Major Competitive Exams
Q. What is the volume of a sphere with radius 5 units? (2021)
A.
500/3π
B.
125/3π
C.
100/3π
D.
200/3π
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Solution
Volume = (4/3)πr³ = (4/3)π(5)³ = (4/3)π(125) = 500/3π cubic units.
Correct Answer: A — 500/3π
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Q. What is the volume of CO2 produced at STP when 2 moles of C2H5OH are completely combusted?
A.
22.4 L
B.
44.8 L
C.
67.2 L
D.
89.6 L
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Solution
C2H5OH + 3O2 → 2CO2 + 3H2O. 2 moles of C2H5OH produce 4 moles of CO2. Volume = 4 * 22.4 L = 89.6 L.
Correct Answer: B — 44.8 L
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Q. What is the volume percent concentration of a solution containing 30 mL of ethanol in 150 mL of solution?
A.
20%
B.
25%
C.
30%
D.
15%
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Solution
Volume percent = (volume of solute / total volume) x 100 = (30 mL / 150 mL) x 100 = 20%.
Correct Answer: B — 25%
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Q. What is the volume percent of ethanol in a solution made by mixing 30 mL of ethanol with 70 mL of water?
A.
30%
B.
70%
C.
50%
D.
20%
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Solution
Volume percent = (volume of solute / total volume) × 100 = (30 mL / (30 mL + 70 mL)) × 100 = 30%.
Correct Answer: A — 30%
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Q. What is the wavelength of a photon with a frequency of 5 x 10^14 Hz?
A.
6 x 10^-7 m
B.
3 x 10^-7 m
C.
1 x 10^-6 m
D.
4 x 10^-7 m
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Solution
Using the formula λ = c/f, where c = 3 x 10^8 m/s, λ = 3 x 10^8 / (5 x 10^14) = 6 x 10^-7 m.
Correct Answer: A — 6 x 10^-7 m
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Q. What is the wavelength of a photon with a frequency of 6 x 10^14 Hz?
A.
5 x 10^-7 m
B.
3 x 10^-7 m
C.
6 x 10^-7 m
D.
1 x 10^-6 m
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Solution
Wavelength λ = c/f = (3 x 10^8 m/s) / (6 x 10^14 Hz) = 5 x 10^-7 m.
Correct Answer: A — 5 x 10^-7 m
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Q. What is the wavelength of an electron accelerated through a potential difference of 100 V? (mass of electron = 9.11 x 10^-31 kg)
A.
1.22 x 10^-10 m
B.
1.23 x 10^-9 m
C.
1.24 x 10^-11 m
D.
1.25 x 10^-12 m
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Solution
The wavelength λ = h/p = h/sqrt(2meV) = (6.63 x 10^-34)/sqrt(2 * 9.11 x 10^-31 * 100) = 1.22 x 10^-10 m.
Correct Answer: A — 1.22 x 10^-10 m
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Q. What is the wavelength of light emitted when an electron transitions from n=3 to n=2 in a hydrogen atom?
A.
410 nm
B.
656 nm
C.
486 nm
D.
434 nm
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Solution
Using the Rydberg formula, the wavelength for the transition from n=3 to n=2 is approximately 434 nm.
Correct Answer: D — 434 nm
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Q. What is the wavelength of light if its frequency is 5 x 10^14 Hz?
A.
6 x 10^-7 m
B.
3 x 10^8 m
C.
5 x 10^-7 m
D.
1 x 10^-6 m
Show solution
Solution
Using the formula λ = c/f, where c = 3 x 10^8 m/s, λ = 3 x 10^8 / (5 x 10^14) = 6 x 10^-7 m.
Correct Answer: C — 5 x 10^-7 m
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Q. What is the wavelength of light if the first-order maximum occurs at an angle of 30° in a double-slit experiment with slit separation of 0.1 mm and a screen distance of 1 m?
A.
300 nm
B.
600 nm
C.
450 nm
D.
750 nm
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Solution
Using d sin θ = mλ, λ = d sin θ/m. For m=1, d=0.1 mm, θ=30°, λ = 0.1 mm * sin(30°) = 0.1 mm * 0.5 = 0.05 mm = 500 nm.
Correct Answer: B — 600 nm
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Q. What is the wavelength of light if the first-order maximum occurs at an angle of 30° in a double-slit experiment with slit separation of 0.1 mm and screen distance of 1 m?
A.
0.5 mm
B.
0.6 mm
C.
0.7 mm
D.
0.8 mm
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Solution
Using d sin θ = mλ, λ = d sin θ/m. For m=1, d=0.1 mm, θ=30°, λ = 0.1 mm * sin(30°) = 0.1 mm * 0.5 = 0.05 mm = 0.5 mm.
Correct Answer: A — 0.5 mm
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Q. What is the wavelength of light if the first-order maximum occurs at an angle of 30° in a double-slit experiment with slit separation of 0.1 mm and distance to screen of 1 m?
A.
0.5 mm
B.
0.3 mm
C.
0.2 mm
D.
0.1 mm
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Solution
Using d sin θ = mλ, λ = d sin θ/m. For m=1, d=0.1 mm, θ=30°, λ = 0.1 mm * sin(30°) = 0.1 mm * 0.5 = 0.05 mm = 0.2 mm.
Correct Answer: C — 0.2 mm
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Q. What is the wavelength of light with a frequency of 5 x 10^14 Hz?
A.
6 x 10^-7 m
B.
3 x 10^8 m
C.
5 x 10^-7 m
D.
1 x 10^-6 m
Show solution
Solution
Using the formula λ = c/f, where c = 3 x 10^8 m/s, λ = 3 x 10^8 / (5 x 10^14) = 6 x 10^-7 m.
Correct Answer: A — 6 x 10^-7 m
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Q. What is the wavelength of light with a frequency of 5 x 10^14 Hz? (Speed of light = 3 x 10^8 m/s)
A.
0.6 m
B.
0.5 m
C.
0.4 m
D.
0.3 m
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Solution
Wavelength λ = c / f = (3 x 10^8 m/s) / (5 x 10^14 Hz) = 0.6 x 10^-6 m = 0.6 µm.
Correct Answer: B — 0.5 m
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Q. What is the weight of a 10 kg mass on Earth (g = 9.8 m/s²)?
A.
9.8 N
B.
10 N
C.
98 N
D.
100 N
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Solution
Weight W = mg = 10 kg * 9.8 m/s² = 98 N.
Correct Answer: C — 98 N
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Q. What is the weight of a 10 kg object on Earth (g = 9.8 m/s²)?
A.
9.8 N
B.
10 N
C.
98 N
D.
100 N
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Solution
Weight W = mg = 10 kg * 9.8 m/s² = 98 N.
Correct Answer: C — 98 N
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Q. What is the weight of a 10 kg object on the surface of a planet where g = 12 m/s²?
A.
120 N
B.
100 N
C.
80 N
D.
60 N
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Solution
Weight W = mg = 10 * 12 = 120 N.
Correct Answer: A — 120 N
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Q. What is the weight of a 10 kg object on the surface of the Earth?
A.
10 N
B.
100 N
C.
1000 N
D.
1 N
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Solution
Weight W = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: B — 100 N
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Q. What is the weight of a 10 kg object on the surface of the Earth? (g = 9.8 m/s²)
A.
9.8 N
B.
19.6 N
C.
98 N
D.
0.98 N
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Solution
Weight W = m * g = 10 kg * 9.8 m/s² = 98 N
Correct Answer: B — 19.6 N
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Q. What is the weight of a 12 kg object on Earth?
A.
12 N
B.
120 N
C.
1.2 N
D.
0.12 N
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Solution
Weight W = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: B — 120 N
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Q. What is the weighted mean of the scores 70, 80, 90 with weights 1, 2, 3 respectively?
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Solution
Weighted mean = (70*1 + 80*2 + 90*3) / (1 + 2 + 3) = 85.
Correct Answer: B — 85
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Q. What is the weighted mean of the scores 70, 80, and 90 with weights 1, 2, and 3 respectively?
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Solution
Weighted mean = (70*1 + 80*2 + 90*3) / (1 + 2 + 3) = 85.
Correct Answer: B — 85
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Q. What is the weighted mean of the scores 80, 90, and 70 with weights 1, 2, and 1 respectively?
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Solution
Weighted mean = (80*1 + 90*2 + 70*1) / (1 + 2 + 1) = 85.
Correct Answer: B — 85
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Q. What is the work done against friction if a box is moved 5 m on a surface with a coefficient of friction of 0.2 and a normal force of 200 N?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Frictional force = μ * N = 0.2 * 200 N = 40 N. Work done = frictional force * distance = 40 N * 5 m = 200 J.
Correct Answer: B — 20 J
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Q. What is the work done against friction when a 5 kg box is pushed 10 m across a surface with a coefficient of kinetic friction of 0.4?
A.
20 J
B.
40 J
C.
50 J
D.
60 J
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Solution
Work done against friction = frictional force * distance = (μk * m * g) * d = (0.4 * 5 kg * 9.8 m/s²) * 10 m = 196 J, approximately 40 J.
Correct Answer: B — 40 J
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Q. What is the work done by a constant force of 10 N acting on an object that moves 5 m in the direction of the force?
A.
10 J
B.
50 J
C.
5 J
D.
25 J
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Solution
Work done (W) = Force (F) * Distance (d) = 10 N * 5 m = 50 J.
Correct Answer: B — 50 J
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Q. What is the work done by a constant force of 10 N acting over a distance of 5 m in the direction of the force?
A.
50 J
B.
10 J
C.
5 J
D.
25 J
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Solution
Work done W = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: A — 50 J
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Q. What is the work done by a constant force of 10 N moving an object 5 m in the direction of the force?
A.
10 J
B.
25 J
C.
50 J
D.
100 J
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Solution
Work done (W) = Force (F) × Distance (d) = 10 N × 5 m = 50 J.
Correct Answer: C — 50 J
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Q. What is the work done by a force of 10 N moving an object 5 m in the direction of the force?
A.
50 J
B.
10 J
C.
5 J
D.
0 J
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Solution
Work done W = Force x Distance = 10 N x 5 m = 50 J.
Correct Answer: A — 50 J
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Q. What is the work done by a force of 5 N moving an object 3 m in the direction of the force?
A.
15 J
B.
5 J
C.
3 J
D.
0 J
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Solution
Work done W = Force × Distance = 5 N × 3 m = 15 J.
Correct Answer: A — 15 J
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