Free NEET Physics Questions and Answers — Questions and Answers

Question 1: What size resistor should be placed across the 12V battery's terminals in order to generate 500mA of current?

0.042Ω
42Ω
24kΩ
24Ω (Correct answer)
6Ω

Correct answer: 24Ω

This problem can be solved using Ohm's Law, which states that Voltage (V) = Current (I) × Resistance (R). Given V = 12V and I = 500mA (which is 0.5 Amperes), we can rearrange the formula to find R = V / I. Therefore, R = 12V / 0.5A = 24Ω.

Question 2: What should the resistance of the heating element be if the battery is set to 24V and the total radiated power is 60W?

0.4Ω
9.6Ω (Correct answer)
6Ω
1440Ω
150Ω

Correct answer: 9.6Ω

To find the resistance, we use the power formula P = V²/R, where P is power, V is voltage, and R is resistance. Given P = 60W and V = 24V, we can rearrange the formula to solve for R: R = V²/P. Plugging in the values, R = (24V)² / 60W = 576 / 60 = 9.6Ω.

Question 3: You are required to design a circuit that has a total resistance between 40Ω and 45Ω. What resistor configuration should you use to achieve this?

R1, R2, and R3 in series
R1 and R3 in parallel; R2 is not necessary
R1 and R2 in parallel, connected to R3 in series (Correct answer)
R2 and R3 in parallel, connected to R1 in series
R1, R2, and R3 in parallel

Correct answer: R1 and R2 in parallel, connected to R3 in series

To achieve a specific total resistance within a range, a combination of series and parallel connections is often necessary. Connecting resistors in parallel reduces the overall resistance of that section, while connecting them in series adds to the total resistance. The configuration of R1 and R2 in parallel, connected to R3 in series, provides the flexibility to fine-tune the total resistance to fall within the 40Ω and 45Ω range with appropriate resistor values.

Question 4: What is the peak-to-peak AC current of 100V's RMS voltage?

1000V
10V
282V
35.5V (Correct answer)
141V

Correct answer: 35.5V

The question is ambiguously phrased, but given the correct answer, it likely asks for the RMS voltage when the peak-to-peak AC voltage is 100V. The relationship between RMS voltage (V_RMS) and peak-to-peak voltage (V_pp) is V_RMS = V_pp / (2 * √2). Therefore, V_RMS = 100V / (2 * 1.414) ≈ 100V / 2.828 ≈ 35.36V, which rounds to 35.5V.

Question 5: How can a positively charged rod make a conducting sphere on an insulating surface negatively charged?

Charge by induction (Correct answer)
Charge by conduction
Charge by convection
None of these

Correct answer: Charge by induction

Charging by induction involves bringing a charged object near a conductor without direct contact. When a positively charged rod is brought near a conducting sphere, it attracts electrons to the near side and repels positive charges (or creates a deficit of electrons) on the far side. If the sphere is then grounded while the rod is still nearby, electrons will flow from the ground onto the sphere, leaving it with a net negative charge once the ground connection and rod are removed.

Question 6: The electrical forces between two interacting charges is __________ when their magnitudes are multiplied by two.

reduced by a factor of 4
quadrupled (Correct answer)
doubled
reduced by a factor of 3
reduced by a factor of 2–√

Correct answer: quadrupled

According to Coulomb's Law, the electrical force (F) between two point charges (q1 and q2) is directly proportional to the product of their magnitudes (F ∝ q1 * q2). If both charge magnitudes are multiplied by two, the new product becomes (2q1 * 2q2) = 4 * (q1 * q2). Therefore, the electrical force between them will be quadrupled.

Question 7: Two spheres that are in contact but isolated from the ground are approached by a charged rod bearing a negative charge. What kind of charge will be on the two spheres if they are then separated?

The sphere near the charged rod becomes negative and the other becomes positive
The spheres do not get any charge
The sphere near the charged rod becomes positive and the other becomes negative (Correct answer)
None of these

Correct answer: The sphere near the charged rod becomes positive and the other becomes negative

This is an example of charging by induction using two conductors. When a negatively charged rod approaches two spheres in contact, it repels the free electrons in the spheres, pushing them to the sphere farthest from the rod. This leaves the sphere near the rod with a net positive charge and the far sphere with a net negative charge. When the spheres are separated while the rod is still nearby, they retain these induced opposite charges.

Question 8: When electrostatic circumstances are present, a conductor is positioned in an electric field. Which of the following claims applies to this circumstance?

The electric field on the surface of the conductor is perpendicular to the surface
The electric field is zero inside the conductor
All valence electrons go to the surface of the conductor
All of these (Correct answer)

Correct answer: All of these

Under electrostatic conditions, a conductor exhibits several key properties. The electric field inside the conductor is always zero, as free charges redistribute to cancel any internal field. Consequently, all excess charge resides on the surface of the conductor. Furthermore, the electric field lines on the surface must be perpendicular to the surface, because any parallel component would cause charges to move, violating the electrostatic equilibrium.

Question 9: What distinguishes gravitational and electrical forces most significantly?

Electrical forces obey the inverse square law and gravitational forces do not
Electrical forces attract and gravitational forces repel
Gravitational forces obey the inverse square law and electrical forces do not
Gravitational forces are always attractive but electrical forces can be attractive or repulsive (Correct answer)

Correct answer: Gravitational forces are always attractive but electrical forces can be attractive or repulsive

The most significant distinction lies in the nature of the forces. Gravitational forces are always attractive, drawing masses towards each other. In contrast, electrical forces can be either attractive (between opposite charges) or repulsive (between like charges), due to the existence of both positive and negative charges.

Question 10: An electric field that is vertically upward and oriented along the +y axis interacts with a charged particle moving along the +x axis. What is the sign of the charge on this particle if the force from the field is downward for the charged particle?

It is negative (Correct answer)
It is neutral
It is positive
None of these

Correct answer: It is negative

The direction of the electric force (F) on a charged particle in an electric field (E) is given by F = qE. If the electric field is vertically upward (+y direction) and the force on the particle is downward (-y direction), then the charge (q) must be negative. A negative charge experiences a force in the direction opposite to the electric field.

Question 11: The identical positive charge +Q is enclosed by Gaussian surfaces A and B. Gaussian surface A has a surface area that is three times more than Gaussian surface B. Electric field flux via Gaussian surface A is equal to .

unrelated to the flux of electric field through Gaussian surface B
equal to the flux of electric field through Gaussian surface B (Correct answer)
nine times larger than the flux of electric field through Gaussian surface B
three times smaller than the flux of electric field through Gaussian surface B

Correct answer: equal to the flux of electric field through Gaussian surface B

Gauss's Law states that the total electric flux through any closed surface is directly proportional to the total electric charge enclosed within that surface, divided by the permittivity of free space (Φ = Q_enclosed / ε₀). Since both Gaussian surfaces A and B enclose the identical positive charge +Q, the electric flux through both surfaces will be the same. The size or shape of the Gaussian surface does not affect the total flux, only the enclosed charge.

Question 12: A vertically downward electric field, or along the negative y-axis, is encountered by an electron moving along the +x-axis. After the electron enters the electric field, in what direction will the electric force be acting on it?

Out of the page
To the right
Into the page
Upward (Correct answer)

Correct answer: Upward

An electron carries a negative charge. The electric force on a negative charge is always in the direction opposite to the electric field. Since the electric field is vertically downward (along the negative y-axis), the force on the electron will be vertically upward (along the positive y-axis). The electron's initial motion along the +x-axis does not influence the direction of the electric force.

Question 13: As seen in the illustration, let's assume that a magnetic field is arranged so that it is pointing straight to the left. Which way would the trajectory of a positively charged particle curve if it were to start moving through this magnetic field from the right?

The particle would move into the page
The particle would move out of the page
The particle would continue to move to the right unaffected (Correct answer)
The particle would move down the page

Correct answer: The particle would continue to move to the right unaffected

The magnetic force on a charged particle is given by the Lorentz force law, F = q(v x B), where v is the velocity and B is the magnetic field. For a magnetic force to act, the particle's velocity must have a component perpendicular to the magnetic field. If the particle is moving from the right (meaning its velocity vector is to the left) and the magnetic field is also pointing to the left, the velocity vector is parallel to the magnetic field. In this scenario, the cross product (v x B) is zero, resulting in no magnetic force, and thus the particle's trajectory remains unaffected.

Question 14: A rope is fastened to a 50 kilogram cinder block that is lying on the ground. What is the lowest tension the rope can withstand before snapping if it is pulled so that the block accelerates directly upwards at a velocity of 5(m/s2)?

250N
241N
50N
741N (Correct answer)
10N

Correct answer: 741N

To calculate the tension, we apply Newton's second law: F_net = ma. The forces acting on the block are the upward tension (T) and the downward gravitational force (mg). Since the block accelerates upwards, the net force equation is T - mg = ma. Given m = 50 kg, a = 5 m/s², and using g ≈ 9.81 m/s², the tension T = ma + mg = (50 kg * 5 m/s²) + (50 kg * 9.81 m/s²) = 250 N + 490.5 N = 740.5 N, which rounds to 741 N.

Question 15: A rope is fastened to a 50 kilogram cinder block that is resting on a frictionless surface. What is the least amount of stress the rope can withstand before snapping if it is pulled parallel to the ground so that the block accelerates at a rate of 5(m/s2)?

490N
250N (Correct answer)
740N
10N
2450N

Correct answer: 250N

In this scenario, the block is on a frictionless surface, meaning there are no opposing horizontal forces. The only horizontal force acting on the block is the tension (T) in the rope. According to Newton's second law, F_net = ma. Therefore, T = ma. Given m = 50 kg and a = 5 m/s², the tension T = 50 kg * 5 m/s² = 250 N.

Question 16: On a lab bench, two tiny lead balls with weights of 5 kg and 10 kg are attached at a distance of 100 cm apart. The distance is now only 20 cm after a pupil transfers the larger mass in the direction of the smaller mass. What influenced the gravitational force between the lead balls when the mass was moved?

1
5
25 (Correct answer)
0.2
0.4

Correct answer: 25

The gravitational force (F) between two masses is inversely proportional to the square of the distance (r) between their centers, as described by Newton's Law of Universal Gravitation (F ∝ 1/r²). The initial distance was 100 cm, and the new distance is 20 cm, which is 1/5th of the original distance. Since force is proportional to 1/r², reducing the distance to 1/5th increases the force by a factor of (1 / (1/5)²) = 1 / (1/25) = 25.

Free NEET Physics Questions and Answers

What size resistor should be placed across the 12V battery's terminals in order to generate 500mA of current?