Test: Physics Minor Mock Test - 3 (March 23) - NEET MCQ

The torque on a current loop depends upon the area of the current loop, when the magnetic field is perpendicular to the plane of the loops the torque has its maximum value,
Τ=I A B
We know I and B for all these cases but A depends upon the geometry. The circle has the greatest area so it should provide the greatest torque.

We want the Voltage difference V_AB.
let A and B be open and not connected to any thing.
There is a current that flows from cell of larger emf to the cell of small emf.
Call that current as  I Amperes.

1.25 V - 0.75 V - I * R2 - I * R1 = 0

I = 0.5 / (R1+R2)
V_AB =  -0.75 - I * R2 =  - 0.75 - 0.5 * R2 / (R1+R2)

= - (0.75 R1 + 1.25 R2) / (R1+R2)

= - ().75 * 1 + 1.25 * 1) / (1 + 1)  volts

=  - 1 volts

As the magnetic field is parallel to the charge, no magnetic Lorentz force will act on it and it will move undeviated with constant velocity.

Step 1: Given data
Area of the current loop, A = 0.01 m²
Electric current flowing in the current loop, I = 100 A
Magnetic field intensity, B = 0.1 T
Angle between the current loop and the magnetic field, θ = 0^∘
Torque acting on the current loop, τ = ?

Step 2: Formula used

τ = I A B sinθ…(a)

Step 3: Calculation of the torque acting on the current-carrying loop
Substituting the given values in equation (a), we get:

τ = 100 A × 0.01 m² × 0.1 T×sin⁡0^∘

τ = 100 × 0.01 × 0.1 × 0

τ = 0

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.

The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.

A current (i) in a magnetic field ( B) experiences a force ( F) given by the equation

 F = i (l × B) 

or F = ilB sin θ,

where l is the length of the wire, represented by a vector pointing in the direction of the current.

According to Ohm’s law, V = IR. Therefore R= V/I

For waves along a string :  

The potential difference between any two points inside the hollow conductor is zero. This means that the potential at all points inside the hollow charged conductor is the same and it is equal to the value of the potential at its surface.

The electric field on the surface of a hollow conductor is maximum and it drops to zero abruptly inside the conductor.

E = - dV/dr

Ohm’s law states I is proportional to V.This holds good at steady temperatures and for the flow of constant current.

We know that in a simple harmonic motion the maximum velocity,
V_max = A⍵
Here A = 50 mm

And ⍵ = 2π / T
= 2π / 2
= π

Hence  V_max = 50 x 10^-3.π
= 0.15 m/s

According to Ohm’s law, V=I R. Therefore R=V/I

   ∴ We get, ω = √3 s^-1
   T = 2π / √3

 Answer :- c

Solution :- When a charged particle moves in a magnetic field, a force is exerted on the moving charged particle. The formula for the force depends on the charge of the particle, and the cross product of the particle's velocity and the magnetic field.

An electroscope is a device that measures the potential difference.
If it is connected in parallel to the capacitor, the potential across it will be equal to the potential across the capacitor, which is equal to the potential across the battery.
On decreasing the battery potential, the potential difference across the electroscope reduces, and hence the reading reduces. 

Energy stored in the capacitor
This is released as heat when the capacitor discharges through the metal block. The quantity of heat = mass xs p.heatx rise in temperature. 
Q = m x s x Δθ = 0.1 x 2.5 x 10² x 0.4 = 10 J
U = Q; 2 x 10⁴ C = 10; C = 5 x 10^-4 F = 500 μF

After one oscillation of a bigger pendulum i.e. 5T/4, ¼ of the total phase is travelled by the smaller pendulum while the bigger is still at initial position.

Thus, the phase difference between two is: ¼ (2π) - 0 = π/2

Periodic time is the time taken for one complete revolution of the particle.

∴ Periodic time (t_p) = 2 π/ω seconds.

We know that,
where v is velocity, T is tension in the string and μ is the mass per unit length
Solving the above equation we get v = 100 m/s
Also, t = d/v = 0.02 secs

Acceleration 

The connecting metal strips include a small resistance in the circuit called the end resistance.On reversing the current, the end resistance tends to get cancelled out.

A device which converts electrical energy into mechanical energy is called an electrical motor.
The working principle of an electric motor depends on the magnetic and electric field interaction.
There are two types of electric motor and they are:
AC motor: Converts alternating current into mechanical power. Linear motor, synchronous motor, and induction motor are the three types of AC motor.
DC motor: Converts direct current into mechanical power. Self-excited motor and separately excited are the two classifications of DC motor.
 

Equation of SHM particle: 
Y=a sinωt
V=aω sinωt
V_max = aω
So the velocity is maximum at mean position

Explanation:

When a negatively charged conductor is connected to the earth, electrons will flow from the conductor to the earth. This is because electrons have a negative charge and they will be repelled from the negatively charged conductor and attracted to the positively charged earth. As electrons flow from the conductor to the earth, the negative charge on the conductor will gradually decrease until it becomes neutral.

• Option A is incorrect because charge flow does occur when a negatively charged conductor is connected to the earth.
• Option B is incorrect because protons have a positive charge and they are not free to move in a conductor.
• Option C is incorrect because electrons flow from the earth to the conductor, not the other way around.

As aluminium and copper are metals, their mobile electrons move under the influence of the external field until they reach the surface of the metal and collect there. External electric fields induce surface changes on metal objects that exactly cancel the field within. Since the field applied is same in both case, the induced charge will be the same.
Hence, the magnitude of induced charge in the aluminium will be the equal to that of copper.

Let the frequencies be f1 and f2
So, f1 = f (vs/vs - v) and f2 = f (vs/vs + v)
ATQ: f1 – f2 = 2 Hz