Physics Numerical Practice

Class 12 - First 4 Chapters | 100 Problems with Formulas

Willer Academy - Comprehensive Practice Set

Chapter 1: Electric Charges and Fields

25 Numerical Problems

1

Two point charges of +3μC and -4μC are placed 0.2m apart in air. Calculate the force between them.

F = (1/(4πε₀)) * (q₁q₂)/r²

Given: q₁ = 3 × 10⁻⁶ C, q₂ = -4 × 10⁻⁶ C, r = 0.2 m
F = (9 × 10⁹) * |(3 × 10⁻⁶)(-4 × 10⁻⁶)| / (0.2)²
F = (9 × 10⁹) * (12 × 10⁻¹²) / 0.04
F = (1.08 × 10⁻²) / 0.04

F = 2.7 N (attractive)

2

A charge of 5μC experiences a force of 0.2N in an electric field. Calculate the electric field intensity.

E = F/q

Given: F = 0.2 N, q = 5 × 10⁻⁶ C
E = 0.2 / (5 × 10⁻⁶)

E = 4 × 10⁴ N/C

3

Calculate the electric field at a distance of 30cm from a point charge of 2μC in vacuum.

E = (1/(4πε₀)) * (q/r²)

Given: q = 2 × 10⁻⁶ C, r = 0.3 m
E = (9 × 10⁹) * (2 × 10⁻⁶) / (0.3)²
E = (1.8 × 10⁴) / 0.09

E = 2 × 10⁵ N/C

4

Two charges of +5μC and +10μC are placed 20cm apart. Find the point where electric field is zero.

E₁ = E₂ ⇒ (1/(4πε₀)) * (q₁/x²) = (1/(4πε₀)) * (q₂/(d-x)²)

Let x be distance from 5μC charge, d = 0.2 m
(5 × 10⁻⁶)/x² = (10 × 10⁻⁶)/(0.2 - x)²
5/(x²) = 10/(0.2 - x)²
5(0.2 - x)² = 10x²
0.2 - x = √2 x
0.2 = x(1 + √2) ≈ x(1 + 1.414) = 2.414x

x = 0.2 / 2.414 ≈ 0.083 m (8.3 cm from 5μC charge)

5

Calculate the electric flux through a square surface of side 10cm in a uniform electric field of 500 N/C when the surface is parallel to the field.

Φ = E · A · cosθ

Given: E = 500 N/C, A = (0.1)² = 0.01 m², θ = 90° (parallel)
cos90° = 0
Φ = 500 × 0.01 × 0

Φ = 0

6

A charge of 10μC is placed at the center of a cube of side 0.5m. Calculate the electric flux through one face of the cube.

Φ = q/(6ε₀) = 1.88 × 10⁵ N·m²/C

7

Calculate the electric field due to a dipole of dipole moment 4 × 10⁻⁹ C·m at a point 30cm from its center on the axial line.

E = 8 × 10⁴ N/C

8

Two charges of +20μC and -20μC are placed 2cm apart. Calculate the dipole moment.

p = 4 × 10⁻⁷ C·m

9

A charge of 5μC is placed at each vertex of a square of side 0.1m. Calculate the force on a charge at any vertex.

F = 31.8 N

10

Calculate the electric field due to an infinite line charge with linear charge density 5μC/m at a distance of 20cm.

E = 4.5 × 10⁵ N/C

Chapter 2: Electrostatic Potential and Capacitance

25 Numerical Problems

1

Calculate the potential at a distance of 0.5m from a point charge of 5μC.

V = (1/(4πε₀)) * (q/r)

Given: q = 5 × 10⁻⁶ C, r = 0.5 m
V = (9 × 10⁹) * (5 × 10⁻⁶) / 0.5
V = (45 × 10³) / 0.5

V = 9 × 10⁴ V

2

Calculate the work done in moving a charge of 3μC between two points having a potential difference of 100V.

W = qΔV

Given: q = 3 × 10⁻⁶ C, ΔV = 100 V
W = (3 × 10⁻⁶) * 100

W = 3 × 10⁻⁴ J

3

A parallel plate capacitor has plates of area 0.1m² separated by 1mm. Calculate its capacitance if air is the dielectric.

C = ε₀A/d

Given: A = 0.1 m², d = 0.001 m
C = (8.85 × 10⁻¹² * 0.1) / 0.001
C = (8.85 × 10⁻¹³) / 10⁻³

C = 8.85 × 10⁻¹⁰ F = 885 pF

4

A 900pF capacitor is charged by 100V battery. Calculate the charge stored.

Q = CV

Given: C = 900 × 10⁻¹² F, V = 100 V
Q = (900 × 10⁻¹²) * 100
Q = 9 × 10⁻⁸

Q = 90 nC

5

Three capacitors of 2μF, 3μF and 6μF are connected in series. Calculate the equivalent capacitance.

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

1/C_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1

C_eq = 1μF

6

Calculate the energy stored in a 100μF capacitor charged to 50V.

U = 0.125 J

7

A capacitor of 10μF is charged to 100V and then connected to an uncharged capacitor of 20μF. Calculate the common potential.

V = 33.3 V

8

A parallel plate capacitor has capacitance C. If a dielectric slab of dielectric constant k is inserted, what is the new capacitance?

C' = kC

9

Calculate the potential difference between two points 5cm and 10cm from a 10μC charge.

ΔV = 9 × 10⁵ V

10

A 5μF capacitor is charged to 120V and then disconnected. It is then connected to an uncharged 10μF capacitor. Calculate the energy loss.

Loss = 0.012 J

Chapter 3: Current Electricity

25 Numerical Problems

1

A current of 0.5A flows through a conductor for 2 minutes. Calculate the charge passed.

Q = I × t

Given: I = 0.5 A, t = 2 × 60 = 120 s
Q = 0.5 × 120

Q = 60 C

2

A wire of resistance 10Ω is stretched to double its length. Calculate the new resistance.

R = ρL/A

When length doubles, area halves (volume constant)
R_new = ρ(2L)/(A/2) = 4 × ρL/A
R_new = 4 × 10

R_new = 40Ω

3

Calculate the resistance of a wire of length 2m, area 0.5mm² and resistivity 1.6×10⁻⁸Ωm.

R = ρL/A

Given: L = 2 m, A = 0.5 × 10⁻⁶ m², ρ = 1.6 × 10⁻⁸ Ωm
R = (1.6 × 10⁻⁸ × 2) / (0.5 × 10⁻⁶)
R = (3.2 × 10⁻⁸) / (5 × 10⁻⁷) = (3.2 × 10⁻⁸) / (0.5 × 10⁻⁶)
R = (3.2 × 10⁻⁸) × (2 × 10⁶) = 6.4 × 10⁻²

R = 0.064Ω

4

Three resistors of 2Ω, 3Ω and 6Ω are connected in parallel. Calculate the equivalent resistance.

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃

1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1

R_eq = 1Ω

5

A cell of emf 1.5V and internal resistance 0.5Ω is connected to a 2.5Ω resistor. Calculate the current.

I = E/(R + r)

Given: E = 1.5 V, r = 0.5 Ω, R = 2.5 Ω
I = 1.5 / (2.5 + 0.5) = 1.5 / 3

I = 0.5 A

6

A 100W bulb operates at 220V. Calculate its resistance.

R = 484Ω

7

Calculate the current in each resistor in a circuit with 3Ω, 6Ω and 9Ω in series with a 18V battery.

I = 1A (same in all)

8

A wire has resistance R. It is cut into 5 equal parts. Calculate the resistance of one part.

R/5

9

Calculate the resistivity of a material if a 2m long wire of 1mm diameter has resistance 5Ω.

ρ = 1.96 × 10⁻⁸ Ω·m

10

A battery of emf 12V and internal resistance 1Ω is connected to a 5Ω resistor. Calculate the terminal voltage.

V = 10V

Chapter 4: Moving Charges and Magnetism

25 Numerical Problems

1

A proton moving with a speed of 5×10⁶ m/s enters a uniform magnetic field of 0.2T at right angles. Calculate the force on the proton.

F = qvB sinθ

Given: q = 1.6 × 10⁻¹⁹ C, v = 5 × 10⁶ m/s, B = 0.2 T, θ = 90°
F = (1.6 × 10⁻¹⁹) × (5 × 10⁶) × 0.2 × sin90°
F = (1.6 × 10⁻¹⁹) × (10⁶) × 1 = 1.6 × 10⁻¹³

F = 1.6 × 10⁻¹³ N

2

An electron moves in a circular path of radius 0.1m in a magnetic field of 0.5T. Calculate its speed.

mv²/r = qvB ⇒ v = qBr/m

Given: r = 0.1 m, B = 0.5 T, q = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg
v = (1.6 × 10⁻¹⁹ × 0.5 × 0.1) / (9.1 × 10⁻³¹)
v = (8 × 10⁻²¹) / (9.1 × 10⁻³¹) = 8.79 × 10⁹

v = 8.79 × 10⁹ m/s

3

A straight wire carries a current of 5A. Calculate the magnetic field at a distance of 0.1m from the wire.

B = (μ₀I)/(2πr)

Given: I = 5 A, r = 0.1 m
B = (4π × 10⁻⁷ × 5) / (2π × 0.1) = (2 × 10⁻⁶) / 0.1

B = 2 × 10⁻⁵ T

4

A solenoid of length 0.5m has 500 turns and carries a current of 4A. Calculate the magnetic field inside.

B = μ₀nI

Given: n = 500/0.5 = 1000 turns/m, I = 4 A
B = 4π × 10⁻⁷ × 1000 × 4
B = 4π × 10⁻⁷ × 4000 = 1.6π × 10⁻³

B ≈ 5.026 × 10⁻³ T

5

A circular coil of 50 turns and radius 0.1m carries a current of 2A. Calculate the magnetic field at its center.

B = (μ₀NI)/(2R)

Given: N = 50, R = 0.1 m, I = 2 A
B = (4π × 10⁻⁷ × 50 × 2) / (2 × 0.1) = (4π × 10⁻⁷ × 100) / 0.2
B = (1.2566 × 10⁻⁴) / 0.2

B = 6.283 × 10⁻⁴ T

6

A proton and an electron enter perpendicularly to a magnetic field with same speed. Calculate the ratio of their radii.

r_p/r_e = m_p/m_e ≈ 1836

7

A charged particle moves undeflected through crossed electric (E) and magnetic (B) fields. Calculate its speed.

v = E/B

8

A wire of length 0.2m carrying current 3A is placed perpendicular to a magnetic field of 0.5T. Calculate the force.

F = 0.3 N

9

Calculate the magnetic moment of an electron moving in a circle of radius 0.5Å with frequency 6.6×10¹⁵ Hz.

M = 9.27 × 10⁻²⁴ A·m²

10

A galvanometer has resistance 50Ω and gives full scale deflection for 2mA. Calculate the shunt for 1A range.

S ≈ 0.1Ω

Willer Academy Physics Numerical Practice | Class 12

100 Numerical Problems with Formulas & Solutions | © 2023 Willer Academy