NCERT Solutions Class 10 Physics - Light, Electricity

Physics Chapters

All Physics topics from Class 10 Science NCERT

Light - Reflection and Refraction

Mirrors, Lenses, Mirror formula, Lens formula, Power

› 11

Electricity

Ohm's law, Resistance, Series/Parallel, Electrical power

› 12

Magnetic Effects of Electric Current

Magnetic field, Electromagnet, Fleming's rules, AC/DC

›

Chapter 9: Light - Reflection and Refraction

Key Formulas:
• Mirror Formula: 1/v + 1/u = 1/f
• Magnification: m = -v/u = h'/h
• Lens Formula: 1/v - 1/u = 1/f
• Power of lens: P = 1/f (in metres) = 100/f (in cm)
• Snell's Law: n₁ sin i = n₂ sin r

Q1. Define the principal focus of a concave mirror.

Solution

The principal focus of a concave mirror is a point on its principal axis where all the light rays, initially parallel to the principal axis, converge after reflection from the mirror.

Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Solution

Given: R = 20 cm
f = R/2 = 20/2 = 10 cm

The focal length is 10 cm.

Q3. Name a mirror that can give an erect and enlarged image of an object.

Solution

A concave mirror can give an erect and enlarged image when the object is placed between the pole and the focus of the mirror.

Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Solution

We prefer a convex mirror because:
• It gives an erect (upright) image
• It gives a diminished (smaller) image
• It provides a wider field of view, allowing the driver to see more traffic behind

Q5. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Solution

Given: u = -10 cm, m = -3 (real image)
m = -v/u
-3 = -v/(-10)
v = -30 cm

The image is located at 30 cm in front of the mirror.

Q6. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Solution

Given: R = +32 cm (convex mirror)
f = R/2 = 32/2 = +16 cm

The focal length is +16 cm.

Q7. A concave mirror produces an image of size double that of the object placed at a distance of 15 cm from it. Find the focal length of the mirror.

Solution

Given: u = -15 cm, m = -2 (real image, inverted, double size)
m = -v/u
-2 = -v/(-15)
v = -30 cm

1/v + 1/u = 1/f
1/(-30) + 1/(-15) = 1/f
-1/30 - 1/15 = 1/f
-1/30 - 2/30 = 1/f
-3/30 = 1/f
f = -10 cm

The focal length is -10 cm (concave mirror).

Q8. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image.

Solution

Given: u = -10 cm, f = +15 cm
1/v + 1/u = 1/f
1/v + 1/(-10) = 1/15
1/v - 1/10 = 1/15
1/v = 1/15 + 1/10
1/v = 2/30 + 3/30
1/v = 5/30
v = +6 cm

The image is located at 6 cm behind the mirror (virtual image).

Q9. A convex lens has a focal length of 10 cm. At what distance from the lens should the object be placed so as to obtain a real and inverted image of double the size?

Solution

Given: f = +10 cm, m = -2
m = v/u → v = -2u

1/v - 1/u = 1/f
1/(-2u) - 1/u = 1/10
-1/(2u) - 1/u = 1/10
-1/(2u) - 2/(2u) = 1/10
-3/(2u) = 1/10
u = -15 cm

The object should be placed at 15 cm from the lens.

Q10. What is the power of a concave lens of focal length 2 m?

Solution

Given: f = -2 m (concave lens)
P = 1/f = 1/(-2) = -0.5 D

The power is -0.5 Dioptre.

Q11. A doctor prescribes a corrective lens of power +1.5 D. Find the focal length of the lens. Is it a converging or diverging lens?

Solution

Given: P = +1.5 D
f = 1/P = 1/1.5 = 0.67 m = 67 cm

The focal length is 67 cm. Since the power is positive, it is a converging (convex) lens.

Q12. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so as to get a sharp image? Find the size of the image.

Solution

Given: h = 7 cm, u = -27 cm, f = -18 cm
1/v + 1/u = 1/f
1/v + 1/(-27) = 1/(-18)
1/v = -1/18 + 1/27
1/v = (-3 + 2)/54
1/v = -1/54
v = -54 cm

m = -v/u = -(-54)/(-27) = -2
h' = m × h = -2 × 7 = -14 cm

The screen should be placed at 54 cm from the mirror. The image size is 14 cm (inverted).

Chapter 11: Electricity

Key Formulas:
• Ohm's Law: V = IR
• Resistance: R = ρl/A
• Series: Rₑ = R₁ + R₂ + R₃
• Parallel: 1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃
• Power: P = VI = I²R = V²/R
• Energy: E = Pt = VIt

Q1. What does an electric circuit mean?

Solution

An electric circuit is a continuous and closed path through which electric current flows. It consists of a source of electricity (battery), wires, and components like resistors, bulbs, and switches. When the circuit is complete (closed), current flows.

Q2. Define the unit of electric current.

Solution

The SI unit of electric current is the Ampere (A). One ampere is defined as the flow of one coulomb of charge per second through any cross-section of a conductor.

Q3. Name the device that helps to maintain a potential difference across a conductor.

Solution

A cell or battery helps to maintain a potential difference across a conductor. A battery is a combination of two or more cells connected together.

Q4. State the factors on which the resistance of a conductor depends.

Solution

Resistance depends on:
• Length (l): R is directly proportional to length
• Cross-sectional area (A): R is inversely proportional to area
• Material: Different materials have different resistivity (ρ)
• Temperature: Resistance generally increases with temperature

Q5. A wire of resistance R is doubled upon itself. What is the new resistance?

Solution

When doubled: l' = l/2, A' = 2A
R' = ρl'/A' = ρ(l/2)/(2A) = ρl/(4A) = R/4

New resistance = R/4

Q6. An electric bulb is connected to a 220V generator. If the current is 0.50A, what is the power of the bulb?

Solution

P = V × I
P = 220 × 0.50
P = 110 W

The power of the bulb is 110 Watts.

Q7. Two lamps, one rated 100 W at 220V, and the other 60 W at 220V, are connected in parallel to electric mains. What is the current drawn from the line if the supply voltage is 220V?

Solution

For 100 W lamp: I₁ = P/V = 100/220 = 0.455 A
For 60 W lamp: I₂ = P/V = 60/220 = 0.273 A

Total current = I₁ + I₂ = 0.455 + 0.273 = 0.727 A

The total current drawn is 0.727 A.

Q8. Calculate the resistance of a metal wire of length 2 m and area of cross-section 1.55 × 10⁻⁶ m². The resistivity of the material is 2.8 × 10⁻⁸ Ωm.

Solution

R = ρl/A
R = (2.8 × 10⁻⁸) × 2 / (1.55 × 10⁻⁶)
R = 5.6 × 10⁻⁸ / 1.55 × 10⁻⁶
R = 3.61 × 10⁻⁸ Ω
R = 0.036 Ω

The resistance is 0.036 Ω.

Q9. An electric heater of resistance 8 Ω draws 15 A current from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Solution

P = I²R
P = (15)² × 8
P = 225 × 8 = 1800 W
P = 1800 J/s

The rate of heat development is 1800 J/s (1800 Watts).

Q10. How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?

Solution

Total resistance needed: Rₑ = V/I = 220/5 = 44 Ω

For n identical resistors in parallel:
Rₑ = R/n
44 = 176/n
n = 176/44 = 4

4 resistors of 176 Ω are needed in parallel.

Q11. Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω

Solution

(i) For 9 Ω:
Two 6 Ω in series = 12 Ω... No.
Better: Two in parallel (3 Ω) + one in series = 3 + 6 = 9 Ω ✓
(R₁ ∩ R₂) + R₃ = (6×6)/(6+6) + 6 = 3 + 6 = 9 Ω

(ii) For 4 Ω:
Three in parallel: 1/Rₑ = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Rₑ = 2 Ω... No.
Two in series (12 Ω) in parallel with one 6 Ω: (12×6)/(12+6) = 72/18 = 4 Ω ✓

(i) Two in parallel + one in series = 9 Ω
(ii) Two in series, all three in parallel = 4 Ω

Q12. What is the commercial unit of electrical energy? Convert it into joules.

Solution

1 kilowatt-hour (kWh) = 1000 W × 3600 s = 3.6 × 10⁶ J

The commercial unit is kilowatt-hour (kWh). 1 kWh = 3.6 × 10⁶ Joules.

Chapter 12: Magnetic Effects of Electric Current

Key Formulas:
• Fleming's Left Hand Rule: For motor effect
• Fleming's Right Hand Rule: For generator effect
• Magnetic field inside solenoid: B = μ₀nI
• Force on current-carrying conductor: F = BIL sinθ
• Transformer: V₁/V₂ = N₁/N₂

Q1. Why does a compass needle get deflected when brought near a bar magnet?

Solution

A compass needle is a small magnet. When brought near a bar magnet, the magnetic field of the bar magnet exerts a force on the compass needle, causing it to deflect. The compass aligns itself along the direction of the magnetic field lines of the bar magnet.

Q2. Draw magnetic field lines around a bar magnet.

Solution

Magnetic field lines emerge from the North pole and enter the South pole outside the magnet. Inside the magnet, they go from South to North, forming closed loops. They never intersect each other. The density of field lines indicates the strength of the magnetic field.

Q3. List the properties of magnetic field lines.

Solution

Properties of magnetic field lines:
• They are closed continuous curves
• They emerge from North pole and enter South pole
• They never intersect each other
• They are tangent to the direction of the magnetic field at any point
• The closer the lines, the stronger the magnetic field
• They are parallel near the centre of a magnet

Q4. Why don't two magnetic field lines intersect each other?

Solution

Two magnetic field lines don't intersect because at the point of intersection, the compass needle would point in two different directions simultaneously, which is impossible. The resultant force at that point would have two different directions, which contradicts the definition of magnetic field direction.

Q5. Consider a circular loop of plane paper lying in a magnetic field. If current is passed through the loop, what will happen to the loop?

Solution

When current passes through the circular loop in a magnetic field, the loop experiences a torque (turning force) and starts to rotate. This is the principle behind the electric motor. The two sides of the loop experience forces in opposite directions (by Fleming's Left Hand Rule), causing rotation.

Q6. What is a solenoid? What are its properties?

Solution

A solenoid is a coil of many turns of insulated copper wire wound in the shape of a cylinder.

Properties:
• When current flows, it behaves like a bar magnet
• The magnetic field lines inside are uniform and parallel
• One end acts as North pole, the other as South pole
• Strength depends on number of turns and current

Q7. What is an electromagnet? What are its uses?

Solution

An electromagnet is a temporary magnet made by winding insulated copper wire around a soft iron core. When current passes through the coil, it becomes magnetic.

Uses:
• In electric bells
• In cranes to lift heavy magnetic materials
• In relays and switches
• In telephones
• In separating magnetic and non-magnetic materials

Q8. What is Fleming's Left Hand Rule? Explain.

Solution

Fleming's Left Hand Rule states: Stretch the forefinger, middle finger, and thumb of your left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the middle finger in the direction of current, then the thumb will point in the direction of the force on the conductor. It is used for electric motors.

Q9. What is the difference between a motor and a generator?

Solution

Motor:
• Converts electrical energy to mechanical energy
• Uses Fleming's Left Hand Rule
• Uses DC power supply
• Has a split-ring commutator

Generator:
• Converts mechanical energy to electrical energy
• Uses Fleming's Right Hand Rule
• Produces AC or DC
• Has slip rings (AC) or split rings (DC)

Q10. Explain the phenomenon of electromagnetic induction.

Solution

Electromagnetic induction is the phenomenon of producing electric current by changing the magnetic field around a conductor. When a conductor moves through a magnetic field (or the magnetic field changes), an induced current is produced. This is the principle behind electric generators and transformers.

Q11. What are the advantages of AC over DC?

Solution

Advantages of AC:
• AC can be transmitted over long distances with less energy loss
• AC voltage can be easily stepped up or down using transformers
• AC is cheaper to generate than DC
• AC is more versatile in use
• DC cannot be easily transmitted over long distances

Q12. What is short circuiting? What are the safety measures?

Solution

Short circuiting occurs when the live wire and neutral wire come in direct contact, causing a large current to flow, which can cause fire.

Safety measures:
• Fuses: Break the circuit when current exceeds safe limit
• MCB (Miniature Circuit Breaker): Automatically switches off
• Proper earthing of appliances
• Using insulated wires
• Not overloading single sockets

Q13. What is the function of an earth wire?

Solution

The earth wire provides a safety measure by conducting the current to the earth in case of any leakage from the metal body of an appliance. It prevents electric shock to users. The earth wire is connected to the metal body of the appliance and goes deep into the earth through a copper plate.