MHT-CET Exam Prep Resources | Soulshift Education

MHT-CET

Chemistry Mathematics Physics

Q. A circuit has a resistance of 10 Ω and a reactance of 10 Ω. What is the impedance of the circuit? (2020)

A. 10 Ω
B. 14.14 Ω
C. 20 Ω
D. 5 Ω

Solution

The impedance Z = √(R² + X²) = √(10² + 10²) = √200 = 14.14 Ω.

Correct Answer: B — 14.14 Ω

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Q. A circuit has a resistance of 10Ω and a reactance of 5Ω. What is the impedance of the circuit? (2021)

A. 5Ω
B. 10Ω
C. √125Ω
D. √(10² + 5²)Ω

Solution

The impedance Z is calculated as Z = √(R² + X²) = √(10² + 5²) = √125Ω.

Correct Answer: D — √(10² + 5²)Ω

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Q. A circuit has a resistance of 10Ω and a reactance of 5Ω. What is the impedance? (2020)

A. 5Ω
B. 10Ω
C. √125Ω
D. √(10² + 5²)Ω

Solution

The impedance Z = √(R² + X²) = √(10² + 5²) = √125Ω.

Correct Answer: D — √(10² + 5²)Ω

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Q. A circuit has a resistance of 4 ohms and a current of 2 A. What is the voltage across the circuit?

A. 2 V
B. 4 V
C. 6 V
D. 8 V

Solution

Using Ohm's Law, V = I × R. Here, I = 2 A and R = 4 ohms. Thus, V = 2 × 4 = 8 V.

Correct Answer: D — 8 V

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Q. A circuit has a total resistance of 12Ω and a current of 1.5A. What is the voltage across the circuit? (2021)

A. 18V
B. 12V
C. 15V
D. 9V

Solution

Using Ohm's law, V = IR = 1.5A * 12Ω = 18V.

Correct Answer: A — 18V

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Q. A circuit has a total resistance of 12Ω and a current of 2A. What is the voltage across the circuit? (2021)

A. 24V
B. 12V
C. 6V
D. 48V

Solution

Using Ohm's law (V = IR), V = 2A * 12Ω = 24V.

Correct Answer: A — 24V

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Q. A circuit has a total resistance of 20Ω and a current of 2A flowing through it. What is the voltage across the circuit? (2023)

A. 30V
B. 40V
C. 20V
D. 10V

Solution

Using Ohm's law, V = I * R = 2A * 20Ω = 40V.

Correct Answer: B — 40V

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Q. A circuit has a total resistance of 5Ω and a current of 2A. What is the voltage across the circuit? (2023)

A. 10V
B. 5V
C. 2V
D. 1V

Solution

Using Ohm's law, V = I * R = 2A * 5Ω = 10V.

Correct Answer: A — 10V

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Q. A circuit has a total resistance of 8Ω and a current of 4A flowing through it. What is the voltage across the circuit? (2021)

A. 32V
B. 16V
C. 8V
D. 4V

Solution

Using Ohm's law, V = I * R = 4A * 8Ω = 32V.

Correct Answer: A — 32V

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Q. A circuit has a total resistance of 8Ω and a current of 4A. What is the voltage across the circuit? (2021)

A. 32V
B. 16V
C. 8V
D. 4V

Solution

Using Ohm's law, V = I * R = 4A * 8Ω = 32V.

Correct Answer: A — 32V

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Q. A circuit has a total voltage of 24V and a total current of 4A. What is the total resistance in the circuit? (2021)

A. 6Ω
B. 8Ω
C. 4Ω
D. 12Ω

Solution

Using Ohm's law, R = V/I = 24V / 4A = 6Ω.

Correct Answer: B — 8Ω

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Q. A circuit has a total voltage of 24V and a total resistance of 8Ω. What is the total current in the circuit? (2020)

A. 3A
B. 2A
C. 4A
D. 6A

Solution

Using Ohm's law, I = V/R = 24V / 8Ω = 3A.

Correct Answer: A — 3A

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Q. A circular loop of wire carries a current. What is the shape of the magnetic field lines around the loop? (2021)

A. Straight lines
B. Concentric circles
C. Ellipses
D. Spirals

Solution

The magnetic field lines around a circular loop of wire carrying current are in the form of concentric circles.

Correct Answer: B — Concentric circles

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Q. A coil has an inductance of 0.5 H and is connected to a 50 Hz AC supply. What is the inductive reactance? (2020)

A. 25.13 Ω
B. 31.42 Ω
C. 50 Ω
D. 100 Ω

Solution

Inductive reactance Xl = 2πfL = 2π(50)(0.5) = 31.42 Ω.

Correct Answer: B — 31.42 Ω

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Q. A coil of wire has 100 turns and is placed in a magnetic field of strength 0.5 T. If the area of the coil is 0.1 m² and the magnetic field is perpendicular to the coil, what is the magnetic flux through the coil? (2022)

A. 5 Wb
B. 0.5 Wb
C. 10 Wb
D. 0.05 Wb

Solution

Magnetic flux (Φ) = B × A × cos(θ). Here, B = 0.5 T, A = 0.1 m², and θ = 0° (cos(0) = 1). Thus, Φ = 0.5 T × 0.1 m² × 1 = 0.05 Wb.

Correct Answer: A — 5 Wb

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Q. A coil of wire has 100 turns and is placed in a magnetic field of strength 0.5 T. If the area of the coil is 0.1 m², what is the maximum magnetic flux through the coil? (2022)

A. 5 Wb
B. 0.5 Wb
C. 10 Wb
D. 1 Wb

Solution

Magnetic flux (Φ) = B × A × N = 0.5 T × 0.1 m² × 100 = 5 Wb.

Correct Answer: A — 5 Wb

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Q. A coil with 100 turns has a magnetic flux of 0.02 Wb passing through it. If the flux changes to 0.01 Wb in 2 seconds, what is the induced EMF? (2022)

A. 0.5 V
B. 1 V
C. 2 V
D. 5 V

Solution

Induced EMF = -N * (ΔΦ/Δt) = -100 * ((0.01 - 0.02) / 2) = 0.5 V.

Correct Answer: B — 1 V

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Q. A coil with 100 turns is placed in a magnetic field that changes from 0.5 T to 1.5 T in 2 seconds. What is the induced EMF in the coil?

A. 50 V
B. 100 V
C. 200 V
D. 400 V

Solution

Using Faraday's law, EMF = -N * (ΔB/Δt) = -100 * ((1.5 - 0.5) / 2) = -100 * (1 / 2) = -50 V. The magnitude is 50 V.

Correct Answer: B — 100 V

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Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force given by Fleming's left-hand rule? (2020)

A. Parallel to the field
B. Opposite to the field
C. Perpendicular to the field
D. In the direction of current

Solution

According to Fleming's left-hand rule, the force is perpendicular to both the magnetic field and the direction of current.

Correct Answer: C — Perpendicular to the field

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Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force? (2020)

A. Parallel to the field
B. Opposite to the field
C. Perpendicular to both current and field
D. In the direction of current

Solution

According to Fleming's left-hand rule, the force on a current-carrying conductor in a magnetic field is perpendicular to both the current and the magnetic field.

Correct Answer: C — Perpendicular to both current and field

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Q. A cyclist accelerates at 3 m/s² for 4 seconds. What is the final velocity if the initial velocity is 2 m/s? (2023)

A. 10 m/s
B. 11 m/s
C. 14 m/s
D. 20 m/s

Solution

Using v = u + at, we have v = 2 m/s + (3 m/s² * 4 s) = 2 + 12 = 14 m/s.

Correct Answer: B — 11 m/s

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the acceleration? (2023)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Acceleration = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the cyclist's acceleration? (2021)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Acceleration a = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A cyclist does 1500 J of work to climb a hill. If he takes 5 minutes, what is his average power output?

A. 5 W
B. 10 W
C. 15 W
D. 20 W

Solution

Power = Work / Time = 1500 J / (5 * 60 s) = 5 W.

Correct Answer: B — 10 W

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Q. A cyclist does 300 J of work to climb a hill. If the cyclist's mass is 75 kg, what is the height of the hill? (2023)

A. 0.4 m
B. 0.5 m
C. 0.6 m
D. 0.8 m

Solution

Potential Energy = mass × g × height; 300 J = 75 kg × 10 m/s² × height; height = 300 J / (75 kg × 10 m/s²) = 0.4 m

Correct Answer: B — 0.5 m

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Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the mass of the cyclist? (g = 9.8 m/s²)

A. 6.12 kg
B. 10 kg
C. 15 kg
D. 20 kg

Solution

Using Work = mgh; m = Work / (gh) = 300 J / (9.8 m/s² * 5 m) = 6.12 kg.

Correct Answer: A — 6.12 kg

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Q. A cyclist does 300 J of work to climb a hill. If the hill is 5 m high, what is the effective weight of the cyclist? (2022)

A. 30 kg
B. 60 kg
C. 90 kg
D. 120 kg

Solution

Weight = Work done / height = 300 J / 5 m = 60 kg

Correct Answer: B — 60 kg

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Q. A cyclist does 300 J of work to climb a hill. If the hill is 5 m high, what is the weight of the cyclist? (2020)

A. 60 kg
B. 30 kg
C. 50 kg
D. 40 kg

Solution

Work = m × g × h. 300 J = m × 9.8 m/s² × 5 m. m = 300 J / (9.8 m/s² × 5 m) = 6.12 kg.

Correct Answer: A — 60 kg

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Q. A cyclist does 600 J of work to climb a hill. If he takes 30 seconds, what is his average power output? (2022)

A. 10 W
B. 20 W
C. 30 W
D. 40 W

Solution

Power = Work / Time = 600 J / 30 s = 20 W.

Correct Answer: B — 20 W

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Q. A cyclist is moving at a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the total kinetic energy?

A. 800 J
B. 900 J
C. 1000 J
D. 1200 J

Solution

KE = 0.5 × m × v² = 0.5 × 75 kg × (15 m/s)² = 0.5 × 75 × 225 = 8437.5 J.

Correct Answer: C — 1000 J

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MHT-CET

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