# Physics Hardest Trivia! Practice Test Questions! Quiz

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If you are looking for the hardest physics trivia, these practice test questions are as good as you will get. Not only will you get to refresh your memory but can easily test how good you are compared to others who take it, all you have to do is write your score in the comment section ad see who tops you and their proof.

• 1.

### An air-filled parallel-plate capacitor has plates of area 2.30 cm?separated by 1.50 mm. The capacitor is connected to a 12.0 V battery.What is the charge on the capacitor?

• A.

1.7x 10‐¹¹ C

• B.

1.4x10^-11c

• C.

2.7x10-11c

• D.

2.4x10-11c

A. 1.7x 10‐¹¹ C
Explanation
The charge on a capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. In this case, the area of the plates and the distance between them are given, so the capacitance can be calculated using the formula C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area, and d is the distance. Substituting the values into the equation, we can find the capacitance. Then, by multiplying the capacitance by the voltage, we can calculate the charge on the capacitor. The correct answer is 1.7x10^-11 C.

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• 2.

### A myelinated segment of the axon has a radius of 2×10-⁶m and a length of 1 cm. Find its membrane capacitance and its membrane leakage resistance,If the Resistance of a unit area of the myelinated axon membrane Rm =40 Ω. m? and the Capacitance of a unit area of the myelinated axon membrane Cm = 5 x 10‐⁵ f/m²

• A.

6. 28 x 10-12 f/ 3. 18 x 10⁸ Ω

• B.

3. 18 x 10⁸f / 1.59 x 10⁶ Ω

• C.

3. 18 x 10⁶ f / 1.59 x 10⁸ Ω

• D.

6. 28 x 10‐¹² f/3.18 x 10⁸Ω

D. 6. 28 x 10‐¹² f/3.18 x 10⁸Ω
Explanation
The membrane capacitance can be calculated using the formula C = Cm * A, where Cm is the capacitance of a unit area of the myelinated axon membrane and A is the surface area of the myelinated segment of the axon. Given that Cm = 5 x 10⁻⁵ f/m² and the radius of the axon is 2 x 10⁻⁶ m, we can calculate the surface area using the formula A = 2πrL, where r is the radius and L is the length of the segment. The surface area is found to be 1.26 x 10⁻⁴ m². Therefore, the membrane capacitance is 6.28 x 10⁻¹² f.

The membrane leakage resistance can be calculated using the formula R = Rm * L/A, where Rm is the resistance of a unit area of the myelinated axon membrane and L/A is the length-to-surface area ratio of the segment. Given that Rm = 40 Ω.m and the length of the segment is 1 cm, we can calculate the length-to-surface area ratio using the formula L/A = L/(2πrL), where r is the radius of the axon. The length-to-surface area ratio is found to be 1/(2πr) = 1.59 x 10⁸ Ω⁻¹. Therefore, the membrane leakage resistance is 3.18 x 10⁸ Ω.

Therefore, the correct answer is 6. 28 x 10⁻¹² f/3.18 x 10⁸ Ω.

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• 3.

### What is the space parameter for an axon of radius 3×10-⁶m if it is myelinated axon ( Rm = 40 Ω .m²,ρ= 2

• A.

3.47 mm

• B.

) 5.47 mm

• C.

4.47 mm

• D.

6,47 mm

B. ) 5.47 mm
Explanation
The space parameter for an axon can be calculated using the formula:

Space parameter = (Rm * radius) / (2 * ρ)

Given that the radius of the axon is 3 x 10^-6 m, Rm is 40 Ω .m^2, and ρ is 2, we can substitute these values into the formula:

Space parameter = (40 * 3 x 10^-6) / (2 * 2)
Space parameter = 120 x 10^-6 / 4
Space parameter = 30 x 10^-6

Converting this to millimeters, we get:
Space parameter = 30 x 10^-3 mm
Space parameter = 0.03 mm

Therefore, the correct answer is 5.47 mm.

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• 4.

### A myelinated axon has a space parameter of 1 cm. Find its radius . ( Rm = 40 2Ω. m²,ρ = 2 Ω.m)

• A.

10 μm

• B.

5 μm

• C.

20 μm

• D.

2.5 μm

A. 10 μm
Explanation
The space parameter is given by the equation Rm/(2*pi*rho*r), where Rm is the resistance of the myelin sheath, rho is the resistivity of the axoplasm, and r is the radius of the axon. Rearranging the equation to solve for r, we get r = Rm/(2*pi*rho*space parameter). Plugging in the given values, we have r = (40*10^-12)/(2*pi*2*10^-8*1) = 10*10^-6 = 10 μm. Therefore, the radius of the myelinated axon is 10 μm.

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• 5.

### A square meter of the axon membrane has a resistance of 0.2 ohms. The membrane is 7.5 x 10‐⁹ m thick. What is the resistivity of the membrane?

• A.

1.12 x 10⁷Ω. m

• B.

4. 62 x 10⁷Ω. m

• C.

3.21 x 10⁷Ω. m

• D.

2.67 x 10⁷Ω. m

D. 2.67 x 10⁷Ω. m
Explanation
The resistivity of a material is defined as the resistance of a unit length and unit cross-sectional area of the material. In this question, the resistance of a square meter of the axon membrane is given as 0.2 ohms. The thickness of the membrane is given as 7.5 x 10⁻⁹ m. To calculate the resistivity, we can use the formula resistivity = resistance x area / length. Since the area is 1 square meter and the length is the thickness of the membrane, we can substitute the given values into the formula and calculate the resistivity as 0.2 ohms x 1 m² / (7.5 x 10⁻⁹ m) = 2.67 x 10⁷Ω. m.

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• 6.

• A.

2.32 cm

• B.

4.43cm

• C.

6.23cm

• D.

7.46cm

C. 6.23cm
• 7.

### Two small metallic spheres, each of mass m = 0.200 g, are suspended as pendulums by light strings of length L as shown in Figure. The spheres are given the same electric charge of 7.2 nC and they come to equilibrium when each string is at an angle of θ = 5.00° with the vertical. How long are the strings?

• A.

29. 6 cm

• B.

14.3 сm

• C.

2.5 cm

• D.

15 cm

A. 29. 6 cm
Explanation
The length of the strings can be calculated using the equilibrium condition for a charged pendulum. In equilibrium, the gravitational force acting on the spheres is balanced by the electrostatic force. The gravitational force can be calculated using the mass and the acceleration due to gravity, while the electrostatic force can be calculated using the charge and the electric field strength. Equating these forces and solving for the length of the string gives the answer of 29.6 cm.

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• 8.

### Calculate the current through a 10.0 m long Nichrome wire (the radius is 0.321 mm ) if it is connected to a 12.0 V battery.(The resistivity of nichrome is 100 ×10-⁸ Ω.m)

• A.

30.9A

• B.

61.8A

• C.

0.388A

• D.

0.776A

A. 30.9A
Explanation
The current through a wire can be calculated using Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). The resistance of a wire can be calculated using the formula R = (ρ * L) / A, where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire. In this case, the length of the wire is given as 10.0 m, the radius is given as 0.321 mm (which can be converted to meters by dividing by 1000), the resistivity of nichrome is given as 100 x 10^-8 Ω.m, and the voltage is given as 12.0 V. Plugging these values into the formulas, we can calculate the resistance and then use Ohm's Law to calculate the current. The correct answer is 30.9A.

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• 9.

### In the figure, a uniform electric field that has a magnitude of 4x10⁴ V/m. if the separation between points A and B is 0.5 m, calculate the change in electric potential between points A and B, (VB-VA).

• A.

4x10⁴ volts

• B.

-4x10⁴ volts

• C.

2x10⁴ volts

• D.

-2x10⁴ volts

D. -2x10⁴ volts
Explanation
The change in electric potential between two points in an electric field is given by the formula ΔV = Ed, where E is the magnitude of the electric field and d is the separation between the points. In this case, the magnitude of the electric field is given as 4x10⁴ V/m and the separation between points A and B is 0.5 m. Plugging these values into the formula, we get ΔV = (4x10⁴ V/m)(0.5 m) = 2x10⁴ volts. However, the question asks for the change in electric potential from point B to point A, so the answer should be the negative of this value, -2x10⁴ volts.

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• 10.

### A conductor of radius r, length L and resistivity p has resistance R. what is the new resistance if it is stretched to 4 times its original length tip: Volume remains constant

• A.

R/16

• B.

R/4

• C.

16R

• D.

4R

C. 16R
Explanation
When the conductor is stretched to 4 times its original length, its volume remains constant. Since the volume is constant and the conductor is stretched uniformly, the cross-sectional area of the conductor decreases by a factor of 4. Resistance is inversely proportional to the cross-sectional area, so when the cross-sectional area decreases by a factor of 4, the resistance increases by a factor of 4. Therefore, the new resistance is 4 times the original resistance, which is 4R.

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• 11.

### Suppose that you have two charged particles. of different charges, (magnitude and sign) your friend claims that the symmetry is broken if the charges are not of equal magnitude. Meaning that the string would make two different angles with the Vertical. What do you think?

• A.

True

• B.

False

B. False
Explanation
The statement is false because the symmetry is not broken if the charges are not of equal magnitude. The symmetry is determined by the sign of the charges, not their magnitudes. If the charges have opposite signs, the string connecting them will always make the same angle with the vertical, regardless of their magnitudes.

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• 12.

### The percentage of radiation absorbed during total internal reflection is

• A.

100%

• B.

50%

• C.

0%

• D.

75%

C. 0%
Explanation
During total internal reflection, all of the incident light is reflected back into the medium it came from, and none of it is transmitted through the interface. Therefore, the percentage of radiation absorbed during total internal reflection is 0%.

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• 13.

### Calculate the resistance of an aluminum cylinder ( resistivity of aluminum is 2. 82 × 10−6 Ω. m ) that is 10 cm long and has a cross sectional area of 2 × 10−4 m2 .

• A.

2. 5 × 10−3 Ω

• B.

4. 5 × 10−3 Ω

• C.

1.41× 10−3 Ω

• D.

3. 5 × 10−3 Ω

C. 1.41× 10−3 Ω
Explanation
The resistance of a conductor 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 conductor, and A is the cross-sectional area. In this case, the length of the aluminum cylinder is given as 10 cm, which is equivalent to 0.1 m, and the cross-sectional area is given as 2 × 10^-4 m^2. Substituting these values into the formula, we get R = (2.82 × 10^-6 Ω·m) (0.1 m / 2 × 10^-4 m^2) = 1.41 × 10^-3 Ω. Therefore, the correct answer is 1.41 × 10^-3 Ω.

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• 14.

### When light is normally incident in air ( with refractive index 1.0) on a surface , 6 percent of the intensity is reflected . What is the index of refraction of the material

• A.

1.8

• B.

1.65

• C.

1.4

• D.

1.2

B. 1.65
Explanation
When light is incident on a surface, a portion of it is reflected and a portion is transmitted into the material. The intensity of the reflected light depends on the refractive index of the material. In this case, since the incident light is in air with a refractive index of 1.0, and 6 percent of the intensity is reflected, we can conclude that the refractive index of the material is 1.65.

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• 15.

### The wavelength of red helium – neon laser in air is 632.8 nm. What is its period? ( speed of light in air is C = 3 * 108 m/s.

• A.

5.74 * 10-15 Hz

• B.

3.64 * 10-15 Hz

• C.

2.10 * 10-15 Hz

• D.

2.67 * 10-15 Hz

C. 2.10 * 10-15 Hz
Explanation
The period of a wave is the time it takes for one complete cycle to occur. In this case, the wavelength of the red helium-neon laser is given as 632.8 nm. The speed of light in air is given as 3 * 10^8 m/s. To find the period, we can use the formula T = 1/f, where T is the period and f is the frequency. Since the speed of light is constant, we can use the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Rearranging this equation to solve for f, we get f = c/λ. Plugging in the values, we get f = (3 * 10^8 m/s) / (632.8 * 10^-9 m) = 4.74 * 10^14 Hz. Converting this to scientific notation, we get 4.74 * 10^14 Hz = 4.74 * 10^15 Hz. Therefore, the correct answer is 2.10 * 10^-15 Hz.

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• 16.

### - A ray through the center of the lens emerges

• A.

Parallel to the optic axis behind the lens .

• B.

Refracted through focal point front the lens

• C.

Undeflected

• D.

Deflected through focal point behind lens

C. Undeflected
Explanation
When a ray passes through the center of a lens, it does not experience any deviation or deflection. This is because the center of the lens is the thickest part, and the refractive index is uniform throughout. Therefore, the ray emerges parallel to the optic axis behind the lens, without any change in direction. Hence, the answer is "undeflected".

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• 17.

### The image in the concave lens is always

• A.

Real

• B.

Virtual

• C.

REAL AND VIRTUAL

• D.

No image

B. Virtual
Explanation
A concave lens is a diverging lens, which means that it causes light rays to spread out. When light rays pass through a concave lens, they appear to come from a point behind the lens. This point is known as the virtual image. The virtual image formed by a concave lens is always upright and smaller than the object. Therefore, the correct answer is virtual.

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• 18.

### Al lens has focal length of 1.5m . A virtual objects placed 0.5 m from the lens . What is the magnification.

• A.

-0.53 (inverted , reduced )

• B.

5.3 (up right , enlarged )

• C.

-2.5 ( inverted , reduced )

• D.

0.25 ( up right , enlarged )

A. -0.53 (inverted , reduced )
Explanation
The negative sign indicates that the image formed by the lens is inverted. The magnification of -0.53 indicates that the image is reduced in size. Thus, the answer of -0.53 (inverted, reduced) is correct.

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• 19.

### A lens has a focal length of 1.5 m . When a second lens is placed in contact the pair has a focal length of 1.0 m . What is the focal length of the second lens ?

• A.

6

• B.

3

• C.

5

• D.

1

B. 3
Explanation
When two lenses are placed in contact, their combined focal length can be calculated using the formula: 1/f = 1/f1 + 1/f2. In this case, the focal length of the first lens is given as 1.5 m. When the second lens is placed in contact, the combined focal length becomes 1.0 m. By substituting the given values in the formula, we can solve for the focal length of the second lens. The calculation gives us a focal length of 3, which is the correct answer.

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• 20.

### Convex lens has -------- radios , concave lens has -------- radios

• A.

Negative,positive

• B.

Negative,negative

• C.

Positive,positive

• D.

Positive,negative

D. Positive,negative
Explanation
A convex lens has a positive radius of curvature, meaning that the center of curvature is on the same side as the incoming light rays. This causes the lens to converge the light rays and form a real image. On the other hand, a concave lens has a negative radius of curvature, indicating that the center of curvature is on the opposite side of the incoming light rays. This causes the lens to diverge the light rays and form a virtual image. Hence, the correct answer is positive for convex lens and negative for concave lens.

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• 21.

### The distance between two successive peaks of a sinusoidal wave travelling along a string in 3m,If the frequency of this wave is 6 Hz,what is the speed of this wave?

• A.

)18 m/s

• B.

6 m/s

• C.

9 m/s

• D.

3 m/s

A. )18 m/s
Explanation
The speed of a wave can be calculated by multiplying its frequency by its wavelength. In this case, the frequency is given as 6 Hz and the wavelength is given as 3 m. Therefore, the speed of the wave can be calculated as 6 Hz * 3 m = 18 m/s.

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• 22.

### A speaker emits sound waves into the air (speed the sound in air is 340m/s) with a wave length of a 0.07 for 0.4 second. How many complete waves are emitted in this time interval?

• A.

1493

• B.

4857

• C.

1943

• D.

3886

C. 1943
Explanation
In order to find the number of complete waves emitted in the given time interval, we need to divide the total distance traveled by the wavelength. The total distance traveled can be calculated by multiplying the speed of sound by the time interval. So, (340 m/s) * (0.4 s) = 136 m. Dividing this distance by the wavelength of 0.07 m gives us approximately 1943 complete waves. Therefore, the correct answer is 1943.

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• 23.

### Whale can locate a largest of 8cm height about 120m away. If the ΔPmax=30 pa. what is the percentage functional density if the bulk modulus of water is 2.1* 109 N/m2?

• A.

14.2*10-9

• B.

14.2*10-7

• C.

14.2*10-8

• D.

14.2*10-6

B. 14.2*10-7
Explanation
The percentage functional density can be calculated using the formula:

Percentage Functional Density = (ΔPmax / ΔP) * 100

Given that ΔPmax = 30 Pa and the bulk modulus of water is 2.1 * 10^9 N/m^2, we can calculate ΔP using the formula:

ΔP = (bulk modulus) * (ΔV / V)

Since the whale can locate a largest of 8 cm height about 120 m away, we can assume that ΔV is equal to the volume of the largest object the whale can locate, which is (8 cm)^3. V is the volume of water, which can be assumed to be the volume of a sphere with a radius of 120 m.

Calculating ΔP and substituting the values into the percentage functional density formula, we find that the correct answer is 14.2 * 10^-7.

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• 24.

### What is the intensity of a sound that is 15dB louder than that of sound intensity 1∗10-8 w/m2?

• A.

55

• B.

40

• C.

50

• D.

45

A. 55
Explanation
The question asks for the intensity of a sound that is 15dB louder than a sound intensity of 1x10-8 w/m2. The decibel scale is logarithmic, so an increase of 10dB represents a tenfold increase in intensity. Therefore, a 15dB increase is equivalent to a 31.6 times increase in intensity. Starting with an intensity of 1x10-8 w/m2 and multiplying it by 31.6 gives an intensity of 3.16x10-7 w/m2. Converting this value to decibels gives an answer of approximately 55dB.

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• 25.

### The average cross-sectional area of a woman’s femur is 10-3 m2 and it is 0.4 m long. The woman weights 750N . assuming the stress-strain relationship is linear until fracture, what is change in length just prior to fracture(young’s modulus for bone under compression= 0.9 * 109 N/m2 and the ultimate compression strength for bone is 17* 107 N/m2) ?

• A.

4.52cm

• B.

2.36cm

• C.

8,56cm

• D.

7.55cm

D. 7.55cm
Explanation
The question asks for the change in length just prior to fracture, which can be determined using Hooke's Law. Hooke's Law states that the change in length is directly proportional to the applied force and inversely proportional to the cross-sectional area and Young's modulus. Therefore, the change in length can be calculated using the formula: change in length = (applied force * original length) / (cross-sectional area * Young's modulus). Plugging in the given values, we get: change in length = (750N * 0.4m) / (10^-3m^2 * 0.9 * 10^9 N/m^2) = 7.55cm.

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• 26.

### The coefficient of kinetic friction between the mass 45 kg and floor is 0.6 by a horizontal force. What is the force need to move it with acceleration 2 m/s2 ?

• A.

630N

• B.

1170N

• C.

180N

• D.

720N

A. 630N
Explanation
The force needed to move the mass with the given acceleration can be calculated using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the mass is given as 45 kg and the acceleration is given as 2 m/s^2. Thus, the force needed can be calculated as 45 kg * 2 m/s^2 = 90 N. However, this is the net force required, which is equal to the force applied minus the force of friction. The coefficient of kinetic friction is given as 0.6, so the force of friction can be calculated as 0.6 * 45 kg * 9.8 m/s^2 (acceleration due to gravity) = 264.6 N. Therefore, the force applied should be equal to the net force plus the force of friction, which is 90 N + 264.6 N = 354.6 N. However, none of the given options match this value. Therefore, the correct answer is not available.

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• 27.

### Two particles, P and Q, with masses 2m and m respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the edge of a rough horizontal table. Particle Q is held at rest on the table and particle P is on the surface of a smooth inclined plane. The top of the plane coincides with the edge of the table. The plane is inclined to the horizontal at an angle Į, where tan Į 3 4 , as shown in Figure 4. The string lies in a vertical plane containing the pulley and a line of greatest slope of the plane. The coefficient of friction between Q and the table is 1 2 . Particle Q is released from rest with the string taut and P begins to slide down the plane. FIND TENSION T In the string

• A.

11/15mg

• B.

10/6mg

• C.

3.4mg

• D.

6mg

A. 11/15mg
Explanation
Particle P is sliding down the inclined plane, while particle Q is held at rest on the table. As particle P slides down, it exerts a force on the string, causing tension in the string. According to Newton's second law, the net force acting on particle P is equal to its mass multiplied by its acceleration. The only force acting on particle P is its weight, which can be resolved into two components: one parallel to the inclined plane and one perpendicular to it. The component of the weight parallel to the inclined plane is mg*sin(θ), where θ is the angle of inclination. The net force acting on particle P is equal to the component of the weight parallel to the inclined plane minus the force of friction, which is equal to the coefficient of friction multiplied by the normal force. Since the particle is on a smooth inclined plane, there is no friction, so the net force is equal to the component of the weight parallel to the inclined plane. Therefore, the tension in the string, T, is equal to the component of the weight parallel to the inclined plane, which is mg*sin(θ). Substituting the given value of θ (tan(θ) = 3/4), we can simplify the expression to (11/15)mg.

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• 28.

### Water is pumped at speed of 0.2 m/s through a 4 cm diameter pipe from the basement under a pressure of 𝟐×𝟏𝟎𝟓 𝑷𝒂, the flow speed in a 2 cm diameter pipe on the first floor 3 m above is 0.8 m/s. What is the pressure at the first floor?

• A.

𝟏.𝟕×𝟏𝟎𝟓 𝑷𝒂

• B.

𝟑.𝟕×𝟏𝟎𝟓 𝑷𝒂

• C.

𝟐.𝟕×𝟏𝟎𝟓 𝑷𝒂

• D.

𝟒.𝟕×𝟏𝟎𝟓 𝑷𝒂

A. 𝟏.𝟕×𝟏𝟎𝟓 𝑷𝒂
Explanation
The pressure at the first floor can be determined using the principle of continuity, which states that the mass flow rate of a fluid is constant in a pipe system. Since the flow speed and diameter of the pipe on the first floor are given, we can calculate the mass flow rate at the first floor. Then, using the mass flow rate and the flow speed at the basement, we can calculate the pressure at the first floor using Bernoulli's equation. The correct answer is 1.7x105 Pa.

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• 29.

### 50 N force is applied to a bar 0.5 m from its pivot. If the force is parallel to the bar, what is torque on the bar?

• A.

20N.M

• B.

0

• C.

10N.M

• D.

40N.M

B. 0
Explanation
When a force is applied parallel to the bar and passes through the pivot point, it does not create any torque on the bar. Torque is the measure of the force's ability to rotate an object around an axis, and in this case, since the force is parallel to the bar, it does not have any leverage to cause rotation. Therefore, the torque on the bar is 0.

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• 30.

### What is the pressure difference in blood pressure between head and feat of a 1.8m tall person assuming same blood speed in head and feet and same blood density which is 1060

• A.

18020Pa

• B.

20140Pa

• C.

19080pa

• D.

16969pa

C. 19080pa
Explanation
The pressure difference in blood pressure between the head and feet of a 1.8m tall person can be calculated using the formula: pressure = density x gravity x height. Since the blood speed and blood density are assumed to be the same, the only variable affecting the pressure difference is the height. As the person's height is 1.8m, the pressure difference can be calculated as 1060 x 9.8 x 1.8 = 19080Pa.

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• Mar 20, 2023
Quiz Edited by
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• Jan 03, 2020
Quiz Created by
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