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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%

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

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

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

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

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

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 screen distance of 1 m?

A. 0.5 mm
B. 0.6 mm
C. 0.7 mm
D. 0.8 mm

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

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 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

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 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

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

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

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

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

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

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

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

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?

A. 80
B. 85
C. 90
D. 75

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?

A. 80
B. 85
C. 90
D. 75

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?

A. 80
B. 85
C. 90
D. 75

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

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

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

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

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

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

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

Solution

Work done W = Force × Distance = 5 N × 3 m = 15 J.

Correct Answer: A — 15 J

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Q. What is the work done by a gas during an isobaric expansion?

A. Zero
B. PΔV
C. ΔU
D. Q

Solution

In an isobaric process, the work done by the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume.

Correct Answer: B — PΔV

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Q. What is the work done by a gas during an isothermal expansion from volume Vi to Vf?

A. nRT ln(Vf/Vi)
B. nR(Tf - Ti)
C. Zero
D. nRT

Solution

The work done by a gas during an isothermal expansion is W = nRT ln(Vf/Vi).

Correct Answer: A — nRT ln(Vf/Vi)

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Q. What is the work done by a gas during expansion against a constant external pressure?

A. PΔV
B. ΔU
C. Q
D. 0

Solution

The work done by a gas during expansion against a constant external pressure is given by W = PΔV, where P is the pressure and ΔV is the change in volume.

Correct Answer: A — PΔV

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