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We are offering TS 10th Class Maths Notes Chapter 13 Probability to learn maths more effectively.

TS 10th Class Maths Notes Chapter 13 Probability

→ Probability :
Probability means the number of occasions that a particular statement (Or) event is likely to occur in a large population of events.

→ Random experiment:
Random experiments (or) tail is an act (or) process that leads to a result, that cannot be predicted in advance.
(or)
An experiment is said to be a random experiment if its outcome cannot be predicted. That is the outcome of an experiment does not obey any rule.

Ex:
(i) If a coin is tossed, we can’t say whether head or tail will appear. So tossing a coin is a random
experiment.
(ii) If we throw a die numbered 1,2,3,4,5,6 on its six faces, it is impossible to say which numbered face will appear. There is no rule to know it. So throwing a die is a random experiment.

→ Equally likely events :
Two or more events are said to be equally likely if each one of them has an equal chance of occurrence.

For example

• When a coin is tossed, the two possible outcomes, head and tail, are equally likely.
• When a die is thrown, the six possible outcomes, 1, 2, 3,4, 5 and 6 are equally likely.

→ Mutually exclusive events :
Two (or) more events are mutually exclusive if the occurrance of each event prevents the every other event.

→ Complementary events :
Consider an event has few outcomes. Event of all other outcomes in the sample survey which are not in the favourable event is called complementary event.

→ Exhaustive events :
All the events are exhaustive if their union is the sample space.

→ Sure events:
The sample space of a random experiments is called sure(or) certain event as any one of its elements will surely occur in any trail of experiment.

→ Impossible event:
An event which will not occur on any account is called an impossible event.

→ Theoretical event:
The theoretical (classical) probability of an event E, written as P(E), is defined as No. of trails in which the events happened
P(E) = \(\frac{\text { No. of trails in which the events happened }}{\text { Total number of trials }}\)
Where we assume that the outcomes of the experiment are equally likely.

→ Elementary event: An event having only one outcome is called an elementary event.
The sum of the probabilities of all the elementary events of an experiment is 1.
The probability of a sure (or certain event) is 1.
The probability of an impossible event is 0.
The probability of an event E is a number P(E) such that 0 < P(E) < 1.
For an event E, P(E) + P(Ē) = 1 where E stands for ‘not E’.
E and Ē are called complementary events. In general, it is true that for an event E,
P(Ē) = 1 – P(E)

→ Sample space : The set of all possible outcomes of an experiment is called a sample space (or) probability space.
Suppose we throw a die once. As it has six faces, every face has equal chance to appear.
Therefore, sample space (s) = {1, 2, 3, 4, 5, 6}
Number of events n(s) = 6
If a coin is tossed, either head (or) tail may appear.
Hence, sample space (s) = {H, T}
Number of events n (s) = 2

→ A cubic dice is a six faced cube, the six faces are marked as 1, 2, 3,4, 5, 6. When such a dice is thrown, any one of the faces come upwards. The number on this face is the outcome of the experiment.

→ Remember the following points :
(a) There are 52 cards in a pack of cards.
(b) Out of these, 26 are red and 26 are black.
(c) Out of 26 red cards, 13 are hearts and 13 are diamonds.
(d) Out of 26 black cards, 13 are spades and 13 are clubs.
(e) Each of four varities (hearts, diamonds, spades, clubs) has an ace (i.e.,) a pack of 52 cards has 4 aces.
Similarly these are 4 kings, 4 queens and 4 jacks.

Flow Chat:

Pierre Simon Laplace:

• The definition of probability was given by Pierre Simon Laplace in 1795.
• Probability theory had its origin in the 16th Century, when an Italian physician and mathematician J.Cardan wrote the first book on the subject, “The Book on Games of Chance”.
• James Bernoulli (1654 – 1705), A. De Moivre(1667-1754) and Pierre Simon Laplace (1749-1827) are among those who made significant contributions to this field.
• In recent years, probability has been used extensively in many areas such as Biology, Economics, Genetics, Physics, Sociology etc.

Telangana SCERT 10th Class Physics Study Material Telangana 10th Lesson Electromagnetism Textbook Questions and Answers.

TS 10th Class Physical Science Solutions 10th Lesson Electromagnetism

Improve Your Learning
I. Reflections on concepts

Question 1.
Are the magnetic field lines closed? Explain.
Answer:
1. The field lines appear to be closed loops but you can’t conclude that lines are closed or open loops by looking at the picture of field lines because we do not know about alignment of lines that are passing through the bar magnet.

2. The direction of a magnetic line at any point gives the direction of the magnetic force on a north pole placed at that point.

3. Since the direction of magnetic field line is the direction of force on a north pole, the magnetic lines always begin from the ‘N’ pole of the magnet and end on the ‘S’ pole of the magnet.

4. Inside the magnet however the direction of magnetic lines is from S-pole of the magnet to the N-pole of the magnet.

5. Thus the magnetic field lines are closed.

Question 2.
See figure, magnetIc lines are shown. What is the direction of the current flowing through the wire?

Answer:
1) The magnetic field lines are in the anti-clockwise direction.
2) The direction of current is vertically upwards. This can be demonstrated with right-hand thumb rule.

Question 3.
A bar magnet with North pole facing towards a coil moves as shown In figure given below. What happens to the magnetic flux passing through the coil?

Answer:
The magnetic flux passing through the coil induces current in the coil. This current is called induced EMF. This induced N EMF is equal to the rate of change of magnetic flux passing through it.

Question 4.
A coil is kept perpendicular to the page. At P, current flows into the page and at Q it comes out of the shown in figure. What is the page as direction of magnetic field due to the coil?

Answer:
To know the direction of magnetic field, we use right-hand rule i. e., ‘when you curl your right-hand fingers ¡n the direction of current, thumb gives the direction of magnetic field’. According to this the direction of magnetic field is as shown in the following figure.

Application of Concepts

Question 1.
The direction of current flowing in a coil Is shown in figure. What type of magnetic pole Is formed at the face that has flow of current as shown in fig.

Answer:
The top surface of the coil shown in the fig. behaves as North pole and direction of magnetic field points towards us.

Question 2.
Why does the picture appear distorted when a bar magnet is brought close to the screen of a television? Explain.
Answer:
Picture on a television screen Is due to motion of the electrons reaching the screen. These electrons are affected by magnetic field of bar magnet. This is due to the fact that the magnetic field exerts a force on the moving charges. This force is called magnetic force. Due to this magnetic force, the picture is distorted when you remove the bar magnet away from the screen, the motion of electron is not affected by the magnetic force and the picture will be normal.

Question 3.
Symbol ‘X’ indicates the direction of a magnetic field into the page. A straight long wire carrying current along its length is kept perpendicular to the magnetic field. What A is the magnitude of force experienced by the wire? In what direction does it act?

Answer:
From the figure, a straight wire carrying current which Is kept perpendicular to a uniform magnetic field B. This ‘B’ is directed Into the page. Let the field be confined to the length L. We know that the electric current means charges in motion. Hence they move with a certain velocity called drift velocity V.

The magnetic force on a single charge is given by F₀ = qv B.
Let total charge inside the magnetic field be Q. So magnetic force on the current carrying wire s given by F=QvB ……………………………. (1)
The time taken by the charge (Q) to cross the field be
t = \(\frac{L}{v} \Rightarrow v=\frac{L}{t} \) ……………………………. (2)
∴ (1) ⇒ F = \(\frac{\mathrm{QLB}}{\mathrm{t}} \Rightarrow \frac{\mathrm{Q}}{\mathrm{t}}(\mathrm{LB}) \) …………………………. (3)
We know that \( \frac{\mathrm{Q}}{\mathrm{t}}\) Is equal to the electhc current in the wire.
I = \(\frac{Q}{t} \)
∴ (3) ⇒ FILB.

The direction of the force: The direction of force can be find by using right-hand rule. Fore finger points towards the velocity of current, middle finger points to the direction of magnetic field (B), then the thumb gives direction of force when the three fingers are stretched in such away that they are perpendicular to each other.

Question 4.
An 8N force acts on a rectangular conductor 20 cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40 A.
Answer:
Formula, F = ILB (or) B = F/IL
Where F = magnetic force = 8N
I = Electrical current in the conductor = 40 A
L, length of conductor = 20 cm. = \(\frac{20}{100}\) = 0.2 mt
∴ B =?
B= \(\frac{8}{40 \times 0.2}=\frac{80}{40 \times 2} \) = 1
Magnetic field induction = 1 tesla.

Question 5.
As shown in figure both coil and bar magnet move in the same direction. Your friend is arguing that there is no change in flux. Do you agree with his statement’ If not what doubts do you have? Frame questions about the doubts you have regarding change in flux.

Answer:
Agree: Yes. I will agree.

1. The induced EMF will not produce when the coil and magnet are moving in the same direction with same velocity.
2. Hence my friend’s argument is correct.

Disagree:

1. What happens if both magnet and coil move in same direction?
2. What happens if both magnet and coil move in opposite direction?
3. What is the direction of the current in the coil?
4. If both move in same direction, is there any linkage of flux with the coil?
5. When N pole is moved towards the coil what is the direction of current?
6. If magnet is reversed, what is the direction of current in the coil?

Question 6.
Give a few applications of Faraday’s law of induction in daily life.
Answer:
The daily life applications of Faraday’s law of induction are :

1. Generation of electricity.
2. Transmission of electricity.
3. Metal detectors in security checking.
4. The tape recorder.
5. Use of ATM cards
6. Induction stoves
7. Transformers
8. Induction coils (spark plugs in automobiles)
9. Break system in railway wheels
10. AC and DC generators
11. Windmills etc.

Question 7.
Explain the working of an electric motor with a neat diagram.
Answer:
1. Consider a rectangular coil kept in uniform magnetic field as shown in figure.
2. Switch on the circuit so that the current flows through the coil. The direction of current is shown in the figure.
3. The sides AB and CD of the coil are always at rigi angles to the magnetic field.
4. According to right-hand rule, at AB the magnetic force acts inward perpendicular to the field of magnet ai
on CD, it acts outward.
5. The top view of coil is shown in the figure.
6. The force on the sides BC and DA varies because they make different angles at different positions of the coil in the field. At BC, magnetic force pulls the coil up and at DA magnetic force pulls it down.
7. The net force acting on AB and on CD is zero because they carry equal currents in the opposite direction.
Similarly, the sum of the forces on sides BC and DA is also zero. So, net force is zero on the coil.
8. But the rectangular coil comes into rotation in clockwise direction because equal and opposite pair of forces acting on the twos sides of the coil.


9. If the direction of current in the coil is unchanged, it rotates, upto half rotation In one direction and the next half In the direction opposite to previous like to and fr0 motion.
10. It the direction of current in the coil is changed the coil will rotate continuously in one and the same direction.
11. To achieve this, brushes B, and B₂ are used.
12. These brushes are connected to the battery. The ends of the coil are connected to slip rings C₁ and C₂ which rotate along with the coil.
13. Initially C₁ is In contact with B₁ and C₂ Is in contact with B₂
14. After half rotation, the brushes come into contact with the other slip rings in such a way that the direction of current through the coil Is reversed. This happens every half-rotation.
15. Thus the direction of rotation of the coil remains the same. This is the principle used in electric motor.
16. In electric motors electrical energy is converted Into mechanical energy.

Question 8.
Explain the working of AC electric generator with a neat diagram.
Answer:
(i) Electric Generator or Dynamo: It is a device which converts mechanical energy into electrical energy.
(ii) Principle: It works on the principle of the electromagnetic induction.
(iii) Construction:
It consists of
(i) Armature coil,
(ii) Brushes,
(iii) Slip rings,
(iv) Strong magnet and
(v) Rotating mechanism (or) motor.
The two A and B of the coil ABCD are connected to the slip rings.

Working:

• When the coil is at rest ¡n vertical position, with side (A) of coil at top position and side (B) at bottom position, no current will be induced in it. Thus current in the coil is zero at this position.
• When the coil is rotated in clockwise direction, current will be induced in it and it flows from A to B, in this position the current increases from zero to a maximum.
• If we continue the rotation of coil current decreases during the second quarter of the rotation and once again becomes zero when coil comes to vertical position with side B at top side and side A at bottom position.
• During the second half of the rotation, current generated follows the same patterns as that in the first half, except that the direction of current is reversed.
• Thus, after every rotation of the current ¡n the respective arm changes, there by generating an alternating current. This device is called A.C. generator.

Question 9.
Explain the working of DC generator with a neat diagram.
Answer:

1. Consider a rectangular coil. Let it be held between the poles of curve-shaped permanent magnets as shown in figure.
2. As the coil rotates the magnetic flux passing through the coil changes.
3. According to the law of electromagnetic induction, an induced current is generated in the coil.
4. If two half-slip rings are connected to the ends of the coil as shown in figure, this generator works as DC generator to produce DC current.

Working:

1. When the coil is in vertical position the induced current generated during the first half rotation, rises from zero to maximum and then falls to zero again.
   

2. As the coil moves further from this position, the ends of the coil go to other slip rings.

3. Hence, during the second half rotation, the current is reversed in the coil itself, the current generated in the second half rotation of the coil is identical with that during the first half of the direct current (DC), for one revolution.

4. Hence, this current ¡s called direct current (DC).

Question 10.
How do you appreciate Faraday’s law, which s the consequence of conservation of energy?
Answer:

1. Faraday’s law of electromagnetic induction, which is the outcome of law of conservation of energy, is very much useful.
2. The law led to the conclusion that mechanical energy can be converted into electrical energy.
3. The fruits of electrical energy, namely, the electric lights, the stoves, electric heaters, electrical irons, the fridge, the electric motor the television, the electronic locomotives and all electric gadgets are serving mankind and making people more comfortable.
4. All these gadgets work on electricity and electrical energy that can be produced by generators on large scale where energy Is created by law of conservation of energy on a very large scale.
5. So, we must appreciate the utilïtanan value of Faraday’s law which is the consequence of conservation of energy.

Question 11.
The value of magnetic field Induction which is uniform is 2T. What is the flux passing through a surface of area 1.5m1 perpendicular to the field?
Answer:
Magnetic Induction of uniform field B = 2T
Flux Φ =?
Surface area A = 1.5 m²
We know B = \(\frac{\varphi}{A} \) ⇒ BA cosθ
Φ= BA
Where B ¡s perpendicular to area is
ΔΦ = 2x 1.5=3webers

Question 12.
Which of the various methods of current generation protects nature well? Give examples to support your answer. )
Answer:
The methods of current generation that protect the nature well:
1. Hydra power plant A power plant that produces electricity by using flowing of water to rotate a turbine Is called hydropower plant.

Advantages:

• It does not produce any environmental pollution.
• It will never get exhausted.
• The construction of dams on rivers helps In controlling floods and in irrigation.

2. Wind generators: A wind generator s used to generate electricity by using wind energy.

Advantages:

• It does not cause any pollution.
• It will never get exhausted.
• Wind energy is available free of cost.

3. Solar cell: Solar cell is a device which converts solar energy directly into electricity.

Advantages:

1. They require no maintenance.
2. They can be set up in remote inaccessible and very sparsely inhabited areas where the laying of usual power transmission is different and expensive.
3. As seawater flows in and out of the tidal barrage during high and low tides It turns the turbines to generate electricity.
4. The energy of moving sea waves can be used to generate electricity.
5. The energy available due to the difference In the temperature of water at the surface of the ocean and at deeper levels is called ocean thermal energy. The ocean thermal energy can be converted into a usable form of energy like electricity.
6. Geothermal energy is the heat energy from hot rocks present inside the earth. It is also used to produce electricity. It is a clean and environmental friendly source of energy.
7. The energy produced during nuclear fission reactions is used for generating electricity at nuclear power plants.

Multiple choice questions

Question 1.
Which of the following converts electrical energy into mechanical energy [ ]
(a) motor
(b) battery
(c) generator
(d) switch
Answer:
(a) motor

Question 2.
Which of the following converts mechanical energy into electrical energy [ ]
(a) motor
(b) battery
(c) generator
(d) switch
Answer:
(c) generator

Question 3.
The magnetic force on a current-carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field is [ ]
(a) 0
(b) ILB
(c) 2ILB
(d) \(\frac{\text { ILB }}{2} \)
Answer:
(b) ILB

Question 4.
One Tesla = [ ]
(a) Newion/Coloumb
(b) Newton / ampere – meter
(c) Ampere I meter
(d) Newton / ampere second
Answer:
(b) Newton/ampere – meter

Question 5.
Magnetic flux [ ]
(a) dyne
(b) Oersted
(c) Guass
(d) Weber
Answer:
(d) Weber

Question 6.
No force works on the conductor carrying electric current when kept [ ]
(a) parallel to magnetic field
(b) perpendicular to magnetic field
(c) in the magnetic field
(d) away from magnetic field
Answer:
(a) parallel to magnetic field

Suggested Experiments

Question 1.
Explain with the help of two activities that current-carrying wire produces magnetic field.
Answer:
Activity – 1.

1. Take a wooden plank and make a hole as shown in adjacent figure. Table
2. Place the plank on the table.
3. Place a retort stand on the plank as shown in figure.
4. Pass 24 gauge copper wire through a hole of the plank and rubber knob of the retort stand In such a way that the wire be arranged In a vertical position and not touch the stand.
5. Connect the two ends of the wire to a battery via switch.
6. Place 6 to lo compass needles In a circular path around the hole so that its centre coincides with the hole. Use 3-volt battery In the circuit.
7. Switch on current.
8. We see the compass needles change the direction In such a way as tangents to the circle.
9. This activity helps us to prove current carrying wire produces magnetic field.

Activity -2
1. Take a thin wooden plank covered with white paper.
2. Make equidistant holes on its surface as shown in adjacent figure.
3. Pass copper wire through the holes. This forms a coil. Join the ends of the coil to a battery through a switch.
4. Switch on the circuit. Current passes through the coil.
5. Now sprinkle iron filings on the surface of the plank around the coil.
6. Give small jerk to It. An orderly pattern of iron filing is seen on the paper.
7. This activity proves that a current-carrying wire develops a magnetic field around it.

Question 2.
How do you verify experimentally that the current-carrying conductor experiences a force when it is kept in a magnetic field?
(Or)
List out the apparatus and experimental procedure for the experiment to observe a current-carrying wire experiences a magnetic force when it is kept in uniform magnetic field.
Answer:
Required apparatus:

• Horse-shoe magnet
• conducting wire
• battery, switch

1. Take a wooden plank.
2. Fix two long wooden sticks on it.
3. These wooden sticks are split at their top ends.
4. A copper wire is passed through splêts of wooden sticks.
5. Connect the wire to 3 volts battery.
6. Close the switch to make the circuit. Current passes through the wire.
7. Bring a horseshoe magnet near the wire.
8. Then a force is experienced on the wire, following right thumb rule.
9. Reverse the polarities of the magnet, then the direction of the force is also reversed.
10. The night hand rule helps the direction of magnetic force exerted by the magnetic field on the current carring wire.

Question 3.
Explain Faraday’s law of induction with the help of activity.
Answer:

1. Connect the terminals of a coil to a sensitive ammeter or galvanometer as shown in the adjacent figure.
2. Now push a bar magnet towards the coil, with its north pole facing the coil.
3. While the magnet is moving towards the coil the needle in galvanometer deflects, showing that a current has been set up in the coil.
4. The needle of galvanometer does not deflect If the magnet is at rest.
5. If the magnet Is moved away from the coil, the needle In the galvanometer again deflects, but ¡n the opposite direction.
6. It means that a current is set up in the coil in the opposite direction.
7. If we use the end of south pole of a magnet instead of north pole in this activity, the experiment works just as described but the reactions are exactly in opposite directions.
8. Further experimentation enables us to understand that the relative motion of the magnet and coil sets up a current in the coil. It makes no difference whether the magnet is moved towards the coil or the coil towards, the magnet.
9. This proves the Faraday’s law: “Whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated In the coil.

Question 4.
What experiment do you suggest to understand Faraday’s law? What Instruments are required? What suggestions do you give to get good results of the experiment? Give precautions also.
Answer:
Aim: To understand Faraday’s law of induction. Materials required: A coil of copper wire, a bar magnet Galvanometer etc.

Procedure:

1. Connect the terminals of a coil to a sensitive galvanometer as shown In the figure.
2. Normally we would not expect any deflections of needle in the galvanometer because there is to be no electromotive force In this circuit.
3. Now if we push a bar magnet towards the coil, with its north pole facing the coil, we observe the needle in the galvanometer deflects, showing that a current Is set up in the coil.
4. The galvanometer does not deflect if the magnet Is at rest.
5. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil, in the opposite direction.
6. If we use end of south pole of a magnet instead of north pole In the above activity, the deflections are exactly reversed.
7. This experiment proves “whenever there is a continuous change of magnetic flux linked with a closed coil, a current Is generated in the coil.

Precautions:

• The coil should be kept on an insulating surface.
• Bar magnet should be of good pole strengths.
• The centre of the Galvanometer scale must be zero.
• The deflections in the galvanometer must be observed while Introducing the bar magnet Into the coil and also while withdrawing It.

Question 5.
How can you verify that a current-carrying wire produces a magnetic field with the help of an experiment?
Answer:
Aim: To demonstrate that a magnetic field is produced around a current-carrying wire.
Apparatus: Copper wire, slits, 3volt battery, key, and connecting wires, thermocol sheets.

Procedure:

• Take a hermocoI sheet and fix two wooden slits of 1 cm at the top of their ends.
• Arrange a copper wire which passes through slits and make a circuit with the elements as shown in the figure.
• Now, keep a magnetic compass below the wire.
• Bring a barmagnet dose to the compass.

Suggested Projects

Question 1.
Collect Information about generation of current by using Faraday’s law.
Answer:
Faraday’s law is useful In generation of current.

1. According to this law, the change in magnetic flux Induces EMF in the coil.
2. He also proposed electromagnetic induction.
3. Electromagnetic induction is a base for generator, which produces electric Current.
4. Transformer also works on the principle of electromagnetic induction, which is helpful In transmission of electricity.
5. Hence Faraday’s law is used in the generation and transmission of current.

Question 2.
Collect information about material required and procedure of making a simple electric motor from Internet and make a simple motor on your Own.
Answer:
Material required:

1. A plastic pipe of 3 or 4 cm. diameter and 5 cm long.
2. A copper wire 30 cm long.
3. A horseshoe magnet.
4. Two corks that exactly fit Into the bore of the plastic pipe.
5. A wooden plank used as base for the whole arrangement.
6. Cello tape and small nails.

Procedure:

1. Wind the copper wire around the plastic pipe.
2. Connect the ends of the wire to the two brushes.
3. Fit in the corks into the hollow of the pipe such that they close either end of the pipe.
4. By means of vertical pieces of wood and the nails arrange the plastic pipe such that it can rotate freely on the pieces of wood. This acts as the armature.
5. Arrange the horseshoe magnet on the wooden base such that the armature lies between both of its poles.
6. Now connect the ends of the copper wire to a sensitive galvanometer.
7. Rotate the armature freely.
8. You can observe reflection in the galvanometer.
9. Thus it is an Improvised electric motor.

Question 3.
Collect information of experiments done by Faraday.
Answer:
Experiment – 1

1. Connect the terminals of a coil to a sensitive galvanometer as shown n the figure.
2. Normally, we would not expect any deflection of needle In the galvanometer because there Is no EMF In the N – circuIt.
3. Now, If we push a bar magnet towards the coil, with N Its north pole facing the coil, the needle In the galvanometer deflects, showing that a current has been set up in the coil, the galvanometer does not deflect if the magnet is at rest.
4. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction,
5. If we use the end of south pole of a magnet Instead of north pole, the results i.e., the deflections in galvanometer are exactly opposite to the previous one.
6. This activity proves that the change in magnetic flux linked with a closed coil, produces current.
7. From this Faradays law of induction can be stated as ‘whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil” This induced EMF is equal to the rate of change of magnetic flux passing through it.

Experiment – 2
1. Prepare a coil of copper wire C₁ and connect the two ends of the coil to a galvanometer.
2. Prepare another coil of copper wire C₂ similar to C₁ and connect the two ends of the coil to a battery via switch.
3. Now arrange the two coils C₁ and C₂ nearby as shown in the figure.
4. Now switch on the coil C₂ We observe a deflection In the galvanometer connected to the coil C₁.
5. The steady current in C₂ produces steady magnetic field. If coil C₂ is moved towards the coil C₁, galvanometer shows a deflection.
6. This indicates that electric current Is induced in coil C₁.

7. When C₂ is moved away, the galvanometer shows a deflection again, but this time in the opposite direction.
8. The deflection lasts as long as coil C₂ in motion.
9. When C₂ is fixed and C₁ is moved, the same effects are observed.
10. This shows the induced EMF due to relative motion between two coils.

TS 10th Class Physical Science Electromagnetism Intext Questions

Page 209

Question 1.
How do they (most of the appliances) work?
Answer:
The electronic motor, generators, electric cranes, rice cookers, electrical irons, washing machines, grinders, fans, and many more appliances work on electricity.

Question 2.
How do electromagnets work?
Answer:
By passing D.C. current through a conductor wound around a core, a magnetic field Is developed around the current-carrying conductor and work as a magnet.

Question 3.
Is there any relation between electricity and magnetism?
Answer:
Yes. A current-carrying conductor develops a magnetic field around it.

Question 4.
Can we produce magnetism from electricity?
Answer:
Yes. By passing electric current (D.C) through an insulated wire wound around an iron rod we can produce magnetism by electricity.

Page 210

Question 5.
Does the needle get deflected by the bar magnet?
Answer:
Yes, the needle gets deflected.

Question 6.
why does the needle get deflected by the magnet?
Answer:
When the N-pole of compass needle and N-pole of magnet come close to each other the needle gets deflected due to repulsion. (or) When the S-pole of compass needle and S-pole of magnet come close to each other (S-pole and S-pole are like poles they repel each other) and the needle gets deflected.

Question 7.
What do you notice?
Answer:
The needle is deflected, as there is magnetic field around the wire.

Question 8.
Is there any change in the position of the compass needle?
Answer:
Yes.

Question 9.
Which force Is responsible for the deflection of the compass needle?
Answer:
The magnetic force developed around the current-carrying wire is responsible for the deflection of compass needle.

Question 10.
Does the current-carrying wire apply a force on the needle?
Answer:
Yes. The current-carrying wire applies a repulsive force on the needle.

Question 11.
What do we call this force?
Answer:
This force is called magnetic force of attraction and repulsion.

Page 211

Question 12.
How was this fleid produced?
Answer:
This field was produced due to the magnetic field developed ¡n the wire by the electric current.\

Question 13.
Can we observe the field of a bar magnet?
Answer:
Yes.

Question 14.
Why does this happen?
Answer:
The needle of the magnetic compass is affected by the bar magnet without any physical contact.

Question 15.
Is there any change in the direction of the needle of the magnetic compass? Why?
Answer:
Yes. The change in direction of needle of magnetic compass is due to the magnetic force developed in the conductor.

Question 16.
What is the nature of force that acts on the needle?
Answer:
The force which acting on the needle from a distance s the magnetic field of the bar magnet.

Question 17.
What do you observe?
Answer:
Almost the needle of compass shows single direction pointing north at far places from bar magnet.

Page 212

Question 18.
What does It mean?
Answer:
It means that strength of field vanes with distance from the bar magnet.

Question 19.
How can we find the strength of the field and direction of the field?
Answer:
The direction of field can be determined by using the compass. To determine the strenath we have to perform activity 3.

Question 20.
What are these curves?
Answer:
These curves are technically called as magnetic field lines,

Page 213

Question 21.
Are these field lines closed loops or open loops?
Answer:
The field lines appear to be closed loops. But we can’t conclude that lines are closed or open loops by looking at the figure of field lines because we do not know about alignment of lines that are passing through the bar magnet.

Question 22.
Can we give certain values to magnitude of the field at every point in the magnetic field?
Answer:
Yes. The number of lines passing through the plane of area ‘A’ perpendicular to the field is called magnetic flux and is deoted by Φ (Phi).

Page 214

Question 23.
What is the flux through unit area perpendicular to the field?
Answer:
It Is equal to Φ/A. (A = Area)

Question 24.
Can we generalize the formula of flux for any orientation of the plane taken in the field?
Answer:
The magnetic flux density denoted by ‘B’ is given by, B = magnetic flux / effective area i.e.; B= Φ /A cos θ)
Where ‘θ’ Is the angle between magnetic field (B) and normal to the plane with area ‘A’.

Question 25.
What is the flux through the plane taken parallel to the field?
Answer:
Since θ is zero between parallel lines. Cos0° = 1 and So, Φ= BA.

Question 26.
What Is the use of Introducing the Ideas of magnetic flux and magnetic flux density?
Answer:
The ideas of magnetic flux and magnetic flux density help us to estimate the strength of the magnetic field.

Question 27.
Are there any sources of magnetic field other than magnets?
Answer:
Yes. The current-carrying conductor develops a magnetic field around It.

Question 28.
Do you know how the old electric calling bells work?
Answer:
Yes. They work on the principle of magnetic effect of electric currents.

Page 215

Question 29.
How do the directions of the compass needles change?
Answer:
They are directed along tangents to the circle.

Question 30.
What is the shape of the magnetic field line around wire?
Answer:
The shape of magnetic line around a current-carrying wire is almost a circle,

Question 31.
What is the direction of magnetic field Induction at any point on the field line?
Answer:
If the current flows vertically in upward direction from the page (plane of paper),the field lines are in anti-clockwise direction.

Page 216

Question 32.
Can you tell the direction of the magnetic field of the coil?
Answer:
Yes. The direction In which compass needle comes to rest Indicates the directton of the field due to the coil. So, the direction of field is perpendicular to the plane of the coil.

Question 33.
Why does the compass needle point In the direction of field?
Answer:
We know that south pole is attracted by the north pole. The needle Is oriented in such a way that Its south pole points towards the north pole of the coil.

Page 217

Question 34.
How do they adjust ¡n such an orderly pattern?
Answer:
A solenoid Is a long wire wound in a closed-packed helix. The magnetic field lines set up by solenoid resemble those of a bar magnet indicating that a solenoid behaves like a bar magnet.

Question 35.
What happens when a current-carrying wire is kept in a magnetic field?
Answer:
When a current-carrying wire is kept in a magnetic field it undergoes orientation.

Question 36.
Do you feel any sensation on your skin?
Answer:
Yes. The hair on my skin rises up when I stand near TV screen.

Question 37.
What could be the reason for that?
Answer:
It is due to the magnetic field produced by electric charges in motion.

Question 38.
Take a bar magnet and bring it near the TV screen. What do you observe?
Answer:
I observed that the picture on the TV screen is not clear. It is distorted.

Question 39.
Why does the picture get distorted?
Answer:
It is because the motion of electrons that form the picture is affected by the magnetic field of bar magnet.

Page 218

Question 40.
Is the motion of electrons reaching the screen affected by magnetic field of the bar magnet?
Answer:
Yes.

Question 41.
Can we calculate the force experienced by a charge moving In a magnetic field?
Answer:
Yes. If the force is F, it is given by the expression, F=qvB
Where, q = Amount of charge
y = Velocity of charged particle perpendicular to the magnetic field.
B = Magnetic field

Question 42.
Can we generalize the equation for magnetic force on charge when there Is an angle ‘q’ between the directions of field ‘B’ and velocity ‘V’?
Answer:
Then force F is given by the formula F = qvB Sin θ.

Question 43.
What Is the magnetic force on the charge moving parallel to a magnetic field?
Answer:
When the charge moves parallel to the magnetic field, the value of θ becomes zero. In the equation F=qvB sin θ. since θ=0, sinθ=0 the value of force F also becomes zero.

Question 44.
What is the direction of magnetic force acting on a moving charge?
Answer:
By applying the “right-hand’ rule we can guess the direction of magnetic force acting on a moving charge is direction of the thumb.

Page 219

Question 45.
What is the direction of force acting on a negative charge moving in a field?
Answer:

1. At first find the direction of force acting on a positive charge, by applying right-hand rule for positive charge.
2. Then reverse the direction and this new direction is the direction of force acting on a negative charge.

Question 46.
What happens when a current-carrying wire ¡s placed in a magnetic field?
Answer:
The current-carrying wire experiences magnetic force when it is kept in magnetic field.

Page 220

Question 47.
Can you determine the magnetic force on a current-carrying wire which is placed along a magnetic field?
Answer:
The magnetic force (F) on a current-carrying wire of length L in a uniform magnetic field ‘B’ where the current through the wire is lis :
It θ is the angle between direction of field and velocity of current and magnetic field, then
F = ILB Sin θ
If the current carrying wire is placed along direction of field θ = 0
∴ F = 0

Question 48.
What Is the force on the wire if its length makes an angle ‘θ’ with the magnetic field?
Answer:
If ‘θ’ is the angle between direction of current and magnetic field, then the force acting on the current-carrying wire is given by F = IL B Sin θ.

Page 221

Question 49.
how could you find Its direction?
Answer:
I could find the direction of this force by applying right-hand rule.

Question 50.
What happens to the wire?
Answer:
The wire is set into deflection.

Question 51.
In which way does lt deflect?
Answer:
The wire deflects upwards according to the right-thumb rule.

Question 52.
Is the direction of deflection observed experimentally same as that of the theoretically expected one?
Answer:
Yes. But it depends on polarities of the horseshoe magnet.

Question 53.
Does the right-hand rule give the explanation for the direction of magnetic force exerted by magnetic field on the wire?
Answer:
The right-hand rule does not help us to explain the reason for deflection of wire.

Question 54.
Can you give a reason for it?
Answer:

1. There exists magnetic field due to external source, ie; horseshoe magnet.
2. When there is current in the wire, it also produces field.
3. These fields overlap and give non-uniform field. This is the reason for it.

Page 222

Question 55.
Does this deflection fit with the direction of magnetic force found by right-hand rule?
Answer:
Yes.

Question 56.
What happens when a current-carrying coil Is placed In uniform magnetic field?
Answer:
The coil is set to motion.

Question 57.
Can we use this knowledge to construct an electric motor?
Answer:
Yes. This is the principle of electric motor which converts electrical energy into kinetic energy.

Question 58.
What is the angle made by AB and CD with magnetic field?
Answer:
AB and CD are at right angles to the magnetic field. (always 900)

Question 59.
Can you draw direction of magnetic force on sides AB and CD?
Answer:
Yes, the direction of magnetic force on sides AB and CD can be determined by applying right hand rule.

Page 223

Question 60.
What are the directions of forces on BC and DA?
Answer:
At BC magnetic force pulls the coil up and at DA magnetic force pulls it down.

Question 61.
What is the net force on the rectangular coil?
Answer:
Net force on the rectangular coil is zero.

Question 62.
Why does the coil rotate?
Answer:
The rectangular coil rotates in clockwise direction because the equal and opposite pair of parallel forces acting on the edges of the coil constitute a couple.

Question 63.
What happens to the rotation of the coil if the direction of current in the coil remains unchanged?
Answer:
If the direction of current in the coil is unchanged, it rotates upto a vertical position and then due to law of conservation of inertia, it rotates further in clockwise direction. But now the sides of the coil experience forces which are in the opposite direction to the previous case. So these forces tend to rotate the wire in anti clockwise direction. As a result, the coil comes, to halt and rotates In anti-clock wire direction.

Question 64.
How could you make the coil to rotate continuously?
Answer:
If the direction of current in coil, after the first half rotation is reversed, the coil continues to rotate in the same direction.

Question 65.
How can we achieve this?
Answer:
To achieve this, brushes B1 and B2 are used.

Page 224

Question 66.
What happens when a coil without current is made to rotate In magnetic field?
Answer:
It a coil without current is made to rotate in magnetic field nothing happens.

Question 67.
How is current produced?
Answer:
The current is produced from the battery to the coil.

Question 68.
What do you notice? (ASI.) (iMark)
Answer:
We notice that the metal ring Is levitated on the coil.

Question 69.
Why is there a difference in behaviour in these two cases?
Answer:
An The A.C. supply changes its direction a number of times in a second. But D.C. is unidirectional current. So there is a difference in the behaviour of the metal ring In these two cases.

Question 70.
What force supports the ring against gravity when it is being levitated?
Answer:
The magnetic force developed in the coil of copper wire supports the ring against gravity when It is being levitated.

Page 225

Question 72.
Could the ring be levitated if D.C. is used?
Answer:
The metal ring is levitated only when the net force on it is zero according to Newton’s second law. So it is not possible If D.C is used.

Question 72.
What is this unknown force acting on the metal ring?
Answer:
The change in polarities at certain intervals at the ends of the solenoid causes the unknown force acting on the metal ring.

Question 73.
What is responsible for the current in the metal ring?
Answer:
The field through the metal ring changes so that flux linked with the metal ring changes and this is responsible for the current in metal ring.

Question 74.
If D.C. is used, the metal ring lifts up and falls down immediately. Why?
Answer:
The flux linked with metal ring is zero when the switch is on. At that instant there should be a change in flux linked with ring. So the ring raises up and falls down. If the switch is off, the metal ring again raises up and falls down. There is no change in flux linked with ring when the switch is off.

Page 226

Question 75.
What could you conclude from the above analysis?
Answer:
The relative motion of the magnet and coil sets up a current in the coil.

Page 227

Question 76.
What is its direction?
Answer:
The direction of the induced current ¡s such that it opposes the change that produced It.

Question 77.
Can you apply conservation of energy for electromagnetic induction?
Answer:
Yes. In electromagnetic induction, mechanical energy is converted into electrical energy and there is neither gain or loss of energy. So the conservation of energy Is maintained.

Page 228

Question 78.
Can you guess what could be the direction of induced current in the coil in such case?
Answer:
Invariably the direction of induced current in the coil must be In anti clockwise direction. In simple terms, when flux Increases through coil, the coil opposes the increase In flux and flux decreases through coil, it opposes the decrease in flux.

Question 79.
Could we get Faraday’s law of induction from conservation of energy?
Answer:
Yes. The mechanical energy used to move crosswire to distance s’ in one second Is converted Into electric energy, such that electronic energy = Φ/Δt and leads to conservation of energy.

Page 229

Question 80.
Can you derive an expression for the force applied on cross wire by the field B?
Answer:
Yes. The force applied F = BIL

Page 231

Question 81.
How could we use the principle of electromagnetic Induction In the case of using ATM card when Its magnetic strip is swiped through a scanner?
Answer:
When the magnetic strip of ATM card coated with Iron oxide moves past the small coil of wire, the magnetic field changes which leads to generation of current and it leads to the operation of the card.

Question 82.
What happens when a coil is continuously rotated In a uniform magnetic field?
Answer:
An Induced current is generated in the coil.

Question 83.
Does it help us to generate the electric current?
Answer:
Yes.

Question 84.
Is the direction of current induced in the coil constant? Does It change?
Answer:
Yes It changes. When the coil is at rest In vertical position, with side (A) of coil at top position and side (B) at bottom position no current will be Induced in It.

Page 232

Question 85.
Can you guess the reason for variation of current from zero to maximum and vice-versa during the rotation of coil?
Answer:
The reason for variation of current from zero to maximum and vice-versa during the second part of the rotation is current generated follows the same pattern as that in the first half except that the direction of current is reversed.

Question 86.
Can we make use of this current? If so, how?
Answer:
Two carbon brushes are arranged in sucti a way that they press the slip rings to obtain current from the coil. When these brushes are connected to external devices like TV, radio, we can make them work with current supplied from ends of carbon brushes.

Page 233

Question 87.
How can we get D.C. current using a generator?
Answer:
By connecting two half slip rings, Instead of a slip ring commutator on either side of the ends of the coil, we can get D.C current.

Question 88.
What changes do we need to make in an A.C. generator to be conected into a D.C. generator?
Answer:
Instead of two slip rings, we use a half slip ring commutator to change A.C. generator Into a D.C. generator.

Think And Discuss

Question 1.
How does a tape recorder which we use to listen songs / speeches work?
Answer:

1. The tape recorder which we use to listen songs/speeches and to record the voices works on the principle of electromagnetic induction.
2. it consists of a piece of plastic tape coated with ferromagnetic substance, iron oxide which is magnetised more in some parts than in others due to varying currents produced by our voice.
3. When the tape moves past as a small coil of wire (it is called head of the tape recorder.), the magnetic field produced by the tape changes which leads to generation of current in the small coil of wire.
4. These varying currents are fed into a loud speaker where they are reproduced as sounds.

Question 2.
How could we use the principle of electro-magnetic induction in the case of using ATM card?
Answer:
When the magnetic strip of ATM card coated with iron oxide moves past as a small coil of wire, the magnetic field changes which leads to generation of current and it leads to the operation of the card.

TS 10th Class Physical Science Electromagnetism Activities

Activity 1

Question 1.
With the help of an activity show the current-carrying wire produces magnetic field.
(or)
How do you prove magnetic field develops around a current-carrying conductor?
Answer:

1. Take a thermocol sheet and fix two thin wooden sticks of height 1cm which have small slit at the top
   of their ends.
2. Arrange a copper wire of 24 gauge which passes through these slits and make a circuit in which the elements1 at 3 volt battery, key and copper; wire are connected in series.
3. Now keep a magnetic compass below the wire.
4. Bring a bar magnet close to the compass.
5. Take a bar magnet far away from the circuit and switch on the circuit. Observe the changes In compass.
6. The compass needle deflects.
7. This deflection Is due to the magnetic field produced by current-carrying wire.

Activity 2

Question 2.
Show that the magnetic field around a bar magnet is three dimentional and its strength and direction varies from place to place.
Answer:

1. Take a sheet of white paper and place it on the horizontal table.
2. Place a bar magnet in the middle of the sheet.
3. Place a magnetic compass near the magnet. It settles to a certain direction:
4. Use a pencil and put dots on the sheet on either side of the needle. Remove the compass. Draw a small line segment connecting the two dots Draw an arrow on it from South Pole of the, needle to North Pole of the needle.
5. Repeat the same by placing the compass needles at various positions on the paper. The compass needle settles in different directions at different positions.
6. This shows that the direction of magnetic field due to a bar magnet varies from place to place.
7. Now take the compass needle to places far away from magnet, on the sheet and observe the orientation of the compass needle in each case.
8. The compass needle shows almost the same direction along north and south at places far from the magnet.
9. This shows that the strength of the field varies with distance from the bar magnet.
10. Now hold the compass a little above the table and at the top of the bar magnet.
11. We observe the deflection in compass needle. Hence we can say that the magnetic field is three dimensional i.e., a magnetic field surrounds its source.
12. From the above activities we can generalize that a magnetic field exists in the region surrounding a bar magnet and is characterized by strength and direction.

Activity 3

Question 3.
Explain the process of tracing magnetic field lines.
Answer:

1. Place a white sheet of paper on a horizontal table.
2. Place a compass in the middle of it. Put two dots on either side of the compass needle. Take it out.
   Draw a line connecting the dots which shows the North and South of the earth.
3. Now place a bar magnet on the line drawn in such a way that its north pole points towards the Magnetic field lines geographic north.
4. Now place the compass at the north pole of the bar magnet. Put a dot at the north pole of the compass needle.
5. Now remove the compass and place it at the dot. It will point in other direction. Again put a dot at the north pole of the compass needle.
6. Repeat the process till you reach the south pole of the bar magnet.
7. Connect the dots from ‘N’ of the bar magnet to S’ of the bar magnet, with a’ free hand ctirve. You will get a curved line.
8. Now select another point from the north pole of the bar magnet. Repeat the process for many points taken near the north pole.
9. You will get different curves as shown in the figure.
10. These lines are called magnetic field lines. They are Imaginary lines.

Activity 4

Question 4.
Describe an activity to find magnetic field due to straight wire carrying current. What Is its direction?
Answer:
1. Take a wooden plank and make a hole as shown in figure, Place this plank on the table.
2. Now place a retort stand on the plank as shown In figure.
3. Pass a 24 gauge copper wire through hole of the plank and rubber knob of the retort stand in such a way that the wire be arranged In a vertical position and not touch the stand.
4. Connect the two ends of the wire to a battery via switch.
5. Place 6 to lo compass needles in a circular path around the hole so that its centre coincides with the hole.
6. Use a 9V battery in the c!rcuet. Swtch on current and it flows through the wire.
7. We notice that they are directed as tangents to the circle.
8. The magnetic field around the wire is circular ¡n shape. This can be verified by sprinkling iron filings around the wire when current flows in the wire.
9. The direction of magnetic field around the wire will be as shown in the following figure.

10. The direction of the magnetic field around the current-carrying wire can be determined by right-hand thumb rule ie., grab the current-carrying wire with your right hand in such a way that thumb is in the direction of current, then the curled fingers show the direction of magnetic field as shown in the figure.

Activity 5

Question 5.
Trace the magnetic field due to circular coil.
Answer:
1. Take a thin wooden plank covered with white paper and make two holes on its surface as shown in the figure.
2. Pass insulated copper wire (24 guage) through the holes and wind the wire 4 to 5 times through holes
such that it looks like a coil.
3. The ends of the wire are connected to terminals of the battery through a switch. Now switch on the
circuit.
4. Place a compass needle on the plank at the centre of the coil. Put dots on either side of the compass needle.
5. Again place compass at one of the dots put other dots further.
6. Do the same till you reach the edge of the plank.
7. Now repeat this for the other side of the coil from the centre. Then draw a line joining the dots. We will get a field line of the circular coil.
8. Do the same for the other points taken In the between the holes. Draw corresponding lines. We will get field lines of the circular coil.
9. The direction of the field due to coil is s determined by using right-hand rule, which states that, when you curl your right-hand fingers in the direction of current, thumb waves the direction of magnetic field.

Activity 6

Question 6.
Find the magnetic field due to a solenoid.
(or)
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid.
Answer:

1. Take a wooden plank covered with a white paper.
2. Make equidistant holes on its surface as shown in the figure.
3. Pass copper wire through the holes. This forms a coil.
4. Join the ends of the coil to a battery through a switch.
5. Switch on the circuit. Current passes through the coil.
6. Now sprinkle Iron filings on the surface of the plank around the coil. Give-a small jerk to it, An orderly pattern Of iron filings is seen on the paper.
7. The long coil is called solenoid. The field of solenoid is shown in the figure.
8. This activity proves that current-carrying solenoid forms magnetic field.

Activity 7

Question 7.
Explain magnetic force on moving charge and current carrying wire.
Answer:
(a) TV Screen Activity:

1. Take a bar magnet and bring it near the TV screen.
2. Then the picture on the screen is distorted.
3. Here the distortion is due to the motion of the electrons reacting the screen are affected by the magnetic field.
4. Now move the bar magnet away from the screen.
5. Then the picture on the screen stabilizes.
6. This must be due to the fact that the magnetic field exerts a torce on moving charges. This force Is called magnetic force.
7. The magnitude of the force is F = Bqv where B Is magnetic induction, q’ is the charge and y Is the velocity of the charged partide.

(b) Procedure:

1. Take a bar magnet near to the TV screen. We observe that electrons are affected by the field produced by the bar magnet. Then the picture Is disturbed.
2. Move the bar magnet away from screen. Then you will get a clear picture.
3. The force on the moving charge is given by F = Baqv.
4. Right hand rule is used when velocity and field are perpendicular to each other

Fore-finger —- Direction of current (V).
Middle finger —- Direcrtion of field
Thumb —- direction of force
This rule is applicable to positive charge.

Activity 8

Question 8.
Explain the result of magnetic force applied on a current-carrying wire by an experiment.
(or)
Why the current carrying straight wire which is kept In a uniform magnetic field, perpendicularly to the direction of the field bends aside P Explain this process with a diagram showing the direction of forces acting on the wire?
Answer:
1. Take a wooden plank. Fix two long wooden sticks on It. These wooden sticks are split at their top ends.
2. A copper wire is passed through these splits and the ends of the wire are connected to a battery
of 3V, through a switch.
3. Close the switch to make the circuit. Current passes through the wire.
4. Now bring a horseshoe magnet near the copper wire as shown in the figure. Observe the deflection of the wire.
5. change polarities of the horseshoe magnet. Again observe the deflection. Repeat this by changing the direction of current in the circuit.
6. When current passes through wire, it produces a magnetic field and this field overlaps with the field by horseshoe magnet and gives a non-uniform field.
7. The field in between north arid south poles of horseshoe magnet is shown in the figure.

8. Let us make a wire passing perpendicular to the paper. Let the current pass through It. It produces magnetic field as shown in the figure.

9. The resultant field will be as shown below.
10. We can see that the direction of the field lines due to wire in upper part (of circular lines) coincides
with the direction of field lines of horseshoe magnet.

11. The direction of field lines by wire in lower part (or circular lines) Is opposite to the direction of the
field lines of horseshoe magnet. The net field in upper part is strong and in lower part it is weak. Hence a non-uniform field is created around the wire.
12. Therefore the wire tries to move to the weaker field region.

Activity 9

Question 9.
Take wooden base. Fix a soft iron cylinder on the wooden base vertically. Wind copper wire around the soft Iron as shown in the figure. Now take a metal ring which is slightly greater in radius than the radius of the soft iron cylinder and insert it through the cylinder on the wooden base. Now explain the behaviour of metal ring when the ends of copper wire are connected to (i) AC source and (ii) DC source.
Answer:
When the ends of copper wire are connected to AC source:

• Connect the two ends of copper wire to AC source and switch on the current.
• The metal ring Levitates due to net force acting on it is zero according to Newton’s second law.

Reason:

1. AC changes both its direction and magnitude in regular intervals.
2. Due to the magnetic field produced by current in the coil, one end of the coil behaves like North pole and the other end behaves like south pole for certain time interval.
3. For the next interval, the coil changes its polarties.
4. Assume that the current flows in clockwise direction in the solenoid as viewed from the top. Then the upper end becomes a south pole.
5. An upward force Is applied on the ring only when the upper side of the ring becomes north pole.
6. It Is only possible when there exists anti clock wise current viewed from the top in the ring.
7. After certain intervals, solenoid changes its polarities, so that the ring should also change its polarities in the same intervals.
8. This is the reason why the metal ring is levitated.

When the ends of copper wire are connected to a DC source:

1. Now connect the ends of copper wire to a DC source.
2. When the current is allowed to flow through the solenoid, It behaves like bar magnet. So the flux is linked to the metal ring when the switch is on.
3. At that instant there is a change in flux linked with the ring. Hence the ring rises up.
4. Thereafter, there Is no change In flux linked with coil, hence it falls down.
5. If the switch is off, the metal ring again lifts up and falls down. In this case also there is change
   in flux linked with ring when the switch is off.

Solving these TS 10th Class Maths Bits with Answers Chapter 10 Mensuration Bits for 10th Class will help students to build their problem-solving skills.

Mensuration Bits for 10th Class

Question 1.
To find out quantity of water in the bottle, we measure
A) surface area
B) total surface area
C) volume
D) base area
Answer:
C) volume

Question 2.
Lateral surface area of a cube is given by
A) 2a²
B) 4a²
C) 6a²
D) a³
Answer:
B) 4a²

Question 3.
Total surface area of a regular circular cylinder is
A) 2πrh
B) πrl
C) 2πr(π + r)
D) 2πr(r + h)
Answer:
D) 2πr(r + h)

Question 4.
The ratio of volumes of a cone and a cylinder whose radii and height are equal is ………………….
A) 3 : 1
B) 1 : 3
C) 1 : 2
D) 1 : 1
Answer:
B) 1 : 3

Question 5.
The diagonal of a cube whose side is ‘a’ units is ……………..
A) a
B) \(\sqrt{2}\) a
C) \(\sqrt{3}\) a
D) 2a
Answer:
C) \(\sqrt{3}\) a

Question 6.
The volume of a sphere of radius ‘r’ is obtained by multiplying its surface area by
A) 4/3
B) r/3
C) 4r/3
D) 3r
Answer:
B) r/3

Question 7.
The total surface area of a solid hemisphere of radius 7cm is
A) 239 π cm²
B) 449 π cm²
C) 221 π cm²
D) 129 π cm²
Answer:
A) 239 π cm²

Question 8.
The curved surface area of a right circular cone of height 15cm and base diameter 16cm is.
A) 144 π cm²
B) 136 π cm²
C) 105 π cm²
D) 120 π cm²
Answer:
B) 136 π cm²

Question 9.
The surface areas of two spheres are in the ratio 1 : 4 then, ratio of their volumes is
A) 1 : 4
B) 2 : 8
C) 1 : 16
D) 1 : 64
Answer:
A) 1 : 4

Question 10.
The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is
A) 19.4 cm³
B) 74.6 cm³
C) 9.7 cm³
D) 8.4 cm³
Answer:
A) 19.4 cm³

Question 11.
The ratio of volume of cone and cylinder of equal diameter and height
A) 3 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 3
Answer:
D) 1 : 3

Question 12.
An iron cylindrical rod has a height 4 times of its radius is melted and cast into spherical balls of the same radius. The number of balls cast is
A) 4
B) 3
C) 2
D) 1
Answer:
D) 1

Question 13.
A cone and a hemi-sphere have equal bases and equal volumes then the ratio of their heights
A) 2 : 1
B) 3 : 1
C) 4 : 1
D) 1 : 1
Answer:
A) 2 : 1

Question 14.
The volume of the greatest cylinder that can be cut form a solid wooden cube of length of edge 14cm is
A) 2156 cm³
B) 1078 cm³
C) 539 cm³
D) 428 cm³
Answer:
A) 2156 cm³

Question 15.
A shuttle cock is a combination of
A) cylinder, sphere
B) sphere, cone
C) cylinder, hemisphere
D) hemisphere, first term cone
Answer:
D) hemisphere, first term cone

Question 16.
T.S.A of a solid hemisphere whose radius is 7cm is …………….. cm².
A) 327 π
B) 144 π
C) 147 π
D) 189 π
Answer:
C) 147 π

Question 17.
If the radius of base of a cylinder is doubled and the height remains unchanged, it’s C.S.A becomes
A) double
B) times
C) half
D) no change
Answer:
A) double

Question 18.
The number of cubes of side 2cm which can be cut from a cube of side 6 cm is
A) 3
B) 18
C) 27
D) 9
Answer:
C) 27

Question 19.
If the diameter of a sphere is’d’ then its volume is
A) \(\frac{1}{6}\) πd³
B) \(\frac{4}{3}\) πd³
C) \(\frac{1}{24}\) πd³
D) \(\frac{1}{3}\) πd³
Answer:
A) \(\frac{1}{6}\) πd³

Question 20.
A cylindrical, a cone and a hemisphere are of equal base and have the same height, then the ratio of their volumes is
A) 3 : 1 : 1
B) 3 : 2 : 1
C) 1 : 2 : 3
D) 1 : 3 : 2
Answer:
A) 3 : 1 : 1

Question 21.
Total surface area of a cube is 216 cm² then its volume is ……………….. cm³.
A) 216
B) 196
C) 212
D) 144
Answer:
A) 216

Question 22.
The total surface area of a cube is 54 cm² then its side is ………………. cm. (A.P. Mar. ’15 )
A) 6
B) 9
C) 12
D) 3
Answer:
D) 3

Question 23.
Base area of a regular cylinder is 154 cm² then its radius is ……………….. (A.P. Mar. ’16, ’15)
A) 49 cm
B) 7 cm
C) 22 cm
D) 14 cm
Answer:
B) 7 cm

Question 24.
If the height and radius of a cone are 1.5 and 8 cm then its slant height = ………………. cm
A) 2.5 cm
B) 7.5 cm
C) 5 cm
D) 10 cm
Answer:
C) 5 cm

Question 25.
Curved surface area of a hemi-sphere = ………………. (A.P. Mar. ’15)
A) πr²
B) \(\frac{1}{3}\)πr²
C) 3πr²
D) 2πr²
Answer:
D) 2πr²

Question 26.
Volume of a cube having 1 cm side is ……………….. (A.P. Mar. ’16, June ’15)
A) 1 cm³
B) 3 cm³
C) 1 cm²
D) 3 cm²
Answer:
A) 1 cm³

Question 27.
Ratio of volumes of two spheres is 8 : 27 then ratio of their curved surface areas is …………….. (A.P. June ’15)
A) 2 : 3
B) 4 : 27
C) 8 : 9
D) 4 : 9
Answer:
C) 8 : 9

Question 28.
Football is in a model of …………………. (A.P. Mar. ’16)
A) circle
B) cylinder
C) Sphere
D) cone
Answer:
C) Sphere

Question 29.
Radius of a cone is ‘r’, height is ‘h’ and its slant height is 7 then which of the following is false ? (A.P. Mar. ’16)
A) always l > h
B) always l > r
C) always r > p
D) l² = r² + h²
Answer:
C) always r > p

Question 30.
Radius, height, slant height of a cone are| r, h, l, then ‘l’ value in terms of r and h is ……………… (T.S. Mar. ’15)
A) \(\sqrt{h^2-r^2}\)
B) \(\sqrt{r^2+h^2}\)
C) \(\sqrt{r^2-h^2}\)
D) \(\sqrt{4 r^2+h^2}\)
Answer:
B) \(\sqrt{r^2+h^2}\)

Question 31.
To calculate the quantity of milk inside a bottle, we need to find out
A) Area
B) Volume
C) Density
D) TSA
Answer:
B) Volume

Question 32.
Sphere, cylinder and cone have same heights and radii, then its ratios of curved surface areas.
A) 4 : 4 : \(\sqrt{5}\)
B) 1 : 1 : \(\sqrt{5}\)
C) \(\sqrt{5}\) : 4 : 4
D) 4 : \(\sqrt{5}\) : 4
Answer:
A) 4 : 4 : \(\sqrt{5}\)

Question 33.
Diagonal of a cuboid is …………….. units.
A) \(\sqrt{l^2+b^2+h^2}\)
B) \(1 \sqrt{\mathrm{b}^2+\mathrm{h}^2}\)
C) \(b \sqrt{h^2+r^2}\)
D) none
Answer:
A) \(\sqrt{l^2+b^2+h^2}\)

Question 34.
The radius of a conical tent is 3 meter and height is 4 meter then its slant height is …………………. meter.
A) 5
B) 725
C) A and B
D) none
Answer:
A) 5

Question 35.
The total surface area of a solid hemisphere of radius 1 unit is
A) 3πr²
B) 2πr²
C) 3π
D) 2π
Answer:
C) 3π

Question 36.
Volume of is cuboid.
A) 16
B) 10
C) 6
D) 12
Answer:
C) 6

Question 37.
The diameter of a metallic sphere is 6 cm and melted to draw a wire of diameter 2cm, then the length of the wire is
A) 48 cm
B) 12 cm
C) 36 cm
D) 24 cm
Answer:
C) 36 cm

Question 38.
A solid sphere of radius r melted and recast into the shape of a solid cone of height r, then radius of the base of the cone is (of equal volume)
A) 2r
B) r
C) 3r
D) 4r
Answer:
A) 2r

Question 39.
If the radii of circular ends of a frustum of a cone are 20 cm and 12 cm and its height is 6cm, then the slant height of the frustum is …………………. cm.
A) 10
B) 6
C) 9
D) 8
Answer:
A) 10

Question 40.
The number of balls, each of radius 1 cm that can be made from a solid sphere of radius 8 cm is
A) 64
B) 216
C) 16
D) 512
Answer:
D) 512

Question 41.
The ratio of volume of two cones is 4 : 5 and the ratio of the radii of their base is 2 : 3 then ratio of their vertical heights is
A) 4 : 5
B) 9 : 5
C) 3 : 5
D) 2 : 5
Answer:
B) 9 : 5

Question 42.
If the ratio of radii of two spheres is 2 : 3 then the ratio of their surface areas is
A) 3 : 2
B) 27 : 8
C) 8 : 27
D) 4 : 9
Answer:
D) 4 : 9

Question 43.
If a cone is cut into two parts by a horizontal plane passing through the mid point of the axis, the ratio of the volumes of the
upper part and the cone is
A) 1 : 2
B) 1 : 4
C) 1 : 6
D) 1 : 8
Answer:
D) 1 : 8

Question 44.
The height of a cylinder is doubled and radius is tripled then its curved surface area will become …………….. times.
A) 7
B) 6
C) 9
D) 12
Answer:
B) 6

Question 45.
Diameter of a sphere which can in-scribe a cube of edge x cm is ……………
A) \(\frac{x}{3}\)
B) \(\frac{x^2}{3}\)
C) \(\frac{x}{\sqrt{3}}\)
D) x\(\sqrt{3}\)
Answer:
D) x\(\sqrt{3}\)

Question 46.
Total surface area of hemisphere of radius r is ………….
A) πr²
B) 2πr²
C) 3πr²
D) none
Answer:
C) 3πr²

Question 47.
Volume of frustrum of a cone is
A) \(\frac{\pi h}{3}\) (R² + r²2 +R.r)
B) \(\frac{\mathrm{h}}{3}\)(R² + r²)
C) \(\frac{\pi h}{3}\)(R² +r²)
D) none
Answer:
A) \(\frac{\pi h}{3}\) (R² + r²2 +R.r)

Question 48.
If the length of each diagonal of a cube is doubled, then its volume become ………………. times.
A) 7
B) 8
C) 9
D) none
Answer:
B) 8

Question 49.
If a right angled triangle is revolved about its hypotenuse then it will form a ………………….
A) double cone
B) triple cone
C) only cone
D) none
Answer:
A) double cone

Question 50.
A solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then radius of small spherical balls is ……………… cm.
A) 10
B) 9
C) 6
D) 5
Answer:
D) 5

Question 51.
In a hollow cuboid box of size 4 × 3 × 2m, the number of solid iron spherical balls of radius 0.5 m that can be packed ……………
A) 71
B) 24
C) 22
D) 16
Answer:
B) 24

Question 52.
If the external and internal radii of a hollow hemispherical bowl are R and r, then its total surface area is ……………….
A) πr² + R²
B) πR² + r²
C) πR² + r
D) π(3R² + r²)
Answer:
D) π(3R² + r²)

Question 53.
Volume of cylinder is …………………. cu. units.
A) πr²h
B) πr²
C) \(\frac{\pi}{r}\)
D) none
Answer:
A) πr²h

Question 54.
Volume of cone is ……………… cu. units.
A) \(\frac{1}{7}\) πr²h
B) \(\frac{1}{2}\) πr²h
C) πr²h
D) \(\frac{1}{3}\) πr²h
Answer:
D) \(\frac{1}{3}\) πr²h

Question 55.
Volume of sphere is ………………… cu.units.
A) \(\frac{4}{3}\) πr²h
B) \(\frac{4}{3}\) πr³
C) \(\frac{1}{3}\) πr³
D) none
Answer:
B) \(\frac{4}{3}\) πr³

Question 56.
Volume of cuboid = …………………… cu.units
A) l²b
B) lbh²
C) lbh
D) none
Answer:
C) lbh

Question 57.
Total surface area of cone is …………………. sq.units.
A) πr² + πrl
B) πr² + πr
C) πr² + πl
D) none
Answer:
A) πr² + πrl

Question 58.
Total surface area of cylinder is ……………………. sq.units.
A) πrh + πr²
B) 2πr + π
C) 2πrh²
D) 2πrh + 2πr²
Answer:
D) 2πrh + 2πr²

Question 59.
Total surface area of hemisphere is ………………… sq.units.
A) \(\frac{\pi r^2}{\mathrm{~h}}\)
B) 4πr²
C) 8πr²h
D) none
Answer:
D) none

Question 60.
Surface area of sphere is ……………… sq.units.
A) \(\frac{\pi r^2}{\mathrm{~h}}\)
B) 4πr²
C) 8πr²h
D) none
Answer:
B) 4πr²

Question 61.
Volume of a cube is ………………… cu.units.
A) 3a³
B) a²h
C) a³
D) none
Answer:
C) a³

Question 62.
The volume of a cube is 216 cm3 then edge is ………………….. cm.
A) 9
B) 10
C) 16
D) 6
Answer:
D) 6

Question 63.
C.S.A of cone = ……………….. sq. units.
A) π²r²l
B) πrl²
C) πr²
D) πrl
Answer:
D) πrl

Question 64.
In a cone, r = 7 cm, h = 10 cm then l = ………………….. cm
A) 12.2
B) 9.2
C) 10.1
D) none
Answer:
A) 12.2

Question 65.
π = …………………
A) \(\frac{22}{7}\)
B) \(\frac{2}{7}\)
C) \(\frac{22}{3}\)
D) none
Answer:
A) \(\frac{22}{7}\)

Question 66.
The volume of a right circular cone with radius 6 cm and height 7 cm is …………………. cm³.
A) 462
B) 264
C) 486
D) none
Answer:
B) 264

Question 67.
A heap of rice is in the form of a cone of diameter 12 m and height 8 m then volume is ……………… m³.
A) 110.53
B) 301.71
C) 310.51
D) none
Answer:
B) 301.71

Question 68.
In a cylinder, r = 8 cm, h = 10 cm, CSA = …………………. cm³.
A) \(\frac{3520}{7}\)
B) \(\frac{1520}{9}\)
C) \(\frac{3310}{41}\)
D) none
Answer:
A) \(\frac{3520}{7}\)

Question 69.
In a hemisphere, r = 1.75 cm then CSA = ……………… cm².
A) 38.5
B) 48.5
C) 93.5
D) none
Answer:
A) 38.5

Question 70.
Volume of cone if r = 2 cm, h = 4 cm is ……………….
A) \(\frac{16}{3}\) π
B) \(\frac{6}{7}\) π
C) \(\frac{18}{31}\) π
D) none
Answer:
A) \(\frac{16}{3}\) π

Question 71.
Surface area of a sphere and cube are equal then the ratio of their volumes is …………………
A) \(\sqrt{\pi}\) : 1
B) \(\sqrt{\pi}\) : \(\sqrt{6}\)
C) π : \(\sqrt{6}\)
D) none
Answer:
B) \(\sqrt{\pi}\) : \(\sqrt{6}\)

Question 72.
In a hemisphere, r = 7 cm then CSA = ………………… cm².
A) 210
B) 308
C) 114
D) 112
Answer:
B) 308

Question 73.
In a cylinder, r = 7 cm then CSA = ………………. cm².
A) 1170
B) 1120
C) 2310
D) 1320
Answer:
C) 2310

Question 74.
Heap of stones is an example of ………………..
A) cylinder
B) cone
C) circle
D) none
Answer:
B) cone

Question 75.
In the figure, l² = ………………

A) h² + r²
B) \(\sqrt{l^2+h^2}\)
C) h² + r
D) h + r²
Answer:
A) h² + r²

Question 76.
Area of equilateral triangle of side ‘a’ units is …………….. sq. units.
A) \(\frac{1}{\sqrt{3}}\) a²
B) \(\frac{4}{\sqrt{3}}\) a²
C) \(\frac{\sqrt{3}}{4}\) a
D) \(\frac{\sqrt{3}}{4}\) a²
Answer:
D) \(\frac{\sqrt{3}}{4}\) a²

Question 77.
Perimeter of square is 20 cm then A = ………….. cm².
A) 12
B) 16
C) 25
D) none
Answer:
C) 25

Question 78.
Diagonal of rectangle is …………… units.
A) \(\sqrt{l^2+b^2}\)
B) \(\sqrt{l+b}\)
C) l + \(\sqrt{b}\)
D) \(\sqrt{l}\) + b
Answer:
A) \(\sqrt{l^2+b^2}\)

Question 79.
Volume of hollow cylinder is …………………
A) πR – r
B) πr² – R
C) πR² – r
D) π(R² – r²)
Answer:
D) π(R² – r²)

Question 80.
……………… gave the symbol π.
A) Euler
B) Pepe
C) Mount
D) None
Answer:
A) Euler

Question 81.
In a cone, (l + r) (l – r) = …………..
A) h²
B) 2h
C) h
D) none
Answer:
A) h²

Question 82.
A cuboid has dimensions 10x8x6 cm then its volume is ……………… cm³.
A) 190
B) 780
C) 680
D) 480
Answer:
D) 480

Question 83.
C.S.A of a cone is 4070 cm² and its diameter is 70 cm then slant height is ………………. cm
A) 27
B) 17
C) 37
D) 16
Answer:
C) 37

Question 84.
The sphere is of radius 2.1 cm then its volume is ………………. cm².
A) 38.08
B) 381.2
C) 83.01
D) none
Answer:
A) 38.08

Question 85.
In l² = h² + r², h = 15, r = 8 then l = ………….
A) 20
B) 17
C) 16
D) 19
Answer:
B) 17

Question 86.
The surface area of a sphere is 616 sq.cm. then its radius is ……………… cm.
A) 16
B) 12
C) 9
D) 7
Answer:
D) 7

Question 87.
In a cone, d = 14 cm, l = 10 cm then CSA = ………………….. cm².
A) 220
B) 140
C) 160
D) none
Answer:
A) 220

Question 88.
In a cube, a = 4 cm then TSA = ……………… cm².
A) 12
B) 70
C) 90
D) none
Answer:
C) 90

Question 89.
Number of edges of a cuboid is ………………
A) 11
B) 16
C) 10
D) 12
Answer:
D) 12

Question 90.
If the diagonals of a rhombus are 10 cm and 24 cm then area is ……………….. cm²
A) 110
B) 814
C) 413
D) 314
Answer:
A) 110

Question 91.
Volume of cone with d as diameter and h as height is …………………. units³.
A) \(\frac{\pi \mathrm{d}^2}{6}\)
B) \(\frac{\pi \mathrm{d}^2 \mathrm{~h}}{12}\)
C) \(\frac{\pi \mathrm{dh}^2}{12}\)
D) none
Answer:
B) \(\frac{\pi \mathrm{d}^2 \mathrm{~h}}{12}\)

Question 92.
The area of the base of a right circular cone is 78.5 cm². If its height is 12 cm then its volume is …………….. cm³.
A) 110
B) 814
C) 413
D) 314
Answer:
D) 314

Question 93.
The volume of a cuboid is 3,36,000 cm³. If; its area is 5,600 cm² then h = ……………… cm
A) 70
B) 60
C) 95.5
D) none
Answer:
B) 60

Question 94.
The volume of cone is 462 cm³, r = 7 cm then h = ……………….. cm.
A) 9
B) 10
C)11
D) none
Answer:
A) 9

Question 95.
The area of equilateral triangle is 36\(\sqrt{3}\) cm² then the perimeter is ……………. cm.
A) 36
B) 63
C) 16
D)10
Answer:
A) 36

Question 96.
Surface area of a cube of side 27 cm is ………………. cm³.
A) 1474
B) 8174
C) 1374
D) 4374
Answer:
D) 4374

Question 97.
The perimeter of an equilateral triangle is 60 cm then its area is ………………….. cm².
A) 149.3
B) 170.1
C) 137.4
D) 173.2
Answer:
D) 173.2

Question 98.
If the diagonal of a cube is 2.5 m then volume is ……………….. m³.
A) 3.01
B) 4.01
C) 8.1
D) none
Answer:
A) 3.01

Question 99.
r² = 1728 then r = ………………..
A) 13
B) 19
C) 10
D) 12
Answer:
D) 12

Question 100.
Number of faces of a cuboid is ………………..
A) 9
B) 10
C) 6
D) 8
Answer:
C) 6

Solving these TS 10th Class Maths Bits with Answers Chapter 11 Trigonometry Bits for 10th Class will help students to build their problem-solving skills.

Trigonometry Bits for 10th Class

Question 1.
If sin θ = \(\frac{\mathrm{a}}{\mathrm{b}}\) then tan θ =

Answer:
(D)

Question 2.
cos² θ + sin² θ is
A) 0
B) 1
C) \(\frac{1}{2}\)
D) θ²
Answer:
B) 1

Question 3.
If sin θ = cos θ, then the value of 2 tan θ + cos²θ
A) 1
B) \(\frac{1}{2}\)
C) \(\frac{5}{2}\)
D) \(\frac{2}{5}\)
Answer:
C) \(\frac{5}{2}\)

Question 4.
If tan θ + sec θ = 8, then sec θ – tan θ is
A) 8
B) \(\frac{1}{8}\)
C) 6
D) 64
Answer:
B) \(\frac{1}{8}\)

Question 5.
The maximum value of sin θ is
A) \(\frac{1}{2}\)
B) \(\frac{\sqrt{3}}{2}\)
C) 1
D) \(\frac{1}{\sqrt{2}}\)
Answer:
C) 1

Question 6.
If tan θ = \(\frac{7}{8}\) then the value of \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
A) \(\frac{7}{8}\)
B) \(\frac{8}{9}\)
C) \(\frac{64}{49}\)
D) \(\frac{49}{64}\)
Answer:
C) \(\frac{64}{49}\)

Question 7.
If 5 tan θ = 4, then the value of \(\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+3 \cos \theta}\) is
A) 0
B) 1
C) \(\frac{1}{7}\)
D) \(\frac{2}{7}\)
Answer:
C) \(\frac{1}{7}\)

Question 8.
\(\frac{\sin \theta}{1+\cos \theta}\) is

Answer:
(C)

Question 9.
The value of \(\frac{2 \tan 30}{1+\tan ^2 30}\) =
A) sin 60°
B) cos 60°
C) tan 60°
D) sin 30°
Answer:
A) sin 60°

Question 10.
The value of sin 45° + cos 45° is
A) \(\frac{1}{\sqrt{2}}\)
B) \(\sqrt{2}\)
C) \(\frac{\sqrt{3}}{2}\)
D) 1
Answer:
B) \(\sqrt{2}\)

Question 11.
If tan θ = 1, the value of \(\frac{5 \sin \theta+4 \cos \theta}{5 \sin \theta-4 \cos \theta}\) is
A) 9
B) 46°
C) 1
D) 0
Answer:
A) 9

Question 12.
tan θ is not defined when θ is
A) 90°
B) 60°
C) 30°
D) 0°
Answer:
A) 90°

Question 13.
If sin θ = \(\frac{\mathrm{a}}{\mathrm{b}}\) then cos θ =
A) \(\frac{\sqrt{a^2-b^2}}{b}\)
B) \(\frac{b}{a}\)
C) \(\frac{\sqrt{b^2-a^2}}{b}\)
D) \(\frac{b-a}{b}\)
Answer:
C) \(\frac{\sqrt{b^2-a^2}}{b}\)

Question 14.
If sin θ = \(\frac{12}{13}\) then tan θ =
A) \(\frac{13}{5}\)
B) \(\frac{5}{12}\)
C) \(\frac{13}{12}\)
D) \(\frac{12}{5}\)
Answer:
D) \(\frac{12}{5}\)

Question 15.
sin θ . sec θ =
A) tan θ
B) cosec θ
C) cot θ
D) sin θ . cos θ
Answer:
A) tan θ

Question 16.
\(\sqrt{1+\cot ^2 \theta}\) =
A) cosec² θ
B) 1 + cot θ
C) sec θ
D) cosec θ
Answer:
D) cosec θ

Question 17.
tan 135° =
A) \(\frac{1}{\sqrt{3}}\)
B) \(\sqrt{3}\)
C) – \(\sqrt{3}\)
D) -1
Answer:
D) -1

Question 18.
\(\sqrt{1+\sin A} \cdot \sqrt{1-\sin A}\) =
A) sin A
B) 1 – sin² A
C) cos A
D) 1
Answer:
C) cos A

Question 19.
sin (90 + θ) =
A) cos θ
B) – cos θ
C) sin θ
D) – sin θ
Answer:
A) cos θ

Question 20.
If tan θ = \(\frac{1}{\sqrt{3}}\), then cos θ =
A) \(\frac{1}{2}\)
B) \(\frac{\sqrt{3}}{2}\)
C) \(\frac{2}{\sqrt{3}}\)
D) \(\sqrt{3}\)
Answer:
B) \(\frac{\sqrt{3}}{2}\)

Question 21.
If cos θ = \(\frac{\sqrt{3}}{2}\) and ‘θ’ is acute, then the value of 4 sin² θ + tan² θ =
A) \(\frac{3}{4}\)
B) 1
C) \(\frac{4}{3}\)
D) \(\frac{5}{3}\)
Answer:
C) \(\frac{4}{3}\)

Question 22.
sin (A – B) = \(\frac{1}{2}\); cos (A + B) = \(\frac{1}{2}\). So, A =
A) 60°
B) 15°
C) 30°
D) 45°
Answer:
D) 45°

Question 23.
sin (-θ) =
A) sin θ
B) cos θ
C) – cos θ
D) – sin θ
Answer:
D) – sin θ

Question 24.
If sin (A – B) = \(\frac{1}{2}\); cos (A + B) = \(\frac{1}{2}\) then B =
A) 15°
B) – sin θ
C) sin θ
D) cos θ
Answer:
A) 15°

Question 25.
If α + β = 90° and α = 2β, then cos² β + sin² β =
A) 1
B) 0
C) \(\frac{1}{2}\)
D) 2
Answer:
A) 1

Question 26.
If tan θ = \(\frac{1}{\sqrt{3}}\) then 7 sin² θ + 3 cos² θ =
A) 1
B) 2
C) 3
D) 4
Answer:
D) 4

Question 27.
cos² 0° + cos² 60° =
A) \(\frac{5}{4}\)
B) \(\frac{2}{\sqrt{3}}\)
C) \(\frac{1}{\sqrt{2}}\)
D) \(\frac{\sqrt{3}}{2}\)
Answer:
A) \(\frac{5}{4}\)

Question 28.
cot (270° – θ) =
A) -tan θ
B) tan θ
C) cot θ
D) – cot θ
Answer:
B) tan θ

Question 29.
sin 240° =
A) \(\frac{1}{\sqrt{2}}\)
B) \(\frac{2}{\sqrt{3}}\)
C) – \(\frac{\sqrt{3}}{2}\)
D) \(\frac{\sqrt{3}}{2}\)
Answer:
C) – \(\frac{\sqrt{3}}{2}\)

Question 30.
\(\frac{1-\tan ^2 30}{1+\tan ^2 30}\) =
A) \(\frac{1}{2}\)
B) 1
C) 0
D) 2
Answer:
A) \(\frac{1}{2}\)

Question 31.
If sin (A + B) = \(\frac{\sqrt{3}}{2}\) ; cos B = \(\frac{\sqrt{3}}{2}\) value of A is
A) 45°
B) 60°
C) 30°
D) 90°
Answer:
C) 30°

Question 32.
cos (A – B) = \(\frac{1}{2}\); sin B = \(\frac{1}{\sqrt{2}}\) measure of A =
A) 15°
B) 105°
C) 90°
D) 60°
Answer:
B) 105°

Question 33.
sin² 45° + cos² 45° + tan² 45° =
A) 2
B) 1
C) 3
D) \(\frac{3}{\sqrt{2}}\)
Answer:
A) 2

Question 34.
sin \(\frac{\pi}{6}\) + cos \(\frac{\pi}{3}\) =
A) \(\frac{2}{\sqrt{3}}\)
B) 2
C) \(\frac{1}{2}\)
D) 1
Answer:
D) 1

Question 35.
If sec θ + tan θ = 4; then cos θ =
A) \(\frac{8}{17}\)
B) \(\frac{4}{17}\)
C) \(\frac{15}{17}\)
D) \(\frac{17}{8}\)
Answer:
A) \(\frac{8}{17}\)

Question 36.
tan 0° =
A) 1
Β) α
C) \(\frac{1}{\sqrt{3}}\)
D) 0
Answer:
D) 0

Question 37.
If sin θ = \(\frac{\mathrm{a}}{\mathrm{b}}\); cos θ = \(\frac{\mathrm{c}}{\mathrm{d}}\) then cot θ =
A) \(\frac{\mathrm{ab}}{\mathrm{cd}}\)
B) \(\frac{\mathrm{bc}}{\mathrm{ad}}\)
C) \(\frac{\mathrm{ca}}{\mathrm{bd}}\)
D) \(\frac{\mathrm{ad}}{\mathrm{bc}}\)
Answer:
B) \(\frac{\mathrm{bc}}{\mathrm{ad}}\)

Question 38.
cos(xy) =
A) cos x sin x + cos y sin
B) cos x sin y + cos y sin y
C) sin x cos y + cos x sin y
D) cos x cos y + sin x sin y
Answer:
D) cos x cos y + sin x sin y

Question 39.
\(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\) =
A) \(\frac{\tan \theta}{1+\tan \theta}\)
B) \(\frac{1}{\tan \theta}\)
C) cot θ
D) sin θ
Answer:
D) sin θ

Question 40.
cos θ:
A) 1 – sin² θ
B) sin² θ + 1
C) \(\frac{\cot \theta}{\ cosec \theta}\)
D) \(\frac{\ cosec \theta}{\ cot \theta}\)
Answer:
C) \(\frac{\cot \theta}{\ cosec \theta}\)

Question 41.
Value of cos 0° + sin 90° + \(\sqrt{3}\) cosec 60° =
A) 2
B) 3
C) 4
D) 1
Answer:
C) 4

Question 42.
π radians =
A) 90°
B) 60°
C) 180°
D) 45°
Answer:
C) 180°

Question 43.
cos⁶θ + sin⁶ θ =
A) 1 + sin³θ cos³ θ
B) 1 – 3sin²θ cos² θ
C) 1 – 3sin³θ cos³ θ
D) 45°
Answer:
C) 1 – 3sin³θ cos³ θ

Question 44.
cos 12 – sin 78 = …………. (A.P. Mar. ’15)
A) 1
B) \(\frac{1}{2}\)
C) 0
D) – 1
Answer:
C) 0

Question 45.
If x = cosec θ + cot θ and y = cosec θ – cot θ, then which of the following is true ………… (A.P Mar.’15)
A) x + y = 0
B) x – y = 0
C) \(\frac{x}{y}\) = 1
D) xy = 1
Answer:
D) xy = 1

Question 46.
cos (A – B) = …… (A.P. Mar.’15)
A) cos A cos B + sin A sin B
B) cos A sin A + cos B sin B
C) sin A sin B – cos A cos B
D) cos A cos B – sin A sin B
Answer:
A) cos A cos B + sin A sin B

Question 47.
cos (90 – θ) = ………….. (A.P. Mar.’15)
A) cos θ
B) sin θ
C) cosec θ
D) tan θ
Answer:
B) sin θ

Question 48.
In ∆ ABC, sin C = \(\frac{3}{5}\) then cos A = (A.P. June’15_
A) \(\frac{3}{5}\)
B) \(\frac{4}{5}\)
C) \(\frac{5}{4}\)
D) \(\frac{5}{3}\)
Answer:
A) \(\frac{3}{5}\)

Question 49.
tan² θ – sec² θ = ………. (A.P. June ’15)
A) 1
B) -1
C) 0
D) α
Answer:
B) -1

Question 50.
sec (90 – A) = …………… (A.P. June
A) cos A
B) cosec A
C) sin A
D) tan A
Answer:
B) cosec A

Question 51.
If cosec θ + cot θ = 5 then cosec θ – cot θ = ………….. (A.P. June ’15)
A) \(\frac{1}{5}\)
B) 5
C) -5
D) – \(\frac{1}{5}\)
Answer:
A) \(\frac{1}{5}\)

Question 52.
If x = 2 sec θ; y = 2 tan θ then x² – y² = ………………. (A.P. June ’15)
A) 0
B) -2
C) 4
D) 2
Answer:
C) 4

Question 53.
If \(\sqrt{3}\) tan θ = 1 then θ = ………. (A.P. June’15)
A) 60°
B) 90°
C) 45°
D) 30°
Answer:
D) 30°

Question 54.
(sec 60) (cos 60) = …………… (A.P. June’15)
A) 1
B) \(\frac{1}{2}\)
C) – 1
D) – \(\frac{1}{2}\)
Answer:
A) 1

Question 55.
sin (60 + 30) = ………………. (A.P. Mar. ’16)
A) 2
B) -2
C) 1
D) 1
Answer:
D) 1

Question 56.
If sin θ + tan θ = \(\frac{1}{2}\), then sin θ – tan θ = …………… (A.P. Mar.’16)
A) 1
B) – 1
C) 2
D) \(\frac{1}{2}\)
Answer:
C) 2

Question 57.
cos (90 – θ) = …………….. (A.P. Mar. ’16)
A) cos θ
B) tan θ
C) cosec θ
D) sin θ
Answer:
D) sin θ

Question 58.
If cot A = \(\frac{5}{12}\) then sin A + cos A = ………… (T.S. Mar. ’15)
A) \(\frac{17}{13}\)
B) \(\frac{12}{13}\)
C) \(\frac{5}{13}\)
D) \(\frac{20}{13}\)
Answer:
A) \(\frac{17}{13}\)

Question 59.
Which of the following is not possible value for sin x ………. (T.S. Mar. ’15)
A) \(\frac{3}{4}\)
B) \(\frac{3}{5}\)
C) \(\frac{4}{5}\)
D) \(\frac{5}{4}\)
Answer:
D) \(\frac{5}{4}\)

Question 60.
If sin cos θ (0 < θ < 90) then tan θ + cot θ = ………….. (T.S Mar. ’16)
A) 2\(\sqrt{3}\)
B) \(\frac{2}{\sqrt{3}}\)
C) 2
D) 1
Answer:
C) 2

Question 61.
If sec θ + tan θ = 3 then sec θ – tan θ = ……………. (T.S. Mar. ’16)
A) \(\frac{1}{3}\)
B) \(\frac{2}{3}\)
C) \(\frac{4}{3}\)
D) \(\frac{5}{3}\)
Answer:
A) \(\frac{1}{3}\)

Question 62.
In ∆ABC, AB = c, BC = a, AC = b and ∠BAC = θ then area of ∆ABC is …………… (θ is acute) (T.S. Mar.’16)
A) \(\frac{1}{2}\) ab sin θ
B) \(\frac{1}{2}\) ca sin θ
C) \(\frac{1}{2}\) bc sin θ
D) \(\frac{1}{2}\) b² sin θ
Answer:
C) \(\frac{1}{2}\) bc sin θ

Question 63.
The value of sin² 60° – sin²30° is ………………..
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) \(\frac{3}{4}\)
D) –\(\frac{1}{2}\)
Answer:
B) \(\frac{1}{2}\)

Question 64.
If cosec θ = 2 and cot θ = \(\sqrt{3}\) p, where θ is an acute angle, then the value of ‘p’ is
A) 2
B) 1
C) 0
D) \(\sqrt{3}\)
Answer:
B) 1

Question 65.
The value of \(\left(\frac{11}{\cot ^2 \theta}-\frac{11}{\cos ^2 \theta}\right)\) is
A) 11
B) 0
C) \(\frac{1}{11}\)
D) -11
Answer:
D) -11

Question 66.
If sec 2A = cosec (A – 27°), where 2A is an acute angle, then the measure of ∠A is
A) 35°
B) 37°
C) 39°
D) 21°
Answer:
C) 39°

Question 67.
\(\frac{1-\sec ^2 A}{{cosec}^2-1}\) = ………………….
A) -tan² A
B) -tan⁴ A
C) 1
D) -sec² A
Answer:
B) -tan⁴ A

Question 68.
If sec θ = 3k and tan θ = \(\frac{3}{\mathrm{k}}\) then (k² – \(\frac{1}{\mathrm{k}^2}\)) = …………….
A) 9
B) 3
C) \(\frac{1}{9}\)
D) 1
Answer:
C) \(\frac{1}{9}\)

Question 69.
If P, Q and R are interior angles of a ∆PQR, then tan \(\left(\frac{P+Q}{2}\right)\) equals.
A) sin \(\left(\frac{\mathrm{R}}{2}\right)\)
B) cos \(\left(\frac{\mathrm{R}}{2}\right)\)
C) cot \(\left(\frac{\mathrm{R}}{2}\right)\)
D) \(\left(\frac{\mathrm{R}}{2}\right)\)
Answer:
C) cot \(\left(\frac{\mathrm{R}}{2}\right)\)

Question 70.
If sin (x – 20)° = cos (3x – 10)°, then ‘x’ is
A) 60
B) 30
C) 45
D) 35.5
Answer:
B) 30

Question 71.
Maximum value of \(\frac{1}{\sec \theta}\), 0° ≤ θ ≤ 90° is
A) 1
B) 2
C) \(\frac{1}{2}\)
D) \(\frac{1}{\sqrt{2}}\)
Answer:
A) 1

Question 72.
The value of cos² 17° – sin²73° is
A) 1
B) \(\frac{1}{3}\)
C) 0
D) -1
Answer:
C) 0

Question 73.
If A = 30°, then sin 2A equals…………..
A) \(\frac{1}{2}\)
B) \(\frac{\sqrt{3}}{2}\)
C) \(\frac{1}{\sqrt{2}}\)
D) 1
Answer:
B) \(\frac{\sqrt{3}}{2}\)

Question 74.
In a right angled ∆ABC, right angle at ‘C’ if tan A = \(\frac{8}{15}\), then the value of cosec² A – 1 is
A) 0
B) \(\frac{64}{225}\)
C) \(\frac{225}{64}\)
D) \(\frac{289}{64}\)
Answer:
C) \(\frac{225}{64}\)

Question 75.
If \(\frac{1}{2}\) tan² 45° = sin²A and ‘A’, acute, then the value of A is
A) 60°
B) 45°
C) 30°
D) 15°
Answer:
B) 45°

Question 76.
sin (45° + θ) cos (45 – θ) =
A) 2 sin θ
B) 0
C) 1
D) 2 cos θ
Answer:
B) 0

Question 77.
If cos 2θ sin 40; here 2θ and 4θ are acute angles, then the value of ‘θ’ is
A) 60°
B) 45°
C) 15°
D) 30°
Answer:
C) 15°

Question 78.
If sin x = cos x, 0 ≤ x ≤ 90°, then x =
A) 30°
B) 90°
C) 0°
D) 45°
Answer:
D) 45°

Question 79.
2 sin θ = sin²θ is true for the value of θ is ………….
A) 0°
B) 45°
C) 30°
D) 60°
Answer:
A) 0°

Question 80.
If sin 45°.cos 45° + cos 60° = tan θ, then the value of θ is
A) 0°
B) 30°
C) 45°
D) 60°
Answer:
C) 45°

Question 81.
From the below figure ON= x; PN = y and Op = r; ∆PON = θ and ∆PON = 90°; sin θ =

Answer:
(D)

Question 82.
cos θ = …………………..
A) \(\frac{x}{r}\)
B) \(\frac{y}{r}\)
C) \(\frac{r}{x}\)
D) \(\frac{y}{x}\)
Answer:
A) \(\frac{x}{r}\)

Question 83.
tan θ =
A) \(\frac{x}{y}\)
B) \(\frac{y}{x}\)
C) \(\frac{r}{x}\)
D) \(\frac{r}{y}\)
Answer:
B) \(\frac{y}{x}\)

Question 84.
sin θ . cosec θ + cos θ . sec θ + tan θ . cot θ =
A) 3
B) 1
C) sinθ.cosθ.tanθ
D) none
Answer:
A) 3

Question 85.
If sinθ. cosec θ = x; then x =
A) 0
B) 1
C) \(\frac{1}{\sin \theta}\)
D) \(\frac{1}{\ cosec \theta}\)
Answer:
B) 1

Question 86.
If sec θ = \(\frac{13}{12}\), then sin θ =
A) \(\frac{5}{13}\)
B) \(\frac{5}{12}\)
C) \(\frac{12}{5}\)
D) \(\frac{12}{13}\)
Answer:
A) \(\frac{5}{13}\)

Question 87.
sin (90 + θ) =
A) cos θ
B) – cos θ
C) sin θ
D) – sin θ
Answer:
A) cos θ

Question 88.
Value of tan² 30° + 2 cot² 60° =
A) \(\frac{2}{3}\)
B) 2
C) 1
D) \(\frac{4}{3}\)
Answer:
C) 1

Question 89.
sin² 75° + cos² 75° =
A) 75
B) 150
C) tan² 75°
D) 1
Answer:
D) 1

Question 90.
sin⁴θ – cos⁴θ =
A) 1
B) cos²θ – sin²θ
C) 2 sin² θ – 1
D) 2 sin²θ
Answer:
C) 2 sin² θ – 1

Question 91.
(1 + tan θ)² =
A) sec² θ
B) sec²θ + 2 tan θ
C) sec²θ + tan²θ
D) sec²θ + tan θ
Answer:
B) sec²θ + 2 tan θ

Question 92.
Expressing tan θ, interms of sec θ.

Answer:
(D)

Question 93.
If 5 sin A = 3; then sec² A – tan² A =
A) \(\frac{9}{25}\)
B) 0
C) \(\frac{25}{9}\)
D) 1
Answer:
D) 1

Question 94.
(sin θ + cos θ)² + (sin θ – cos θ)² =
A) 2 sin²θ+ cos²θ
B) 2
C) 2 sin²θ + 4 cos²θ
D) 2 sin²θ
Answer:
B) 2

Question 95.

In ∆ABC, ∠B = 90°; ∠C = 0. From the figure, tan θ =
A) \(\frac{8}{17}\)
B) \(\frac{15}{8}\)
C) \(\frac{8}{15}\)
D) \(\frac{17}{15}\)
Answer:
B) \(\frac{15}{8}\)

Question 96.
Value of cos 0° + sin 90° + \(\sqrt{2}\) sin 45°
A) 0
B) 2 + \(\sqrt{2}\)
C) 4
D) 3
Answer:
D) 3

Question 97.
Value of 3 sin² 45° + 2cos² 60° =
A) 2
B) 4
C) 32
D) 1 1/2
Answer:
A) 2

Question 98.
Value of cos 240° =
A) \(\frac{1}{2}\)
B) –\(\frac{\sqrt{3}}{2}\)
C) – \(\frac{1}{2}\)
D) none
Answer:
C) – \(\frac{1}{2}\)

Question 99.
If tan θ + cot θ = 2; then tan²θ + cot²θ =
A) 4
B) 2
C) 6
D) 1
Answer:
B) 2

Question 100.
Value of tan 60° – tan 30°
A) \(\frac{1}{\sqrt{3}}-\sqrt{3}\)
B) \(\frac{1}{\sqrt{3}}\)
C) \(\frac{2 \sqrt{3}}{3}\)
D) \(\frac{\sqrt{3}}{3}\)
Answer:
C) \(\frac{2 \sqrt{3}}{3}\)

Question 101.
If sin θ = cos θ, then θ =
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
B) 45°

Question 102.
(1 + tan² 60°)² =
A) 1
B) 2
C) 4
D) 16
Answer:
D) 16

Question 103.
cos (270° – θ)
A) – cos θ
B) – sin θ
C) sin θ
D) cos θ
Answer:
B) – sin θ

Question 104.
When 0° ≤ θ ≤ 90°; the maximum value of sin θ + cos θ is
A) \(\sqrt{2}\)
B) \(\frac{1}{\sqrt{2}}\)
C) 1
D) 2
Answer:
A) \(\sqrt{2}\)

Question 105.
In right angle ∆ABC; ∠B = 90°; tan C = \(\frac{5}{12}\) then the length of hypotenuse is
A) 16
B) 13
C) 21
D) 17
Answer:
B) 13

Question 106.
If A, B are acute angles; sin (A – B) = \(\frac{1}{2}\) sin A = \(\frac{1}{2}\) then B =
A) \(\frac{\pi}{3}\)
B) \(\frac{\pi}{5}\)
C) \(\frac{\pi}{6}\)
D) \(\frac{\pi}{12}\)
Answer:
D) \(\frac{\pi}{12}\)

Question 107.
In ∆ABC, a = 3; b = 4; c = 5 then cos A =
A) \(\frac{3}{5}\)
B) \(\frac{3}{4}\)
C) \(\frac{5}{3}\)
D) \(\frac{4}{5}\)
Answer:
D) \(\frac{4}{5}\)

Question 108.

If sin C = \(\frac{3}{5}\) then cos A =
A) \(\frac{3}{5}\)
B) \(\frac{4}{5}\)
C) \(\frac{5}{4}\)
D) \(\frac{5}{3}\)
Answer:
A) \(\frac{3}{5}\)

Question 109.
sec A. \(\sqrt{1-\sin ^2 A}\) =
A) cos A
B) sec A
C) 0
D) 1
Answer:
D) 1

Question 110.
cot(270° – θ) =
A) -tan θ
B) tan θ
C) cot θ
D) – cot θ
Answer:
B) tan θ

Question 111.
\(\frac{\sin 18^{\prime \prime}}{\cos 72^{\prime \prime}}\) =
A) 1
B) \(\frac{1}{4}\)
C) 0
D) ∞
Answer:
A) 1

Question 112.
If π < θ < \(\frac{3 \pi}{2}\) then θ lies in
A) first quadrant
B) second quadrant
C) third quadrant
D) fourth quadrant
Answer:
C) third quadrant

Question 113.
If sinθ . cosθ = k; then sin θ + cos θ =
A) K²
B) K² – 1
C) \(\sqrt{2 K^2-1}\)
D) \(\sqrt{1+2 \mathrm{~K}}\)
Answer:
D) \(\sqrt{1+2 \mathrm{~K}}\)

Question 114.
If tan² 60° + 2 tan² 45° = x tan 45°; then x =
A) 0
B) 5
C) 1
D) 2
Answer:
B) 5

Question 115.
sin³θ cos θ . cos³ θ . sin θ =
A) sin θ + cos θ
B) sin θ cos θ
C) sin θ
D) cos θ
Answer:
B) sin θ cos θ

Question 116.
\(\frac{\sqrt{1+\tan ^2 \theta}}{\sqrt{1+\cot ^2 \theta}}\) =
A) sin θ
B) cos θ
C) tan θ
D) cot θ
Answer:
C) tan θ

Question 117.
sin² 47° + sin² 43° =
A) 0
B) ∞
C) 1
D) can not be determined
Answer:
C) 1

Question 118.
sec (360° – θ) =
A) cos θ
B) sec θ
C) cosec θ
D) cot θ
Answer:
B) sec θ

Question 119.
If sin θ + cos θ = \(\sqrt{2}\); then value of θ =
A) 0°
B) 30°
C) 45°
D) 60°
Answer:
C) 45°

Question 120.
(1 + cot²45°)²: =
A) 4
B) 2
C) 1
D) \(\sqrt{2}\)
Answer:
A) 4

Question 121.
\(\frac{{\ cosec}^2 \theta}{\cot \theta}\) – cot θ =
A) cot θ
B) cosec θ
C) sec θ
D) tan θ
Answer:
D) tan θ

Question 122.
If the following, which are in geometric progression ?
A) sin 30°, sin 45°, sin 60°
B) sec 30°, sec 45°, sec 60°
C) tan 30°, tan 45°, tan 60°
D) cos 45°, cos 60°, cos 90°
Answer:
C) tan 30°, tan 45°, tan 60°

Question 123.
\(\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}\) =
A) sec x + tan x
B) sec x- tan x
C) 2 tan x
D) 2 sec x
Answer:
A) sec x + tan x

Question 124.
\(\frac{1}{\sec ^2 A}+\frac{1}{{cosec}^2 A}\) =
A) 2
B) 1
C) tan2A + cos2A
D) 0
Answer:
B) 1

Question 125.
If 4 sin 30° . sec 60° = x tan 4°; then x =
A) 0
B) 1
C) 3
D) 4
Answer:
D) 4

Question 126.
Value of sin 60° cos 30° + cos 60°. sin 30°
A) \(\frac{1}{2}\)
B) 1
C) \(\frac{\sqrt{3}}{2}\)
D) \(\frac{2}{\sqrt{3}}\)
Answer:
B) 1

Question 127.
Value of cos 60°. cos 30° + sin 60°. sin 30°
A) \(\frac{\sqrt{3}}{2}\)
B) \(\frac{1}{\sqrt{2}}\)
C) 1
D) \(\frac{2}{\sqrt{3}}\)
Answer:
A) \(\frac{\sqrt{3}}{2}\)

Question 128.
\(\frac{\tan 45^{\prime \prime}}{\ cosec 30^{\prime \prime}}+\frac{\sec 60^{\prime \prime}}{\cot 45^{\prime \prime}}\) =
A) 1 1/2
B) 1
C) 2
D) 2 1/2
Answer:
D) 2 1/2

Question 129.
Value of cos 75° =
A) sin 15°
B) – sin 15°
C) cos 15°
D) \(\frac{\sqrt{3}}{2}\)
Answer:
A) sin 15°

Question 130.
tan θ . cot θ = sec θ . x; then x =
A) cos θ
B) sec θ
C) tan θ
D) cot θ
Answer:
A) cos θ

Question 131.
(1 + tan²A) (1 – sin² A) =
A) sec²A
B) cos²A
C) 1
D) 1 – sin²A + tan²A
Answer:
C) 1

Question 132.
tan 240° =
A) \(\frac{1}{\sqrt{3}}\)
B) \(\sqrt{3}\)
C) –\(\sqrt{3}\)
D) –\(\frac{1}{\sqrt{3}}\)
Answer:
B) \(\sqrt{3}\)

Question 133.
\(\frac{\sin ^4 A-\cos ^4 A}{\sin ^2 A-\cos ^2 A}\) =
A) sin² A – cos² A
B) 1
C) sin² A
D) cos²
Answer:
B) 1

Question 134.
If sin θ = \(\frac{1}{2}\); then cos \(\frac{3 \theta}{2}\)
A) \(\frac{1}{\sqrt{2}}\)
B) \(\frac{\sqrt{3}}{2}\)
C) \(\frac{1}{2}\)
D) \(\frac{2}{\sqrt{3}}\)
Answer:
A) \(\frac{1}{\sqrt{2}}\)

Question 135.
cos \(\left(\frac{3}{2}+\theta\right)\) =
A) cos θ
B) sin θ
C) – sin θ
D) sec θ
Answer:
B) sin θ

Question 136.
\(\frac{\tan \theta \cdot \sqrt{1-\sin ^2 \theta}}{\sqrt{1-\cos ^2 \theta}}\)
A) sin θ
B) cos θ
C) sec θ
D) 1
Answer:
D) 1

Question 137.
If sec θ + tan θ = \(\frac{1}{5}\), then sin θ =
A) \(\frac{5}{13}\)
B) \(\frac{12}{13}\)
C) \(\frac{13}{12}\)
D) \(\frac{5}{13}\)
Answer:
B) \(\frac{12}{13}\)

Question 138.
(sec 45° + tan 45°) (sec 45° – tan 45°) =
A) 1
B) 0
C) 2
D) 2\(\sqrt{2}\)
Answer:
A) 1

Question 139.
If the angle in a triangle are in the ratio of 1 : 2 : 3 then the smallest angle in radius is
A) \(\frac{\pi}{3}\)
B) \(\frac{\pi}{6}\)
C) \(\frac{2\pi}{3}\)
D) \(\frac{\pi}{2}\)
Answer:
B) \(\frac{\pi}{6}\)

Question 140.
\(\sqrt{{\ cosec}^2 \theta-\cot ^2 \theta}\) =
A) cosec θ – cot θ
B) cosec θ + cot θ
C) 1
D) 0
Answer:
C) 1

Question 141.
tan A, in terms of sin A is

Answer:
(D)

Question 142.
\(\frac{1}{1-\sin \theta}\) + \(\frac{1}{1+\sin \theta}\) =
A) 2 tan² θ
B) 2 sec² θ
C) 2 cosec² θ
D) 2 cot² θ
Answer:
B) 2 sec² θ

Question 143.
If cot² θ = 3; then cosec θ =
A) 4
B) 2
C) 3
D) 1
Answer:
B) 2

Question 144.
If cos θ = – cos θ, then θ is
A) 60°
B) 45°
C) 30°
D) 90°
Answer:
B) 45°

Question 145.
If A is acute and tan A = \(\frac{1}{\sqrt{3}}\); then sin A = ……………….
A) \(\frac{1}{2}\)
B) 1
C) \(\sqrt{2}\)
D) \(\frac{\sqrt{3}}{2}\)
Answer:
A) \(\frac{1}{2}\)

Question 146.
If sin θ = \(\frac{\mathrm{a}}{\mathrm{b}}\); cos θ = \(\frac{\mathrm{c}}{\mathrm{d}}\) ;then tan θ =
A) \(\frac{\mathrm{bc}}{\mathrm{ad}}\)
b) \(\frac{\mathrm{ac}}{\mathrm{bd}}\)
C) \(\frac{\mathrm{ab}}{\mathrm{cd}}\)
D) \(\frac{\mathrm{ad}}{\mathrm{bc}}\)
Answer:
D) \(\frac{\mathrm{ad}}{\mathrm{bc}}\)

Question 147.
If sin A = \(\frac{1}{\sqrt{2}}\) ; then tan A =
A) 3
B) 4
C) 1
D) \(\sqrt{2}\)
Answer:
C) 1

Question 148.
If sec θ = cosec θ; then value of θ =
A) \(\frac{\pi}{2}\)
B) \(\frac{\pi}{4}\)
C) \(\frac{\pi}{6}\)
D) \(\frac{\pi}{3}\)
Answer:
B) \(\frac{\pi}{4}\)

Question 149.
The radius of a circle is ‘r’; an arc of length ‘l’ is making an angle θ, at the centre of the circle, then θ =
A) \(\frac{\mathrm{l}}{\mathrm{r}}\)
B) \(\frac{\mathrm{r}}{\mathrm{l}}\)
C) lr
D) l + r
Answer:
A) \(\frac{\mathrm{l}}{\mathrm{r}}\)

Question 150.
tan (A + B) =

Answer:
(B)

Question 151.
tan(A – B) =

Answer:
(C)

Question 152.
\(\frac{\sqrt{\ cosec^2 \theta-1}}{\ cosec \theta}\) =
A) 1+ sec θ
B) cosec θ + cot θ
C) cos θ
D) tan θ
Answer:
C) cos θ

Question 153.
\(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\) =
A) tan θ
B) cosec θ
C) cot θ
D) sec θ
Answer:
A) tan θ

Question 154.
\(\frac{\sqrt{\sec ^2 \theta-1}}{\sec \theta}\) =
A) cosec θ
B) sin θ
C) cosec θ – cot θ
D) 2 sec θ
Answer:
B) sin θ

Question 155.
If x = 2 cosec θ; y = 2 cot θ; then x² – y² =
A) 4
B) 0
C) 1
D) 2
Answer:
A) 4

Question 156.
If cos(A + B) = θ, cos B = \(\frac{\sqrt{3}}{2}\); then A is
A) 60°
B) 180°
C) 15°
D) 115°
Answer:
A) 60°

Question 157.
\(\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}\) =
A) tan θ
B) cos θ
C) sec θ
D) cot θ
Answer:
A) tan θ

Question 158.
\(\frac{1}{\sqrt{1+\tan ^2 \theta}}\) =
A) sin θ
B) cos θ
C) sec θ
D) cosec θ
Answer:
B) cos θ

Question 159.
cos θ . tan θ =
A) cos θ
B) cot θ
C) sin θ
D) cos² θ
Answer:
C) sin θ

Question 160.
sin 225° =
A) \(\frac{1}{\sqrt{2}}\)
B) –\(\frac{1}{\sqrt{2}}\)
C) \(\frac{\sqrt{3}}{2}\)
D) \(\frac{2}{\sqrt{3}}\)
Answer:
B) –\(\frac{1}{\sqrt{2}}\)

Question 161.
sec² 33° – cot² 57° =
A) 0
B) 1
C) -1
D) \(\frac{1}{2}\)
Answer:
B) 1

Question 162.
sin 180° =
A) 0
B) 1
C) -1
D) ∞
Answer:
A) 0

Question 163.
If cos θ = \(\frac{1}{2}\) ; then cos \(\frac{\theta}{2}\) =
A) \(\frac{1}{4}\)
B) \(\frac{\sqrt{3}}{2}\)
C) \(\frac{1}{2}\)
D) 1
Answer:
B) \(\frac{\sqrt{3}}{2}\)

Question 164.
sin (A + B).cos (A – B) + sin (A – B). cos (A + B) =
A) cos 2A
C) sin 2A
B) cos 2B
D) sin 2B
Answer:
B) cos 2B

Question 165.
Value of cos 60° . cos30° – sin 60° . sin 30° =
A) 1
B) 0
C) \(\frac{1}{4}\)
D) \(\frac{1}{2}\)
Answer:
B) 0

Question 166.
cos 300° =
A) \(\frac{\sqrt{3}}{2}\)
B) 1
C) 0
D) \(\frac{1}{2}\)
Answer:
D) \(\frac{1}{2}\)

Question 167.
If sin (A + B) = 1; sin B = \(\frac{1}{2}\) ; then A =
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
C) 60°

Question 168.
\(\sqrt{\sec ^2 A+\ cosec^2 A}\) =
A) cos A + sin A
B) sec A + cosec A
C) tan A+ cot A
D) 1
Answer:
C) tan A+ cot A

Question 169.
If \(\sqrt{3}\) tan θ = 1; then θ =
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
A) 30°

Question 170.
\(\sqrt{\ cosec ^2 \theta-\sin ^2 \theta-\cos ^2 \theta}\) =
A) cot θ
B) tan θ
C) sec θ
D) cosec θ
Answer:
A) cot θ

Question 171.
sin 750° =
A) \(\frac{1}{\sqrt{2}}\)
B) \(\frac{1}{2}\)
C) 1
D) \(\frac{\sqrt{3}}{2}\)
Answer:
B) \(\frac{1}{2}\)

Question 172.
Cosec 60° × cos 90° =
A) ∞
B) \(\frac{2}{\sqrt{3}}\)
C) 0
D) 3
Answer:
C) 0

Question 173.
sin 81° = ………………
A) cos 9°
B) cos 81°
C) – cos 9°
D) can not be possible to determine without tables
Answer:
A) cos 9°

Question 174.
If sin θ = \(\frac{11}{15}\) ; then cos θ =
A) \(\frac{\sqrt{26}}{7}\)
B) \(\frac{2 \sqrt{26}}{3}\)
C) \(\frac{2 \sqrt{26}}{15}\)
D) none
Answer:
C) \(\frac{2 \sqrt{26}}{15}\)

Question 175.
If tan θ = \(\sqrt{3}\), then sec θ = ……………….
A) 2
B) -2
C) 4
D) 5
Answer:
A) 2

Question 176.
If 3 cot θ = 5, then \(\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+3 \cos \theta}\) = ………………..
A) -1
B) 1
C) 7
D) 0
Answer:
D) 0

Question 177.
(1 + tan²θ) cos²θ = …………….
A) 1
B) 0
C) 8
D) 14
Answer:
A) 1

Question 178.
(sec²θ – 1) (1 – cosec²θ) = ………………
A) 2
B) -1
C) 3
D) -4
Answer:
B) -1

Question 179.
cot²θ – \(\frac{1}{\sin ^2 \theta}\)
A) 4
B) -3
C) 2
D) -1
Answer:
D) -1

Question 180.
If cos θ . sin θ = \(\frac{1}{2}\) ; then θ = …………….
A) 1
B) -1
C) 3
D) 4
Answer:
A) 1

Question 181.
If cos θ = -cos θ; then θ in radian measure is ………….
A) π^c
B) \(\frac{\pi^c}{2}\)
C) \(\frac{\pi^c}{3}\)
D) \(\frac{\pi^c}{7}\)
Answer:
C) \(\frac{\pi^c}{3}\)

Question 182.
If sin A = \(\frac{3}{5}\) ; then sin (90 + A) = ………….
A) \(\frac{4}{5}\)
B) \(\frac{5}{4}\)
C) \(\frac{1}{3}\)
D) \(\frac{2}{3}\)
Answer:
A) \(\frac{4}{5}\)

Question 183.
\(\sqrt{(\sec \theta+1)(\sec \theta-1)}\) = ……………..
A) cot θ
B) tan θ
C) cos θ
D) sin θ
Answer:
B) tan θ

Question 184.
\(\sqrt{\sec ^2 \theta-\tan ^2 \theta+\cot ^2 \theta}\) = …………………..
A) -cos θ
B) 1
C) sec θ
D) cosec θ
Answer:
D) cosec θ

Question 185.
cos 150° = …………….
A) –\(\frac{\sqrt{3}}{2}\)
B) – \(\sqrt{3}\)
C) –\(\frac{1}{2}\)
D) none
Answer:
A) –\(\frac{\sqrt{3}}{2}\)

Question 186.
sin² 75° + cos² 75° = ………………..
A) 3
B) 2
C) 4
D) 1
Answer:
D) 1

Question 187.
sin 240° + sin 120° = …………..
A) 0
B) -1
C) 3
D) none
Answer:
A) 0

Question 188.
If cosec θ + cot θ = 2; then cosec θ – cot θ = ………………
A) -1
B) 2
C) \(\frac{1}{2}\)
D) 3
Answer:
C) \(\frac{1}{2}\)

Question 189.
If sec A + tan A = \(\frac{1}{3}\); then sec A – tan A = ……………
A) 4
B) 1
C) -3
D) 3
Answer:
D) 3

Question 190.
sin 30° + cos 60° = ………………
A) 1
B) 4
C) 3
D) none
Answer:
A) 1

Question 191.
(cos A + sin A)² + (cos A – sin A)² = ………………..
A) 1
B) 2
C) 4
D) none
Answer:
B) 2

Question 192.
sin 450° = ……………….
A) 4
B) 2
C) -1
D) none
Answer:
D) none

Question 193.
cos(A + B) = ……………..
A) cos A cos B – sin A sin B
B) cos A sec B – sin A sin B
C) cos A cos B + sin A sec B
D) none
Answer:
A) cos A cos B – sin A sin B

Question 194.
tan(A – B) = ………….
A) tan A – cos B
B) tan B – tan A
C) \(\frac{\tan B-\tan A}{1+\tan A+\tan B}\)
D) none
Answer:
C) \(\frac{\tan B-\tan A}{1+\tan A+\tan B}\)

Question 195.
tan (360 – θ) = ……………….
A) sin θ
B) sec θ
C) tan θ
D) -tan θ
Answer:
D) -tan θ

Question 196.
The value of tan 75° = …………..
A) 2 + \(\sqrt{3}\)
B) 2 – \(\sqrt{3}\)
C) \(\sqrt{3}\) – 1
D) none
Answer:
A) 2 + \(\sqrt{3}\)

Question 197.
cos 110°. cos 70° – sin 110°.sin 70° = ……………..
A) 4
B) 1
C) -1
D) 3
Answer:
C) -1

Question 198.
Express tan θ, interms of sin θ = ………………
A) \(\frac{\cos \theta}{1-\sin \theta}\)
B) \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
C) \(\frac{\sin \theta}{1+\cos \theta}\)
D) none
Answer:
B) \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)

Question 199.
cosec 60° . sec 60° = ……………….
A) \(\frac{1}{2}\)
B) \(\frac{\sqrt{3}}{4}\)
C) \(\frac{4}{3}\)
D) \(\frac{4}{\sqrt{3}}\)
Answer:
D) \(\frac{4}{\sqrt{3}}\)

Question 200.
sin(A + B) =
A) sin A cos B + cos A sin B
B) sin A – cos B sin B
C) sin A cos B – sin A sin B
D) none
Answer:
A) sin A cos B + cos A sin B

Question 201.
cos (180 – θ) = ……………..
A) – cos θ
B) cos θ
C) sec θ
D) none
Answer:
A) – cos θ

Question 202.
sin 2A = ………………
A) 2 sin A cos A
B) cos A sin A
C) sin² A
D) cos² A
Answer:
A) 2 sin A cos A

Question 203.
\(\sqrt{\ cosec ^2 \theta-\sin ^2 \theta-\cos ^2 \theta}\) = ……………….
A) -tan θ
B) -sin θ
C) sec θ
D) cot θ
Answer:
D) cot θ

Question 204.
(1 – sec²θ) (1 – cosec²θ) = ……………….
A) 3
B) -1
C) 4
D) 1
Answer:
D) 1

Question 205.
If cos θ = \(\frac{3}{5}\); then cos (-θ) = …………….
A) \(\frac{-3}{5}\)
B) \(\frac{3}{5}\)
C) \(\frac{1}{5}\)
D) \(\frac{1}{4}\)
Answer:
B) \(\frac{3}{5}\)

Question 206.
2 sin 45°. cos 45° = …………….
A) 1
B) 4
C) -1
D) none
Answer:
A) 1

Question 207.
If cot = x; then cosec θ = ………….
A) \(\sqrt{2 x+1}\)
B) \(\sqrt{1+x}\)
C) \(\sqrt{x^2+1}\)
D) none
Answer:
C) \(\sqrt{x^2+1}\)

Question 208.
In the figure, AB = ………………

A) 16 \(\sqrt{3}\)
B) 10 \(\sqrt{3}\)
C) 9\(\sqrt{3}\)
D) 20\(\sqrt{3}\)
Answer:
D) 20\(\sqrt{3}\)

Question 209.
sin 45°.cos 45° + \(\sqrt{3}\) sin 60° = ………..
A) 2
B) -2
C) 3
D) none
Answer:
A) 2

Question 210.
\(\frac{\sqrt{\sec ^2 A-1}}{\sec A}\) = ………………
A) sec A
B)-cos A
C) cos A
D) sin A
Answer:
D) sin A

Question 211.
\(\frac{\sqrt{\ cosec ^2 \theta-1}}{\ cosec \theta}\) = …………..
A) -sin θ
B)-cos θ
C) cos θ
D) none
Answer:
C) cos θ

Question 212.
If α + β = 90° and α = 2β; then cos² α + sin² β = ………………
A) \(\frac{-1}{2}\)
B) -1
C) 2
D) \(\frac{1}{2}\)
Answer:
D) \(\frac{1}{2}\)

Question 213.
tan 30° + cot 30° = ………….
A) \(\frac{4}{\sqrt{3}}\)
B) \(\frac{4}{3}\)
C) \(\frac{\sqrt{3}}{4}\)
D) none
Answer:
A) \(\frac{4}{\sqrt{3}}\)

Question 214.
\(\sqrt{\tan ^2 \theta+\cot ^2 \theta+2}\) = ………………..
A) tan θ – cos θ
B) tan θ
C) tan θ + cot θ
D) tan θ – cot θ
Answer:
C) tan θ + cot θ

Question 215.
If sin θ = \(\frac{\mathrm{a}}{\mathrm{b}}\); then tan θ = …………….

Answer:
(D)

Question 216.
tan θ = \(\frac{1}{\sqrt{3}}\) ; cos θ = ……………
A) \(\frac{\sqrt{3}}{2}\)
B) \(\sqrt{3}\)
C) \(\frac{2}{\sqrt{3}}\)
D) \(\frac{1}{2}\)
Answer:
A) \(\frac{\sqrt{3}}{2}\)

Question 217.
If 5 sin A = 3; sec²A – tan²A = ……………..
A) 3
B) -1
C) 4
D) 1
Answer:
D) 1

Question 218.
sin (-θ) = …………………..
A) cos θ
B) -tan θ
C) sec θ
D) -sin θ
Answer:
D) -sin θ

Question 219.
cos (-θ) = ………………..
A) sec θ
B) -cos θ
C) cos θ
D) 1
Answer:
C) cos θ

Question 220.
sin (180 – θ) = ………………
A) cos θ
B) sin θ
C) tan θ
D) 0
Answer:
B) sin θ

Question 221.
cos (270 – θ)
A) cos θ
B) sin θ
C) – sin θ
D) none
Answer:
C) – sin θ

Question 222.
tan (360 – θ) = ……………
A) -tan θ
B) tan θ
C) sec θ
D) cos θ
Answer:
A) -tan θ

Question 223.
cosec (270 – θ) = ……………..
A) sec θ
B) -sec θ
C) tan θ
D) none
Answer:
C) tan θ

Question 224.
sec (90 + θ) = …….
A) tan θ
B) cosec θ
C) – cos θ
D) – cosec θ
Answer:
D) – cosec θ

Question 225.
cos 240° = ……………..
A) \(\frac{-1}{2}\)
B) -1
C) \(\frac{2}{3}\)
D) -3
Answer:
A) \(\frac{-1}{2}\)

Question 226.
sin 420° = ………….
A) – \(\frac{\sqrt{3}}{2}\)
B) 1
C) \(\frac{\sqrt{3}}{2}\)
D) \(\frac{2}{\sqrt{2}}\)
Answer:
C) \(\frac{\sqrt{3}}{2}\)

Question 227.
tan 750° = ……………..
A) \(\frac{-1}{\sqrt{3}}\)
B) \(\sqrt{3}\)
C) -1
D) \(\frac{1}{\sqrt{3}}\)
Answer:
D) \(\frac{1}{\sqrt{3}}\)

Question 228.
cosec 300° = ………………
A) \(\frac{-2}{\sqrt{3}}\)
B) \(\frac{1}{\sqrt{3}}\)
C) \(\frac{-1}{2}\)
D) –\(\frac{1}{\sqrt{3}}\)
Answer:
A) \(\frac{-2}{\sqrt{3}}\)

Question 229.
sec 240° = …………..
A) 3
B) -1
C) 2
D) -2
Answer:
D) -2

Question 230.
sin² 47° + sin² 43° = ……………
A) 1
B) -1
C) 3
D) none
Answer:
A) 1

Question 231.
If cosec θ – cot θ = 4, then cosec θ + cot θ = ……………….
A) 1
B) \(\frac{1}{2}\)
C) \(\frac{-1}{4}\)
D) \(\frac{1}{4}\)
Answer:
D) \(\frac{1}{4}\)

Question 232.
tan θ in terms of cosec θ = ………………….
A) \(\frac{1}{\sqrt{\ cosec ^2 \theta-1}}\)
B) \(\frac{1}{\sqrt{1+\ cosec \theta}}\)
C) \(\frac{1}{1+\tan ^2 \theta}\)
D) none
Answer:
A) \(\frac{1}{\sqrt{\ cosec ^2 \theta-1}}\)

Question 233.
\(\frac{\ cosec ^2 \theta}{\cot \theta}\) – cot θ = …………………
A) cot θ
B) sec θ
C) tan θ
D) none
Answer:
C) tan θ

Question 234.
\(\frac{1-\tan ^2 30^{\prime \prime}}{1+\tan ^2 30^{\prime \prime}}\) = ………………
A) -2
B) \(\frac{-1}{2}\)
C) 1
D) \(\frac{1}{2}\)
Answer:
D) \(\frac{1}{2}\)

Question 235.
\(\frac{1}{\sec ^2 A}+\frac{1}{\ cosec ^2 A}\) = ………………
A) -1
B) 1
C) 3
D) 4
Answer:
B) 1

Question 236.
tan θ. cot θ = sec θ. x; then x = …………………….
A) cos θ
B) – cos θ
C) tan θ
D) none
Answer:
A) cos θ

Question 237.
If sin (A + B) = \(\frac{\sqrt{3}}{2}\) ; cos B = \(\frac{\sqrt{3}}{2}\) then A = ……………….
A) 70°
B) 45°
C) 60°
D) 30°
Answer:
D) 30°

Question 238.
cos \(\left(\frac{3 \pi}{2}+\theta\right)\) = …………….
A) tan θ
B) cos θ
C) -sin θ
D) sin θ
Answer:
D) sin θ

Question 239.
sec θ + tan θ = \(\frac{1}{2}\); then sin θ = …………….
A) \(\frac{2}{13}\)
B) \(\frac{1}{13}\)
C) \(\frac{12}{13}\)
D) \(\frac{13}{2}\)
Answer:
C) \(\frac{12}{13}\)

Question 240.
If sec = cosec θ; then the value of θ in radians = …………………
A) \(\frac{\pi^c}{2}\)
B) \(\frac{\pi^c}{4}\)
C) \(\frac{\pi^c}{3}\)
D) \(\frac{\pi^c}{12}\)
Answer:
B) \(\frac{\pi^c}{4}\)

Question 241.
Maximum value of sin θ + cos θ = ……………..
A) 3
B) \(\sqrt{3}\)
C) 2
D) \(\sqrt{2}\)
Answer:
D) \(\sqrt{2}\)

Question 242.
Maximum value of cos θ = …………………
A) 1
B) -1
C) 2
D) 0
Answer:
A) 1

Question 243.
Minimum and maximum value of tan θ = ……………
A) (-∞, ∞)
B) (- ∞, 0)
C) (3, 2)
D) (1, -1)
Answer:
A) (-∞, ∞)

Question 244.
\(\frac{\sin ^4 \theta-\cos ^4 \theta}{\sin ^2 \theta-\cos ^2 \theta}\) = ……………….
A) 2
B) -1
C) 1
D) none
Answer:
C) 1

Question 245.
cos 0° + sin 90° + \(\sqrt{3}\) cosec 60° = ………………
A) 0
B) -1
C) 3
D) 4
Answer:
D) 4

Question 246.
\(\left|\begin{array}{l}
\tan \theta \sec \theta \\
\sec \theta \tan \theta
\end{array}\right|\) = …………………
A) -1
B) -4
C) 1
D) none
Answer:
C) 1

Question 247.
If cosec θ + cot θ = 3, then cosec θ – cot θ = …………….
A) \(\frac{1}{2}\)
B) \(\frac{1}{3}\)
C) \(\frac{-1}{3}\)
D) none
Answer:
B) \(\frac{1}{3}\)

Question 248.
\(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\) = ……………..
A) -cos θ
B) -sin θ
C) sin θ
D) sin² θ
Answer:
C) sin θ

Question 249.
\(\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}\) = ……………
A) -cot θ
B) tan θ
C) sec θ
D) none
Answer:
B) tan θ

Question 250.
x = 2 cosec θ; y = 2 cot θ; x² – y² = …………….
A) 4
B) -1
C) -3
D) 2
Answer:
A) 4

Question 251.
\(\sqrt{\sec ^2 A+\ cosec ^2 A}\) = ……………….
A) tan A – cos A
B) tan A + cos A
C) 1
D) tan A + cot A
Answer:
D) tan A + cot A

Question 252.
sin 81° = ……………….
A) cos 9°
B) cos 20°
C) tan 9°
D) none
Answer:
A) cos 9°

Question 253.
\(\frac{\sqrt{\sec ^2 \theta-1}}{\sec \theta}\) = ………………
A) – tan θ
B) cos θ
C) sin θ
D) none
Answer:
C) sin θ

Question 254.
If sin A = \(\frac{1}{\sqrt{2}}\); then tan A = ……………
A) 4
B) 3
C) -1
D) 1
Answer:
D) 1

Question 255.
sin 225° = ………………
A) \(\frac{-1}{\sqrt{2}}\)
B) \(\sqrt{2}\)
C) \(\frac{-1}{2}\)
D) 1
Answer:
A) \(\frac{-1}{\sqrt{2}}\)

Question 256.
cos (x – y) = ………………
A) cos x cos y + sin x sin y
B) cos x – sin x sin y
C) cos x cos y – 1
D) all
Answer:
A) cos x cos y + sin x sin y

Question 257.
\(\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\) = ………………..
A) sec⁴ θ
B) sec θ
C) \(\frac{\sec ^2 \theta}{2}\)
D) 2 sec² θ
Answer:
D) 2 sec² θ

Question 258.
sin⁴ θ – cos⁴ θ = ………………..
A) 2 sec² θ – 1
B) sec² θ + 1
C) sec² θ – 3
D) none
Answer:
A) 2 sec² θ – 1

Question 259.
tan θ is not defined if θ = ……………..
A) 0°
B) 70°
C) 90°
D) 20°
Answer:
C) 90°

Question 260.
If cosec θ = \(\frac{25}{7}\) ; then cot θ = …………….
A) \(\frac{4}{7}\)
B) \(\frac{7}{24}\)
C) \(\frac{4}{23}\)
D) \(\frac{24}{7}\)
Answer:
D) \(\frac{24}{7}\)

Question 261.
tan 26°. tan 64° = …………..
A) 1
B) -1
C) 3
D) 7
Answer:
A) 1

Question 262.
If tan 2A = cot (A – 18°) where 2A is an acute angle then A = ……………
A) 116°
B) 20°
C) 16°
D) 36°
Answer:
D) 36°

Question 263.
sin(90 – Φ) = ……………
A) cos Φ
B) sin Φ
C) – cos Φ
D) 0
Answer:
A) cos Φ

Question 264.
In ∆ABC, sin \(\left(\frac{B+C}{2}\right)\) = ………………
A) cos \(\frac{A}{2}\)
B) cos \(\frac{C}{2}\)
C) tan \(\frac{A}{2}\)
D) 1
Answer:
A) cos \(\frac{A}{2}\)

Question 265.
\(\frac{\sec 35^{\prime}}{\ cosec 55^{\prime \prime}}\) = …………………
A) -3
B) 8
C) 4
D) 1
Answer:
D) 1

Question 266.
sin \(\frac{\pi^c}{4}\) + cos 45° = ……………….
A) 2
B) \(\sqrt{2}\)
C) -1
D) 0
Answer:
B) \(\sqrt{2}\)

Question 267.
sec 0° = ………………..
A) -1
B) 1
C) 0
D) 7
Answer:
B) 1

Question 268.
sec θ – tan θ = \(\frac{1}{n}\) then sec θ + tan θ = ………………
A) -n
B) -1
C) n
D) none
Answer:
C) n

Question 269.
If 3 tan A = 4 then cos A = ……………..
A) \(\frac{1}{2}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{7}\)
D) none
Answer:
D) none

Question 270.
From the figure, sin C = …………..

Answer:
(A)

Question 271.
cos² θ = ……………….
A) 1 + sin² θ
B) 1 – sin² θ
C) 1 – sin θ
D) 1 + cos θ
Answer:
B) 1 – sin² θ

Question 272.
sec θ is not defined if θ = …………………
A) 0°
B) 90°
C) 30°
D) 45°
Answer:
B) 90°

Question 273.
(1 + tan² 60)² = ………………..
A) 1
C) 16
B) 10
D) 12
Answer:
B) 10

Question 274.
Reciprocal of cot A = ……………..
A) sin A
B) sin² A
C) sec² A
D) tan A
Answer:
D) tan A

Question 275.
sin A = cos B then A + B = …………….
A) 20°
B) 70°
C) 90°
D) none
Answer:
C) 90°

Question 276.
Trigonometry was introduced by ……………….
A) Cantor
B) Cayley
C) Hipparchus
D) none
Answer:
C) Hipparchus

Question 277.
If tan A = \(\frac{3}{4}\) then sec² A – tan² A = ………………..
A) 4
B) 3
C) -1
D) 1
Answer:
D) 1

Question 278.
tan² Φ sec² Φ = ………………..
A) -1
B) 1
C) 3
D) 0
Answer:
A) -1

Question 279.
In the figure, tan X = ……………..

Answer:
(B)

Question 280.
If sin θ = cos 66° then θ = ………………
A) 30°
B) 24°
C) 36°
D) 48°
Answer:
B) 24°

Question 281.
If sec = \(\frac{\mathrm{X}}{\cos \theta}\) then X = ……………….
A) \(\frac{1}{2}\)
B) 0
C) -1
D) 1
Answer:
D) 1

Question 282.
Which of the following is not the value of sin θ ?
A) 1
B) \(\frac{3}{4}\)
C) \(\frac{4}{3}\)
D) \(\frac{1}{2}\)
Answer:
B) \(\frac{3}{4}\), C) \(\frac{4}{3}\)

Question 283.
Which of the following is not correct?
A) cos 0°
B) sin 90° = 0
C) tan 45° = cot 45°
D) both A and B
Answer:
D) both A and B

Question 284.
(sec A + tan A) (1 = sin A) = ……………….
A) sec A
B) sin A
C) cosec A
D) cos A
Answer:
D) cos A

Question 285.
If sec θ + tan θ = X then cosec θ = ……………..

Answer:
(C)

Question 286.
From the adjacent figure, \(\frac{c}{a}=\frac{29}{21}\) represents

A) cos θ
B) cosec θ
C) cot θ
D) sin θ
Answer:
B) cosec θ

Question 287.
From the adjacent figure, value of ‘sin²A + cos²A’

Answer:
(D)

We are offering TS 10th Class Maths Notes Chapter 2 Sets to learn maths more effectively.

TS 10th Class Maths Notes Chapter 2 Sets

→ A set is a well-defined collection of objects.
E.g.: The collection of instruments in a Mathematics Instrument Box.
The collection of districts in Andhra Pradesh.

→ An object belonging to a set is known as an element of the set.
For example, if B = {1, 2, 3, 4, 5}
1, 2, 3, 4, 5 are the elements of set B.

→ We use upper case letters A, B, C,…., X, Y, Z etc., to denote a set while the elements of a set are represented by small letters a, b, c,…. etc.

→ If ‘x’ is an element of set A.
We say that ‘x’ belongs to ‘A’ and we write xe A.
If Y is not element of set B.
We say that ‘x does not belongs to B’ and we write xg B.

→ A set can be represented in two ways.

• Roster (or) Tabular form
• Set builder form.

→ Roster form :

• In roster form, all the elements of a set are list the elements being separated by commas and are enclosed within braces (curly brackets).
• For example, the set of all even positive integers less than 20 is described in the roaster form as {2, 4, 6, 8, 10,12,14,16,18}.
• The set of all vowels in english alphabet can be written as V = {a, e, i, o, u}
• It may be noted while writing the set in roaster form, an element is not generally repeated.
• For example, the set of letters forming the word ‘FOLLOWER’ is written as {F, O, L, W, E, R}.

→ Set builder form :
In set builder form, we use a symbol x (or any other symbol like the letters y, z etc.,) for the elements of the set. After that we write either a colon or a vertical line. Then we write the characteristic property possessed by the elements of the set. Lastly we enclose the description within braces.
Let A = (2, 3, 5, 7,11,13,17}. This is the set of all prime numbers less than 19. It can be represented in set builder form as {x/x is a prime number less than 19}

→ Null set:
A set which does not contain any element is called the empty set or the null set or the void set. The empty/ null set is denoted by the symbol Φ or {}.
E.g.: Let P = {x : 1 < x < 2 and x e N}
P is an empty set because there is no natural number between 1 and 2.

→ Finite set:
A set is called a finite set if it is possible to count the number of elements in it.
C = {x, y, z} and A = {1, 2, 3, 4, 5}

→ Infinite set:
A set is called an infinite set if the number of elements in it is not finite (i.e.,) we cannot count the number of elements in it.
For example, E = {x : x is a multiple of 3}
0 = {x/x is an odd natural number}

→ Cardinal number of a set:
The number of elements in a set is called the cardinal number of the set.
If A = {1, 2, 3, 4, 5}, n(A) = 5 If B = {5, 6, 7}, n(B) = 3 But n(Φ) = 0

→ Universal set; subset:
If we consider all the students in a school as universal set, then the students in any class of that school will be a subset of it.
The subset is denoted by the symbol: ⊂
A set S is a subset of a set R, if every element of S is an element of R.
i. e., S ⊂ R
S ⊂ R whenever x ∈ S, then x ∈ R.
If U = {a, b, c,…., x, y, z} and V = (a, e, i, o, u}, then V ⊂ U.
If we consider the alphabet in English as universal set, then the set of vowels is a subset of it.

→ Equal sets :
Two sets A and B are said to be equal if every element of A is also an element of B and if every element of B is also an element of A.
In other words, two sets A and B are said to be equal if they have exactly the same elements.
We write A = B
E.g.: A = {5, 8,10,11}, B = {10, 5, 11, 8}
Then A = B

→ Equivalent sets:
Two finite sets A and B are said to be equivalent if they have the same number of elements. We write A = B
For example, let A = {x, y, z} and B = {2, 4, 6} then A and B are equivalent sets.

→ Difference of sets :
Let A and B are two sets. The difference of sets A and B, in the same order, is the set of elements that belong to A but not to B.
We write it as A – B and read as ‘A’ difference B’.
A – B = {x: x ∈ A and x ∉ B}

→ Union of sets :
Let A and B be any two sets. The union of A and B is the set that contains all the elements of A and also the elements of B, the common elements being taken only once.

Symbolically,
A ∪ B = {x : x ∈ A or x ∈ B}
E.g.: If A = {1, 3, 5, 7} and B = {5, 7, 9,11} then A ∪ B = {1, 3, 5, 7, 9,11}
We read A ∪ B as ‘A union B’.

→ Intersection of sets :
Let A and B be any two sets. The intersection of sets A and B is the set of all elements which are common to both A and B.

Symbolically,
A ∩ B = {x : x ∈ A and X ∈ B}
E.g.: If A = {2,4, 6, 8} and B = {6, 8,10,12} then A ∩ B = {6, 8}
We read A ∩ B as ‘A intersection B’.

→ Disjoint sets:
Two sets A and B are said to be disjoint, if A ∩ B = Φ

i. e., there is no common element in the sets.
E.g.: A = {1, 2, 3, 4, 5} and B = {6, 7, 8}
Here, A and B have no elements in common.
∴ A and B are called disjoint sets.

→ Venn diagrams:
Venn diagrams are a convenient way of showing operations between sets. Most of the ideas about sets and their properties can be visualised by means of diagrams. These diagrams are known as venn diagrams because they are named after John Venn. In these diagrams, the universal set is represented by the interior of a rectangle and its subsets are represented by the interior of circles.
(i) n(A) + n(B) – n(A ∪ B) = n(A ∩ B)
(ii) n(A) + n(B) – n(A ∩ B) = n(A ∪ B)
E.g.:
(i) If n(A) = 7, n(B) = 8, n(A ∩ B) = 3, find n(A ∪ B).
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 7 + 8 – 3 = 12
(ii) If n(A) = 14, n(B) = 5, n(A ∪ B) = 16, find n(A ∩ B).
n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 14 + 5 – 16
= 19 – 16 = 3

Important Formula:

• Null Set. P = {x: 1 < x < 2 and x ∈ N}
• A set is called Finite Set. If it is possible to count.
• A set is called Infinite Set. If it is not possible to count.
• The number of elements in a set is called the cardinal number of the set.
• Two sets A and B are said to be equal if they have the same elements, i.e A = B
• Symbol of subset is ⊂.

Flow Chat:

George Cantor(1845 – 1918A.D):

• George Cantor was a German mathematician, best known as the inventor of set theory, which has become a fundamental theory in mathematics.
• Cantor established the importance of one-to-one correspondence between the numbers of two sets, defined infinite and well-ordered sets and proved that the real numbers are “more numerous” than the natural numbers.
• He defined the cardinal and ordinal numbers and their arithmetic.
• Cantor’s work is of great philosophical interest, a fact of which he was well aware.

Solving these TS 10th Class Maths Bits with Answers Chapter 12 Applications of Trigonometry Bits for 10th Class will help students to build their problem-solving skills.

Applications of Trigonometry Bits for 10th Class

Question 1.
If a pole 6 m high casts a shadow 2 \(\sqrt{3}\) m long on the ground, then the sun’s angle of elevation is
A) 60°
B) 45°
C) 30°
D) 90°
Answer:
A) 60°

Question 2.
If the angle of elevation of a tower from a distance of 100 m from its foot is 60°then the height of the tower is …………. m.
A) 100\(\sqrt{3}\)
B) \(\frac{100}{\sqrt{3}}\)
C) 50 \(\sqrt{3}\)
D) \(\frac{50}{\sqrt{3}}\)
Answer:
A) 100\(\sqrt{3}\)

Question 3.
The height of a tower is 10 m. The length of its shadow when Sun’s altitude is 45° is …………….. m.
A) 10
B) 20
C) 10\(\sqrt{3}\)
D) 50
Answer:
A) 10

Question 4.
The length of the shadow of a tower on the plane ground is \(\sqrt{3}\) times the height of the tower, the angle of elevation of sun is
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
A) 30°

Question 5.
The ratio of the length of a rod and it’s shadow is 1 : \(\sqrt{3}\) then the angle of elevation of the Sun is
A) 45°
B) 30°
C) 75°
D) 90°
Answer:
B) 30°

Question 6.
If two towers of height X and Y subtend angles of 30° and 60°respectively at the centre of the line joining their feet, then X : Y is equal to
A) 1 : 3
B) 1 : \(\sqrt{3}\)
C) 3 : 1
D) \(\sqrt{3}\) : 1
Answer:
A) 1 : 3

Question 7.
A wall of 8 m long casts a shadow 5 m long at the same time a tower casts a shadow 50m long, then the height of tower is
A) 20 m
B) 40 m
C) 80 m
D) 200 m
Answer:
C) 80 m

Question 8.
If the Sun’s angle of elevation is 60°, then a pole of height 6 m will cast a shadow of length ………….. m.
A) \(\sqrt{3}\)
B) 5\(\sqrt{3}\)
C) 6\(\sqrt{3}\)
D) 2\(\sqrt{3}\)
Answer:
D) 2\(\sqrt{3}\)

Question 9.
A pole of 12 m high casts a shadow 4\(\sqrt{3}\)m on the ground, then the Sun’s angle of elevation is
A) 60°
B) 120°
C) 45°
D) 30°
Answer:
A) 60°

Question 10.
If the height and length of the shadow of a man are the same then the angle of elevation of the Sun is
A) 60°
B) 45°
C) 90°
D) 120°
Answer:
B) 45°

Question 11.
What is the angle of elevation of the top of a temple of height 10 m at a point whose distance from the base of the tower is 10\(\sqrt{3}\) m ?
A) 30°
B) 60°
C) 45°
D) 90°
Answer:
A) 30°

Question 12.
The length of the shadow of 5m height tree whose angle of elevation of the Sun is 30° is?
A) 5 m
B) \(\sqrt{3}\) m
C) 5\(\sqrt{3}\) m
D) 10 m
Answer:
C) 5\(\sqrt{3}\) m

Question 13.
From the top of a 10m height tree the angle of depression of a point on the ground is 30° then the distance of the point from the foot of the tree is
A) 10 m
B) 10\(\sqrt{3}\) m
C) \(\frac{10}{\sqrt{3}}\) m
D) 5\(\sqrt{3}\) m
Answer:
B) 10\(\sqrt{3}\) m

Question 14.
Ladder ‘x’ meters long is laid against a well making an angle ‘0’ with the ground. If we want to directly find the distance between the foot of ladder and foot of the wall, which trigonometrical ratio should be considered ?
A) sin θ
B) cos θ
C) tan θ
D) cot θ
Answer:
B) cos θ

Question 15.
Top of a building was observed at an angle of elevation ‘α’ from a point, which is at distance ‘d’ meters from the foot of the building. Which trigonometrical ratio should be considered for finding height of buildings.
A) tan α
B) sin α
C) cos α
D) sec α
Answer:
A) tan α

Question 16.
In the given figure, the value of angle θ is

A) 30°
B) 60°
C) 45°
D) 90°
Answer:
A) 30°

Question 17.
The given figure shows the observation of point ‘C’ from point A. The angle of depression from A is

A) 30°
B) 45°
C) 90°
D) 75°
Answer:
A) 30°

Question 18.
If the length of the shadow of a tower is \(\frac{1}{\sqrt{3}}\) times the height of the tower, then the angle of elevation of the sun is ……………..
A) 30°
B) 45°
C) 60°
D) 75°
Answer:
C) 60°

Question 19.
A tower is 50 m high. Its shadow is x m shorter when the sun’s altitude is 45° then when it is 30°, then x = ………… cm
A) 105
B) 20
C) 10
D) 100
Answer:
D) 100

Question 20.
The length of the string of a kite flying at 100 m above the ground with the elevation of 60° is ………….
A) \(\frac{200}{\sqrt{3}}\)
B) \(\frac{20}{\sqrt{3}}\)
C) \(\frac{291}{\sqrt{3}}\)
D) none
Answer:
A) \(\frac{200}{\sqrt{3}}\)

Question 21.
A player sitting on the top of a tower of height 40 m observes the angle of depression of a ball lying on the ground is 60 The distance between the foot of the tower and ball is …………… m.
A) 20
B) \(\frac{80}{\sqrt{61}}\)
C) \(\frac{40}{\sqrt{3}}\)
D) \(\frac{40}{\sqrt{6}}\)
Answer:
C) \(\frac{40}{\sqrt{3}}\)

Question 22.
If the ratio of height of a tower and the length of its shadow on the ground is \(\sqrt{3}\) :1, then the angle of elevation of the sun is ……………….
A) 80°
B) 60°
C) 70°
D) 100°
Answer:
B) 60°

Question 23.
The angle of depression of the top of a tower at a point 100 m from the house is 45°, then the height of the tower is …………. m.
A) 18.1
B) 16.3
C) 36.6
D) 26.7
Answer:
C) 36.6

Question 24.
An object is placed above the observer’s horizontal, we call the angle between the line of sight and observer’s horizontal is ……………..
A) angle of elevation
B) angle of depression
C) point
D) none
Answer:
A) angle of elevation

Question 25.
Angle of elevation of the top of a building from a point on the ground is 30. Then the angle of depression of this point from the top of the building is …………………
A) 65°
B) 60°
C) 70°
D) 30°
Answer:
D) 30°

Question 26.
What change will be observed in the angle of elevation as we move away from the object ?
A) increase
B) decrease
C) can’t be determined
D) none
Answer:
A) increase

Question 27.
An object is placed below the observer’s horizontal, then what is the angle between line of sight and observer’s horizontal ?
A) angle of elevation
B) angle of depression
C) can’t be determined
D) none
Answer:
B) angle of depression

Question 28.
What change will be observed in the angle of elevation as we approach the foot of the tower ?
A) 0
B) 60°
C) Data not correct
D) none
Answer:
D) none

Question 29.
In the figure given below, the imaginary line through the object and eye of the observer is called …………………

A) line of sight
B) angle of depression
C) angle of elevation
D) none
Answer:
A) line of sight

Question 30.
In the figure given below, a man on the top of cliff observers a boat coming towards him. Then 6 represents the angle of …………….

A) depression
B) elevation
C) equal
D) none
Answer:
A) depression

Question 31.
In the figure given below, if AB = 10 m and AC = 20 m, then θ = ………………..

A) 60°
B) 30°
C) 70°
D) none
Answer:
B) 30°

Question 32.
A pole 6 m high casts a shadow 2\(\sqrt{3}\) m long on the ground, then the sun’s elevation is …………….
A) 70°
B) 20°
C) 80°
D) 60°
Answer:
D) 60°

Question 33.
In the figure given below, if AB = CD = 10\(\sqrt{3}\) m then BC = ………………

A) 90
B) 60
C) 40
D) None
Answer:
C) 40

Question 34.
In the figure given below, if AB = 10\(\sqrt{3}\) m, then CD = …………….. (take \(\sqrt{3}\) = 1.732)

A) 7.32
B) 8.14
C) 3.1
D) 1.92
Answer:
A) 7.32

Question 35.
In the figure given below, if AD = 7\(\sqrt{3}\) m, then BC = ………………. m.

A) 13
B) 19
C) 28
D) None
Answer:
C) 28

Question 36.
The length of the shadow of a tree is 7 m high, when the sun’s elevation is …………………..
A) 45°
B) 60°
C) 70°
D) 90°
Answer:
A) 45°

Question 37.
If two tangents inclined at an angle of 60 are drawn to a circle of radius 3 cm, then length of tangent is equal to …………. m.
A) 4\(\sqrt{3}\)
B) 2\(\sqrt{91}\)
C) \(\sqrt{3}\)
D) 3\(\sqrt{3}\)
Answer:
D) 3\(\sqrt{3}\)

Question 38.
The angle formed by the line of sight with horizontal, when the point being viewed is above the horizontal level is called
A) angle of elevation
B) angle of depression
C) point
D) none
Answer:
A) angle of elevation

Question 39.
cot² B – Cosec² B = ………………
A) 0
B) – 1
C) 1
D) 2
Answer:
B) – 1

Question 40.
\(\frac{\tan \theta}{\sec \theta}\) = ……………….
A) – cos θ
B) sin θ
C) – tan θ
D) none
Answer:
B) sin θ

Question 41.
A boy observed the top of an electrical pole to be at angle of elevation of 60° when the observation point is 8 m away from the foot of the pole then the height of the pole is ……………… m.
A) 18\(\sqrt{3}\)
B) 14
C) 7\(\sqrt{3}\)
D) 8\(\sqrt{3}\)
Answer:
D) 8\(\sqrt{3}\)

Question 42.
Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60 what is the distance between you and the object ?
A) 9
B) 7\(\sqrt{3}\)
C) 12\(\sqrt{3}\)
D) None
Answer:
D) None

Question 43.
Sin \(\frac{\pi^{\mathrm{c}}}{2}\) = ……………….
A) 4
B) 3
C) 1
D) -1
Answer:
C) 1

Question 44.
Domain of sin θ = ………………..
A) R
B) R – {30°}
C) N
D) None
Answer:
D) None

Question 45.
tan \(\frac{\pi^{\mathrm{c}}}{4}\) = ……………….
A) 2
B) 3
C) -1
D) 1
Answer:

Question 46.
cot 15° = ………………
A) 2 + \(\sqrt{3}\)
B) 2 – \(\sqrt{3}\)
C) \(\sqrt{2}\)
D) \(\sqrt{3}\) – 1
Answer:
A) 2 + \(\sqrt{3}\)

Question 47.
A + B = 180° then cos A + cos B = ………………
A) 4
B) 1
C) 0
D) none
Answer:
C) 0

Question 48.
sin 15° = ……………….
A) \(\frac{\sqrt{3}}{9 \sqrt{2}}\)
B) \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
C) \(\frac{\sqrt{3}+1}{2}\)
D) none
Answer:
B) \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

Question 49.
tan A = \(\frac{\mathrm{n}}{\mathrm{n}+1}\), tan B = \(\frac{\mathrm{n}}{2\mathrm{n}+1}\), A + B = …………..
A) 4
B) 3
C) -1
D) 1
Answer:
D) 1

Question 50.
The angle of elevation of tower at a point 40 m apart from it is cot^-1 \(\left(\frac{3}{5}\right)\). Obtain the height of the tower.
A) \(\frac{200}{3}\) m
B) \(\frac{100}{3}\) m
C) \(\frac{210}{17}\) m
D) none
Answer:
A) \(\frac{200}{3}\) m

Question 51.
A ladder 20 m long is placed against a vertical wall of height 10 m, then the distance between the foot of the ladder and the wall is …………………. m.
A) 7\(\sqrt{3}\)
B) 20\(\sqrt{3}\)
C) 30\(\sqrt{3}\)
D) none
Answer:
C) 30\(\sqrt{3}\)

Question 52.
sin 18° = ………………
A) \(\frac{\sqrt{5}}{4}\)
B) \(\frac{\sqrt{5}-1}{4}\)
C) \(\frac{1+\sqrt{3}}{2}\)
D) \(\frac{\sqrt{3}-1}{4}\)
Answer:
B) \(\frac{\sqrt{5}-1}{4}\)

Question 53.
In the below figure x = …………………. cm.

A) 10
B) 12
C) 13
D) 19
Answer:
A) 10

Question 54.
cot (90 – A) = ………………
A) 3 tan A
B) sin A
C) cot A
D) tan A
Answer:
D) tan A

Question 55.
cos⁴ A – sin⁴ A = …………….
A) sin² A
B) cos² A
C) cos 2A
D) cos 3A
Answer:
C) cos 2A

Question 56.
If cosec θ + cot θ = k then cos θ ……………..
A) \(\frac{k^2-1}{k^2+1}\)
B) \(\frac{k^2}{k^2-1}\)
C) \(\frac{k^2+1}{k}\)
D) none
Answer:
D) none

Question 57.
x = (sec θ + tan θ), y = (sec θ – tan θ) then xy ………………
A) -1
B) 0
C) 1
D) -2
Answer:
C) 1

Question 58.
tan 15° = ……………..
A) \(\frac{\sqrt{3}}{\sqrt{3}+1}\)
B) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
C) \(\frac{\sqrt{3}-1}{2}\)
D) none
Answer:
B) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

Question 59.
cosec θ = ……………….
A) \(\sqrt{1+\cot ^2 \theta}\)
B) \(\sqrt{\cot ^2 \theta-1}\)
C) \(\sqrt{1+\sin \theta}\)
D) \(\sqrt{\cot \theta-1}\)
Answer:
A) \(\sqrt{1+\cot ^2 \theta}\)

Question 60.
x = a sin θ, y = a cos θ then x² + y² = ………………
A) \(\frac{a}{3}\)
B) \(\frac{a}{2}\)
C) a
D) a²
Answer:
D) a²

Question 61.
Example of a Pythagorean Triplet is ………………
A) 5, 12, 13
B) 5, 10, 11
C) 8, 9, 11
D) none
Answer:
A) 5, 12, 13

Question 62.
sec² A = …………….
A) 1 – tan² A
B) 1 + tan² A
C) cot² A
D) none
Answer:
B) 1 + tan² A

Question 63.
\(\frac{1}{\cos \theta}\) – cos θ = ………………
A) tan θ . sin θ
B) sec θ . cos θ
C) tan θ . cot θ
D) none
Answer:
A) tan θ . sin θ

Question 64.
sin θ = cos θ, θ ∈ Q₁ then θ = …………..
A) \(\frac{\pi^c}{2}\)
B) \(\frac{\pi^c}{3}\)
C) \(\frac{2 \pi^c}{3}\)
D) \(\frac{\pi^c}{4}\)
Answer:
D) \(\frac{\pi^c}{4}\)

Question 65.
72° = …………………
A) \(\frac{\pi^c}{2}\)
B) \(\frac{\pi^c}{3}\)
C) \(\frac{2 \pi^c}{5}\)
D) \(\frac{\pi^c}{5}\)
Answer:
C) \(\frac{2 \pi^c}{5}\)

Question 66.
sin² 105° + cos² 105° = ……………….
A) 1
B) 0
C) 9
D) 10
Answer:
A) 1

Question 67.
sin 45° (cos 45°) = ………………..
A) 1
B) \(\frac{1}{2}\)
C) 3
D) none
Answer:
B) \(\frac{1}{2}\)

Question 68.
cos 40° = 0.76 then sin 50² = ………………..
A) 0.76
B) 7.6
C) 76.6
D) none
Answer:
A) 0.76

Question 69.
At a point 15 m away from the base of a 15 m high pole, the angle of elevation of the top is …………………
A) 30°
B) 45°
C) 60c
D) 90°
Answer:
B) 45°

Question 70.
When the length of the shadow of a person is equal to his height, then the elevation of source of light is …………
A) 15°
B) 30°
C) 45°
D) 60°
Answer:
C) 45°

Question 71.
The angle of elevation of top of a tree is 30. On moving 20 m nearer, the angle of elevation is 60. The height of the tree is
A) 15\(\sqrt{3}\) m
B) 2\(\sqrt{3}\) m
C) 10\(\sqrt{3}\) m
D) 5\(\sqrt{3}\) m
Answer:
C) 10\(\sqrt{3}\) m

Question 72.
The ratio of length of a pole and its shadow is 1 :\(\sqrt{3}\).The angle of elevation is
A) 90°
B) 60°
C) 45°
D) 30°
Answer:
D) 30°

Question 73.
The upper part of a treee is broken by wind and makes an angle of 30° with the ground and at a distance of 21 m from the foot of the tree. Find the total height of the tree.
A) 30\(\sqrt{3}\) m
B) 21 m
C) 30 m
D) 21\(\sqrt{3}\) m
Answer:
D) 21\(\sqrt{3}\) m

Question 74.
From a bridge 25 m high, the angle of depression of a boat is 45°. Find the horizontal distance of the boat from the bridge.
A) 25\(\sqrt{3}\) m
B) 25 m
C) \(\frac{25}{\sqrt{3}}\)
D) 45 m
Answer:
B) 25 m

Question 75.
A tower makes an angle of elevation equal to the angle of depression from the top of a cliff 25 m height. Find the height of the tower.
A) 25 m
B) 75 m
C) 5m
D) 50 m
Answer:
D) 50 m

Question 76.
When the angle of elevation of a pole is 45°, the length of the pole and its shadow are
A) equal
B) length > shadow
C) shadow > length
D) none of the above
Answer:
A) equal

Question 77.
In a rectangle, if the angle between a diagonal and a side is 30, and the length of the diagonal is 6 cm, the area of the rectangle is
A) 18 cm²
B) 9 cm²
C) 18\(\sqrt{3}\) cm²
D) 9\(\sqrt{3}\) cm²
Answer:
D) 9\(\sqrt{3}\) cm²

Question 78.
Two posts are 15 m and 25 m high and the line joining their tops make an angle of 45° with the horizontal. The distance between the two posts is
A) 15 m
B) 25 m
C) 18 m
D) 10 m
Answer:
D) 10 m

Question 79.
An electric pole 20 m high stands up right! on the ground with the help of steel wire to its top and affixed on the ground. If the steel wire makes 60° with the horizontal ground, find the length of steel wire.
A) 60\(\sqrt{3}\) m
B) 20 m
C) 60 m
D) \(\frac{20}{\sqrt{3}}\) m
Answer:
D) \(\frac{20}{\sqrt{3}}\) m

Question 80.
A building casts a shadow of length 50\(\sqrt{3}\) m when the sun is 30° about the horizontal. The height of the building is
A) 30 m
B) 40 m
C) 50 m
D) 60 m
Answer:
C) 50 m

Question 81.
When the angle of elevation of a light! changes from 30° to 45°, the shadow of pole becomes 100\(\sqrt{3}\) m less. The height of the pole is
A) 30 m
B) 120 m
C) 75 m
D) 100 m
Answer:
D) 100 m

Question 82.
From the top of a building 50 m from horizontal, the angle of depression made by a car is 30°. How far is the car from the building ?
A) \(\frac{50}{\sqrt{3}}\)
B) 50\(\sqrt{3}\) m
C) 150 m
D) 30\(\sqrt{3}\) m
Answer:
B) 50\(\sqrt{3}\) m

Question 83.
From the top of a building with height 30°(\(\sqrt{3}\) + 1) m two cars make angles of depression of 45° and 30° due east. What is the distance between two cars ?
A) 30 m
B) 60 m
C) 45 m
D) 75 m
Answer:
B) 60 m

Question 84.
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retires 40 m from the bank, he finds the angle to be 30°. The breadth of the river is
A) 10 m
B) 15 m
C) 20 m
D) 25 m
Answer:
C) 20 m

Question 85.
A ladder of 10 m length touches a wall at a height of 5 m. The angle made by it with the horizontal is
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
A) 30°

Question 86.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance between the top of the tree and the ground is 10 m. Find the height of the tree.
A) 10 m
B) 30\(\sqrt{3}\) m
C) 10\(\sqrt{3}\) m
D) 30 m
Answer:
C) 10\(\sqrt{3}\) m

Question 87.
The angle of elevation of a cloud from a point 200 m above the lake is 30° and the angle of depression of its reflection in the lake is 60°. The height of the cloud above the lake is
A) 100 m
B) 200 m
C) 300 m
D) 400 m
Answer:
D) 400 m

Question 88.
An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60. After a flight of 10 seconds, its angle of elevation is observed to be 30 from the same point on the ground. Find the speed of the aeroplane.
A) 415.7 km/h
B) 215.3 km/h
C) 700 km/h
D) none of the above
Answer:
A) 415.7 km/h

Question 89.
If AB = 4m, and AC = 8m, then the angle of elevation of A as observed from C is
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
A) 30°

Question 90.
If a pole of height 6 m casts a shadow 2\(\sqrt{3}\) m long on the ground, then the sun’s elevation is
A) 30°
B) 60°
C) 45°
D) 90°
Answer:
B) 60°

Question 91.
Find the elevation of the sun at the moment when the length of the shadow of a tower is just equal to its height.
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
B) 45°

Question 92.
If the shadow of a tree is \(\frac{1}{\sqrt{3}}\) times the height of the tree, then the angle of elevation of the sun is
A) 30°
B) 45°
C) 60°
D) 90°
Answer:
C) 60°

Telangana SCERT 10th Class Physics Study Material Telangana 9th Lesson Electric Current Textbook Questions and Answers.

TS 10th Class Physical Science 9th Lesson Questions and Answers Electric Current

Improve Your Learning
I. Reflections on concepts

Question 1.
Explain how electron flow causes electric current with Lorentz-Orude’s theory of electrons.
Answer:
1. Lorenti and Drude, scientists of 19th Century proposed that conductors like metals contain large number of free electrons while the positive ions are fixed in their locations. The arrangement of the positive ions is aimed lattice.
2. Assume that a conductor is an open circuit. The electrons move randomly in lattice space of a conductor as shown in the following figure.

3. When the electrons are in random motion, they can move in any direction.

4. Hence if we imagine any cross-section as In above figure, the number of electrons, crossing the cross-section of a conductor from left to right in one second is equal to that of electrons passing the cross-section from right to left in one second and the nett charge moving along conductor through any cross-section Is zero when the conductor is in open circuit.

5. When the ends of the conductor are connected to a source (say, battery) through a bulb, the bulb glows because energy transfer takes place from battery to the bulb.

6. As the electrons are responsible for transfer of energy from battery to the bulb, they must have an ordered motion.

7. When the electrons are in ordered motion there will be a net charge crossing any cross-section of the conductor.

8. This ordered motion of electrons Is called electric current.

Question 2.
Write the difference between potential difference and emf.
Answer:

Question 3.
How can you verify that the resistance of a conductor is temperature dependent?
Answer:

1. Take a bulb and measure the resistance of the bulb using a multimeter in open circuit. Note the value of the resistance.
2. Now connect the bulb in a circuit and switch on the circuit.
3. After a few minutes the bulb gets heated.
4. Now measure the resistance of the bulb again with multimeter.
5. The value of resistance of the bulb in second instance is more than the resistance of the bulb in open circuit.
6. Here the increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.
7. Thus the value of resistance of a conductor depends on the temperature.

Question 4.
What do you mean by electric shock? Explain how it takes place.
Answer:

• If we touch live wire of 240V which gives 0.0024 A of current flows through the body the function g of organs inside the body gets disturbed.
• This disturbance inside the body is felt as electric shock.
• If the current flow continues further, it damages the tissues of the body which leads to decrease ¡n resistance of the body.
• When this current flows for a longer time, damage to the tissues increases and there by the resistance of human body decreases further.
• Hence, the current through the human body will increase.
• If this current reaches 0.07 A, it affects the functioning of the heart.
• If this current passes through the heart for more than one-second t could be fatal.

Question 5.
Draw a circuit diagram for a circuit in which two resistors A and B are connected in series with a battery and a voltmeter is connected to measure the potential difference across the resistor A.
Answer:
A circuit diagram in which two resistors A and B are connected ¡n series with a battery and a voltmeter is connected to measure the p.d. across the resistor A.

Question 6.
In the below figure, the potential at A is ……………………… when the potential at B is zero.

Answer:
Apply Kirchoff’s loop rule V_A-5 x 1 – 2—V_b=0
V_A-5 – 2-0 =0 ⇒V_A = 7V
The potential at A = 7V when the potential at B = O

Question 7.
How does a battery work? Explain.
Answer:
1. A battery consists of two metal plates (electrodes) and a chemical (electrolyte).
2. This electrolyte consists positive and negative ions which move In opposite directions.
3. This electrolyte exerts a force called chemical force (F_c) to make the ions move in a specified direction.
4. Positive Ions move towards one plate and accumulate on that. As a result this plate becomes positively charged (Anode).

5. Negative Ions move to another plate and accumulate on that. As a result of this the plate becomes negatively charged (Cathode).
6. This accumulation continues till both plates are sufficiently charged.
7. But the ions experience another force called electric force (F_e) when sufficient number of charges accumulated on the plates.
8. The direction of F_e is opposite to F_c and magnitude depends on the amount of charge accumulated on the plates.
9. The accumulation of charges on plates is continuous till F_e becomes equal to F_c Now there will not be any motion due to balance of F_e and F_c.

10. The new battery that we buy from the shop is under the Influence of balanced forces. This Is the reason for the constant RD. between the terminals of a battery.

11. When a conducting wire Is connected to the terminals of the battery, a RD. is created between the ends of the conductor which sets up an electric field through-out the conductor.

12. The large number of electrons In the conductor, near the positive terminal of the battery are attracted by It and start to move towards positive terminal. As a result the amount of positive charge on this plate decreases. So F_e becomes weaker than F_c and F_c pulls negative ions from anode towards cathode.

13. The negative terminal pushes one electron into the conductor because of stronger repulsion between negative terminal and negative ion.

14. Hence. the total number of electrons in the conductor remains constant during the current flow. The above-said process continues till F_e = F_c.

Question 8.
Explain Kirchhoff’s laws with examples.
Answer:
Kirchhoffs laws:
1. The junction law: At any junction point in a circuit where the current can divide, the sum of the currents in the junction must equal to the sum of the currents leaving the junction.
This means that there is no accumulation of electric charges at any junction in a circuit. Eg: ‘P Is the junction
I₁, I_4, and I₆ are the currents into the junction.
I₂, I_3, and I₅ are the currents leaving the junction.
According to Kirchhoff’s junction law
I₁+I₄+I₆ = I₂+I₃+I₅

2) The loop law: The algebraic sum of the increases and decreases in potential difference across various components of a closed circuit loop must be zero.
Eg: For the loop ACDBA
-V₂+I₂R₂-I₁R₁ +V₁=O
For the loop EFDCE
– (I₁ + I₂) R₃ -I₂R₂ + V₂ = O
For the loop EFBAE
-(I₁+I₂)R₃-I₁R₁+V₁=O

Question 9.
Deduce the expression for the equivalent resistance of three resistors connected In series.
Answer:
Two or more resistors are said to be connected in series if the current flowing through one, also flows through the others.
In series combination, we know that
1. The same current passes through the resistors.
2. The potential difference across combination of resistors is equal to the sum of the voltages across the individual resistors.
Connect the circuit as shown n the figure.
The cell connected across the series combination of 3 resistors maintains a potential difference (v) across the combination. The current through the combination is I.


Let us replace the combination of 3 resIstors by a single resistor R_eq such that current does not change.
R_eq is given by Ohm’s law as
R_eq = \( \frac{V}{I}\)
⇒V=IR_eq

The potential differences V₁, V₂, V₃ across the resistors R₁, R₂ and R₃
respectively are given by Ohm’s law as
V₁ = IR₁, V₂ = IR₂, V = IR₃
Since the resistances are connected in series
V= V₁ +V₂+V₃
IR_eq = IR₁ + IR₂ + IR₃
I (R_eq) = I (R₁ + R₂ + R₃)
⇒ R_eq = R₁ + R₂ + R₃
Similarly, for n resistors connected in series,
R_eq = R₁ + R₂ + R₃+ ………………………. +R_n.

Question 10.
Deduce the expression for the equivalent resistance of three resIstors connected In parallel.
Answer:
If resistances are connected in such a way that the same potential difference gets applied across each of them, they are said to be connected In parallel.

For a parallel combination, we know that,

1. The total current flowing Into the combination is equal to the sum of the currents passing through the individual resistors ⇒ I = I₁+ I₂+ I₃
2. The potential difference remains constant V₁ = V₂ = V₃ = V.
   Connect the circuit as shown in the figure.
3. The cell connected across 3 resistors maintains the same potential difference across each resistor.
4. The current I gets divided at A into 3 parts I₁ I₂ and I₃ which flows through R₁, R_2, and R₃ respectively.
5. Let us replace the combination of resistors by an equivalent resistance R_eq such that potential difference across the circuit does not change.
6. The equivalent resistance R_eq = \(\frac{V}{I} \Rightarrow I=\frac{V}{R_{e q}}\)
7. The currents I₁,I₂,I₃ across R₁,R₂ and R3 are given by I₁ = \(\frac{V}{R_1}\), I₂ = \(\frac{V}{R_2} \), I₃ = \(\frac{V}{R_3} \),

Since the resistors are in parallel,

Question 11.
What is the value of 1KWH in joules?
Answer:
1 KWH = (1000 J/s) (60 × 60s) = 3600 × 1000J = 3.6 × 10⁶ So, 1KWH is equal to 3.6 x 10⁶ Joules.

Question 12.
Silver is a better conductor of electricity than copper. Why do we use copper wire for conduction of electricity?
Answer:
Reasons:

• Copper has low resistivity. When electricity is passed through copper wires, the power losses in the form of heat are very small.
• Cost of copper versus that of silver metal, copper is less expensive.
• Copper has flexibility and resistance to breakage.
• Copper is cheaply available than silver.

Application of concepts

Question 1.
Explain overloading of household circuit.
Answer:

1. Electricity enters our homes through two wires called lines. These lines have low resistance and the potential difference between the wires is usually about 240 V.
2. All electrical devices are connected in parallel in our home. Hence, the potential drop across each device is 240 V.
3. Based on the resistance of each electric device, it draws some current from the supply. Total current drawn from the mains is equal to the sum of the currents passing through each device.
4. If we add more devices to the household circuit the current drawn from the mains also increases.
5. This leads to overheating and may cause a fire. This is called “overloading”.

For example:
If we switch on devices, such as heater shown in the figure, from the mains exceeds the maximum limit 20 A.

Question 2.
Why do we use fuses in household circuits?
Answer:

1. The fuse consists of a thin wire of low melting point.
2. To prevent damages due to overloading we connect an electric fuse to the household circuit.
3. When the current in the fuse exceeds 20A, the wire will heat up and melt.
4. The circuit then becomes open and prevents the flow of current into the household circuit. So all the electric devices are saved from damage that could be caused by overload.
5. Thus, we can save the household wiring and devices by using fuses.

Question 3.
Two bulbs have ratings 100W, 220V, and 60W, 220V. Which one has the greater resistance?
Answer:
Resistance of first bulb R₁ = \(\frac{V^2}{P}=\frac{220 \times 220}{100}\) = 484 Ω
Resistance of second bulb R₂ = \(\frac{V^2}{P}=\frac{220 \times 220}{60}=\frac{4840}{6}\) = 806.6Ω
∴ The bulb rated 60W, 220V has higher resistance.

Question 4.
Why don’t we use series arrangement of electrical appliances like bulb, Television, fan and others in domestic circuits?
Answer:

1. We have seen that in a series circuit, the current is constant throughout the electric circuit.
2. But it is obviously impracticable to connect an electric bulb and an electric heater in series because they need currents of widely different values to operate properly.
3. Another major disadvantage of a series circuit is that when one component fails, the circuit is broken and none of the other components works.

Question 5.
Are the headlights of a car connected in series or parallel? Why?
Answer:
The headlights of a car are connected in parallel.
Reason:

• When they are connected in parallel, same voltage (P.D) will be maintained in the two lights.
• If one of the lights damaged, the other will work without any disturbance.

Question 6.
Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?
Answer:
We should connect the electric appliances in parallel to household circuit because

• Each appliance gets the full voltage.
• The parallel circuit divide the current through the appliances. Each appliance gets proper current depending on its resistance.
• If one appliance is switched on/off others are not affected.

If appliances are connected in series the following disadvantages are arised:

• The same current will flow through all the appliances, which is not desired.
• Total resistance becomes large and the current gets reduced.
• We cannot use independently on/off switches with individual appliances.
• All appliances have to be used simultaneously even if we don’t need them.

Question 7.
If the resistance of your body is 10000012, what would be the current that flows in your body when you touch the terminals of a 12V battery?
Answer:
Resistance of the body (R) 1,00,000 Ω
Potential difference of the battery (V) =12V
Current that flows in the body (I) = ?
According to Ohm’s law, \(\frac{V}{I}\) = R
⇒ I = \(\frac{V}{R}=\frac{12}{1,00,000}\) = 0.00012 A

Question 8.
A wire of length 1m and radius 0.1 mm has a resistance of 100W. Find the resistivity of the material.
Answer:
Resistance of wire, R = 100 Ω
Radius of wire, r = 0.1mm = 1 x 10^-4 m
Length of wire, l = 1m
Formula for resistivity of wire: ρ = \(\frac{R A}{l}=\frac{R \pi r^2}{l} \)
Substituting the given values, ρ = \(\frac{22}{7} \times 10^2 \times \frac{10^{-4} \times 10^{-4}}{l} \) = \(\frac{22}{7} \times 10^{-6} \) ohm—meter
= 3.14 × 10^-6 ohm-mt

Question 9.
Why do we consider tungsten as a suitable material for making the filament of a bulb?
(Or)
What is the reason for using tungsten as a filament in electric bulb?
Answer:
Tungsten has a high resistivity value (5.60 ×10^-8 m) and a high melting point (3422°C). So the filament of a bulb is usually made of tungsten. Its high resistivity enables the filament to become red hot soon and then it produces white heat to emit light. Its high melting point keeps it in a solid state and also prevents oxidation.

Question 10.
How can you appreciate the role of a small fuse in house wiring circuit in preventing damage to various electrical appliances connected to the circuit?
Answer:

1. The fuse consists of a thin wire of low melting point. When the current in the fuse exceeds $20 \mathrm{~A}$, the wire will heat up and melt.
2. The circuit then becomes open and prevents the flow of current into the household circuit. So all the electric devices are saved from change that could be caused by overload.
3. Thus we can save the house holding wiring and devices by using fuses.
4. So we should appreciate the role of fuse in preventing damage to electrical appliances in household circuits.

Question 11.
Uniform wires of resistance 100c are melted and recast into wire of length double that of the original. What would be the resistance of the new wire formed?
Answer:
Before recasting,
Resistance R₁ = 100 Ω
length (l₁)= l (say)
After recasting
Resistance R₂ = ?
length (‘l₂) = 2l

We know that R α l,
\(\frac{R_1}{R_2}=\frac{l_1}{l_2}\)

Higher Order Thinking Questions

Question 1.
Imagine that you have three resistors of 30 Ω each. How many resultant resistances can be obtained by connecting these three in different ways. Draw the relevant diagrams.
Answer:
Let R₁ = 30Ω, R₂ = 30Ω, R₃ = 30Ω
We get different resistors by different combinations as shown below.

Question 2.
A house has 3 tube lights, two fans and a Television. Each tube light draws 40W. The fan draws 80W and the Television draws 60W. On the average, all the tube lights are kept on for five hours, two fans for 12 hours and the television for five hours every day. Find the cost of electric energy used in 30 days at the r ate of Rs. 3.00 per Kwh.
Answer:
Power consumption by tube lights in a day = 40W x 3 x 5H = 600 WH
Power consumption by fans in a day = 80W x 2 x 12H = 1920 WH
Power consumption by television in a day = 60W x 1 x 5H = 300 WH
Total power consumption in a day = 600 + 1920 + 300 = 2820 WH
Power consumption for 30 days = 2820 x 30 = 84600 WH

Rate of 1 KWH = Rs 3/- Total consumption = 84.60 x Rs.3 = Rs. 253.80

Question 3.
Observe the circuit and answer the questions given below:
(i) Are resIstors 3 and 4 In series?
(ii) Is the battery in series with any resistor?
(iii) What is the potential drop across the resistor 3?
(iv) What is the total emf in the circuit if the potential drop across resistor 1 is 6V?
Answer:
(i) No. Resistors 3 and 4 are not in series. They are in parallel.
(ii) No.
(iii) As resistors 3 and 4 are in parallel, same potential difference will be allowed
through them. Hence the potential drop across resistor 3 is 8V.
(iv) Total emf=V₁+V₂+V₃+V₄=6V+14V+8V+8V=36V.

Multiple choice questions

Question 1.
A uniform wire of resistance 50 Q is cut into five equal parts. These parts are now connected in parallel. Then the equivalent resistance of the combination is : ( )
(A) 2 Q
(B) 12 Q
(C) 250 Q
(D) 6250 Q
Answer:
(A) 2 Q

Question 2.
A charge is moved from a point A to a point B. The work done to move unit charge during this process is called ( )
(A) potential at A
(B) potential at B
(C) potential difference between A&B
(D) current from A to B
Answer:
(C) potential difference between A&B

Question 3.
Joule / coulomb is the same as ( )
(A) 1-watt
(B) 1-volt
(C) 1 ampere
(D) 1-ohm
Answer:
(B) 1-volt

Question 4.
The resistors of values 2 Ω, 4 Ω, and 6 Ω are connected in series. The equivalent resistance in the circuit is ( )
(A) 2 Ω
(B) 4 Ω
(C) 12 Ω
(D) 6 Ω
Answer:
(C) 12 Ω

Question 5.
The resistors of values 3 Ω, 6 Ω, and 18 Ω are connected in parallel. The equivalent resistance in the circuit is ( )
(A) 12 Ω
(B) 36 Ω
(C) 18 Ω
(D) 1.8 Ω
Answer:
(D) 1.8 Ω

Question 6.
The resistors of values 6 Ω, and 6 Ω are connected in series and 12 Ω are connected in parallel. The equivalent resistance of the circuit is ( )
(A) 24 Ω
(B) 6 Ω
(C) 18 Ω
(D) 2.4 Ω
Answer:
(B) 6 Ω

Question 7.
The current in the wire depends ( )
(A) only on the potential difference applied
(B) only on the resistance of the wire
(C) on both of them
(D) none of them
Answer:
(C) on both of them

Suggested Experiments

Question 1.
State Ohm’s law. Suggest an experiment to verify it and explain the procedure.
Answer:
A. Ohm’s Law: The potenbal difference between the ends of a conductor is directly proportional toe the electric current passing through it at constant temperature.
Verification:
Aim: To verify Ohm’s law or to show that -=Co,,ctani
Materials required: 5 dry cells of 1.5V each, conducting wires, an ammeter,
a voltmeter, Manganin wire of length 10cm, LED and Key.

Procedure:
1. Connect a circuit as shown in figure.
2. Solder the connecting wires to the ends of the Manganin wire.
3. Close the key.
4. Note the readings of current from Baltery Key ammeter and potential difference from volt meter in the following table.

5. Now connect 2 cells (in series) instead of one cell in the circuit. Note the values of ammeter and voltmeter and record them in the above table.
6. Repeat the same for three cells, four cells, five cells respectively.
7. Record the values of V and I corresponding to each case in the table.
8. Find \(\frac{V}{I} \) for each set of values.
9. We notice that \(\frac{V}{I} \)  is a constant.
V ∝ I ⇒ \(\frac{V}{I} \)= Constant
This constant is known as resistance of the conductor, denoted by R.
⇒ \(\frac{V}{I} \) = R
∴ Ohm’s law is verified.

Question 2.
How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross-section area and temperature?
Answer:
1. Collect Iron spokes of different lengths with the same cross-sectional area.
2. Make a circuit as shown in the figure.
3. Connect one of the iron spokes between P and Q.
4. Measure the value of the current using the ammeter connected to the circuit and note in your notebook.
5. Repeat this for other lengths of the iron spokes. Note the corresponding values of currents in your notebook as shown below.

1. We observe that current decreases as the length of the spoke increases.
2. We also know that resistance increases as current decreases.
3. Hence the resistance of iron spoke increases as its length increases.
4. We conclude that the resistance of a conductor is directly proportional to its length for a constant-potential difference and constant cross-sectional area.

R ∝ l

Suggested Projects

Question 1.
a. Take a battery and measure the potential difference. Make a circuit and measure the potential difference when the battery is connected in the circuit. Is there any difference in potential difference of battery?
Answer:

1. The potential difference across the terminals of a battery when it is not connected in any circuit is called the Electromotive force of battery or emf of battery.
2. As soon as the battery is connected to an external circuit, there will be a current through the battery as well as the external circuit.
3. Due to this current flowing through the battery, there will be a voltage drop inside the battery because of the internal resistance of the battery itself.
4. Hence when external circuit is connected, the voltage appeared across the terminal of the battery is somewhat less than the open circuit voltage of the battery.
5. This is because of voltage drop due to internal resistance ¡nside the battery.

b. Measure the resistance of a bulb (filament) in open circuit with a multimeter. Make a circuit with elements such as bulb, battery of 12V and key in series. Close the key. Then again measure the resistance of the same bulb (filament) for every 30 seconds. Record the Observations in a proper table. What can you conclude from the above results?
Answer:
Materials required: a bulb, 12 v battery, key, and multimeter.

Procedure:

1. Measuretheresistanceofabulb in open circuit.
2. Connect the bulb, battery and a key in series in a circuit as shown in fig.
3. Close the key and measure the resistance of a bulb for every 30 seconds with a multimeter and note down them in the following table.

Observations: Resistance of the bulb in open circuit = 4.3 Q.

Conclusion:

1. From the above observations, it is clear that the resistance of a bulb (filament increases as the time increases.
2. This Is because, as the current passes through the filament of a bulb, filament gets heated up and its temperature increases.
3. As the temperature of the filament increases, its 20 resistance also increases
4. So, the resistance of a conductor depends upon its temperature.

Question 2.
Calculate the resistance of venous bulbs that you use at your home and find which one is having higher / lower resistance value. Write the report on your observations.
Answer:
We are using following types of bulbs in my bouse,

1. In candescent bulb (100 W)
2. Fluorescent tube lights (40 W)
3. CFL lamp (20 W)
4. LED bulb (10 W)

Conclusion:

• From the above observations, it is clear that the LED bulb of low wattage has higher resistance.
• So, it is clear that the resistance of a electrical appliance is more ¡f its wattage is less.

Question 3.
Collect the information and prepare a report on power consumption in your home/school.
Answer:
In my house, we are using the following electric appliances.
Tubelights (40 W) – 3 (Using daily each for 8 hours)
Fans(80 W) – 3 (UsIng daily each for 10 hours)
Television (60 W) – 1 (UsIng daily each for S hours)
Electric heater (1000 W) – 1 (Using daily each for 30 mm)

Calculation of power consumption :

Total electric energy consumed Per Day = 4160 WH

Conclusion:
Total electric energy consumed for one month (30 days) = 4160 x 30 = 1248000
Total electric energy consumed for one month in KWH = \(\frac{124800}{1000} \) = 124.8 KWH
∴ We are consuming nearly 125 KWH (units) of electric energy in my house in a month.

TS 10th Class Physical Science Electric Current Intext Questions

Page 176

Question 1.
What do you mean by electric current?
Answer:
The flow of electrons in a particular direction is called electric current.

Question 2.
Which type of charge (+ve or -ve) flows through an electric wire when it is connected in an electric circuit?
Answer:
Electrons carry nagative charge. So negative charge flows through circuit.

Question 3.
Is there any evidence for the motion of charge in daily life situations?
Answer:

1. Lightning, which is observed in the sky at the time of heavy rain is an evidence for the motion of charge in the atmosphere.
2. When we put the switch of an electric lamp ‘on’, the bulb glows. It is also evidence to motion of charge.

Question 4.
Does motion of charge always lead to electric current?
Answer:
Yes.

Question 5.
What do you notice in activity 1?
Answer:
The bulb glows.

Page 177

Question 6.
Can you predict the reasons for the bulb not glowing in situations 2 & 3?
Answer:

1. In situation 2, the source of current, namely battery is removed from circuit. So the bulb does not glow.
2. In situation 3, nylon wires do not conduct electricity. Nylon is a non-conductor. So the bulb does not glow.

Question 7.
Why do all materials not act as conductors?
Answer:
The materials in which electrons do not move freely do not act as conductors.

Question 8.
How does a conductor transfer energy from source to bulb?
Answer:
The electrons in a conductor move randomly in lattice space of conductor. These electrons transfer energy from source to bulb.

Question 9.
What happens to the motion of electrons when the ends of the conductor are connected to the battery?
Answer:
When the ends of the conductor are connected to the battery, the transfer of charged particles takes place from battery to bulb and again to the battery. As the circuit is complete and closed the bulb glows.

Page 178

Question 10.
Why do electrons move in specified direction?
Answer:
When the conductor Is connected to a battery, a uniform electric field is set up throughout the conductor. This field makes the electrons move towards positive end.

Question 11.
In which direction do the electrons move?
Answer:
The free electrons in the conductor are accelerated by the electric field and move in a direction opposite to the direction of the field.

Question 12.
Do the electrons accelerate continuously?
Answer:
No

Question 13.
Do they move with constant speed?
Answer:
The electrons collide with lattice ions, lose energy and may even come to rest at every collision.

Page 179

Question 14.
Why does a bulb glow immediately when we switch on?
Answer:
When we switch on any electric circuit, irrespective of length of the connecting wire, an electric field is set up throughout the conductor instantaneously, due to the potential difference of the source connected to the circuit. This electric field makes all the electrons move in a specified direction simultaneously. So the bulb glows immediately.

Question 15.
How can we decide the direction electric current?
Answer:
The direction of electric current is determined by the signs of the charge (q) and drift speed (y).

Page 180

Question 16.
How can we measure electric current?
Answer:
We can measure electric current, using an Ammeter.

Question 17.
Where do the electrons get energy for their motion from?
Answer:
The field exerts a force on the charge (electrons) The free charges accelerate the electric field, if the free charges ar electrons, then the direction of electric force on them ‘s opposite to the direction of electric field. It means that the electric field does some work to move free charges in a specified direction.

Question 18.
Can you find the work done by the electric force?
Answer:
Work done by the electric force on a free charge q ‘s given by W = EJ.

Page 181

Question 19.
What Is the direction of electric current In terms of potential difference?
Answer:
In terms of potential difference, the direction of electric current is from positive terminal to the negative terminal.

Question 20.
Do positive charges move In a conductor? Can you give an example of this?
Answer:
In electrolytic positive charges move towards of negative electrode.

Question 21.
How dosas battery maintain a constant potential difference between Ita terminals?
Answer:
The accumulation of charge on plates continues till the electric force F becomes equal to chemical force F, At this situation, the potential difference between the terminals Is maintained constant.

Question 22.
Why does the battery discharge when ita positive and negative terminals are connected through s conductor?
Answer:
When a conducting wire is connected to the terminals of the battery, a potential difference Is created between the ends and it sets up an electric field throughout the conductor. The electrons near the positive terminal of the battery are attracted by it and start moving towards positive terminal. As a result, the amount of positive charge on the plate decreases and the F becomes weaker than F. So the battery becomes discharged.

Page 182

Question 23.
What happens when the battery is connected in a circuit?
Answer:
When a conducting wire is connected to the terminals of the battery, a potential difference is created between the ends of the conductor. This potential difference sets up an electric field throughout the conductor and Its direction is from positive terminal to negative terminal In the conductor.

Page 183

Question 24.
How can we measure potential difference or emf?
Answer:
Generally, a voltmeter s used to measure potential difference or emf.

Page 184

Question 25.
Is there any relation between emf of battery and drift speed of electrons in the conductor connected to a battery?
Answer:
The ratio of emf and drift speed of electrons Is constant for some materials at constant temperatures.

Page 186

Question 26.
Can you guess the reason why the ratio of V and I in case of LED Is not constant?
Answer:
LED (Light Emitting Diode) is made up of semiconducting material. It is non Ohmic material and so the ratio of V and I in case of LED is not constant.

Question 27.
Do all materials obey Ohm’s law?
Answer:
No. Some materials such as silicon, germanium etc. do not obey Ohm’s law.

Question 28.
Can we classify the materials based on Ohm’s law?
Answer:
Yes. Based on Ohms’s law materials are classified into three categories.
They are:

• Ohmic materials,
• Non-Ohmic materials and
• Semiconductors.

Question 29.
What is resistance?
Answer:
Resistance of a conductor is the obstruction to the motion of electrons in a conductor.

Question 30.
Is the value of resistance the same for all materials?
Answer:
No. Silver and copper have least resistance value. Other materials such as iron, aluminum etc. have little higher resistance values. Tungsten has a very high resistance value.

Question 31.
Is there any application of Ohm’s law in daily life?
Answer:
Ohm’s law has a wide application in daily life:

1. We use materials like copper which are ohmic conductors to make household electrical wiring and in Industries.
2. Semiconductors which find an extensive application in modern electronic devices such as TV, DVD, Computers etc., are made up of non-ohmic materials.
3. The fuse, a device which protects household electrical appliances from high-voltage electric currents, is also an application of Ohm’s law.

Question 32.
What causes electric shock in the human body-current or voltage?
Answer:
It Is the electric current that causes electric shock in the human body. When 0.0024 Amperes of current flows through human body the functioning of organs inside the body gets disturbed. This disturbance inside the body Is felt as electric shock. ¡f the current flow continues further, It damages the tissues of the body which leads to decrease in resistance of the body.

Page 187

Question 33.
Do you know the voltage of mains that we use in our household circuits?
Answer:
The voltage of mains that we use in our household circuits is 240V. Usually, it varies between 220V and 240V.

Question 34.
What happens to our body if we touch live wire of 240V?
Answer:
The current passing through our body when we touch a live wire of 240V is given by, I= \(\frac{240}{100000}\) = 0.024 A
When this quantity of current flows through the body the functioning of organs inside the body gets disturbed. This disturbance inside the body is felt as electric shock and damages the tissues of the body.
When this current flows for a longer time, damage to tissues increases and resistance of body decreases.
A current of 0.07A, effects the functioning of heart and it may lead to fatal consequences.

Page 188

Question 35.
Why does not a bird get a shock when it stands on a high-voltage wire?
Answer:

1. There are two parallel transmission lines on electric poles.
2. The p.d. between the two lines is 240 V throughout their lengths.
3. When the bird stands on a high-voltage wire, there is no potential difference between the legs of the bird because it stands on a single wire.
4. If the two lines are connected across by a conducting device, then only current flows between the wires.
5. As the bird stands on a single wire no current passes through its body.
6. So, it does not feel any electric shock.

Page 189

Question 36.
What could be the reason for an increase in the resistance of the bulb when current flows through it?
Answer:

1. As the bulb glows it gets heated.
2. The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb since resistance is temperature dependent.

Question 37.
What happens to the resistance of a conductor If we increase its length?
Answer:
The resistance of a conductor increases with the increase in its length.i.e., R α l

Page 190

Question 38.
Does the thickness of a conductor influence its resistance’?
Answer:
Yes. The resistance of a conductor decreases with Increase in Its thickness. i.e., R α\(\frac{1}{A}\) (A = thickness or cross-section area of conductor)

Page 192

Question 39.
How are electric devices connected in circuits?
Answer:
Electric devices are connected in circuits either is series combination or an parallel combination.

Page 193

Question 40.
What do you notice in activity 6?
Answer:
In series connection of resistors, there Is only one path for the flow of current in the circuit. If the current in the entire circuit is I, It Is the same current lin the parts also.

Question 41.
What do you mean by equivalent resistance?
Answer:
If the current drawn by a resistor is equal to the current drawn by the combination of resistors, then the resistor is called as equivalent resistor.

Question 42.
What happens when one of the resistors In series breaks down?
Answer:
When one of the resistors In senes combination breaks down, the circuit becomes open and flow of current cannot take place In the circuit.

Question 43.
Can you guess in what way household wiring has been done?
Answer:
The household wiring has been done in parallel combination because
(i) the equivalent resistance of the parallel combination is less than the resistance of each of the resistors and (ii) through one of the resistors in parallel combination is cut off the other resistors continue to work.

Page 194

Question 44.
How much current is drawn from the battery?
Answer:
Measure the current (1) drawn from the battery using the ammeter and it Is 1.5 amps.

Question 45.
Is it equal to Individual currents drawn by the resistors?
Answer:
Yes, the current drawn from the battery‘s equal to the sum of individual currents drawn by the individual resistors (here bulbs).
That is, I=I₁ +I₂+I₃+ ……………………….. .

Page 199

Question 46.
You might have heard the sentence like “this month we have consumed 100 units of current”. What does unit mean?
Answer:
The electric appliances that we use n our daily life consume electric energy.
This energy Is measured in units.
i.e., 1 unit = 1 K.W,H. (Kilo Watt hour)
⇒ 1 K.W.H = 1000 W.H.

Question 47.
A bulb Is marked “60w and 120V”. What do these values Indicate?
Answer:
60 W and 120 V’ marked on a bulb indicates that If the bulb is connected to 120 volts mains, it will be able to convert 60 Watts of electrical energy into heat or light in one second.

Page 200

Question 48.
What is the energy lost by the charge in 1 Sec?
Answer:
The energy lost by the charge is equal to W/t where W = Work done and ‘t’ is the time in seconds.

Page 201

Question 49.
What do you mean by overload?
Answer:

1. Electricity enters our homes, through two wires called lines.
2. These line wires have low resistance and the p.d. of the wires is usually 240V.
3. These two line wires run throughout the household circuit to which we connect various appliances such as bulbs, fans, TV, refrigerator, air cooler etc.
4. These appliances are connected in parallel combination.
5. It we add more devices to the household circuit the current drawn from the mains abnormally increases. This is called overload.

Question 50.
Why does It cause damage to electric appliances?
Answer:

1. The maximum current that we can draw from the mains is 20A.
2. When the current drawn from the mains Is more than 20A, overheating occurs and may cause a fire.
3. It also causes damage to electrical appliances.

Question 51.
What happens when this current Increases greatly?
Answer:
When the current drawn from the mains is more than 20A, overheating occurs and may cause breaking of fire. This is called overloading and causes the damage of electrical appliances. Sometimes it may lead to fire accidents

Page 202

Question 52.
How can we prevent damage due to overloading?
Answer:
To prevent damage due to overloading an electric fuse is connected in the household circuit.

Think And discuss

Question 1.
What do you mean by short circuit?
Answer:
Short circuit means a connection across an electric circuit with a very low resistance, by an insulation failure etc. Current passes through this by pass.

Question 2.
Why does a short circuit damage electric wiring and devices connected to it?
Answer:
As the current takes short cut, which results in heating or burning which damages the wiring and devices connected to it.

TS 10th Class Physical Science Electric Current Activities

Activity 1

Question 1.
Write an activity to check when a bulb glows in a circuit.
Answer:
Aim: To check when a bulb glows In a circuit.
Materials required:

1. A bulb
2. a battery
3. a switch
4. insulated copper wire

Procedure (1):

• Take a bulb, a battery, a switch and few insulated copper wires.
• Connect the ends of the copper wires to the terminals of the battery through the bulb and switch.
• Now switch on the circuit.