# PHYS 104 : Principles Of Physics! Trivia Questions Quiz

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Physics helps us understand how the world around us works. Do you consider yourself a know it all when it comes to physics and where the study began, for example, the common formulas used? If you are not sure this Phys 104: principles of physics trivia questions quiz will come in handy to refresh your memory. Try it out!

• 1.

### A lamp emits yellow light of a wavelength of 600 nm in the air (Refractive index n=1). What would be the wavelength under the water (Refractive Index n=1.33)?

• A.

600 nm

• B.

451 nm

• C.

350 nm

• D.

700 nm

B. 451 nm
Explanation
When light passes from one medium to another, its wavelength changes according to the refractive index of the two mediums. In this case, the light is passing from air (refractive index of 1) to water (refractive index of 1.33). The wavelength of the light decreases when it enters a medium with a higher refractive index. Therefore, the wavelength of the yellow light (600 nm) will decrease to a shorter wavelength under water. The correct answer is 451 nm.

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• 2.

### The diagram shows a particle with positive charge +Q and a particle with negative -Q located at the same distance from a point p. The direction of the total electric field at point p is:

• A.

Up

• B.

Down

• C.

Right

• D.

Left

A. Up
Explanation
The positive charge +Q and the negative charge -Q create electric fields that point away from them. Since the positive charge is above point p and the negative charge is below point p, the electric fields from both charges will add up and point upwards at point p. Therefore, the direction of the total electric field at point p is up.

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• 3.

### A block shown is pulled across the horizontal surface at a constant speed by the force shown. If M = 5 kg, F = 14 N andθ = 370 , what is the coefficient of kinetic friction between the block and the horizontal surface?. Consider g = 10 m/s2

• A.

0.33

• B.

0.27

• C.

0.44

• D.

0.50

B. 0.27
Explanation
The coefficient of kinetic friction can be determined using the equation Ff = μk * Fn, where Ff is the force of kinetic friction, μk is the coefficient of kinetic friction, and Fn is the normal force. In this case, the force of kinetic friction is equal to the force applied to the block (F = 14 N). The normal force can be calculated as Fn = M * g, where M is the mass of the block (5 kg) and g is the acceleration due to gravity (10 m/s^2). Plugging in the values, we get Fn = 5 kg * 10 m/s^2 = 50 N. Therefore, the coefficient of kinetic friction is μk = Ff / Fn = 14 N / 50 N = 0.28.

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• 4.

### Instead of describing fluids In terms of mass and forces, we use the ... and the...

• A.

Volume, mass

• B.

Newton, Pascal

• C.

Mass, pressure

• D.

Density, pressure

D. Density, pressure
Explanation
Fluids are described in terms of density and pressure rather than mass and forces. Density is a measure of how much mass is contained in a given volume of fluid, while pressure is the force exerted by the fluid per unit area. By using density and pressure, we can better understand and analyze the behavior of fluids, such as their flow, buoyancy, and stability.

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• 5.

### When the distance between two charges is cut in half, the electric force between the charges is

• A.

Increased by a factor of two

• B.

Increased by a factor of four

• C.

Decreased by a factor of two

• D.

Decreased by a factor of four

B. Increased by a factor of four
Explanation
When the distance between two charges is cut in half, the electric force between the charges is increased by a factor of four. This is because the electric force between charges is inversely proportional to the square of the distance between them. Therefore, if the distance is halved, the square of the distance becomes four times smaller, resulting in the electric force being four times stronger.

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• 6.

### The slope of the straight line shown on the graph is called the

• A.

Bulk modulus.

• B.

Young's modulus.

• C.

Compressibility.

• D.

Spring constant.

B. Young's modulus.
Explanation
The slope of a straight line on a graph represents the ratio of the change in the dependent variable to the change in the independent variable. In this case, the straight line on the graph represents the relationship between stress and strain, which is described by Young's modulus. Young's modulus is a measure of the stiffness or elasticity of a material, indicating how much it deforms under a given amount of stress. Therefore, the correct answer is Young's modulus.

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• 7.

### The strain in an object subjected to stress is the ...

• A.

Shear

• B.

Fractional deformation

• C.

Elastic limit

• D.

Hook's law

B. Fractional deformation
Explanation
The correct answer is "fractional deformation" because when an object is subjected to stress, it undergoes a change in shape or size, which is known as deformation. Fractional deformation refers to the ratio of the change in length or size to the original length or size of the object. This term is commonly used to describe the amount of strain experienced by the object.

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• 8.

### What is the normal force on the mass M = 7 kg in the figure if F = 60 N and  = 30°? Consider g = 10 m/s2

• A.

100 N

• B.

40 N

• C.

70 N

• D.

30 N

A. 100 N
Explanation
The normal force is equal to the component of the gravitational force acting perpendicular to the surface. In this case, the mass is on a flat surface, so the normal force is equal to the weight of the mass. The weight is calculated by multiplying the mass by the acceleration due to gravity, which is 10 m/s^2. Therefore, the normal force is 7 kg * 10 m/s^2 = 70 N.

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• 9.

### The coefficient of static friction is usually

• A.

Greater than the coefficient of kinetic friction

• B.

Greater than 1

• C.

Negative

• D.

Equals to zero

A. Greater than the coefficient of kinetic friction
Explanation
The coefficient of static friction is usually greater than the coefficient of kinetic friction because it takes more force to overcome the initial static friction and start an object's motion than to keep it in motion. The coefficient of static friction represents the maximum amount of friction that can be applied to prevent an object from moving, while the coefficient of kinetic friction represents the amount of friction experienced by an object in motion. Therefore, the coefficient of static friction is generally higher than the coefficient of kinetic friction.

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• 10.

### A rubber band is stretched by 1.0 cm when a force of 0.35 N is applied to each end. If instead a force a force of 0.70 N is applied to each end, estimate how far the rubber band will stretch from its un-stretched length

• A.

0.25 cm

• B.

0.5 cm

• C.

1.0 cm

• D.

2.0 cm

D. 2.0 cm
Explanation
When the force applied to the rubber band is doubled from 0.35 N to 0.70 N, it can be inferred that the stretch of the rubber band will also double. This is because the stretch of a rubber band is directly proportional to the force applied to it. Therefore, if a force of 0.35 N causes a stretch of 1.0 cm, then a force of 0.70 N will cause a stretch of 2.0 cm.

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• 11.

### You hold a piece of wood in one hand and a piece of iron in the other. Both pieces have the same volume, and you hold them fully underwater at the same depth. At the moment you let them go, which one experiences the greater buoyancy force?

• A.

The piece of wood.

• B.

The piece of iron.

• C.

Both experience the same buoyant force.

• D.

C. Both experience the same buoyant force.
Explanation
Both the piece of wood and the piece of iron experience the same buoyant force because buoyant force depends on the volume of the object displaced, not the material it is made of. Since both pieces have the same volume and are fully submerged at the same depth, they displace the same amount of fluid, resulting in equal buoyant forces acting on them.

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• 12.

### A beam of light of wavelength 550 nm traveling in a medium of refractive index 2.33 is incident on a slab of transparent material of refractive index 1.66. Find the ratio sinθ1 sinθ2 , where θ1 is the angle of incidence and θ2 is the angle of refraction.

• A.

1.4

• B.

2

• C.

0.71

• D.

3.99

C. 0.71
Explanation
The ratio sinθ1/sinθ2 is equal to the ratio of the velocities of light in the two mediums, which is equal to the ratio of the refractive indices of the two mediums. In this case, the refractive index of the first medium is 2.33 and the refractive index of the second medium is 1.66. Therefore, the ratio sinθ1/sinθ2 is equal to 2.33/1.66, which simplifies to approximately 0.71.

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• 13.

### Find the critical angle for water-air interface (nwater = 1.33, nair = 1).

• A.

90 degree

• B.

60 degree

• C.

30 degree

• D.

48.8 degree

D. 48.8 degree
Explanation
The critical angle for a water-air interface can be found using the formula: critical angle = arcsin(n2/n1), where n1 is the refractive index of the first medium (water) and n2 is the refractive index of the second medium (air). In this case, n1 = 1.33 and n2 = 1. Plugging these values into the formula, we get: critical angle = arcsin(1/1.33) = 48.8 degrees. Therefore, the correct answer is 48.8 degrees.

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• 14.

### A beam of light of wavelength λ1 traveling from an optically dense medium (n1= 1.8) to an optically rare medium (n2 = 1). Find the ratio λ1:λ2, where λ2 is the wavelength of light in the optically rare medium.

• A.

0.56

• B.

1.8

• C.

0.8

• D.

2.8

A. 0.56
Explanation
The ratio of λ1 to λ2 can be found using the formula: n1λ1 = n2λ2, where n1 and n2 are the refractive indices of the two media, and λ1 and λ2 are the wavelengths of light in the respective media. In this case, n1 = 1.8 and n2 = 1. Since λ1 and λ2 are inversely proportional to n1 and n2, the ratio of λ1 to λ2 will be the inverse of the ratio of n1 to n2. Therefore, the ratio of λ1 to λ2 is 1.8:1, which simplifies to 0.56.

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• 15.

### A beam of light travels fastest in

• A.

Glass

• B.

Water

• C.

Plastic

• D.

Air

D. Air
Explanation
Air is the correct answer because it has the lowest refractive index among the given options. The refractive index determines the speed of light in a medium, with a lower refractive index indicating a faster speed. Air has a refractive index close to 1, while glass, water, and plastic have higher refractive indices. Therefore, light travels fastest in air compared to the other mediums mentioned.

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• 16.

### The wavelength of red helium-neon laser in air is 632.8 nm. What is its frequency?. Speed of light in air is c = 3 x 108 m/s.

• A.

4.74 x 10^14 Hz

• B.

3.74x 10^14 Hz

• C.

2.74x 10^14 Hz

• D.

1.74x 10^14 Hz

A. 4.74 x 10^14 Hz
Explanation
The frequency of a wave can be calculated using the equation f = c/λ, where f is the frequency, c is the speed of light, and λ is the wavelength. In this case, the wavelength of the red helium-neon laser is given as 632.8 nm, which is equal to 632.8 x 10^-9 m. Plugging this value into the equation, we get f = (3 x 10^8 m/s) / (632.8 x 10^-9 m) = 4.74 x 10^14 Hz. Therefore, the correct answer is 4.74 x 10^14 Hz.

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• 17.

### If a liquid has an index of refraction of 1.461, what is the speed of light in this liquid? Speed of light in air is c = 3 x 108 m/s.

• A.

2.88 × 10^8 m/s

• B.

2.05× 10^8 m/s

• C.

1.25× 10^8 m/s

• D.

1.46× 10^8 m/s

B. 2.05× 10^8 m/s
Explanation
The speed of light in a medium is given by the equation v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the index of refraction of the medium. In this case, the index of refraction is given as 1.461. Plugging this value into the equation, we get v = (3 x 10^8 m/s) / 1.461 = 2.05 x 10^8 m/s. Therefore, the speed of light in this liquid is 2.05 x 10^8 m/s.

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• 18.

### A certain source of sound waves radiates uniformly in all directions. At a point P, which is a distance of 20 m from the source, the intensity level is 51 dB. What is the total power output of the source?. The threshold of hearing is I0 = 1 x 10−12W/m2

• A.

4.52 × 10^−4 W

• B.

6.33× 10^−4 W

• C.

8.36× 10^−4 W

• D.

3.16× 10^−4 W

B. 6.33× 10^−4 W
Explanation
The intensity level of sound is given by the equation L = 10log(I/I0), where L is the intensity level, I is the intensity of the sound wave, and I0 is the threshold of hearing. In this case, we are given the intensity level L = 51 dB and I0 = 1 x 10^-12 W/m^2. We can rearrange the equation to solve for I: I = I0 * 10^(L/10). Plugging in the given values, we get I = (1 x 10^-12) * 10^(51/10) = 6.33 x 10^-4 W/m^2. Since the sound waves radiate uniformly in all directions, the total power output of the source is equal to the intensity multiplied by the surface area of a sphere with a radius of 20 m. Therefore, the total power output is (6.33 x 10^-4 W/m^2) * (4π(20^2)) = 6.33 x 10^-4 W.

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• 19.

### A light ray initially in water (nwater = 1.33) enters a transparent medium at an angle of incidence of 370, and the transmitted ray is refracted at an angle of 250. Calculate the speed of light in the transparent medium.

• A.

2.58 × 10^8 m/s

• B.

1.58× 10^8 m/s

• C.

1.24× 10^8 m/s

• D.

2.24× 10^8 m/s

B. 1.58× 10^8 m/s
Explanation
The speed of light in a medium can be calculated using the formula v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium. In this question, the incident angle is given as 37° and the refracted angle is given as 25°. Using Snell's law, we can calculate the refractive index of the medium as n = sin(37°) / sin(25°) = 1.33. Plugging this value into the formula v = c/n, we can calculate the speed of light in the medium as v = c/1.33. Therefore, the correct answer is 1.58 × 10^8 m/s.

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• 20.

### Calculate the axoplasm resistance of a 1 cm long segment of myelinated axon of radius 2 μm if the axoplasm resistivity is 2 Ω.m.

• A.

1.59 x 10^9 Ω

• B.

2.5 x 10^8 Ω

• C.

3.18 x 10^9 Ω

• D.

2.6 x 10^7 Ω

A. 1.59 x 10^9 Ω
Explanation
The axoplasm resistance can be calculated using the formula R = (ρ * L) / A, where R is the resistance, ρ is the resistivity, L is the length of the segment, and A is the cross-sectional area of the segment. In this case, the length is given as 1 cm and the radius is given as 2 μm. To calculate the cross-sectional area, we use the formula A = π * r^2, where r is the radius. Plugging in the values, we get A = π * (2 μm)^2 = 4π μm^2. Converting the units to meters, we get A = 4π * 10^-12 m^2. Now we can plug in the values into the resistance formula: R = (2 Ω.m * 1 cm) / (4π * 10^-12 m^2). Simplifying this expression gives us R = 1.59 x 10^9 Ω.

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• 21.

### An unmyelinated segment of axon has a radius of 5 μm and a length of 1 cm. find its membrane capacitance if the capacitance per unit area of the membrane, cm = 1x10−2 Fm−2

• A.

3.1 x10^−12 F

• B.

3.1 x10^−9 F

• C.

3.1 x10^−6 F

• D.

3.1 x10^−8 F

B. 3.1 x10^−9 F
Explanation
The membrane capacitance of the unmyelinated segment of the axon can be calculated using the formula C = cm * A, where C is the capacitance, cm is the capacitance per unit area of the membrane, and A is the cross-sectional area of the axon. The cross-sectional area can be calculated using the formula A = π * r^2, where r is the radius of the axon.

Given that the radius is 5 μm (5 * 10^-6 m) and the length is 1 cm (0.01 m), we can calculate the cross-sectional area as A = π * (5 * 10^-6)^2 = 7.85 * 10^-10 m^2.

Substituting the values into the formula, C = (1 * 10^-2) * (7.85 * 10^-10) = 7.85 * 10^-12 F.

Therefore, the correct answer is 3.1 x 10^-12 F.

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• 22.

### What is the ratio (I2:I1) between the intensities of sound waves of two identical speakers if the difference in their sound levels is 3.0 dB. The threshold of hearing is I0 = 1 x 10−12W/m2.

• A.

3

• B.

2

• C.

0.5

• D.

4

B. 2
Explanation
The difference in sound levels of 3.0 dB corresponds to a ratio of intensities of 10^(3.0/10) = 2. Therefore, the ratio of the intensities of the sound waves between the two identical speakers is 2, meaning that the intensity of sound wave I2 is twice the intensity of sound wave I1.

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• 23.

### The frequency of a wave is doubled when the wavelength remains the same. What happens to the speed of the wave?

• A.

• B.

Remains unchanged

• C.

It doubles

• D.

It’s cut to one-half

C. It doubles
Explanation
When the frequency of a wave is doubled while the wavelength remains the same, the speed of the wave doubles as well. This is because the speed of a wave is determined by the product of its frequency and wavelength. Since the frequency has doubled, the speed of the wave also doubles to maintain the same wavelength.

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• 24.

### A ring weighs 6.327 x 10^-3 N in air and 6.033 x 10^-3 N when submerged in water (ρwater = 1000kg/m3). What is the density of the ring?

• A.

3.2 x 10^4 kg.m^-3

• B.

2.15x 10^4 kg.m^-3

• C.

4.3x 10^4 kg.m^-3

• D.

5.2x 10^4 kg.m^-3

B. 2.15x 10^4 kg.m^-3
Explanation
The density of an object can be calculated using the formula density = mass/volume. In this case, the mass of the ring is given by the weight divided by the acceleration due to gravity. The volume of the ring can be calculated by dividing the weight loss (difference in weight in air and water) by the weight of an equal volume of water. By substituting the given values into the formula, we can calculate the density of the ring to be 2.15x10^4 kg.m^-3.

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• 25.

### Calculate the gauge pressure at the eardrum of a diver at a depth of 1000 m in seawater (ρseawater = 1030 kg/m3). The atmospheric pressure is 1 atm = 1.013x10^5 Pa.

• A.

10.30 x10^6Pa

• B.

5.15 x10^6Pa

• C.

20.6 x10^6Pa

• D.

10.4 x10^6Pa

A. 10.30 x10^6Pa
Explanation
The gauge pressure at the eardrum of a diver at a depth of 1000 m in seawater can be calculated using the formula P = P₀ + ρgh, where P is the gauge pressure, P₀ is the atmospheric pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the atmospheric pressure is given as 1.013x10^5 Pa, the density of seawater is 1030 kg/m^3, and the depth is 1000 m. Plugging these values into the formula, we get P = 1.013x10^5 + (1030)(9.8)(1000) = 10.30 x10^6 Pa. Therefore, the correct answer is 10.30 x10^6 Pa.

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• 26.

### One end of a meter stick (متريه مسطره ) is pinned on a table. Two forces, F1, F2 are applied in such a way that the net torque is zero, if F2 = 6 N. find the magnitude of F1.

• A.

1.5 N

• B.

6 N

• C.

3 N

• D.

0.5 N

A. 1.5 N
Explanation
Since the net torque is zero, it means that the clockwise torque is equal to the counterclockwise torque. The torque is calculated by multiplying the force by the distance from the pivot point. Since one end of the meter stick is pinned on the table, the distance for F1 is 1 meter. Therefore, if F2 is 6 N, F1 must be half of that, which is 3 N.

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• 27.

### A vacuum-filled parallel plate capacitor of a plate area1 m^2. If its two plates are separated by 1 cm and the potential difference between them is 100 V. Determine the magnitude of the charge on each plate. Permittivity of free space is ε0 = 8.85 × 10^−12 C^2/N.m^2

• A.

8.84 x 10^−8 C

• B.

3.6x 10^−8 C

• C.

7.5 x 10^−6 C

• D.

2.7 x 10^−8 C

A. 8.84 x 10^−8 C
Explanation
The magnitude of the charge on each plate can be determined using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. In this case, the capacitance can be calculated using the formula C = ε0A/d, where ε0 is the permittivity of free space, A is the plate area, and d is the separation between the plates. Plugging in the given values, we get C = (8.85 x 10^-12 C^2/N.m^2)(1 m^2)/(0.01 m) = 8.85 x 10^-8 F. Then, using the formula Q = CV, we find Q = (8.85 x 10^-8 F)(100 V) = 8.85 x 10^-6 C. However, since there are two plates, the magnitude of the charge on each plate is half of this value, which is 8.84 x 10^-8 C.

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• 28.

### A bat can hear sound at frequencies up to 120000 Hz. The wavelength of sound in air at this frequency is

• A.

0.887 cm

• B.

0.287 cm

• C.

5.356 cm

• D.

6.757 cm

B. 0.287 cm
Explanation
The wavelength of sound in air can be calculated using the formula: wavelength = speed of sound / frequency. Since the frequency given is 120,000 Hz and the speed of sound in air is approximately 343 meters per second, we can calculate the wavelength as follows: wavelength = 343 m/s / 120,000 Hz. Converting the units, we get the wavelength as 0.00286 meters or 0.287 cm.

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• 29.

### A sphere of weigh 3.0 x 10^-3 N is suspended from a string. A steady horizontal force F pushes the sphere so that the string makes an angle θ = 370 with the vertical as shown in the figure. The magnitude of F is

• A.

4.3 x 10^-3 N

• B.

2.3  x 10^-3 N

• C.

1.5  x 10^-3 N

• D.

5.5  x 10^-3 N

B. 2.3  x 10^-3 N
Explanation
The magnitude of the force F can be determined using trigonometry. Since the string makes an angle of 37° with the vertical, we can use the sine function to find the vertical component of the force. The weight of the sphere (3.0 x 10^-3 N) is equal to the vertical component of the force. Therefore, the magnitude of F is equal to the weight divided by the sine of the angle, which is 2.3 x 10^-3 N.

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• 30.

### An office window has dimensions 3.4 m by 2.1 m. As a result of the passage of a storm, the air pressure outside the window drops to 0.96 atm, but the pressure inside the window is held at 1.0 atm. What net force pushes out on the window?

• A.

3.89 x 10^4 N

• B.

4.89 x 10^4 N

• C.

2.89 x 10^4 N

• D.

5.89 x 10^4 N

C. 2.89 x 10^4 N
Explanation
The net force that pushes out on the window can be calculated using the formula: Force = Pressure x Area. The pressure inside the window is 1.0 atm, which is equal to 1.013 x 10^5 N/m^2. The area of the window is 3.4 m x 2.1 m = 7.14 m^2. Plugging these values into the formula, we get: Force = (1.013 x 10^5 N/m^2) x (7.14 m^2) = 7.25 x 10^5 N. Therefore, the correct answer is 7.25 x 10^5 N, which is equivalent to 2.89 x 10^4 N.

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