Engineering Entrance Exam Preparation Resources

Engineering Entrance

MHT-CET

Q. If 50 mL of a 3 M solution is diluted to 200 mL, what is the new molarity? (2020)

A. 0.75 M
B. 1.5 M
C. 2 M
D. 3 M

Solution

Using M1V1 = M2V2, (3 M)(50 mL) = M2(200 mL) => M2 = (3 M × 50 mL) / 200 mL = 0.75 M.

Correct Answer: A — 0.75 M

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Q. If 500 J of heat is added to a system and 200 J of work is done by the system, what is the change in internal energy of the system?

A. 300 J
B. 500 J
C. 700 J
D. 200 J

Solution

According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.

Correct Answer: A — 300 J

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Q. If 500 J of heat is added to a system and it does 200 J of work, what is the change in internal energy? (2022)

A. 300 J
B. 200 J
C. 500 J
D. 700 J

Solution

According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.

Correct Answer: A — 300 J

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Q. If 5x = 25, what is the value of x? (2019)

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

Solution

5x = 25, so x = 25/5 = 5.

Correct Answer: A — 5

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Q. If a 10 kg object is dropped from a height of 20 m, what is the velocity just before it hits the ground? (Assume g = 10 m/s²) (2019)

A. 10 m/s
B. 20 m/s
C. 14 m/s
D. 15 m/s

Solution

Using v² = u² + 2gh, v = √(0 + 2 × 10 m/s² × 20 m) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

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Q. If a 10 kg object is moving with a speed of 2 m/s, what is its kinetic energy?

A. 10 J
B. 20 J
C. 40 J
D. 80 J

Solution

KE = 0.5 × m × v² = 0.5 × 10 kg × (2 m/s)² = 0.5 × 10 × 4 = 20 J.

Correct Answer: B — 20 J

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Q. If a 10 kg object is raised to a height of 5 m, what is the work done against gravity? (g = 9.8 m/s²)

A. 490 J
B. 980 J
C. 150 J
D. 250 J

Solution

Work = m * g * h = 10 kg * 9.8 m/s² * 5 m = 490 J.

Correct Answer: B — 980 J

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Q. If a 10 N force is applied at an angle of 60 degrees to the horizontal while moving an object 3 m, what is the work done? (2020)

A. 15 J
B. 30 J
C. 25 J
D. 5 J

Solution

Work done = Force × Distance × cos(θ) = 10 N × 3 m × cos(60°) = 10 N × 3 m × 0.5 = 15 J

Correct Answer: A — 15 J

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Q. If a 10 N force is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done? (2020)

A. 20 J
B. 40 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance × cos(θ) = 10 N × 4 m × cos(60°) = 10 N × 4 m × 0.5 = 20 J

Correct Answer: C — 30 J

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Q. If a 1000 kg car is moving at a speed of 15 m/s, what is its kinetic energy?

A. 112,500 J
B. 225,000 J
C. 150,000 J
D. 75,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (15 m/s)² = 112,500 J

Correct Answer: B — 225,000 J

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Q. If a 1000 kg car is moving with a speed of 15 m/s, what is its kinetic energy?

A. 112,500 J
B. 225,000 J
C. 75,000 J
D. 150,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (15 m/s)² = 112,500 J

Correct Answer: B — 225,000 J

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Q. If a 1000 W motor runs for 2 hours, how much work does it do? (2000)

A. 7200 J
B. 2000 J
C. 3600000 J
D. 100000 J

Solution

Work = Power × Time = 1000 W × (2 hours × 3600 s/hour) = 1000 W × 7200 s = 7200000 J

Correct Answer: C — 3600000 J

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Q. If a 100Ω resistor is connected in series with a 200Ω resistor, what is the total resistance? (2023)

A. 300Ω
B. 100Ω
C. 200Ω
D. 400Ω

Solution

In series, R_total = R1 + R2 = 100Ω + 200Ω = 300Ω.

Correct Answer: A — 300Ω

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Q. If a 10Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2022)

A. 3.33Ω
B. 5Ω
C. 7.5Ω
D. 15Ω

Solution

For parallel resistors, 1/R_eq = 1/R1 + 1/R2 = 1/10 + 1/5 = 1/10 + 2/10 = 3/10. Therefore, R_eq = 10/3 = 3.33Ω.

Correct Answer: A — 3.33Ω

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Q. If a 10Ω resistor is connected in parallel with a 5Ω resistor, what is the total resistance? (2022)

A. 3.33Ω
B. 5Ω
C. 7.5Ω
D. 15Ω

Solution

For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 = 1/10 + 1/5 = 1/10 + 2/10 = 3/10. Therefore, R_eq = 10/3 = 3.33Ω.

Correct Answer: A — 3.33Ω

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Q. If a 1500 kg car is moving at a speed of 30 m/s, what is its kinetic energy?

A. 675,000 J
B. 450,000 J
C. 225,000 J
D. 900,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1500 kg × (30 m/s)² = 675,000 J

Correct Answer: B — 450,000 J

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Q. If a 15Ω resistor is connected in parallel with a 30Ω resistor, what is the equivalent resistance? (2021)

A. 10Ω
B. 7.5Ω
C. 5Ω
D. 12Ω

Solution

For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 = 1/15 + 1/30 = 2/30 + 1/30 = 3/30. Therefore, R_eq = 30/3 = 10Ω.

Correct Answer: B — 7.5Ω

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Q. If a 15Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2023)

A. 3.75Ω
B. 4Ω
C. 5Ω
D. 10Ω

Solution

1/R_eq = 1/15 + 1/5 = 1/15 + 3/15 = 4/15; R_eq = 15/4 = 3.75Ω.

Correct Answer: A — 3.75Ω

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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground (g = 10 m/s²)? (2023)

A. 10 m/s
B. 20 m/s
C. 30 m/s
D. 40 m/s

Solution

Using v = √(2gh), for h = 10 m, v = √(2 * 10 m/s² * 10 m) = √200 = 20 m/s.

Correct Answer: B — 20 m/s

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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground? (g = 9.8 m/s², h = 20 m) (2023)

A. 14 m/s
B. 19.6 m/s
C. 20 m/s
D. 28 m/s

Solution

Using v² = u² + 2gh, where u = 0, we find v = √(2 * 9.8 * 20) = 19.6 m/s.

Correct Answer: B — 19.6 m/s

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Q. If a 2 kg object is lifted to a height of 10 m, what is the potential energy gained by the object? (g = 9.8 m/s²)

A. 19.6 J
B. 98 J
C. 39.2 J
D. 78.4 J

Solution

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.

Correct Answer: B — 98 J

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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (g = 9.8 m/s²)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.

Correct Answer: A — 58.8 J

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Q. If a 2 kg object is moving with a velocity of 15 m/s, what is its kinetic energy?

A. 225 J
B. 150 J
C. 300 J
D. 450 J

Solution

Kinetic Energy = 1/2 * m * v^2 = 1/2 * 2 kg * (15 m/s)^2 = 225 J.

Correct Answer: A — 225 J

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Q. If a 2 kg object is moving with a velocity of 3 m/s, what is its kinetic energy?

A. 9 J
B. 6 J
C. 3 J
D. 12 J

Solution

Kinetic Energy = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.

Correct Answer: A — 9 J

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Q. If a 3 kg object is moving with a speed of 4 m/s, what is its kinetic energy?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

KE = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 0.5 × 3 × 16 = 24 J.

Correct Answer: B — 24 J

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Q. If a 3 kg object is subjected to a net force of 12 N, what is its acceleration? (2023)

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

Solution

Acceleration a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: A — 4 m/s²

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Q. If a 4 kg object is lifted to a height of 2 m, what is the work done against gravity? (2021)

A. 40 J
B. 80 J
C. 20 J
D. 60 J

Solution

Work done = mass × g × height = 4 kg × 10 m/s² × 2 m = 80 J

Correct Answer: B — 80 J

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Q. If a 4 kg object is moving with a speed of 5 m/s, what is its kinetic energy?

A. 50 J
B. 40 J
C. 100 J
D. 80 J

Solution

Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 0.5 × 4 × 25 = 50 J.

Correct Answer: D — 80 J

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Q. If a 5 kg object is dropped from a height of 10 m, what is its potential energy just before it hits the ground? (g = 9.8 m/s²)

A. 490 J
B. 980 J
C. 245 J
D. 2450 J

Solution

Potential Energy (PE) = m × g × h = 5 kg × 9.8 m/s² × 10 m = 490 J.

Correct Answer: B — 980 J

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Q. If a 5 kg object is dropped from a height of 10 m, what is its potential energy just before it hits the ground? (2021)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

PE = m × g × h = 5 kg × 9.8 m/s² × 10 m = 490 J.

Correct Answer: B — 100 J

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Engineering Entrance

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