ELECTRICITY

Multiple choice questions. (ONE MARKS)

1. The resistance of a conductor is 27 Ω. If it is cut into three equal parts and connected in
parallel, then its total resistance is (April-2019)
(A) 6 Ω (B) 3 Ω (C) 9 Ω (D) 27 Ω
R1 = 9 Ω R2 = 9 Ω R3 = 9 Ω
1 1 1 1 1 1 1 3 1
= + + = + + = =
Rp R1 R2 R3 9 9 9 9 3
Rp = 3 Ω
2. A piece of metallic wire of resistance R is cut into 3 equal parts. These parts are then
connected in parallel. If the total resistance of this combination is R1, then the value of R: R1 is
(June-2019)
(A) 1:3 (B) 9:1 (C) 9:1 (D) 3:1
R R R
R1 = Ω R2 = Ω R3 = Ω
3 3 3
1 1 1 1 1 3 3 3 1 9 R 9
1
= + + 1
= + + 1
= 1
= R: R1 = 9: 1
R R1 R2 R3 R R R R R R R 1

3. An electric lamp whose resistance is 30 Ω and a conductor of 6 Ω resistances are connected in

series to 9V battery as shown in the figure. The total current flowing in the circuit is
(April-2021)
(A) 4 A (B) 36 A (C) 0.25 A (D) 0.6 A

R1 = 30 Ω R2 = 6 Ω Rs = 30 + 6 = 36 Ω V = 9V
V 9 1
I = = = = 0.25 A
R 36 4

4. A device used to change the resistance in the electric circuit is (April-2021)

(A) Voltmeter (B) ammeter (C) galvanometer (D) rheostat
5. ‘Ohm’ is the SI unit of (April-2021)
(A) Electric potential difference (B) resistance (C) electric current (D) electric charge
6. The device used to produce electricity is (April-2022)
(A) Galvanometer (B) Electric generator (C) Ammeter (D) Electric motor.
7. The correct formula that shows the relationship between potential difference, electric current
and resistance in an electric circuit is (April-2022)
𝑅 𝐼 𝑽
(A) I = (B) I = VR (C) V = (D) R =
𝑉 𝑅 𝑰
8. The SI unit of resistivity is (June-2022)
(A) Ohm (B) volt (C) watt (D) ohm-meter.
9. The device used to measure the rate of the current in a circuit is (April-2023)
(A) Ammeter (B) Voltmeter (C) Galvanometer (D) Battery
10. A device that converts electrical energy into mechanical energy is (June-2023)
(A) Electric generator (B) Electric motor (C) Galvanometer (D) Voltmeter
11. In an electric circuit to get an equivalent resistance Rs four resistors of 2 Ω each are first
connected in series. Later to get an equivalent resistance of Rp the same resistors are
connected in parallel. Then the ratio of Rs / Rp is (April-2024)
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 (A) 16: 1 (B) 2: 1 (C) 4: 1 (D) 8: 1
R1 = 2 Ω R2 = 2 Ω R3 = 2 Ω R4 = 2 Ω
Rs = 2 + 2 + 2 + 2 = 8 Ω
1 1 1 1 1 1 1 1 1 4 2
= + + + = + + + = =
Rp R1 R2 R3 R4 2 2 2 2 2 1

1 1
=2 Rp =
Rp 2
Rs 8 16
= = Rs:Rp = 16:1
Rp 1/2 1
12. SI unit of electric charge is (Aug-2024)
(A) Coulomb (B) ampere (C) joule (D) volt
Answer the following questions. (ONE MARKS)
1. What is the SI unit of potential difference? Name the device used to measure the potential
difference. (Mar-2020)
SI unit of potential difference is volt (V).
Voltmeter is used to measure the potential difference.
2. Write the symbols of the following components used in an electric circuit. (April-2023)
i) Rheostat ii) Wires crossing without joining
i) ii)

3. Draw the symbol diagram of rheostat used in electric circuit. (June-2023)

4. Write the symbols of the following components used in an electric circuit : (April-2024)
i) Combination of two cells ii) Wires crossing without joining.
i) ii)

5. Can an electric heater of 2kW be connected to a domestic circuit rated 15 A and has a
potential difference of 220V? Support your answer. (April-2024)
Can be connected
Because the rate of electric circuit is less than 15 A.
6. Write the symbols of the following components used in an electric circuit. (June-2024)
i) Wires crossing without joining ii) Voltmeter
i)
ii)

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 7. Write the symbols of the following components used in an electric circuit. (Aug-2024)
i) A rheostat ii) A wire joint.
i) ii)

Answer the following questions. (TWO MARKS)

1. Draw the diagram of an electric circuit in which the resistors R1, R2 and R3 are connected in
parallel including an ammeter and a voltmeter and mark the direction of the current.
(April-2019)

2. It is advantageous to connect electric devices in parallel instead of connecting them in series.

Why? (April-2019)
➢ The appliances connected in series need currents of widely different values to operate properly.
➢ In a series circuit, if component fails, the circuit is broken and none of the component works.
➢ But in parallel circuits current divides through electrical gadgets.
➢ This is helpful particularly when each gadget has different resistance and requires different
current to operate properly.
3. According to Joule’s law of heating, mention the factors on which heat produced in a resistor
depends. According to this law write the formula used to calculate the heat produced.
(April-2019)
The heat produced in a resistor is
i) Directly proportional to the square of the current for a given resistance.
ii) Directly proportional to the resistance for a given circuit.
iii) Directly proportional to the time for which the current flows through the resistor.
H = I2Rt
4. An electric refrigerator rated 400 W is used for 8 hours a day. An electric iron box rated 750
W is used for 2 hours a day. Calculate the cost of using these appliances for 30 days, if the cost
of 1 kWh is Rs. 3/-. (April-2019)
E = nPt
The total energy consumed by refrigerator in 30 days = 400 x 8 x 30 = 96000 Wh = 96 kWh
The total energy consumed by iron box in 30 days = 750 x 2 x 30 = 45000 Wh = 45 kWh
The total energy consumed by refrigerator and iron box = 96 kWh + 45 kWh = 141 kWh
The sum of bill amount for 141 kWh at rate of Rs. 3 per 1kWh = 141 x 3 = Rs. 423

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 5. A bulb is marked 220 V and 40 W. Calculate the current flowing through the bulb and its
resistance. (June-2019)
V = 220V P = 40 W
P= VI
40= 220×I
I= 40/220=2/11 A = 0.19(approx.)
Now for resistance
V= IR
220=2/11×R
R=220×11/2=110×11=1210 ohm.
6. Observe the given circuit diagram. Calculate the total resistance and the total current flowing
through the circuit. (Sep-2020)
Here, R1 = 5 Ω, R2 = 4 Ω, R3 = 12 Ω, V =24 V.
Total resistance of the circuit RT =?
Total current flowing through the circuit, I =?
1 1
Total resistance of the circuit RT = R1 + +
R2 R3
1 1
=5+ +
4 12
3+1 4 1
=5+ =5+ =5+
12 12 3
=5+3
=8Ω
V 24
Total current flowing through the circuit I= = = 3A
R 8

7. The resistivity of manganese wire of length 1 m is 1·84 × 10− 6 Ω m at 20°C. If the diameter of
the wire is 3 × 10− 4 m, what will be the resistance of the wire at that temperature? (Mar-2020)
Resistivity 𝜌 = 1.84 x 10-6 Ω m Length l = 1 m diameter d = 3 x 10-4 m radius r =3/2 x 10-4
22 3
Area of cross section A = 𝜋r2 = x ( x 10−4 )2
7 2
22 9
= x x 10-8
7 4
99
= x 10-8 m2
14
𝜌x𝑙 1.84 𝑋 10−6 𝑋 1 𝑋 14
Resistance R = =
𝐴 99 𝑋 10−8
25.76 𝑋 102
=
99
2576
=
99
= 26.02 Ω

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 8. Observe the given circuit: Calculate the total resistance in the circuit and the total current
flowing in the circuit. (Mar-2020)
R1 = 2 Ω R2 = 4 Ω R3 = 4 Ω V = 6V
In parallel connections:
1 1 1 1 1 1 1 4+2+2 8
= + + = + + = = =1Ω Rp = 1 Ω
Rp R1 R2 R3 2 4 4 8 8
Total resistance in the circuit R = Rp + R4 = 1 + 5 = 6 Ω
V 6
Current I = = = 1A
R 6
9. Draw the schematic diagram of an electric circuit comprising of electric cell, electric bulb,
ammeter and plug key. (April-2022)

10. An electric bulb with a resistance of 50 Ω is connected to 10 V battery in an electric circuit.

Calculate the electric current flowing through the electric bulb and electric power of the bulb.
(June-2022)
R= 50 Ω V = 10 V
V 10
I = = = 0.2 A
R 50
Electric current flowing through the electric bulb is 0.2 A.
P = VI = 10 x 0.2 = 2 W
Power of bulb is 2 W.
11. 1000 J of heat is produced each 2 seconds in a 5 Ω resistor. Find the potential difference
across the resistor. (June-2023)
H = 1000 J R = 5 Ω t=2s
2
H = I Rt
H 1000 1000
I2 = = = = 100
Rt 5x2 10
I = √100 = 10 A
Potential difference across the resistor
V = IR = 10 x 5 = 50V
12. A wire of given material having length ‘l’ and area of cross-section ‘A’ has a resistance of ‘4
Ω’. Find the resistance of another wire of the same material having length ‘l/2’ and area of
cross-section ‘2A’. (June-2023)
R=4Ω
For first wire
𝑙
Resistance of a wire R1 = 𝜌
𝐴
𝑙
4=𝜌
𝐴
For second wire 𝑙1 = l/2 𝐴1 = 2A
𝑙 𝑙 𝑙
R1 = 𝜌 1 = 𝜌 =𝜌
𝐴1 2 x 2A 4𝐴

R1 = 𝜌 ( 𝐴𝑙 ) ( 14 )
1
R1 = 4 x =1Ω
4
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 Resistance of another wire is 1 Ω
13. Give reason: a) the tungsten is used in filaments of electric lamps.
b) In domestic circuits the electric devices are not connected in series. (Aug-2024)
a) The melting point and resistivity of tungsten are very high. It does not melt readily at a high
temperature. The electric lamps glow at very high temperature. Hence tungsten is mainly used as
heating element of electric bulbs.
b) In domestic electric circuits, the different electric components need widely different electric
current values to operate properly. When one component fails, the circuit is broken and none of the
components works. Therefore the series arrangement is not used for domestic electric circuits.
14. Observe the following electric circuit: When a wire of resistance ‘R’ Ω is connected between
‘X’ and ‘Y’, then the ammeter reading is 3A. If ‘R’ Ω resistance is replaced by ‘2R’ Ω in the
same circuit, what would be the reading in ammeter? Give scientific reason for your answer.
(Aug-2024)
If R = R Ω I = 3 A V = IR = 3 x R = 3R
If R Ω is replaced by 2R Ω
I = V/R = 3R / 2R = 3/2 = 1.5 A
The ammeter reading is 1.5 A
Reason: If the resistance is doubled, the current gets halved.
Answer the following questions. (THREE MARKS)
1. State Joule’s law of heating. Explain the working of electric filament bulb. (Mar-2020)
The heat produced in a resistor is
i) Directly proportional to the square of the current for a given resistance.
ii) Directly proportional to the resistance for a given circuit.
iii) Directly proportional to the time for which the current flows through the resistor.
H = I2Rt
A strong metal with high melting point such as tungsten is used for making bulb filaments.
The bulbs are usually filled with chemically inactive nitrogen and Argon gases to prolong the life of
the filament.
Most of the power consumed by the filament appears as heat, but a small part of it is in the form of
light radiated.
2. State Ohm’s law. How ammeter and voltmeter should be connected in electric circuit? What is
the use of these instruments, in the circuit? (Mar-2020)
The potential difference V across the ends of a given metallic wire in an electric circuit is directly
proportional to the current flowing through it at constant temperature.
Ammeter should be connected in series and Voltmeter should be connected in parallel in the circuit.
Ammeter is used to measure current.
Voltmeter is used to measure potential difference.
3. State Ohm’s law. On which factors does the resistance of a conductor depend? Mention the SI
unit of electric power. (April-2023)
The potential difference (V) across the ends of a given metallic wire in an electric circuit is directly
proportional to the current (I) flowing through it, provided its temperature remains the same.
V = IR
The resistance of a conductor depends on:
i) Its length ii) its area of cross-section iii) the nature of its material iv) temperature.
The SI unit of electric power is watt

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 4. State Joule’s law of heating. How is fuse connected in the circuits? Name the metal used in the
filament and the gas filled in electric bulb. (April-2023)
The heat produced in a resistor is
i) Directly proportional to the square of the current for a given resistance.
ii) Directly proportional to the resistance for a given circuit.
iii) Directly proportional to the time for which the current flows through the resistor. H = I2Rt
A fuse is connected in series with the live wire of an electrical circuit.
The metal used in the filament is tungsten and the gas is filled in the bulb is argon.
5. The resistors R1, R2 and R3 have the values 10 Ω, 20 Ω and 60 Ω respectively, which have been
parallelly connected to a battery of 24 V in an electric circuit. Then calculate the following.
i) The current flowing through each resistor ii) The total current in the circuit
iii) The total resistance of the circuit. (April-2023)
R1 = 10 Ω R2 = 20 Ω R3 = 60 Ω V = 24V
V 24
i) I1 = = = 2.4 A
R1 10

V 24
I2 = = = 1.2 A
R2 20

V 24
I3 = = = 0.4 A
R3 60

ii) I = I1 + I2 + I3 = 2.4 + 1.2 + 0.4 = 4 A

1 1 1 1 1 1 1 6+3+1 10 1
iii) = + + = + + = = =
Rp R1 R2 R3 10 20 60 60 60 6
Rp = 6 Ω
6. 200J of heat is produced each second in an 8Ω resistance. Find the potential difference across
the resistor. (April-2024)
H = 200 J R=8Ω t=1s
H = I2Rt
H 200
I2 = = = 25
Rt 8x1
I = √25 = 5 A
Potential difference across the resistor
V = IR = 5 x 8 = 40V
7. An electric refrigerator rated 300W operates 6 hours in a day. What is the cost of the energy to
operate it for 30 days at Rs. 7.00 per kWh? (April-2024)
The total energy consumed by refrigerator in 30 days = 300 x 6 x 30 = 54000 Wh = 54 kWh
The cost of energy for 54 kWh at rate of Rs. 7 per 1kWh = 54 x 7 = Rs. 378
Answer the following questions. (FOUR MARKS)
1. (i) Define electric potential difference. How is ammeter connected in an electric circuit?
(ii) Explain the application of heating effect of electric current in an electric bulb and the fuse
used in an electric circuit. (June-2019)
i) The amount of work done to move a unit positive charge from one point to another in an electric
circuit.
Ammeter connected in series in an electric circuit.
ii)When electric current passes through a very thin , high resistance tungsten filament of an electric
bulb, the filament becomes white hot and emits light .

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 if a current larger than the specified values flows thin ,its temperature increases enough to melt it,
hence the circuit breaks and appliance is protected from damage
2. (i)State Ohm’s law
(ii) Explain the factors on which the resistances of a conductor depend. (June-2019)
i) The potential difference (V) across the ends of a given metallic wire in an electric circuit is
directly proportional to the current (I) flowing through it, provided its temperature remains the same.
V = IR
ii) The resistance of a conductor depends on:
i) Its length ii) its area of cross-section iii) the nature of its material iv) temperature.
3. What is the meaning of the statement “The potential difference between two points is 1 V”?
Name the device used to measure potential difference. What is resistance of a conductor?
What is electric power? Write three formulae used to find electric power. (Sep-2020)
• If 1 Joule (1J) of work is done to move a charge of 1 Coulomb (1 C) from one point to another point
in a current carrying conductor, the potential difference between the two points is 1 volt (1 V).
• The device used to measure it is voltmeter.
• The property of a conductor to resist the motion of electric charges flowing through it is called
resistance of a conductor.
• The rate at which electric energy is dissipated or consumed in an electric circuit is called electric
power.
I2
• Three formulae used to find electric power are P = VI P= P = V2 R
R
4. a) What are the advantages of connecting electrical devices in parallel in an electric circuit
instead of connecting them in series?
b) How are ammeter and voltmeter connected in an electric circuit? What are their functions?
(April-2022)
a) Parallel circuit divides the current through the electrical devices connected. This is helpful
particularly when each device has different resistance and requires different current to operate
properly. But in a series circuit when one component fails the current is broken and none of the
components works.
b) An ammeter is connected in series and voltmeter is connected in parallel.
Ammeter measures the amount of current flowing through in a circuit.
Voltmeter measures the potential difference between two points in a circuit.
5. a) State Joule’s law of heating. Name any two devices that work on the application of this law.
b) Why are the alloys like nichrome used in electrical heating devices? (June-2022)
a) The heat produced in a resistor is
i) Directly proportional to the square of current for a given resistance.
ii) Directly proportional to the resistance for a given current.
iii) Directly proportional to the time for which the current flows through the resistor.
H = I2Rt
The devices that work on this law are Electric Toaster, Electric Oven, Electric Kettle, Electric Bulb,
and Electric Fuse.
b) Resistivity of alloys is more than / higher than that of metals.
Alloys do not oxidize (burn) readily at high temperature. Alloys have high melting point.
6. a) State Ohm’s law. In domestic electric circuit electrical appliances are not connected in series.
Why?
b) Write the factors on which resistance of a conductor depends. (June-2022)
a) The potential difference (V) across the ends of a given metallic wire in an electric circuit is
directly proportional to the current (I) flowing through it, provided its temperature remains the same.
V = IR
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 • In a series circuit the current is constant throughout the electric circuit due to this all electrical
appliances not possible to work at the same value.
• In a series connection, when one component fails, the circuit is broken.
b) The resistance of a conductor depends on:
i) Its length ii) its area of cross-section iii) the nature of its material iv) temperature.
7. a) A bread-toaster rated 350 W is used for 15 hours a day. An electric iron box rated 250 W is
used for 5 hours a day. Calculate the cost of using these appliances for 30 days, if the cost of 1
kWh is Rs. 4.
b) In which method the resistors R1 and R2 could be connected so that the equivalent resistance
of that electric circuit becomes low? What is the change in the value of current in the circuit by
this type of connection? (June-2023)
a) The total energy consumed by bread toaster in 30 days = 350 x 15 x 30 = 157500Wh = 157.5
kWh
The total energy consumed by iron box in 30 days = 250 x 5 x 30 = 37500 Wh = 37.5 kWh
The total energy consumed by bread toaster and iron box = 157.5 kWh + 37.5 kWh = 195 kWh
The sum of bill amount for 195 kWh at rate of Rs. 4 per 1kWh = 195 x 4 = Rs. 780
b) If the resistor is connected in parallel, the resistance of R1 and R2 will be lowest.
The value of current in the circuit increases.
8. a) What is resistance of a conductor? On what factors does the resistance of a conductor
depend?
b) It is advantageous to connect electrical devices in parallel instead of connecting them in
series. Why? Explain. (June-2024)
a) Resistance of a conductor is a property that resists the flow of electric charges in the conductor.
The resistance of a conductor depends on:
i) Its length ii) its area of cross-section iii) the nature of its material iv) temperature.
b) Parallel circuit divides the current through the electrical devices connected. This is helpful
particularly when each device has different resistance and requires different current to operate
properly. But in a series circuit when one component fails the current is broken and none of the
components works.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 MEGNITIC EFFECT OF ELECTRIC CURRENT
Multiple choice questions. (ONE MARKS)
1. Observe the diagram. The magnetic poles represented by P and Q respectively are(Sep-2020)
(A) South ( S ) and south ( S ) (B) north ( N ) and south ( S )
(C) north ( N ) and north ( N ) (D) south ( S ) and north ( N ).

2. The magnetic field lines inside a solenoid are in the form of parallel straight lines. The reason
for this is, the magnetic field inside the solenoid is (April-2021)
(A) Very high (B) uniform (C) zero (D) produced by electric current
3. In Fleming’s right hand rule, the middle finger indicates the direction of (April-2022)
(A) Induced electric current (B) magnetic field
(C) Motion of the conductor (D) mechanical force.
4. Imagine, you are holding a straight current carrying conductor as per the right hand thumb
rule. If the thumb is upward, then the direction of the field lines of the magnetic field is
(June-2023)
(A) Downward (B) upward (C) anti-clockwise (D) clockwise
5. The magnetic field inside a long straight solenoid carrying current. (June-2024)
(A) Is the same at all points (B) is zero
(C) decreases as we move towards its end (D) increases as we move towards its end
6. In Fleming’s left hand rule the middle finger represents the direction of (Aug-2024)
(A) Magnetic field (B) current (C) movement of conductor (D) induced current
Answer the following questions. (ONE MARKS)
1. Observe the given figure. What type of current is induced in the coil by doing the experiment
related to this figure? Give reason for your answer. (Mar-2020)
Alternating current
Because the deflections in the galvanometer would just be opposite to
the previous.

2. Suggest any two measures to avoid overloading in domestic circuits. (Sep-2020)

➢ Live and neutral wires should not come into direct contact.
➢ There should not be any short-circuit in the circuit.
➢ Too many appliances should not be connected to a single socket.
➢ Should always use quality wires and good quality electrical appliances.
3. Magnetic field lines do not intersect each other. Why? (April-2022)
At the point of intersection the compass needle would point towards two directions which is not
possible.
4. What are the reasons for occurring overload in an electric circuit? (June-2022)
Accidental hike in the supply voltage
Connecting too many appliances to a single socket
When live wire and neutral wire come into direct contact.
5. What does the thumb indicate in the right hand thumb rule? (April-2023)
The thumb indicates the direction of current
6. Observe the figure and mention the direction of the force acting on the current carrying
conductor AB. Name the rule that helped you to find the direction of the force. (June-2023)
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 The direction of the force acting on the current carrying conductor is
perpendicular to its length and the magnetic field.
Fleming’s left hand rule helped to find the direction of the force.

Answer the following questions. (TWO MARKS)

1. Observe the given figure: If the key connected to Coil-2 is
plugged, in which of the other two coils more current is
induced? Why? (June-2024)
➢ More current is induced in Coil-1.
➢ Coil-1 has more number of turns than Coil-3
➢ As the number of turns increases the current induced also
increases.
2. Observe the given figures and answer the questions that follow :
i) Which of the above figures shows the correct direction of magnetic field?
ii) Name and state the rule that helped to choose the correct figure. (Aug-2024)
i) Figure (b) shows the correct direction of magnetic field.
ii) Right hand thumb rule helps to choose the correct figure.
Statement: Holding a current carrying straight conductor in
right hand such that the thumb indicates the direction of current
and the fingers wrapping around the conductor indicates
direction of field lines of the magnetic field.
3. Placing a ‘fuse’ in electric circuits is essential. Why?
Explain. (Aug-2024)
Fuse is a safety device which protects electric circuits and electric appliances by stopping the flow
of any unduly high electric current. It is a piece of wire made of a metal or an alloy of appropriate
melting point, placed in series with the device. If current larger than the specified value flows, the
temperature of fuse wire increases. This melts the fuse wire and breaks the circuit. Thus placing of
fuse is must in electric circuits.
Answer the following questions. (THREE MARKS)
1. What are the functions of an earth wire? It is necessary to connect the electric appliances
having metallic body to the earth wire in domestic electric circuit. Why? Explain. (April-2022)
Functions of the earth wire:
This is used as a safety measure for appliances have a metallic body in domestic circuit
This provides a low resistance conducting path for the current
Any leakage of current in the appliances keeps its potential to that of the earth and the user may not
get a severe electric shock.
2. List the properties of the magnetic field due to the flow of electric current in a solenoid. What
are the two methods of increasing magnetic field in a solenoid? (June-2022)
The magnetic field in a current carrying solenoid is similar to that of magnetic field produced in a
bar magnet.
The magnetic field is uniform inside the solenoid.
The two methods to increase magnetic field in a solenoid.
i) By increasing the number of turns of the coil.
ii) By increasing the current flowing through solenoid.
3. Observe the given diagram: Explain the experiment related to this diagram. What conclusions
can be drawn from this experiment? (June-2023)

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 • Take two different coils of copper wire say 100 and 50
turns respectively. Insert them over a non-
conducting cylindrical roll.
• Connect the coil-1 in series with battery and plug key,
coil-2 with galvanometer.
• When the key is plugged in, needle of the
galvanometer deflects and return to zero. This
indicates the presence of current in the coil-2.
• Disconnect coil-1 from battery. Needle of the galvanometer deflects in the opposite direction of the
current.
Conclusion:
• Changing electric current in coil-1 induces current in coil-2. This is electromagnetic induction.
• This is due to the change in the magnetic field.
4. In domestic circuits,
i) What are the reasons for overloading?
ii) Explain the working of earth wire. (April-2024)
i ) * Accidental hike in the supply of voltage
* Connecting too many appliances to a single socket
* When live wire and neutral wire come into direct contact.
ii) Functions of the earth wire:
* This is used as a safety measure for appliances have a metallic body in domestic circuit.
* This provides a low resistance conducting path for the current.
* Any leakage of current in the appliances keeps its potential to that of the earth and the user may
not get a severe electric shock.
5. A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar
magnet is (April 2024)
i) Pushed into the coil? ii) Withdrawn from inside the coil? iii) Held stationary inside the coil?
i) The needle of the galvanometer deflects.
ii) The needle of the galvanometer deflects in the direction opposite to the first
iii) The needle of the galvanometer does not deflect needle indicates zero.
Answer the following questions. (FOUR MARKS)
6. (i) How does overload and short-circuit occur in an electric circuit? Explain. What is the
function of fuse during this situation?
(ii) Mention two properties of magnetic field lines. (April-2019)
i) * Overload and short circuit occur in an electric circuit if
Accidental hike in the supply voltage
Connecting too many appliances to a single socket
When live wire and neutral wire come into direct contact.
* Fuse is a safety device which protects electric circuits and electric appliances by stopping the flow
of any unduly high electric current. It is a piece of wire made of a metal or an alloy of appropriate
melting point, placed in series with the device. If current larger than the specified value flows, the
temperature of fuse wire increases. This melts the fuse wire and breaks the circuit.
ii) Properties of magnetic field lines.
➢ Field lines emerge from North Pole of a magnet and merge at South Pole.
➢ Inside the magnet the direction of the field lines is from its south pole to North Pole.
➢ Magnetic field lines are closed curves.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 ➢ Magnetic field is stronger where the field lines are crowded.
➢ No two field-lines are found to cross each other.
7. Observe the given diagram. Explain the experiment related to this
diagram. What conclusions can be drawn from this experiment?
(Sep-2020)
➢ The ends of the copper coil (AB) are connected to a galvanometer.
The north pole of the bar magnet (NS) is moved inside the coil.
Induced current is produced in the coil and hence the needle of the
Galvanometer shows momentary deflection in one direction.
➢ When the north pole of the magnet is withdrawn from the coil, the needle of the galvanometer is
deflected in the opposite direction.
➢ When the magnet is held stationary inside the coil, the deflection of the galvanometer drops to zero
(shows no deflection).
➢ When the magnet is moved inside the coil with greater force, galvanometer shows greater deflection
and when the magnet is moved with smaller force, the galvanometer shows smaller deflection.
When the magnet is stationary and the coil is moved towards / away from the magnet, galvanometer
show deflection.
Conclusions that can be drawn from this experiment:
Motion of the magnet with respect to the coil produces an induced potential difference, which sets up
an induced electric current in the circuit.
8. a) What is solenoid? Write the properties of the magnetic field lines formed around a current
carrying solenoid.
b) What is alternating current? Electric appliances having metallic body are connected to
earth wire. Why? (April-2023)
a) A coil of many circular turns of insulated copper wire wrapped closely in the shape of cylinder is
called a solenoid.
The magnetic field in a current carrying solenoid is similar to that of magnetic field produced in a
bar magnet.
The magnetic field is uniform inside the solenoid.
b) A type of current that reverse direction and changes its magnitude at regular intervals of time.
Earth wire is used as a safety measure for appliances have a metallic body in domestic circuit
This provides a low resistance conducting path for the current
Any leakage of current in the appliances keeps its potential to that of the earth and the user may not
get a severe electric shock.
9. a) State the right hand thumb rule. Write any two properties of the magnetic field lines.
b) What is solenoid? How can this be converted into an electromagnet? (April-2024)
Right hand thumb rule:
a) When you are holding a current carrying conductor such that the thumb points towards the
direction of current then your fingers will wrap around the conductor in the direction of the field
lines of the magnetic field.
Properties of the magnetic field lines:
*Emerge from North Pole and merge at the South Pole. *Closed curves
*Never intersect each other *Have magnitude and direction.
b) Solenoid: A coil of many circular turns of insulated copper wire wrapped closely in the shape of
cylinder is called a solenoid.
Current carrying solenoid can be used to magnetize a piece of magnetic material like soft iron when
placed inside the coil.
10. a) Explain an experiment of drawing magnetic field lines around a bar magnet with the help of
a compass needle.
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 b) Mention two properties of magnetic field lines. (June-2024)
a) Drawing magnetic field lines around a bar magnet using a compass needle:
➢ Place a bar magnet on a white paper and mark the boundary of the magnet
➢ Place the compass needle near the north pole of the magnet. The south pole of the compass needle
directs towards the north pole of the magnet. Mark it with a point.
➢ Move the needle to a new position such that it’s South Pole occupies the position previously
occupied by its north pole. Mark it with a point.
➢ In this way proceed step by step till you reach the south pole of the magnet.
➢ Join the points marked on the paper by a small curve.
➢ This curve represents a field line.
b) Properties of magnetic field lines:
➢ Field lines emerge from North Pole of a magnet and merge at South Pole.
➢ Inside the magnet the direction of the field lines is from its south pole to North Pole.
➢ Magnetic field lines are closed curves.
➢ Magnetic field is stronger where the field lines are crowded.
➢ No two field-lines are found to cross each other.
11. a) Explain an experiment to show that a current carrying conductor experiences the force in a
magnetic field.
b) How is a simple electric motor converted into a commercial motor? (June-2024)
a) Take a small aluminum rod and suspend it horizontally using connecting wires.
Place a strong horse-shoe magnet in such a way that rod lies between the two poles with the
magnetic field directed upwards.
Connect the aluminum rod in series with a battery, a key and a rheostat.
Now pass the current through the aluminum rod in one particular direction.
The rod displaces towards one side.
Reverse the direction of current flowing through the rod. The rod displaces towards the opposite side.
Hence a current carrying conductor experiences a force perpendicular to its length in a magnetic field.
b) By replacing permanent magnet with an electromagnet.
By increasing the number of turns of the conducting wire in the current-carrying coil.
By using a soft iron core on which the coil is wounded.
12. Explain Faraday’s experiment related to the electromagnetic induction. (Aug-2024)
Faraday made an important breakthrough by discovering how a moving magnet can be used to
generate electric current.
• Take a coil of wire having a large number of turns.
• Connect the ends of the coil to a galvanometer
• Take a strong bar magnet and move its north pole towards the one end of the coil
• There is a momentary deflection in the needle of the galvanometer. This indicates the presence of a
current in the coil. The moment, when the motion of magnet stops, the deflection becomes zero.
• Withdraw the North Pole away from the coil. The needle of galvanometer is deflected towards left
showing the current is now set up in the opposite direction to the first.
• Keeping the North Pole stationary near the coil and when coil is moved towards the North Pole, we
see the galvanometer needle deflects towards right. Similarly the needle moves toward left when the
coil is moved away.
• When the coil is kept stationary with respect to magnet, the deflection drops to zero.
• It is thus clear from the experiment that motion of magnet with respect to coil produces an induced
potential difference which sets up an induced electric current in the circuit.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 LIGHT: REFLECTION AND REFRACTION
Multiple choice questions. (ONE MARKS)
1. To obtain a diminished image of an object from a concave mirror, position of the object should
be (F = principal focus, C = center of curvature, P = pole) (April-2019)
(A) Between C and F (B) beyond C
(C) Between P and F (D) at F
2. Identify the emergent ray in the given figure. (June-2019)
(A) CD (B) BC (C) AB (D) IJ

3. An object is kept at the center of curvature of a concave mirror. The position and nature of the
image formed is (Mar-2020)
(A) Between F and C and inverted (B) behind the mirror and erect
(C) Between F and P and erect (D) at the center of curvature and inverted.
4. The image of the English letter “ ” in convex mirror looks like (Sep-2020)

(A) (B) (C) (D)

5. The focal length of a lens is + 0.50 m. The
power of the lens and type are (April-2021)
(A) + 2.0 D and convex lens (B) + 2.0 D and concave lens
(C) – 2.0 D and concave lens (D) – 2.0 D and convex lens
6. The nature and the size of the image formed when an object is kept between the principal focus
F1 and optical center O of a convex lens are (April-2021)
(A) Virtual, erect and enlarged (B) real, inverted and small size
(C) Virtual, inverted and small size (D) real, inverted and enlarged
7. Observe the following table: Material medium P Q R S
In which material medium the speed of light is
very high? (April-2021) Refractive index 1.52 1.44 2.42 1.33
(A) Q (B) P (C) S
(D) R
8. One property of a convex lens among the following is that, it (April-2021)
(A) Diverges the light rays (B) is thicker at the edges and thinner at the middle
(C) Forms real and erect image (D) is thinner at the edges and thicker at the middle

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 9. To get diminished and real image of an object from a convex lens, the object should be placed
(A) At principal focus F1 (B) between principal focus F1 and 2F1
(C) Beyond 2F1 (D) between principal focus F1 and optical center O. (April-2022)
10. The correct statement among the following related to the concave lens is (June-2022)
(A) Converges the light rays (B) diverge the light rays
(C) Forms inverted image (D) forms real image
11. The focal Length of a lens is + 0.50 m. The power of the lens and type is (April-2023)
(A) + 2.0 D and concave lens (B) + 2.0 D and convex lens
(C) – 2.0 D and concave lens (D) – 2.0 D and convex lens
12. A light ray enters to rarer medium from a denser medium. Then the speed of that light ray
(April-2023)
(A) Decreases and bends towards the normal (B) increases and bends away from the normal
(C) Decreases and bends away from the normal (D) increases and bends towards the normal
13. A mirror forms an erect and enlarged image of an object. Then the type of the mirror and the
nature of the image respectively are (June-2023)
(A) Convex mirror and virtual image (B) concave mirror and real image
(C) Plane mirror and real image (D) concave mirror and virtual image
14. To get virtual and erect image by a convex lens, an object is to be placed (June-2024)
(A) Beyond 2F1 (B) betweenF1 and 2F1
(C) at focusF1 (D) between focusF1 and optical center O
15. Type of the mirror used in vehicles as rear view mirror is (Aug-2024)
(A) plane mirror (B) concave mirror (C) convex mirror (D) Plano concave mirror
Answer the following questions. (ONE MARKS)
1. Convex mirror is commonly used as rear-view mirror in vehicles. Why? (April-2019)
Because it gives erect image. It also helps the driver to view large area.
2. What is the center of curvature of a spherical mirror? (June-2019)
The reflecting surface of a spherical mirror forms a part of a sphere. This sphere has a center. This
point is called the center of curvature of the spherical mirror.
3. Observe the given incomplete diagram. Complete the diagram by drawing refracted rays and
show the image formed. (Sep-2020)

4. Mention the SI unit of power of lens. (April-2022)

The SI unit of power of lens is diopter.
5. Calculate the power of convex lens with a focal length of + 0·5 m. (June-2022)
f = + 0.5 m p = 1/f = 1/ 0.5=+2.0D

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 Answer the following questions. (TWO MARKS)
1. The focal length of a concave lens is 30 cm. At what distance should the object be placed from
the lens so that it forms an image at 20 cm from the lens? (April-2019)
f = - 30 cm v = - 20 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −3+2 −1
= - = - = + = =
𝑢 𝑣 𝑓 − 20 − 30 − 20 30 60 60
−30
u= = - 60 cm object distance is 60 cm
2
2. Draw the ray diagram to show the formation of image by a convex lens when the object is at
2F1. [F1: Principal focus] (June-2019)
AB = object
A1B1 = image

3. Object distance and image distance of a lens are –30 cm and –10 cm respectively. Find the
magnification and decide the type of lens used and nature of the image. (Sep-2020)
u = -30 cm v = - 10 cm
𝑣 −10
Magnification m = = = 0.33
𝑢 −30
Concave lens is used and nature of the image is virtual and erect and diminished.
4. An object is placed at 25 cm in front of a concave mirror of focal length 15 cm. At what
distance from the mirror should a screen be placed in order to obtain a sharp image?
(April-2022)
u = - 25cm f = - 15 cm
1 1 1
= +
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −5+3 −2
= - = - = + = =
𝑣 𝑓 𝑢 −15 −25 −15 25 75 75
−75
v= = - 37.5 cm
2
The screen should be placed at a distance of 37·5 cm, in front of the concave mirror.
5. A concave lens has focal length of 15 cm. At what distance should the object from the lens be
placed so that it forms an image at 10 cm from the lens? (April-2022)
f = - 15 cm v = - 10 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −3+2 −1
= - = - = + = =
𝑢 𝑣 𝑓 −10 −15 −10 15 30 30
u = - 30 cm
The object is placed at a distance of 30 cm from the concave lens.
6. Light enters from air to benzene having refractive index 1.50. Calculate the speed of light in
benzene. ( Speed of light in air is 3 × 108 ms− 1 ) (April-2023)
8 −1
Refractive index of benzene m = 1.50 speed of light in air c = 3 × 10 ms
𝑐
m=
𝑣
𝑐 3 X 108
v= = =2 x 108 ms-1 the speed of light in benzene is 2 x 108 ms-1
𝑚 1.50

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 7. A concave lens has focal length of 12 cm. At what distance should the object from the lens be
placed so that it forms an image at 9 cm from the lens? (April-2023)
f = - 12 cm v = - 9 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −4+3 −1
= - = - = + = =
𝑢 𝑣 𝑓 −9 −12 −9 12 36 36
u = - 36 cm
The object is placed at a distance of 36 cm from the concave lens.
8. A person who has a defect of the eye as shown in the below figure purchases a spectacle having
lens of – 2.0D power. Is this lens suitable to rectify the eye defect of that person? Analyze.
(April-2024)
• This lens is not suitable for that person’s defect of eye.
• Light rays from a close by object are focused at a point
behind the retina.
• Therefore it is farsightedness and it is corrected by using a
convex lens of appropriate power in spectacle.
• Lens of – 2·0 D power is a concave lens and that does not
rectify this defect.
9. State two laws of reflection of light. (June-2024)
i) The angle of incidence is equal to the angle of reflection.
ii) The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in
the same plane
Answer the following questions. (THREE MARKS)
1. Draw the ray diagrams for the image formation in a convex lens when an object is placed
(i) at focus F1 (ii) beyond 2F1. (April-2019)

2. A concave lens has focal length 30 cm. At what distance should the object be placed from the
lens so that it forms an image at 20 cm from the lens? Also, find the magnification produced by
the lens. (June-2019)
f = - 30 cm v = - 20 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −3+1 −2
= - = - = + = =
𝑢 𝑣 𝑓 − 20 − 30 − 20 30 30 30
−30
u= = - 15 cm
2
𝑣 −20 4
Magnification m = = = = 1.33
𝑢 −15 3

3. An object is kept on the principal axis of a concave mirror of focal length 12 cm. If the object is
at a distance of 18 cm from the mirror, calculate the image distance. Determine the nature of
the image formed by calculating the magnification produced by the mirror. (Mar-2020)
f = - 12cm u = - 18 cm
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 1 1 1
= +
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −3+2 −1
= - = - = + = =
𝑣 𝑓 𝑢 −12 −18 −12 18 36 36
v = -36 cm
Image distance is 36 cm
−𝑣 −(−36)
Magnification m = = =-2
𝑢 −18
The image formed is inverted real and enlarged.
4. A doctor prescribes a corrective lens of power – 0·5 D to a person. Find the focal length of the
lens. Is this lens diverging or converging? Give reason. How does the property of this lens can
be used to correct eye defects? (Mar-2020)
P = -0.5D
f = 1/p = 1/-0.5=-2 m
• This lens is diverging because power of a lens is negative means the given lens is concave.
• The diverging property of the lens is used to correct myopia. In myopic eye the image of a distant
object is formed in front of the retina.
• A concave lens of suitable power diverges the light rays and brings the image back on the retina.
5. Draw the ray diagram when the object is kept between F1 and 2F1 of the convex lens. With the
help of the diagram mention the position and nature of the image formed. (F1: Principal focus
of the lens) (Mar-2020)
Position of the image is beyond 2F2
Nature of the image is enlarged and
Real and inverted.

6. a) State the laws of refraction of light.

b) In the given figure, AB is the incident ray, BC is the refracted ray and MN is the normal at
the point of incidence. Which medium is more denser? Why? (Sep-2020)
a) Laws of refraction of light:
• The incident ray, the refracted ray and the normal to the interface of
two transparent media at the point of incidence, all lie in the same
plane.
• The ratio of sine of angle of incidence to the sine of angle of refraction
is a constant for the light of a given colour and for the given pair of
media.
• If i is the angle of incidence and r is the angle of refraction, then,
sin 𝑖
= constant.
sin 𝑟
b) Medium 1 is more denser, because When a ray of light travels from rarer medium to denser
medium, it always bends towards the normal.

7. a) Differentiate between convex mirror and concave mirror.

b) Define the principal focus of a convex lens. (Sep-2020)
Convex mirror Concave mirror

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 Reflecting surface is curved outwards Reflecting surface is curved inwards

Always forms virtual and erect images Forms real and inverted images. (Except the case

when object is kept between P and F )

Diverges the light rays Converges the light rays.

b) The rays of light falling on a convex lens parallel to the principal axis, after refraction from the
lens converge to a point on the principal axis. This point on principal axis is called the ‘principal
focuses of the convex lens.
8. Draw the ray diagram to show the image formation by a convex lens, when the object is kept at
2F1 of the lens. With the help of the ray diagram mention the position and nature of the image
formed. [F1: Principal focus of the lens] (April-2022)
Position of the image is at 2F2
Nature of the image is real and inverted
Size of the image same as object

9. a) State the two laws of refraction of light.

b) “The refractive index of diamond is 2·42.” Write the meaning of this statement. (June-2022)
a) Laws of refraction of light:
• The incident ray, the refracted ray and the normal to the interface of two transparent media at the
point of incidence, all lie in the same plane.
• The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for the light of a
given colour and for the given pair of media.
sin 𝑖
• If i is the angle of incidence and r is the angle of refraction, then, = constant.
sin 𝑟
b) The ratio of the speed of light in air and the speed of light in diamond is 2·42.
10. Draw the ray diagram of image formation when the object is kept at 2F1 of the convex lens.
With the help of the ray diagram, mention the position and nature of the image formed.
[F1: Principal focus of the lens] (June-2022)
Position of the image is at 2F2
Nature of the image is real and inverted
Size of the image same as object

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 11. Draw the ray diagram of image formation when the object is kept between C and F of the
concave mirror. With the help of the ray diagram mention the position and the nature of the
image formed. [F: Principal focus of the mirror, C: Centre of curvature of mirror] (June-2022)
Position: beyond C
Nature: real and inverted

12. Draw the ray diagram for the image formation in a convex lens when the object is placed
beyond 2F1. Mention the position and nature of the image formed. [F1 : Principal focus of the
lens ]
(April-2023)
Position: between F2 and 2f2
Nature: real and inverted

13. What is meant by the ‘aperture’ of a spherical mirror? Mention the four uses of a concave
mirror. (June-2023)
Diameter of the mirror’s reflecting surface.
Uses of concave mirror.
i) Used in torches, search-lights
ii) Used in vehicles head lights
iii) Used as shaving mirrors
iv) The dentists used to test / examine teeth of patients
v) Used in solar furnace.
14. a) What is meant by the power of a lens? Write the formula used to find the power of a lens.
What is the SI unit of power of a lens?
b) If the focal lengths of two lenses ‘A’ and ‘B’ are + 0·50 m and – 0·40 m respectively.
Mention the types of these lenses in the same order. (June-2023)
a) The power of lens is defined as the reciprocal of the focal length.
1
𝑃= p= power of lens f= focal length
𝑓
The SI unit of Power of a lens is diopter (D).
b) Lens A has convex lens and lens B has concave lens
15. Draw the ray diagram for the image formation by a convex lens, when the object is placed at
2F1. With the help of the ray diagram mention the position and the nature of the image
formed. [F1: Principal focus of the lens] (June-2023)
Position of the image is at 2F2
Nature of the image is real and inverted
Size of the image same as object

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 16. Draw the ray diagram for the image formation in a convex lens when the object is placed
beyond 2F1. With the help of the ray diagram mention the position and the nature of the image
formed. [F1: Principal focus of the lens] (June-2023)
Position: between F2 and 2f2
Nature: real and inverted

17. Draw the ray diagram of image formation when the object is kept at 2F1 of the convex lens.
With the help of ray diagram mention the position and the nature of the image formed.
(F1: Principal focus of the lens) (April-2024)
Position of the image is at 2F2
Nature of the image is real and inverted
Size of the image same as object

18. A concave lens has focal length of 25 cm. At what distance should the object from the lens be
placed so that it forms an image at 20 cm from the lens? Find the magnification of the image
produced by the lens. (June-2024)
f = -25 cm v = -20 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −5+4 −1
= - = - = + = =
𝑢 𝑣 𝑓 −20 −25 −20 25 100 100
u = -100 cm
Object distance is 100 cm
𝑣 −20 1
Magnification m = = = = 0.2
𝑢 −100 5
19. A concave lens has focal length of 30 cm. At what distance should the object from the lens be
placed so that it forms an image at 20 cm from the lens? (Aug-2024)
f = - 30 cm v = - 20 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −3+2 −1
= - = - = + = =
𝑢 𝑣 𝑓 − 20 − 30 − 20 30 60 60
−30
u= = - 60 cm object distance is 60 cm
2
20. a) Find the focal length of a convex mirror whose radius of curvature is 6 cm.
b) Find the power of convex lens of focal length 0.2 m. (Aug-2024)
a) Radius of curvature R = 6 cm
Focal length f = R / 2 = 6 / 2 = 3 cm
b) Focal length f = 0.2 m
Power of convex lens p = 1 / f = 1/0.2 = +5.0D

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 21. Draw the ray diagram for the image formation in a convex lens when the object is placed
between 2F1 and F1. Mention the position and nature of the image formed.[F1 : Principal focus
of the lens ] (Aug-2024)
Position of the image is beyond 2F2
Nature of the image is real and inverted

Answer the following questions. (FOUR MARKS)

1. a) What is refraction of light? State two laws of refraction of light.
b) What is refractive index of light? “The refractive index of diamond is 2·42.” What is the
meaning of this statement? (April-2022)
a) The bending of light when it travels from one medium into another is called refraction of light.
Laws of refraction of light:
• The incident ray, the refracted ray and the normal to the interface of two transparent media at the
point of incidence, all lie in the same plane.
• The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for the light of a
given colour and for the given pair of media.
sin 𝑖
If i is the angle of incidence and r is the angle of refraction, then, = constant.
sin 𝑟
b) The ratio of speed of light in air and the speed of light in medium is called refractive index.
The ratio of speed of light in air and the speed of light in diamond is 2·42.
2. a) Write any four uses of concave mirror.
b) An object is placed at a distance of 15 cm on the principal axis in front of a concave lens
with a focal length of 10 cm. Find the image distance. (June-2022)
a) Uses of concave mirror.
i) Used in torches, search-lights
ii) Used in vehicles head lights
iii) Used as shaving mirrors
iv) The dentists used to test / examine teeth of patients
v) Used in solar furnace.
b) u = - 15 cm f = - 10 cm
1 1 1
= -
𝑓 𝑣 𝑢
1 1 1 1 1 1 1 −3−2 −5
= + = + = - = =
𝑣 𝑓 𝑢 −10 −15 −10 15 30 30
v = - 30/5 = - 6 cm
3. a) Define focal length, principal axis and aperture of the spherical lens.
b) State two laws of refraction of light. (April-2023)
a) Focal length (f): the distance between the pole and the principal focus of a spherical mirror.
Principal axis: imaginary line joining the pole and center of the curvature.
Aperture: Diameter of the mirror’s reflecting surface.
b) Laws of refraction of light:
• The incident ray, the refracted ray and the normal to the interface of two transparent media at the
point of incidence, all lie in the same plane.
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 • The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for the light of a
given colour and for the given pair of media.
sin 𝑖
If i is the angle of incidence and r is the angle of refraction, then, = constant.
sin 𝑟
4. a) State two laws of reflection of light.
b) Write any two differences between concave mirror and convex mirror. (April-2024)
a) Laws of reflection of light:
• The angle of incidence is equal to the angle of reflection
• The incident ray, the normal to the mirror at the point of incidence and the reflected ray all lie in the
same plane.
b) Differences between concave mirror and convex mirror.
Convex mirror Concave mirror

Reflecting surface is curved outwards Reflecting surface is curved inwards

Always forms virtual and erect images Forms real and inverted images. (Except the case

when object is kept between P and F )

Diverges the light rays Converges the light rays.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 THE HUMAN EYE AND THE COLOURFUL WORLD
Multiple choice questions. (ONE MARKS)
1. The change that occurs in the eye to see the distant objects clearly is (April-2019)
(A) Focal length of the eye lens decreases (B) Curvature of the eye lens increases
(C) Focal length of the eye lens increases (D) ciliary muscles of the eye contract
2. The sky as seen from the surface of the moon appears dark because, (Sep-2020)
(A) Only a little of the blue and violet colors are scattered
(B) All the colors are absorbed by the atmosphere present in the moon
(C) All the colors are scattered (D) Atmospheric particles needed to scatter the light are not
present.
3. Right statement regarding the color of the scattered sunlight and the size of scattering
atmospheric particles is (April-2024)
(A) Small particles scatter red color (B) big particles scatter blue color
(C) Big particles scatter violet color (D) too larger particles scatter all colors equally
4. The color that is least scattered by fog and smoke is (June-2024)
(A) Orange (B) blue (C) red (D) violet
5. Identify the wrong statement among the following statements regarding refraction and
dispersion of light. (June-2024)
(A) Stars twinkle (B) Sky appears blue to an astronaut flying at very high altitudes
(C) The sun is visible to us about two minutes before the actual sunrise (D) Planets do not twinkle
Answer the following questions. (ONE MARKS)
1. Observe the given figure. Name the eye defect indicated in the figure and also mention the lens
used to correct this defect. (April-2019)
Eye defect indicated in the figure is Myopia or near-sightedness
Concave lens used to correct this defect.

2. What is the function of pupil of the human eye? (June-2019)

Control and regulate the amount of light entering the eye
6. A student sitting in the last bench has difficulty in reading the blackboard writing. Which is the
defect of vision the student has? How can it be corrected? (Sep-2020)
A student has myopia of near-sightedness
It is corrected using concave lens
3. Observe the below figure showing the refraction of light through a glass prism. Name the angle
represented as ∠x and give reason for the formation of that angle. (June-2024)
∠x is an angle of deviation.
The particular shape of the prism makes the emergent
ray bends at an angle to the direction of the incident ray

4. What is spectrum of white light? (Aug-2024)

The band of the coloured components of a white light beam is called spectrum of white light.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 Answer the following questions. (TWO MARKS)
1. What is dispersion of light? Mention the color that bends the least and the color that bends the
most when light undergoes dispersion through a prism. (April-2019)
Phenomenon of splitting up of white light into 7 colours is known as dispersion of light.
The color that bends the least is red and the color that bends the most is violet.
2. Mention any four phenomena that can be observed due to atmospheric refraction of light on the
earth. (April-2019)
i) Apparent star position ii) Twinkling of stars
iii) Advanced sun rise and delayed sunset. iv) Rainbow formation
3. (i) What is Tyndall effect?
(ii) Name the color that bends the least and the color that bends the most when white light is
dispersed by a prism. (June-2019)
i) When a beam of light passes through a colloidal solution, the path of light becomes visible due to
the scattering of light by the colloid particles. This is known as Tyndall effect.
ii) The color that bends the least is red and the color that bends the most is violet.
4. (i)What is meant by the power of accommodation of the eye?
(ii) What are the far point and near point of the human eye with normal vision? (June-2019)
i) The ability of the eye lens to focus near and far objects clearly on the retina by adjusting its focal
length.
The near point is about 25 cm
The far point of the eye is at infinity
5. What is hypermetropia or far-sightedness? Name the type of lens used to correct it.
(June-2019)
A person cannot see nearby object clearly, but can see distant object clearly
Corrected by using convex lens
6. What is spectrum of white light? Name any two phenomena that occur in the atmosphere due
to the refraction of light. (April-2024)
The band of coloured components of a light beam formed by the splitting of light through glass
prism is called spectrum of light.
Phenomena that occur due to the refraction of light:
Twinkling of stars, Advanced sunrise and delayed sunset, Rainbow formation.
7. What is cataract of eye? What are the near point and far point of the human eye with normal
vision? (April-2024)
Sometimes, the crystalline lens of people at old age becomes milky and cloudy. This condition is
called cataract.
The near point is about 25 cm the far point of the eye is at infinity
Answer the following questions. (THREE MARKS)
1. Draw the diagram to show the recombination of the spectrum of white light and label the
following parts. (Sep-2020)
a) The ray of light that bends the most b) The ray of light that bends the least.
a) The ray of light that bends the most is red.
b) The ray of light that bends the least is violet.

2. Draw the ray diagrams that show :

(June-2024)
SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE
 i) Near point of hypermetropic eye ii) Hypermetropic eye
iii) Correction for hypermetropic eye.

Answer the following questions. (FOUR MARKS)

1. Explain the experiment conducted by Newton to show that white light contains seven colors.
Sun appears red in color during sunrise but appears white at noon. Explain with the reasons.
(Mar-2020)
Isaac Newton was the first to use a glass prism to obtain the spectrum of sun light.
He tried to split the colours of the spectrum of white light further by using similar prism. He could
not get any more colours.
He then placed a second prism (identical) in an inverted position with respect to the first prism. This
allowed all the colours of the spectrum to pass through the second prism.
He found a beam of white light emerging from the other side of the second prism. This observation
gave Newton the idea that the sun light is made up of seven colours.
When sun is near the horizon, the light travels longer distance, due to this most of the blue light
scattered away by the particles. The light of longer wave length red light will reach our eye and sky
appears red.
At noon sun is over head and sunlight would travel shorter distance relatively through the
atmosphere .hence at noon sun appears white as only little of the blue and violet colour are scattered.
2. a) How does the lens of human eye accommodate to see the nearby objects and the distant
objects? Explain.
b) Explain the formation of rainbow in the nature. (June-2023)
a) The eye lens is composed of a fibrous, jelly like material. Its curvature can be modified to some
extent by the ciliary muscles. The change in curvature of the eye lens can thus change its focal
length.
When the muscles are relaxed, the lens becomes thin, thus focal length increases. This enables us to
see distant objects clearly.
When ciliary muscles are contract, this increases the curvature of the eyes lens. The eye lens
becomes thicker. Consequently, the focal length of the eye lens decreases. This enables us to see
nearby objects clearly.
b) Rainbow is a spectrum of sun light in nature
It is formed due to the dispersion of sunlight by small water droplets.
When sunlight enters water droplets its gets refracted and dispersed, reflect it internally and finally
refract it again when it comes out of the raindrop. Due to the dispersion of light and internal
reflection, different colours reach the observer’s eye.
3. a) How does the eye accommodate to see far and near objects?
b) Why do stars twinkle? Explain. (Aug-2024)
a) The eye lens is composed of a fibrous, jelly like material. Its curvature can be modified to some
extent by the ciliary muscles. The change in curvature of the eye lens can thus change its focal
length.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE

 When the muscles are relaxed, the lens becomes thin, thus focal length increases. This enables us to
see distant objects clearly.
When ciliary muscles are contract, this increases the curvature of the eyes lens. The eye lens
becomes thicker. Consequently, the focal length of the eye lens decreases. This enables us to see
nearby objects clearly.
b) The twinkling of star is due to atmospheric refraction of sunlight.
Since the stars are very distant, they approximate point sized sources of light. As the path of rays of
light coming from the stars goes on varying slightly, the apparent position of the stars fluctuates and
the amount of starlight entering the eye flickers — the star sometimes appears brighter, and at some
other time, fainter, which is the twinkling effect.

SAVITHA V , ADARSHA VIDYALAYA CHALLAKERE