The comprehensive approach in SCERT Class 10 Physics Solutions Chapter 6 Electromagnetic Induction in Daily Life Important Questions with Answers ensure conceptual clarity.
SSLC Physics Chapter 6 Important Questions Kerala Syllabus
Electromagnetic Induction in Daily Life Class 10 Important Questions
Question 1.
When there is a change in magnetic flux linked with a conductor, an emf is induced in it. Name the phenomenon?
Answer:
Electromagnetic Induction
Question 2.
The secondary coil of a transformer has double turns than its primary coil. If the voltage applied in the primary coil is 25 V, what will the voltage be in the secondary?
(25 V, 50 V, 2 V, 12.5 V)
Answer:
50 V
Question 3.
What is the energy conversion taking place in a windmill?
Answer:
Wind Energy → Mechanical Energy → Electrical Energy
Question 4.
Statement: A current is induced in a coil when a magnet is moved towards or away from it
Reason: The/elative motion between the coil and the magnet changes the magnetic flux through the coil
(a) Both statement and reason are true; the reason explains the statement.
(b) Both statement and reason are true; but the reason does not explain the statement.
(c) Both the statement and the reason are incorrect.
(d) Statement is incorrect; reason is correct.
Answer:
(a) Both statement and reason are true; the reason explains the statement.
Question 5.
Some characteristics of stepup transformers is given below. Select the correct statements suitable for stepup transformers.
(a) Secondary voltage is greater than primary voltage.
(b) Thick wires are used in the secondary.
(c) Primary voltage is greater than secondary voltage.
(d) The current in the primary coil is greater than that in the secondary coil.
(i) (a), (b) and (d)only
(ii) (a) and (d) only
(iii) (a) and (c) only
(iv) (b) only
Answer:
(ii) (a) and (d) only
Question 6.
Write any two precautions to be taken to avoid electric shock.
Answer:
- Never handle electric equipment or operate switches when the hands are wet.
- Do not fly kites near electric lines.
Question 7.
The primary and secondary currents of a transformer are 2 A and 4 A respectively. The output voltage of this transformer is 100 V.
(a) Identify the transformer.
(b) Calculate the primary voltage of the transformer.
Answer:
(a) Stepdown Transformer
(b) VP × Ip = Vs × Is
∴ VP = (Vs × Is) / Ip
= (100 × 4) / 2
= 200 V
Question 8.
The current in the secondary coil of a transformer with no power loss is 5 A and that is the primary is 0.5 A.
(a) What type of transformer is this?
(b) If the input voltage of this transformer is 240 V calculate the output voltage.
Answer:
(a) Step down transformer
(b) IS = 5 A
IP = 0.5 A
VP = 240 V
VP = ?
IS × VS = IP × VP
VS = \(\frac{V_P \times I_P}{I_S}\) = \(\frac{0.5 \times 240}{5}\) = 24 V
Question 9.
Fuse, MCB, ELCB, and RCCB are used to ensure safety in household electric circuits.
(a) What is the difference between the functioning of Fuse and MCB?
(b) What is the advantage of MCB over Fuse?
(c) What is the function of ELCB and RCCB in electric circuits?
Answer:
(a) Fuse – In fuse, fuse wire melts due to excess current. It works based on the heating effect of electric current.
MCB – It has an internal switch that gets tripped when there is excess current. It works based on the magnetic effect and the heating effect of electrical energy.
(b) In MCB, after solving the problem, the switch can be turned on. But in the fuse, the melted fuse wire is to be replaced by another suitable wire. Moreover, MCB is more sensitive than fuse. MCB is simple to resume the supply.
(c) ELCB helps to break the circuit automatically when there is leakage of current due to insulation failure or short circuit, it protects the user from electric shock. RCCB also breaks the circuit, so that there’is no harm due to electric shock. RCCB ensures more safety.
Question 10.
A transformer has 100 turns in the primary and 1000 turns in the secondary.
(a) Which coil of this transformer is made using thick wire? Give reason.
(b) Explain/how electrical energy is transferred from the primary to the secondary of the transformer
Answer:
(a) Primary coil is made using thick wire.
(Hint: Number of turns is less in the Primary here. So it is a step-up transformer. So thick wire is used in the primary coil)
When the thick wire is used, the resistance of the coil can be minimized. There are two benefits to this.
(i) Overheating of the coil can be prevented.
(ii) Energy loss can be minimized.
(b) It is through mutual induction power is transferred from the primary coil to the secondary coil.
When the strength or direction of the current in one of the two adjacent coils changes, the magnetic flux around it changes. As a result, an emf is induced in the secondary coil. This phenomenon is mutual induction.
Question 11.
AC generators are used in power stations in our Country.
(a) What is the voltage produced by the generators in our power stations?
(b) What do you mean by transmission loss?
(c) Explain how it is minimized.
Answer:
(a) 11000 V (11 kV)
(b) When electrical energy is transmitted through conducting wires, energy is wasted in the form of heat.
(c) As H = I2Rt, for reducing H, I and R are to be decreased.
As P = VI, For decreasing I without affecting the power, voltage V is to be increased before transmission.
Question 12.
A schematic diagram of a generator is given:
(a) Which type of generator is this? (AC/DC)
(b) Name the parts of this generator marked as
(c) State the working principle of this device.
1, 2, 3, 4,
1. ________________
2. ________________
3. ________________
4. ________________
(c) State the working principle of this device.
Answer:
(a) AC Generator
(b) 1. Field magnet
2. Armature
3. Slip rings
4. Brush
(c) Electromagnetic induction: Whenever there is a change in the magnetic flux linked with a coil, an emf is induced in the coil. This phenomenon is electromagnetic induction.
Question 13.
Observe the circuit of household electrification.
(a) Which device is used to measure the electrical energy consumed in household circuits?
(b) Write any two advantage of connecting the devices in parallel in a household circuit.
(c) Write the function of ELCB.
Answer:
(a) Watt-hour meter.
(b) Devices work according to the marked power.
Devices can be controlled using switches as needed.
(c) ELCB helps to break the circuit automatically whenever there is a current leak due to insulation failure or other reasons.
Question 14.
A transformer without power loss works at an input of 250 V. A current of 1 A flows through the secondary coil when an electrical device of power 50 W is connected to the secondary.
(a) Which type of transformer is used here?
(b) What is the working principle of a transformer?
(c) Calculate the current through the primary.
Answer:
(a) Ps = Vs × Is
50 = Vs × 1
Vs = 50 V
Since Vs is less than Vp, it is a step-down transformer.
(b) Mutual Induction
(c) Vp × Ip = Vs × Is
Vp = 250 V, Vs = 50 V
Is = 1 A
Ip = (Vs × Is) / Vp = (50 × 1) / 250 = 0.2 A
Question 15.
Analyze the following schematic diagram and answer the questions.
(a) Name the device.
(b) Which are the main parts of this device?
(c) Illustrate the graphic representation of the output when the field magnet is rotated by keeping the armature stationary.
Answer:
(a) DC Generator.
(b) Armature, field magnet, split ring commutator, and brushes.
(c) 