Engineering Entrance
Q. A chord of a circle is 12 cm long and is 5 cm away from the center. What is the radius of the circle? (2020)
A.
10 cm
B.
13 cm
C.
15 cm
D.
12 cm
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Solution
Using Pythagoras theorem: r² = (5)² + (6)² = 25 + 36 = 61; r = √61 ≈ 7.81 cm.
Correct Answer: B — 13 cm
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Q. A circle has a diameter of 20 cm. What is its circumference? (2022)
A.
62.8 cm
B.
40 cm
C.
31.4 cm
D.
20 cm
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Solution
Circumference = πd = π * 20 = 62.8 cm.
Correct Answer: A — 62.8 cm
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Q. A circle has a radius of 5 cm. What is the length of an arc that subtends a central angle of 60 degrees? (2021)
A.
5.24 cm
B.
10.47 cm
C.
3.14 cm
D.
6.28 cm
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Solution
Arc length = (θ/360) * 2πr; = (60/360) * 2π * 5 = 10.47 cm.
Correct Answer: B — 10.47 cm
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Q. A circle has a radius of 5 cm. What is the length of an arc that subtends an angle of 60 degrees at the center? (2020)
A.
5.24 cm
B.
3.14 cm
C.
5.00 cm
D.
10.47 cm
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Solution
Arc length = (θ/360) * 2πr = (60/360) * 2 * π * 5 = 5.24 cm.
Correct Answer: A — 5.24 cm
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Q. A circle has an area of 154 cm². What is the radius? (2019) 2019
A.
7 cm
B.
14 cm
C.
21 cm
D.
28 cm
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Solution
Area = πr². Therefore, r = √(Area/π) = √(154/3.14) ≈ 7 cm.
Correct Answer: B — 14 cm
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Q. A circle has an area of 78.5 cm². What is its radius? (Use π = 3.14) (2019)
A.
5 cm
B.
7 cm
C.
10 cm
D.
12 cm
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Solution
Area = πr². Therefore, r² = Area / π = 78.5 cm² / 3.14 = 25. r = √25 = 5 cm.
Correct Answer: B — 7 cm
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Q. A circle is inscribed in a square of side 10 cm. What is the area of the circle? (2019)
A.
78.5 cm²
B.
100 cm²
C.
50 cm²
D.
25 cm²
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Solution
Radius = 10/2 = 5 cm; Area = πr² = π * 5² = 78.5 cm².
Correct Answer: A — 78.5 cm²
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Q. A circle is inscribed in a square of side 10 cm. What is the area of the circle? (Use π = 3.14) (2020)
A.
78.5 cm²
B.
50 cm²
C.
100 cm²
D.
25 cm²
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Solution
Radius of the circle = side/2 = 10 cm / 2 = 5 cm. Area = πr² = 3.14 * 5² = 78.5 cm².
Correct Answer: A — 78.5 cm²
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Q. A circle is inscribed in a square of side 8 cm. What is the area of the circle? (2022)
A.
50.24 cm²
B.
64 cm²
C.
25.12 cm²
D.
32 cm²
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Solution
Radius = 8/2 = 4 cm; Area = πr² = π * 4² = 50.24 cm².
Correct Answer: A — 50.24 cm²
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Q. A circuit has a 9V battery and two resistors of 3Ω and 6Ω in series. What is the current flowing through the circuit? (2022)
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Solution
Total resistance R = 3Ω + 6Ω = 9Ω. Current I = V/R = 9V / 9Ω = 1A.
Correct Answer: B — 2A
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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
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Solution
The impedance Z = √(R² + X²) = √(10² + 5²) = √125Ω.
Correct Answer: D — √(10² + 5²)Ω
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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
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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
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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 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
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Solution
Power = Work / Time = 1500 J / (5 * 60 s) = 5 W.
Correct Answer: B — 10 W
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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
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Solution
Power = Work / Time = 600 J / 30 s = 20 W.
Correct Answer: B — 20 W
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Q. A cyclist travels at a speed of 10 m/s for 5 seconds. What distance does he cover? (2023)
A.
25 m
B.
50 m
C.
75 m
D.
100 m
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Solution
Distance = speed * time = 10 m/s * 5 s = 50 m.
Correct Answer: B — 50 m
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Q. A cylinder rolls down an incline without slipping. If its mass is M and radius is R, what is the acceleration of its center of mass? (2020)
A.
g/2
B.
g/3
C.
g/4
D.
g/5
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Solution
For rolling without slipping, a_cm = (2/3)g sin(θ).
Correct Answer: B — g/3
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Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)
A.
10 cm height, 10 cm radius
B.
5 cm height, 15.87 cm radius
C.
8 cm height, 12.5 cm radius
D.
12 cm height, 8.33 cm radius
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Solution
Using the formula for surface area and volume, the optimal dimensions are found to be 5 cm height and approximately 15.87 cm radius.
Correct Answer: B — 5 cm height, 15.87 cm radius
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Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022) 2022
A.
10, 10
B.
5, 20
C.
8, 15
D.
6, 18
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Solution
Using the formula for surface area and volume, the optimal dimensions are found to be radius = 5 and height = 20.
Correct Answer: C — 8, 15
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Q. A disc of radius R and mass M is rotating about its axis with an angular velocity ω. What is its rotational kinetic energy? (2020)
A.
(1/2)Iω²
B.
(1/2)Mω²
C.
Iω
D.
Mω²
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Solution
Rotational kinetic energy K.E. = (1/2)Iω². For a disc, I = (1/2)MR², thus K.E. = (1/4)MR²ω².
Correct Answer: A — (1/2)Iω²
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Q. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied, what is the angular acceleration? (2021)
A.
τ/I
B.
I/τ
C.
ω/τ
D.
Iω
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Solution
From Newton's second law for rotation, τ = Iα, thus α = τ/I.
Correct Answer: A — τ/I
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Q. A force of 10 N is applied to an object, causing it to accelerate at 2 m/s². What is the mass of the object? (2020)
A.
5 kg
B.
10 kg
C.
15 kg
D.
20 kg
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Solution
Using F = ma, mass m = F/a = 10 N / 2 m/s² = 5 kg.
Correct Answer: A — 5 kg
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Q. A force of 10 N is applied to move an object 5 m. What is the work done? (2023)
A.
50 J
B.
10 J
C.
5 J
D.
20 J
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Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: A — 50 J
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Q. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².
Correct Answer: B — 4 m/s²
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Q. A force of 50 N is applied to move an object 4 m. How much work is done?
A.
100 J
B.
150 J
C.
200 J
D.
250 J
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Solution
Work = Force * Distance = 50 N * 4 m = 200 J.
Correct Answer: C — 200 J
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Q. A gas expands from 2 L to 5 L at a constant pressure of 1 atm. How much work is done by the gas? (2023)
A.
3 L·atm
B.
5 L·atm
C.
2 L·atm
D.
1 L·atm
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Solution
Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.
Correct Answer: A — 3 L·atm
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Q. A gas expands isothermally at 300 K and absorbs 600 J of heat. What is the work done by the gas? (2023)
A.
600 J
B.
300 J
C.
900 J
D.
0 J
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Solution
In an isothermal process, the work done by the gas is equal to the heat absorbed. Therefore, work done = 600 J.
Correct Answer: A — 600 J
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Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure of the gas is 100 kPa, what is the work done by the gas? (2020)
A.
0 kJ
B.
10 kJ
C.
20 kJ
D.
30 kJ
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Solution
Work done (W) = P * ΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 10 kJ.
Correct Answer: B — 10 kJ
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Q. A light ray strikes a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection? (2021)
A.
30 degrees
B.
60 degrees
C.
90 degrees
D.
45 degrees
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Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is also 30 degrees.
Correct Answer: A — 30 degrees
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Q. A machine does 1200 J of work in 60 seconds. What is its efficiency if the input energy is 2000 J? (2022)
A.
60%
B.
70%
C.
80%
D.
90%
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Solution
Efficiency = (Output Work / Input Energy) * 100 = (1200 J / 2000 J) * 100 = 60%.
Correct Answer: A — 60%
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