induced in a coil due to its rotation Magnetic field.
Answer:
(i) Right hand thumb rule
(ii) Flemings left hand rule
(iii) Fleming’s right hand rule.

Question 16.
Explain the underlying principle and working of an Electric generator by drawing a labelled diagram. What is The Function of brushes?
Answer:
Principle : It is based on the principle Electromagnetic induction, Which states in simple form As “whenever a conductor is Rotated mechanically in a Magnetic field, there is a Brushes Change in the magnetic lines of force within it and an Induced current is generated In the conductor”. The Direction of induced current is given by Fleming’s right hand rule.

Construction : It consists of a rotating armature ABCD containing coils of wire, pole pieces, brushes and a commutator. The two ends of the armature ABCD are connected to two metallic Rings S₁ and S₂ the two brushes B₁ and B₂ which are connected To a galvanometer G are in contact with the rings S₁ and S₂ Respectively.

Working : When the armature coil ABCD rotates in the magnetic field provided by the strong field magnet, it cuts the magnetic lines of force. Thus the changing magnetic field produces induced current in the coil. The current flows out through the brush B, in one direction in the first half of the revolution and through the Brush B, in the next half revolution in the reverse direction. This Process is repeated. Therefore, induced current produced is of alternating nature. Such a current is called alternating current.

Question 17.
When does an electric short circuit occur?
Answer:
If the plastic insulation of the live wire and neutral wire gets torn, than the two wires touch each other. The touching of the live wire and neutral wire directly is known as short circuiting. It occurs by the direct touching of live and neutral wires.

Question 18.
What is the function of an earth wire ? Why is it necessary to earth metallic appliances?
Answer:
Function : A very high current flows through the earth Wire and the fuse of household wiring blows out or melts and it cuts of the power supply. We earth the metallic appliances to save ourselves from electric shocks. If by chance, the live wire touches. The metal part of the electric appliance, which has been earthed, then the current passes directly to the earth through the earth wire. It does not need our body to pass the current and thus we do not get an electric shock.

Class 10 Science Chapter 13 Magnetic Effects of Electric Current Textbook Activities

Activity 13.1 (Page 223)

• Take a straight thick copper wire and place it between the points X and Y in an electric circuit, as shown in Fig.

Fig. : Compass needle is deflected on passing an electric current through a metallic conductor

• place a small compass near to this copper wire. See the position of its needle.
• Pass the current through the circuit by inserting the key into the plug.

Question 1.
Observe the change in the position of the compass needle.
Answer:
We see that the needle is deflected.

Question 2.
What does it mean ?
Answer:
It means that the electric current through the copper wire has produced a magnetic effect. Thus we can say that electricity and magnetism are linked to each other.

Activity 13.2 (Page 223)

• Fix a sheet of white paper on a drawing board using some adhesive material.
• Place a bar magnet in the centre of it,
• Sprinkle some iron filings uniformly around the bar magnet (Fig. ). A salt- sprinklei may be used for this purpose.
• Now tap the board gently.

Question 1.
What do you observe?
Answer:

The iron filings arrange themselves in a pattern as shown Fig. Why do the iron filings arrange in such a pattern? What does this pattern demonstrate? The magnet exerts its influence in the region surrounding it. Therefore the iron filings experience a force. The force thus exerted makes iron filings to arrange in a pattern. The region surrounding a magnet, in which the force of the magnet can be detected, is said to have a magnetic field. The lines along which the iron filings align themselves represent magnetic field lines.

Activity 13.3 (Page 224)

• Take a small compass and a bar magnet.
• Place the magnet on a sheet of white paper fixed on a drawing board, using some adhesive material.
• Mark the boundary of the magnet.

Fig. : (a) Drawing a magnetic field line with the help of a compass needle

• Place the compass near the north pole of the magnet. How does it behave? The south pole of the needle points towards the north pole of the magnet. The north pole of the compass is directed away from the north pole of the magnet.

• Mark the position of two ends of the needle.
• Now move the needle to a new position such that its south pole occupies the position previously occupied by its north pole.
• In this way, proceed step by step till you reach the south pole of the magnet as shown in Fig. (a).
• Join the points marked on the paper by a smooth curve. This curve represents a field line.
• Repeat the above procedure and draw as many lines as you can. You will get a pattern shown in Fig. (b). These lines represent the magnetic field around the magnet. These are known as magnetic field lines.
• Observe the deflection in the compass needle as you move; it along a field line. The deflection increases as the needle is moved towards the poles.

Observations : Magnetic field is a quantity that has both direction and magnitude. The direction of the magnetic field is taken to be the direction in which a north pole of the compass needle moves inside it. Therefore it is taken by convention that the field lines emerge from north pole and merge at the south pole (note the arrows marked on the field lines in Fig. (b). Inside the magnet, the direction of field lines is from its south pole to its north pole. Thus the magnetic field lines are closed curves.

The relative strength of the magnetic field is shown by the degree of closeness of the field lines. The field is stronger, that is, the force acting on the pole of another magnet placed is greater where the field lines are crowded (see Fig. (b).

No two field-lines are found to cross each other. If they did, it would mean that at the point of intersection, the compass needle would point towards two directions, which is not possible.

Activity 13.4 (Page 226)

• Take a long straight copper wire, two or three cells of 1.5 V each, and a plug key. Connect all of them in series as shown in Fig. (a).
• Place the straight wire parallel to and over a compass needle.
• Plug the key in the circuit.
• Observe the direction of deflection of the north pole of the needle. If the current flows from north to south, as shown in Fig. (a), the north pole of the compass needle would move towards the east.
• Replace the cell connections in the circuit as shown in Fig. (b). This would result in the change of the direction of current through the copper wire, that is, from south to north.
• Observe the change in the direction of deflection of the needle. You will see that now the needle moves in opposite direction, that is, towards the west [Fig.] It means that the direction of magnetic field produced by the electric current is also reversed.

Observations:


Fig.: A simple electric circuit in which a straight copper wire is placed parallel to and over a compass needle. The deflection in the needle becomes opposite when the direction of the current is reversed.

Activity 13.5 (Page 226)

• Take a battery (12 V), a variable resistance (or a rheostat), an ammeter (0-5 A), a plug key, connecting wires and a long straight thick copper wire.
• Insert the thick wire through the centre, normal to the plane of a rectangular cardboard. Take care that the cardboard is fixed and does not slide up or down.
• Connect the copper wire vertically between the points X and Y, as shown in Fig. (a), in series with the battery, a plug and key.
• Sprinkle some iron filings uniformly on the cardboard. (You may use a salt sprinkler for this purpose.)
• Keep the variable of the rheostat at a fixed position and note the current through the ammeter.

• Close the key so that a current flows through the wire. Ensure that the copper wire placed between the points X and Y remains vertically straight.
• Gently tap the cardboard a few times. Observe the pattern of the iron filings. You would find that the iron filings align themselves showing a pattern of concentric circles around the copper wire (Fig. a).
• What do these concentric circles represent? They represent the magnetic field lines.
• Flow can the direction of the magnetic field be found? Place a compass at a point (say P) over a circle. Observe the direction of the needle. The direction of the north pole of the compass needle would give the direction of the field lines produced by the electric current through the straight wire at point P. Show the direction by an arrow.

Fig. : (a) A pattern of concentric circles indicating the field lines of a magnetic field around a straight conducting wire. The arrows In the circles show the direction of the field lines (b) A close up of the pattern obtained

Does the direction of magnetic field lines get reversed if the direction of current through the straight copper wire is reversed? Check it.

Observations: What happens to the deflection of the compass needle placed at a given point if the current in the copper wire is changed? To see this, vary the current in the wire. We find that the deflection in the needle also changes. In fact, if the current is increased, the deflection also increases. It indicates that the magnitude of the magnetic field produced at a given point increases as the current through the wire increases.

What happens to the deflection of the needle if the compass is moved away from the copper wire but the current through the wire remains the same? To see this, now place the compass at a farther point from the conducting wire (say at point Q). What change do you observe? We see that the deflection in the needle decreases. Thus the magnetic field produced by a given current in the conductor decreases as the distance from it increases. From Fig. it can be noticed that the concentric circles representing the magnetic field around a current-carrying straight wire become larger and larger as we move away from it.

Activity 13.6 (Page 229)

• Take a rectangular cardboard having two holes. Insert a circular coil having large number of turns through them, normal to the plane of the cardboard.
• Connect the ends of the coil in series with a battery, a key and a rheostat, as shown in Fig. (a).
• Sprinkle iron filings uniformly on the cardboard.
• Plug the key.
• Tap the cardboard gently a few times. Note the pattern of the iron filings that emerges on the cardboard.

Fig. : (a) Magnetic field produced by a current-carrying circular coil

Observations : A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder is called a solenoid. The pattern of the magnetic field lines around a current-carrying solenoid is shown in Fig. (h). Compare the pattern of the field with the magnetic field around a bar magnet (Fig.). Do they look similar? Yes, they are similar. In fact, one end of the solenoid behaves as a magnetic north pole, while the other behaves as the south pole. The field lines inside the solenoid are in the form of parallel straight lines This indicates that the magnetic field is the same at all points inside the solenoid. That is, the field is uniform inside the solenoid.

Fig. : (b) Field lines of the magnetic field through and around a current carrying solenoid.

Activity 13.7 (Page 230)

• Take a small aluminium rod AB (of about 5 cm). Using two connecting wires suspend it horizontally from a stand, as shown in Fig.

Fig. : A current carrying rod, AB, experiences a force perpendicular to its length and the magnetic field

• Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the north pole of the magnet vertically below and south pole vertically above the aluminium rod (Fig.).
• Connect the aluminium rod in series with a battery, a key and a rheostat.
• Now pass a current through the aluminium rod from end B to end A.
• What do you observe? It is observed that the rod is displaced towards the left. You will notice that the rod gets displaced.
• Reverse the direction of current flowing through the rod and observe the direction of its displacement. It is now towards the right.
• Why does the rod get displaced?

Observation: The displacement of the rod in the above activity suggests that a force is exerted on the current-carrying aluminium rod when it is placed in a magnetic field. It also suggests that the direction of force is also reversed when the direction of current through the conductor is reversed. Now change the direction of field to vertically downwards by interchanging the two poles of the magnet. It is once again observed that the direction of force acting on the current-carrying rod gets reversed. It shows that the direction of the force on the conductor depends upon the direction of current and the direction of the magnetic field. Experiments have shown that the displacement of the rod is largest (or the magnitude of the force is the highest) when the direction of current is at right angles to the direction of the magnetic field. In such a condition we can use a simple rule to find the direction of the force on the conductor.

Activity 13.8 (Page 233)

• Take a coil of wire AB having a large number of turns.
• Connect the ends of the coil to a galvanometer as shown in Fig.
• Take a strong bar magnet and move its north pole towards the end B of the coil. Do you find any change in the galvanometer needle?

Fig. : Moving a magnet towards a coll sets up a current in the coll circuit, as Indicated by deflection in the galvanometer needle

• There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.
• Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first.
• Place the magnet stationary at a point near to the coil, keeping its north pole towards the end B of the coil. We see that the galvanometer needle deflects toward the right when the coil is moved towards the north pole of the magnet. Similarly the needle moves toward left when the coil is moved away.
• When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero. What do you conclude from this activity?

Observations : A galvanometer is an instrument that can detect the presence of a current in a circuit. The pointer remains at zero (the centre of the scale) for zero current flowing through it. It can deflect either to the left or to the right of the zero mark depending on the direction of current.

We can also check that if you had moved south pole of the magnet towards the end B of the coil, the deflections in the galvanometer would just be opposite to the previous case. When the coil and the magnet are both stationary, there is no deflection in the galvanometer. It is, thus, clear from this activity that motion of a magnet with respect to the coil produces an induced potential difference, which sets up an induced electric current in the circuit.

Activity 13.9 (Page 235)

• Take two different coils of copper wire having large number of turns (say 50 and 100 turns respectively). Insert them over a non-conducting cylindrical roll, as shown in Fig (You may use a thick paper roll for this purpose.)
• Connect the coil-1, having larger number of turns, in series with a battery and a plug key. Also connect the other coil-2 with a galvanometer as shown.

• Plug in the key. Observe the galvanometer. Is there a deflection in its needle? You will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil- 2.
• Disconnect coil-1 from the battery. You will observe that the needle momentarily moves, but to the opposite side. It means that now the current flows in the opposite direction in coil-2.

Observations: In this activity we observe that as soon as the current in coil-1 reaches either a steady value or zero, the galvanometer in coil-2 shows no deflection.

From these observations, we conclude that a potential difference is induced in the coil-2 whenever the electric current through the coil-1 is changing (starting or stopping). Coil-1 is called the primary coil and coil-2 is called the secondary coil. As the current in the first coil changes, the magnetic field associated with it also changes. Thus the magnetic field lines around the secondary coil also change. Hence the change in magnetic field lines associated with the secondary coil is the cause of induced electric current in it. This process, by which a changing magnetic field in a conductor induces a current in another conductor, is called electromagnetic induction. In practice we can induce current in a coil either by moving it in a magnetic field or by changing the magnetic field around it. It is convenient in most situations to move the coil in a magnetic field.

Class 10 Science Chapter 13 Magnetic Effects of Electric Current Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
In household circuits is a fuse wire connected in Series or in parallel.
Answer:
Series.

Question 2.
What is an electric generator ?
Answer:
A device which converts mechanical energy into electrical energy.

Question 3.
What are the various components of an electric motor?
Answer:

• Magnet
• brushes
• coil
• commutator.

Question 4.
What is meant by back e.m.f. in a motor ?
Answer:
Back e.m.f. means induced e.m.f. in a motor.

Question 5.
What is electric motor ?
Answer:
A device which converts electric energy into mechanical energy.

Short Answer Type Questions

Question 1.
What is the source of magnetic field ?
Answer:
The origin of magnetic field is electrical. The electrons That revolve in an atom act as tiny current loops are responsible for generating the magnetic field.

Question 2.
What is the basic difference between electric lines of force and magnetic lines of force.
Answer:
The electric lines of force originate from positive charge End as negative charge hence there is discontinuous curve. On the other hand, magnetic lines of force are dosed loops because Isolated magnetic poles do not exist.

Question 3.
Give some ways to increase the magnitude of induced current.
Answer:
The ways to increase the magnitude of induced current are as follows:

• Magnitude of induced current can be increased by Increasing the strength of used magnetic field.
• It can be increased if the speed of movement of the conduction in the magnetic field is increased.
• It can also be increased if the conductor is taken in the form of rectangular coil of many turns of insulated wire.

Long Answer Type Question

Question 1.
Why electric voltage is stepped up for transmission?
Answer:
Mathematically, Power = voltage × current.

When the electric voltage is stepped up, the potential differences increases and current decreases so that the power remains the same When there is transmission of electricity over large distances. Through the wires, the wires get heated and the heat thus produced is in the atmosphere. Heat produced is I²R where 1 is the current flowing through the wire. When the voltage is stepped up, the current I that flows through the wire gets decreased and hence I²R become small and there would be small heat loss or power less. During transmission. Hence to minimize the power loss during transmission, the electric voltage is stepped up.

Multiple Choice Questions

Question 1.
Magnetic field produced at the Centre of a current-carrying Circular wire is
(a) directly proportional to the square of the radius of the circular wire.
(b) directly proportional to the radius of the circular wire
(c) Inversely proportional to the square of the radius of the inversely Circular wire
(d) Inversely proportional to the radius of the circular wire
Answer:
(d) Inversely proportional to the radius of the circular wire

Question 2.
Fuse in an electric circuit acts as a
(a) Current multiplier
(b) Voltage multiplier
(c) Power multiplier
(d) Safety device
Answer:
(d) Safety device

Question 3.
Who invented generator ?
(a) Oersted
(b) Coulomb
(c) Rutherford
(d) Faraday
Answer:
(d) Faraday

Question 4.
The frequency of household supply of a.c. is
(a) 45 Hz
(b) 50 Hz
(c) 60 Hz
(d) 100 Hz
Answer:
(b) 50 Hz

Question 5.
The value of magnetic dip at magnetic
(a) 0°
(b) 30°
(c) 90°
(d) 180°
Answer:
(c) 90°

These NCERT Solutions for Class 10 Science Chapter 12 Electricity Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Electricity NCERT Solutions for Class 10 Science Chapter 12

Class 10 Science Chapter 12 Electricity InText Questions and Answers

In-text Questions (Page 200)

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path of an electricity current is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
Ampere is the unit of current. When one coulomb of charge flow per second in the circuit, it is known as one ampere.
1A = \(\frac{1 \mathrm{C}}{1 \mathrm{~S}}\)

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
We know that, charge on one electron = 0.6 × 10^-19 C
Let 1 C contain ‘n’ electrons
so, 1 C = n × 0.6 × 10^-19 C
or

In-text Questions (Page 202)

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Battery or Cell.

Question 2.
What is meant by saying that the potential difference between two points is 1V?
Answer:
The potential difference between two points in a current carrying conductor is said to be one volt when one joule of work is done to move a charge of one coulomb from one point to the other.

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
We know that, the amount of energy or work is given by the formula.
W = VQ
= 6 V × 1 C
= 6 V – C = 6 J

In-text Questions (Page 209)

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends upon two factors.
(i) Length of the conductor ; Resistance of a conductor is directly proportional to its length i.e.,
R ∝ 1
(ii) Area of cross section of the conductor :
The resistance of conductor is directly proportional to its area of cross section.

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Current will flow more easily through a thick wire because the resistance of a wire is inversely proportional to its area of cross section. Thick wire has a large surface area than a thin wire.

Question 3.
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
Let ‘R’ be the resistance, and ‘V’ be the potential difference than ‘I’ current flow through the wire, so from Ohm’s Law
V = iR ……….(i)
When potential difference is half of its initial value than, from Ohm’s Law
\(\frac{\mathrm{V}}{2}\) = i₁ R …….(ii)
Now from equation (i) and (ii) we get

The current become half of its initial value (i).

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The resistivity of an alloy is generally higher than that of its constituent metals but alloys do not oxidise readily at high temperature so they are commonly used in elecricity toasters and electric irons.

Question 5.
Use the data in Table 12.2 to answer the following-
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
(a) Iron because its resistivity is lower than mercury.
(b) Silver is the best conductor because it has lowest resistivity.
Table 12.2 s Electrical resistivity of some substances at 20°C

In-text Questions (Page 213)

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:

Question 2.
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 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:

According to Ohm’s Law
V = IR
since R = R₁ + R₂ + R₃
= 5Ω + 8Ω + 12Ω
R= 25 W
And V = 6 volt
6 volt = 1 × 25 Ohm
I = \(\frac{6 \text { volt }}{25 \mathrm{Ohm}}\) = 0.24 Ampere
V¹ = -MR
V¹ = 0.24 \(\frac{\text { volt }}{\mathrm{Ohm}}\) × 12 ohm
V¹ = 2.88 volt
So the reading in the ammeter = 0.24 Ampere
And reading of the volmeter = 2.88 volt.

In-text Questions (Page 216)

Question 1.
Judge the equivalent resistance when the following are connected in parallel.
(A) 1 Ω and 10⁶ Ω
(b) 1 Ω and 10³ Ω, and 10⁶ Ω.
Answer:
(a) Let R is the equivalent resistance

(b) Let ‘R’ is the equivalent resistance.

Question 2.
An electric lamp of 100 Ω, a toaster of resistance 50 Ω?, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
In series the current is constant throughout the electric circuit Thus, it is obviously impracticable to connect an electric bulb and and an electric heater in series because they require currents of widely different values to operate properly. Another major disadvantages of a series circuit is that when one component fails the circuit is broken and none of the components works. On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has different resistance and require different current to operate properly.

Question 4.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of
(a) 4 Ω
(b) 1 Ω?
Answer:
If 3 Ω and 6 Ω are in series than resultant resistant resistant say (R) can be calculated as

And if R and 2 Ω in series then resultant resistor (R¹) can be calculated as
R¹ = R + 2Ω
= 2Ω + 2Ω
R¹ = 4 Ω

(b) If all the resistance are in parallel than resultant resistance (R) can be calculated as follows:

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω ?
Answer:
(a) When the four resistances are in parallel, the resultant resistance will be lowest.
Let the resultant resistance be ‘R’ so

(b) When the four resistance are in series, the resultant resistance will be highest.
Let the resultant resistant be ‘R’ so.
R₁ = 4Ω + 8Ω + 12Ω + 24Ω
R₁ = 48 Ω

In-text Questions (Page 2018)

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The cord of an electric heater not glow because it is less resistive while the heating element glow because it is purely resistive.

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
We know that the heat generated in time is
H = V I t
Now I = \(\frac{\mathrm{Q}}{\mathrm{t}}=\frac{96000 \mathrm{C}}{60 \times 60 \mathrm{Sec}}\)
= 26.67 Amp.
now, H = V I t
= 50 volt × 26.67 Amp. × 3600 Sec
= 4800600 J

Question 3.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 sec.
Answer:
Given I = 5 A
R = 20 W
t = 30 S
H = ?
H = I² R t
= (5 A)² × 20 Ω × 30 Sec
= 25 A² × 20 Ω × 30Sec
= 15000 J

In-text Questions (Page 220)

Question 1.
What determines the rate at energy is delivered by a current ?
Answer:
It is electric power (p) given by the formula,
P = V I = I²R = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Given I = 5 A
V = 220 V
since P = V × I
= 220 V × 5 A
= 1100 V – A
P = 1100 W
Energy consumed in ‘2’ hours
= V × I × t
= 220 volt × 5 A × 2 × 60 × 60 sec
= 7920000 J

Class 10 Science Chapter 12 Electricity Textbook Questions and Answers

Page no. 221

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 ration R/R’ is
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electric power in a circuit ?
(a) I²R
(b) IR²
(c) V I
(d) \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
Answer:
(b) IR²

Question 3.
An electric bulb is rated 230 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 an electric circuit. 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:
A voltmeter is connected in the circuit in parallel 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 W ? How much does the resistance change if the diameter is doubled ?
Answer:
Given
Diameter = 0.5 mm
so radius (r) = \(\frac {0.5}{2}\) mm
= \(\frac{0.5}{2 \times 10^{3}}\) metre
Resistivity (ρ) = 1.6 × 10 10^-8 Ωm
R = 10 W
A = πr²
l = ?
We know that


Second Case

Question 7.
The values 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 the resistor.
Answer:
Resistance (R) = tan Q = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)

Question 8.
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.
Answer:
By Ohm’s Law

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.5 Ω, and 12 Ω respectively. How much current would flow through the 12 Ω resistor ?
Answer:
Let the resultant resistor be R.
So, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
R = 13.4 Ω
By Ohm’s Law
V = I × R
I = \(\frac{V}{R}=\frac{9 V}{13.4} \mathrm{Ohm}\)
= 0.671 A.
In series current remain same in all the resistors. So current in 12 Ω resistor will be 0.671 Amp.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line ?
Answer:
By Ohm’s Law
V= I × R
R = \(\frac{\mathrm{V}}{\mathrm{I}}=\frac{220 \mathrm{Volt}}{5 \text { Ampere }}\)
= 44 Ohm
Let ‘n’ resistors are connected in parallel.

So the no. of resistors will be ‘4’.

Question 11.
Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (b) 4 Ω.
Answer:
(i)

Two resistance in parallel and their resultant should be in series with the third resistance.

Two resistance in series and their resultant should be parallel with the third resistance.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 Ω. 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 ?
Answer:
The resistance of an electric bulb can be calculated as follows :
P = \(\frac{\mathrm{V}_{2}}{\mathrm{R}}\)
R = \(\frac{220 \times 220}{10}\) = 22 × 220
= 4840 Ohm.
If the allowable current is 5 A and voltage is 220 V than resistance
R¹ = \(\frac{V}{I}=\frac{220 \mathrm{~V}}{5 \mathrm{Amp}}\)
= 44 Ohm.
Let the number of lamps are ‘n’, so

So the number of lamps will be 110.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series or in parallel. What are the currents in the three cases ?
Answer:
Case I when the coils are in series,
Let resultant resistance be R, so
R = 24 Ω + 24 Ω = 48 Ω
Now, I = \(\frac{\mathrm{V}}{\mathrm{R}}=\frac{220 \mathrm{~V}}{48 \Omega}\) = 4.58 A
Case when the coil ‘A’ and ‘B’ are in parallel, let the resultant resistance be ‘R’

Question 14.
Compare the power used in the 2 Ω resistor in each of the following circuit (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
(i) The resultant resistance (R)
= 1 Ω + 2 Ω = 3 Ω
= (R₁)(R₂)

(ii) When a battery in parallel with 12 Ω and 2 Ω resistors, then voltage remain same in both the resistance. Current will be different.
So power (P) = \(\frac{V^{2}}{R}=\frac{(4)^{2}}{2}\)
= \(\frac {16}{2}\) = 8 J
∴ \(\frac{P}{P^{1}}=\frac{8 J}{8 J}\)
P : P¹ = 1 : 1

Question 15.
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 ?
Answer:
Resistance of bulb rated 100 W at 220 say R1

for the bulb rated 60 W at 220 V, say resistance is R₂
So, R₂ = \(\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{220 \times 220}{60}\)
= 806.6 Ohm
When R₁ and R₂ are in series, the resultant resistance say R, so

Question 16.
Which uses more energy, a 250 W.T.V. set in 1 hr, or a 1200 W toaster in 10 minutes ?
Answer:
Energy used by T.V. in one second = 250 J
So energy used by T.V. in 1 hr i.e., 3600 sec
= 250 × 3600 J = 900000 J
Energy used by a toaster in one second = 1200 J
So energy used by a toaster in 10 minutes i.e., 600 sec
= 1200 × 600 J = 720000 J
Hence T.V. will used more energy than a toaster.

Question 17.
An electric heater of resistance 8 W draws 15 A from the service mains 2 hours. Calculate the rate of which heat is developed in the heater.
Answer:
The rate at which heat is developed is called power.
So P = V × I
= \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
Since V = I × R
So, = I² × R
= (15A)² × 8 Ohm
= 225 A² × 8 Ohm
= 1800 J/Sec

Question 18.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps ?
(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 ?
(c) Why is the series arrangement not used for domestic circuits.
(d) How does the resistance of a wire vary with its area on cross-section ?
(e) Why are copper and aluminium wires usually employed for electricity transmission ?
Answer:
(a) The resistivity of tungsten does not change with high temperature. It has very high melting point. It can operate upto 2700° C. So tungsten is almost used exclusively for filament of electricity lamps.
(b) Because alloys do not oxidise on heating at high temperature.
(c) Because when light are in series if one bulb fuses or one tube light stops working, the circuit breaks and all light go off.
(d) The resistance of a given wire is inversely proportional to area of cross-section of the wire.
(e) Because copper and aluminium have low resistivity that is why they are good conductors of electricity so they are usually employed for electricity transmission.

Class 10 Science Chapter 12 Electricity Textbook Activities

Activity 12.1 (Page 203)

• Set up a circuit as shown in Fig,, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)

• First use only one cell as the source in the circuit. Note the reading in the ammeter 1, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.
• Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
• Repeat the above steps using three cells and then four cells in the circuit separately.

Question 1.
Calculate the ratio of V to I for each pair of potential difference V and current J.
Answer:

Question 2.
Plot a graph between V and I, and observe the nature of the graph.
Answer:
In this Activity, you will find that approximately the same value for V/I is obtained in each case. Thus the V-I graph is a straight line that passes through the origin of the graph, as shown in Fig. Thus, V/I is a constant ratio.

Activity 12.1 (Page 205)

• Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.
• Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig.

• Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit. ]
• Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
• Now repeat the above step with the 10 W bulb in the gap XY.
• Are the ammeter readings differ for different components connected in the gap XY? What do the above observations indicate?
• You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

Question 1.
In this Activity we observe that the current is different for different components. Why do they differ?
Answer:
Certain components offer an easy path for the flow of electric current while the others resist the flow. We know that motion of electrons in an electric circuit constitutes an electric current. The electrons, however, are not completely free to move within a conductor. They are restrained by the attraction of the atoms among which they move. Thus, motion of electrons through a conductor is retarded by its resistance. A component of a given size that offers a low resistance is a good conductor. A conductor having some appreciable resistance is called a resistor. A component of identical size that offers a higher resistance is a poor conductor. An insulator of the same size offers even higher resistance.

Activity 12.3 (Page 206)

• Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length 1 [say, marked (1)] and a plug key as shown in Fig.
• Now, plug the key. Note the current in the ammeter.
• Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 21 [marked (2) in the Fig.].
• Note the ammeter reading.
• Now replace the wire by a thicker nichrome wire, of the same length 1 [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.

Fig. : Electric circuit to study the factors on which the resistance of conducting wires depends

• Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig.] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)] Note the value of the current.
• Notice the difference in the current in all cases.
• Does the current depend on the length of the conductor?
• Does the current depend on the area of cross-section of the wire used?

It is observed that the ammeter reading decreases to one-half when the length of the wire is doubled. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit. A change in ammeter reading is observed when a wire of different material of the same length and the same area of cross-section is used. On applying Ohm’s law, we observe that the resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material. Precise measurements have shown that resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A).

Activity 12.4 (Page 210)

• Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Fig. You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω etc., and a battery of 6 V for performing this Activity.
• Plug the key. Note the ammeter reading.
• Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.

Question 1.
Do you find any change in the value of current through the ammeter?
Answer:
Observation : We observe that the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in a series combination of resistors the current is the same in every part of the circuit or the same current through each resistor.

Activity 12.5 (Page 211)

• In Activity 12.4, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Fig. 12.8 of text-book.

• Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Fig.
• Plug the key and measure the potential difference across the first resistor. Let it be V₁.
• Similarly, measure the potential difference across the other two resistors, separately. Let these values be V₂ and V₃ respectively.
• Deduce a relationship between V, V₁, V₂ and V₃.

We observe that the potential difference V is equal to the sum of potential differences V₁, V₂, and V₃. That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is,
V = V₁ + V₂ + V₃

Activity 12.6 (Page 213)

• Make a parallel combination, X’. of three resistors having resistances R₁, R₂, and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig. Also connect a voltmeter in parallel with the combination of resistors.
• Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig.).

• Take out the plug from the key: Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R₁, as shown in Fig. Note the ammeter reading, I₁.

• Similarly, measure the currents through R₂ and R₃. Let these be I₂ and I₃, respectively.

Question 1.
What is the relationship between I, I₁, I₂ and I₃?
Answer:
Observation : It is observed that the total current I, is equal to the sum of the separate currents through each branch of the combination.
I = I₁ +I₂ + I₃

Class 10 Science Chapter 12 Electricity Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Why resistance becomes more in series combination ?
Answer:
Because resistance is directly proportional to the length in series.

Question 2.
Why resistance becomes less in parallel combination ?
Answer:
Because resistance is inversely proportional to the length in parallel.

Question 3.
What is the difference between kilowatt and kilo hour ?
Answer:
Kilowatt is the unit of electric power whereas kilowatt hour is the unit of electric power.

Question 4.
What is the main property of charge ?
Answer:
Unlike charges attract each other while like charges repel each other.

Question 5.
What are the special property of a heating wire ?
Answer:
High resistance and high melting point.

Short Answer Type Questions

Question 1.
What is the resistance of an electric arc lamp, if the lamp uses 20 A when connected to a 220 Volt line ?
Answer:
Here, V = 220 V
I = 20 A
Since R = \(\frac{\mathrm{V}}{\mathrm{I}}\)
Or R = \(\frac{220}{22}\) = 11 Ohm

Question 2.
If you connect three resistors having values 2Ω, 3Q and 5 Q in parallel, then will the value of the total resistance be less than 2 Ω, or greater than 5 Ω, or lie between 2 Ω and 5 Ω ? Explain.
Answer:
Since the resistors are connected in parallel.

This show that the resistance is less than 2.

Long Answer Type Questions

Question 1.
How much work is done during flow of current for a given time in a wire ?
Answer:
As the conductor offers resistance to the flow of current, work is done continuously to move the electrons or maintain the flow of current.

The work done in carrying charge 0 through potential difference is given by
W = Q × V
Q = I × t
W = I × t × V
According to the Ohm’s law,
V = 1R
On putting the value of V in equation (i), we get
W = I²Rt.
Since, energy is the capacity or ability to do work, therefore we can also say that H [electrical energy] = I²Rt.

Multiple Choice Questions

Question 1.
The element of an electric heater is made by:
(a) Nichrome
(b) Aluminium
(c) Tungsten
(d) Platinium
Answer:
(a) Nichrome

Question 2.
In V-I, graph the slope of the line gives:
(a) Current
(b) Potential difference
(c) Resistance
(d) Reciprocal of potential difference
Answer:
(c) Resistance

Question 3.
The resistance of a straight conductor does not depend on its:
(a) Resistivity
(b) Materials
(c) Length
(d) Shape of cross section
Answer:
(d) Shape of cross section

Question 4.
The unit of electric charge is :
(a) Coulomb
(b) Volt
(c) Ohm
(d) Ampere
Answer:
(a) Coulomb

Question 5.
The unit of resistance of:
(a) Ohm
(b) Volt
(c) Ampere
(d) Coulomb
Answer:
(a) Ohm

These NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
We know that

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
We know that

Question 3.
Evaluate:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
(ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
(i) We know that

(ii) We know that
sin 25° cos 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25° = 1 (sin² θ + cos² θ)

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ ) =
(A) 0
(B) 1
(C) 2
(D) -1

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)
(A) sec² A
(B) -1
(C) cot² A
(D) tan² A
Solution:
(i) We know that
9 sec² A – 9 tan² A = 9(sec² A – tan² A)
= 9 x 1 = 9
Correct option is (B)

(ii) We know that
(1 + tan θ + sec θ) (1 + cot θ – cosec θ )

So, the correct option is (C)

(iii) We have given,

So, the correct option is (D)

(iv) We have given,

So, the correct option is (D)

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:

(vi) L.H.S
\(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Multiply both numerator and denominator by \(\sqrt{1+\sin A}\)

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A.
Solution:
L.H.S.
(sin A + cosec A)² (cos A + sec A)²
= (sin² A + cosec² A + 2sin A + cosec A) + (cos²A + sec² A + 2cos. A sec A)
= sin² A + cosec² A + ²) + (cos² A + sec² A + 2) [∵sin A cosec A = 1 and cos A sec A = 1]
= (sin² A + cos² A) + 4 + sec² A + cosec² A
= 5 (1 + tan² A) + 1 + cot² A) [∵ sec² A = (1 + tan² A) and cosec² A = (1 + cot² A)]
= 7 tan² A + cot² A = R.H.S.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
[Hint: Simplify LHS and RHS separately]
Solution:
L.H.S.