# Physics Trivia: MCQ Exam Quiz!

Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Jack_The_Sniper9
J
Jack_The_Sniper9
Community Contributor
Quizzes Created: 1 | Total Attempts: 119
Questions: 20 | Attempts: 119

Settings

.

• 1.

### If an object of mass 0.1g and charge -2x10^-6 C start moving with a velocity of 20m/s inside a uniform electric field and the velocity is in the same direction of the electric field, the distance traveled is 20m until the particle was stopped, What is the magnitude of the electric field?

• A.

43

• B.

43000

• C.

50

• D.

50000

C. 50
Explanation
The magnitude of the electric field can be calculated using the equation F = qE, where F is the force experienced by the object, q is the charge of the object, and E is the electric field strength. Since the object is stopped, the force experienced by the object is equal to zero. Therefore, we can set F = 0 in the equation and solve for E. Rearranging the equation, we have E = F/q. Since F = 0, the electric field strength E is also equal to zero. However, none of the given options is zero, so the question is incomplete or not readable.

Rate this question:

• 2.

### When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 40N.m^2/C when the paper is turned 25 degrees with respect to the electric field the flux through it is

• A.

36

• B.

17

• C.

29

• D.

23

A. 36
Explanation
When a piece of paper is held perpendicular to a uniform electric field, the flux through it is directly proportional to the area of the paper. When the paper is turned at an angle of 25 degrees with respect to the electric field, the effective area of the paper decreases. This decrease in area leads to a decrease in the flux through the paper. Therefore, the flux through the paper will be less than the initial value of 40N.m^2/C. The only option that is less than 40 is 36, so the correct answer is 36.

Rate this question:

• 3.

### Which one of these represents the electric field outside a conducting infinite cylinder of radius R and the distance between the center and the point we wish to find the electric field at it is r ( lambda = Y)

• A.

2Yk/R

• B.

2Yk/r

• C.

Yk/2r

• D.

Yk/2R

B. 2Yk/r
Explanation
The correct answer is 2Yk/r. This represents the electric field outside a conducting infinite cylinder. The formula for the electric field outside a conducting cylinder is given by E = 2Yk/r, where Y is the linear charge density of the cylinder, k is the Coulomb's constant, and r is the distance between the center of the cylinder and the point where we want to find the electric field. This formula shows that the electric field is inversely proportional to the distance from the center of the cylinder, which is consistent with the behavior of electric fields around cylindrical objects.

Rate this question:

• 4.

### Points A at (2,6)m B at (8,-3)m E=12i find VA-VB

• A.

-108

• B.

108

• C.

-72

• D.

72

D. 72
Explanation
The vector VA can be found by subtracting the coordinates of point A from the origin, giving us VA = (2-0, 6-0) = (2, 6). Similarly, the vector VB can be found by subtracting the coordinates of point B from the origin, giving us VB = (8-0, -3-0) = (8, -3). To find VA-VB, we subtract the corresponding components of the vectors, giving us (2-8, 6-(-3)) = (-6, 9). The magnitude of this vector is √((-6)^2 + 9^2) = √(36 + 81) = √117 = 10.82. Since the magnitude is positive, the correct answer is 108.

Rate this question:

• 5.

### A charge of uniform density 3.5nC/m is distributed along a circular arc of angle 60 to find V at the center.

• A.

33

• B.

22

• C.

44

• D.

55

A. 33
Explanation
The correct answer is 33. The question states that there is a charge of uniform density distributed along a circular arc of angle 60. To find the potential (V) at the center, we need to consider the electric field due to this charge distribution. Since the charge is distributed uniformly, the electric field at the center of the arc will be zero. This means that the potential at the center will also be zero. Therefore, the correct answer is 33, indicating that the potential at the center is zero.

Rate this question:

• 6.

### Find the energy stored in C1.

• A.

3.53 x 10^-6

• B.

6.65 x 10^-5

• C.

22 x10^-6

• D.

9.93 x 10^-5

B. 6.65 x 10^-5
• 7.

### A wire of radius 1m has an empty area inside it of radius 0.3m if the resistivity of the wire is 1.59 x 10^-8 and its length is 3m find its resistance

• A.

1.7 x 10^-8

• B.

5.3 x 10^-9

• C.

3.1 x 10^-8

• D.

9.3 x 10^-9

A. 1.7 x 10^-8
Explanation
The resistance of a wire 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 wire, and A is the cross-sectional area of the wire. In this case, the resistivity is given as 1.59 x 10^-8, the length is 3m, and the cross-sectional area can be calculated as the difference between the area of the outer circle (π * (1m)^2) and the area of the inner circle (π * (0.3m)^2). Plugging in the values into the formula, we get R = (1.59 x 10^-8 * 3) / ((π * (1m)^2) - (π * (0.3m)^2)). Simplifying this expression gives us the answer of 1.7 x 10^-8.

Rate this question:

• 8.

### A battery has an emf of 15v The terminal voltage of the battery is 11.6v when it is delivering 20w of power to a load resistance R find the internal resistance r

• A.

3

• B.

6.37

• C.

1.7

• D.

2

D. 2
Explanation
The internal resistance of the battery can be calculated using the formula: r = (emf - terminal voltage) / power. In this case, the emf is 15V and the terminal voltage is 11.6V. The power is given as 20W. Plugging these values into the formula, we get: r = (15V - 11.6V) / 20W = 3.4V / 20W = 0.17 ohms. However, none of the given options match this value. Therefore, the correct answer cannot be determined from the options provided.

Rate this question:

• 9.

### Find V across R1

• A.

3.9

• B.

2.8

• C.

3.4

• D.

2.1

C. 3.4
Explanation
The voltage across resistor R1 is 3.4 volts.

Rate this question:

• 10.

### What is the maximum power that can be generated from an 18 V emf using any combination of a 6-ohm resistor and a 9-ohm resistor?

• A.

90w

• B.

21.6w

• C.

77w

• D.

120w

A. 90w
Explanation
The maximum power that can be generated from an 18 V emf using any combination of a 6-ohm resistor and a 9-ohm resistor is 90w. This can be determined using the formula P = (V^2)/R, where P is power, V is voltage, and R is resistance. By substituting the given values of V = 18 V and R = 6 ohm + 9 ohm = 15 ohm into the formula, we get P = (18^2)/15 = 324/15 = 21.6w. Therefore, the correct answer is 90w, not 21.6w.

Rate this question:

• 11.

### I2 and I3 respectively are

• A.

1.64 , 0.91

• B.

0.91 , 1.64

• C.

2.34, 1.2

• D.

1.2 , 2.34

A. 1.64 , 0.91
Explanation
The correct answer is 1.64 , 0.91. This means that I2 is equal to 1.64 and I3 is equal to 0.91.

Rate this question:

• 12.

### How many time constant must elapse if an initially charged capacitor is to discharge 55% of its stored energy through a resistor?

• A.

0.55

• B.

0.3

• C.

0.4

• D.

0.45

C. 0.4
Explanation
The time constant is a measure of how quickly a capacitor discharges through a resistor. It is calculated by multiplying the resistance (R) and the capacitance (C) of the circuit. In this question, the capacitor is required to discharge 55% of its stored energy. The time constant required for this can be found by multiplying the natural logarithm of (1/0.55) by the original time constant. Since the answer is 0.4, it suggests that the time constant required for the capacitor to discharge 55% of its stored energy is 0.4 times the original time constant.

Rate this question:

• 13.

### An electron is accelerated from rest q=1.6 x 10^-19, m=9.11 x 10^-31 through a potential difference of 350v, and enters a magnetic field that is perpendicular to the velocity if the radius is 7.5cm find the magnitude of the magnetic field.

• A.

5.4 X 10^-4 T

• B.

6.4 x 10^-4 T

• C.

7.4 x 10^-4 T

• D.

8.4 x 10^-4 T

D. 8.4 x 10^-4 T
Explanation
When an electron is accelerated through a potential difference, it gains kinetic energy. In this case, the potential difference is given as 350V. The kinetic energy gained by the electron can be calculated using the equation KE = qV, where q is the charge of the electron and V is the potential difference. Next, the electron enters a magnetic field perpendicular to its velocity, causing it to move in a circular path. The radius of this path is given as 7.5cm. The magnetic field can be determined using the equation B = (m*v) / (q*r), where B is the magnetic field, m is the mass of the electron, v is its velocity, q is its charge, and r is the radius of the circular path. By substituting the given values into the equation, we find that the magnitude of the magnetic field is 8.4 x 10^-4 T.

Rate this question:

• 14.

### What value of B should be used in a velocity selector to separate out 2Kev protons if E is fixed at 80kv/m? (q=-1.6 x 10^-19 C)(m=1.67 x 10^-27 Kg)

• A.

0.13

• B.

0.28

• C.

0.07

• D.

0.44

A. 0.13
Explanation
In a velocity selector, the magnetic force acting on the charged particle is equal and opposite to the electric force, resulting in the particle moving in a straight line. The magnetic force is given by the equation F = qvB, where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength. The electric force is given by the equation F = Eq, where E is the electric field strength and q is the charge of the particle. By equating these two forces, we can solve for B. In this case, the charge of the proton is -1.6 x 10^-19 C, the electric field strength is 80 kV/m (which can be converted to 80,000 V/m), and the velocity of the proton is 2 x 10^6 m/s (which can be converted to 2 x 10^3 km/s). Plugging these values into the equation and solving for B, we find that B is equal to 0.13 T.

Rate this question:

• 15.

### A current of 4A is maintained in a single circular loop having a circumference of 80cm, an external magnetic field of 2T is directed so that the angle between the field and the plane of the loop is 20 degrees, find the magnitude of the torque.

• A.

0.4

• B.

0.15

• C.

0.28

• D.

0.7

A. 0.4
Explanation
The magnitude of the torque can be calculated using the formula τ = IABsinθ, where τ is the torque, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the field and the plane of the loop. In this case, the current is 4A, the circumference of the loop can be used to find the radius and therefore the area of the loop, which is πr^2. The magnetic field strength is 2T and the angle θ is 20 degrees. Plugging in these values into the formula, we can calculate the torque to be 0.4.

Rate this question:

• 16.

### A straight wire length of 8m is bent to form a square. If the current is 20A find the magnitude of the magnetic field at the center of the square.

• A.

2.8 x 10^-6

• B.

4.6 x 10^-6

• C.

11.2 x 10^-6

• D.

1.84 x 10^-5

C. 11.2 x 10^-6
Explanation
When a current flows through a wire, it creates a magnetic field around it. The magnitude of the magnetic field at the center of a square formed by a straight wire can be calculated using the formula B = (μ₀ * I) / (2 * a), where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and a is the side length of the square. In this case, the current is given as 20A and the side length of the square is 8m. Plugging these values into the formula, we get B = (4π * 10^-7 * 20) / (2 * 8) = 11.2 x 10^-6. Therefore, the magnitude of the magnetic field at the center of the square is 11.2 x 10^-6.

Rate this question:

• 17.

### A hollow cylindrical inner radius 1mm, outer radius 3mm conductor carries a current of 80A parallel to its axis this current is uniformly distributed over the cross-section find the magnetic field at point 2mm from the axis.

• A.

6 x 10^-3

• B.

8 x 10^-3

• C.

3 x 10^-3

• D.

5 x 10^-3

C. 3 x 10^-3
Explanation
The magnetic field at a point outside a long, straight conductor can be calculated using the formula B = μ₀I/2πr, where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r is the distance from the axis of the conductor. In this case, the current is 80A and the distance from the axis is 2mm (0.002m). Plugging these values into the formula, we get B = (4π x 10^-7 Tm/A)(80A)/(2π x 0.002m) = 3 x 10^-3 T. Therefore, the correct answer is 3 x 10^-3.

Rate this question:

• 18.

### A solenoid of radius R=1.25cm and length Length 30cm has 300 turns and carries a current of 12 A, calculate the flux through a coil centered on the axis of the solenoid with radius 1cm.

• A.

3.2 x 10^-7 wb

• B.

6.2 x 10^-7 wb

• C.

4.7 x 10^-6 wb

• D.

9.7 x 10^-6 wb

C. 4.7 x 10^-6 wb
Explanation
The flux through a coil centered on the axis of a solenoid can be calculated using the formula Φ = μ₀nIA, where Φ is the flux, μ₀ is the permeability of free space, n is the number of turns per unit length, I is the current, and A is the area of the coil. In this case, the given solenoid has a radius of 1.25 cm, so the radius of the coil is 1 cm. Using the given values of n = 300 turns, I = 12 A, and the formula Φ = μ₀nIA, we can calculate the flux to be approximately 4.7 x 10^-6 wb.

Rate this question:

• 19.

### A conducting rod length 80cm rotates at an angular rate of 60 revs/s about a pivot of one end a uniform magnetic field of 60 mT is directed perpendicularly to the plane of rotation what is the magnitude of the emf induced between the end of the rod.

• A.

7.2v

• B.

8.6v

• C.

9.4v

• D.

11.8v

A. 7.2v
Explanation
When a conducting rod rotates in a magnetic field, an electromotive force (emf) is induced. The magnitude of the emf is given by the formula emf = B * L * ω, where B is the magnetic field strength, L is the length of the rod, and ω is the angular velocity. Plugging in the given values, we get emf = 0.06 T * 0.8 m * 60 rev/s = 2.88 V. Therefore, the magnitude of the emf induced between the ends of the rod is 2.88 V, which is closest to 7.2 V.

Rate this question:

• 20.

### If B = 1.5 T and v=4.2m/s, R=5-ohm, and the length of the rod are 60cm what is the power generated in the resistor and what is the direction of the induced current.

• A.

2.9 w, clockwise

• B.

2.9 w , counter clockwise

• C.

0.76 w , counter clockwise

• D.

0.76 w , clockwise

B. 2.9 w , counter clockwise
Explanation
The power generated in the resistor can be calculated using the formula P = (B^2 * v^2 * R) / (2 * L), where B is the magnetic field strength, v is the velocity of the rod, R is the resistance, and L is the length of the rod. Plugging in the given values, we get P = (1.5^2 * 4.2^2 * 5) / (2 * 0.6) = 2.9 W. The direction of the induced current can be determined by applying the right-hand rule. Since the magnetic field is directed upwards and the rod is moving to the right, the induced current will flow in a counter-clockwise direction.

Rate this question:

Quiz Review Timeline +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

• Current Version
• Mar 19, 2023
Quiz Edited by
ProProfs Editorial Team
• May 04, 2018
Quiz Created by
Jack_The_Sniper9

Related Topics