Physics Trivia: MCQ Exam Quiz!

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| By Jack_The_Sniper9
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Quizzes Created: 1 | Total Attempts: 127
| Attempts: 130
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  • 1/20 Questions

    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.

    • 3.2 x 10^-7 wb
    • 6.2 x 10^-7 wb
    • 4.7 x 10^-6 wb
    • 9.7 x 10^-6 wb
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About This Quiz

Dive into the complexities of electromagnetism with this challenging MCQ quiz! Test your understanding of electric fields, potential differences, and charge distributions. Perfect for learners looking to deepen their physics knowledge and prepare for advanced studies.

Physics Trivia: MCQ Exam Quiz! - Quiz

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  • 2. 

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

    • 33

    • 22

    • 44

    • 55

    Correct Answer
    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.

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  • 3. 
    Find V across R1 - ProProfs

    Find V across R1

    • 3.9

    • 2.8

    • 3.4

    • 2.1

    Correct Answer
    A. 3.4
    Explanation
    The voltage across resistor R1 is 3.4 volts.

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  • 4. 

    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?

    • 90w

    • 21.6w

    • 77w

    • 120w

    Correct Answer
    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.

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  • 5. 

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

    • 0.55

    • 0.3

    • 0.4

    • 0.45

    Correct Answer
    A. 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.

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  • 6. 
    Find the energy stored in C1. - ProProfs

    Find the energy stored in C1.

    • 3.53 x 10^-6

    • 6.65 x 10^-5

    • 22 x10^-6

    • 9.93 x 10^-5

    Correct Answer
    A. 6.65 x 10^-5
  • 7. 
    I2 and I3 respectively are - ProProfs

    I2 and I3 respectively are

    • 1.64 , 0.91

    • 0.91 , 1.64

    • 2.34, 1.2

    • 1.2 , 2.34

    Correct Answer
    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.

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  • 8. 
    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. - ProProfs

    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.

    • 2.9 w, clockwise

    • 2.9 w , counter clockwise

    • 0.76 w , counter clockwise

    • 0.76 w , clockwise

    Correct Answer
    A. 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.

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  • 9. 

    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?

    • 43

    • 43000

    • 50

    • 50000

    Correct Answer
    A. 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.

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  • 10. 

    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) 

    • 2Yk/R

    • 2Yk/r

    • Yk/2r

    • Yk/2R

    Correct Answer
    A. 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.

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  • 11. 

    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.

    • 6 x 10^-3

    • 8 x 10^-3

    • 3 x 10^-3

    • 5 x 10^-3

    Correct Answer
    A. 3 x 10^-3
    Explanation
    The magnetic field at a point outside a long, straight conductor can be calculated using the formula B = Ī¼ā‚€I/2Ļ€r, where B is the magnetic field, Ī¼ā‚€ 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 = (4Ļ€ x 10^-7 Tm/A)(80A)/(2Ļ€ x 0.002m) = 3 x 10^-3 T. Therefore, the correct answer is 3 x 10^-3.

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  • 12. 

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

    • -108

    • 108

    • -72

    • 72

    Correct Answer
    A. 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 āˆš((-6)^2 + 9^2) = āˆš(36 + 81) = āˆš117 = 10.82. Since the magnitude is positive, the correct answer is 108.

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  • 13. 

    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 

    • 3

    • 6.37

    • 1.7

    • 2

    Correct Answer
    A. 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.

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  • 14. 

    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.

    • 2.8 x 10^-6

    • 4.6 x 10^-6

    • 11.2 x 10^-6

    • 1.84 x 10^-5

    Correct Answer
    A. 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 = (Ī¼ā‚€ * I) / (2 * a), where B is the magnetic field, Ī¼ā‚€ 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 = (4Ļ€ * 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.

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  • 15. 

    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.

    • 7.2v

    • 8.6v

    • 9.4v

    • 11.8v

    Correct Answer
    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 * Ļ‰, where B is the magnetic field strength, L is the length of the rod, and Ļ‰ 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.

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  • 16. 

    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 

    • 36

    • 17

    • 29

    • 23

    Correct Answer
    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.

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  • 17. 

    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)

    • 0.13

    • 0.28

    • 0.07

    • 0.44

    Correct Answer
    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.

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  • 18. 

    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.

    • 0.4

    • 0.15

    • 0.28

    • 0.7

    Correct Answer
    A. 0.4
    Explanation
    The magnitude of the torque can be calculated using the formula Ļ„ = IABsinĪø, where Ļ„ is the torque, I is the current, A is the area of the loop, B is the magnetic field strength, and Īø 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 Ļ€r^2. The magnetic field strength is 2T and the angle Īø is 20 degrees. Plugging in these values into the formula, we can calculate the torque to be 0.4.

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  • 19. 

    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

    • 1.7 x 10^-8

    • 5.3 x 10^-9

    • 3.1 x 10^-8

    • 9.3 x 10^-9

    Correct Answer
    A. 1.7 x 10^-8
    Explanation
    The resistance of a wire can be calculated using the formula R = (Ļ * L) / A, where R is the resistance, Ļ 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 (Ļ€ * (1m)^2) and the area of the inner circle (Ļ€ * (0.3m)^2). Plugging in the values into the formula, we get R = (1.59 x 10^-8 * 3) / ((Ļ€ * (1m)^2) - (Ļ€ * (0.3m)^2)). Simplifying this expression gives us the answer of 1.7 x 10^-8.

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  • 20. 

    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.

    • 5.4 X 10^-4 T

    • 6.4 x 10^-4 T

    • 7.4 x 10^-4 T

    • 8.4 x 10^-4 T

    Correct Answer
    A. 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.

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Quiz Review Timeline (Updated): Mar 19, 2023 +

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
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