Physics: CUET Mock Test - 2 - CUET MCQ

A scalar quantity is a quantity with magnitude only but no direction. But a vector quantity possesses both magnitude and direction. An electric field has a very specific direction (away from a positive charge or towards a negative charge). Hence electric field is a vector quantity. Moreover, we have to use a vector addition for adding two electric fields.

Concept:

The expression for torque (τ) experienced by an electric dipole of dipole moment (P) in an external uniform electric field (E) is given by:
τ = P x E
where
τ is the torque vector,
P is the electric dipole moment vector, and
E is the external electric field vector.
The symbol "x" represents the cross-product between the two vectors.

• The direction of the torque is perpendicular to both the dipole moment vector and the external electric field vector and follows the right-hand rule.
• The magnitude of the torque is given by:
• |τ| = P E sinθ
• where θ is the angle between the dipole moment vector and the external electric field vector.

The correct answer is option (4)

Calculation:

The given circuit can be rearranged as

Where the resistance of each resistors are 2 ohms and voltage applied is 2 V.
We can see that arm resistances R₁ ,R₂,R3,R4

R₁R3 = R₂R₄

the above circuit is a wheat stone bridge.

The current flowing through the middle branch would be zero.

Net resistance "R" will be = (2 + 2) II (2 + 2)

R = 2 ohm.

The current from the battery would be

I = V/R

V = 2 Volts.

So, I = 2/2 = 1 A

I = 1 A.

The correct answer is option (4).

Calculation:
We know that the voltmeter is always connected in parallel.
The voltmeter resistance is given as 150 Ω and is connected across AB.
This means 150 Ω is connected in parallel with 100 Ω .
The resistance across AB will be a parallel combination of 150 and 100 Ω resistors.
The resistance across AB = (150) II (100)
= 150*100/(150 + 100)
= 60 Ω.
The net resistance of the circuit would be series combination of 60 and 100 Ω .
R = 60 + 100
= 160 Ω.
Current I = V/R
I = 40/160
= 0.25.
The voltage across AB will be the voltmeter reading which is equal to I*R.
Voltmeter reading = 0.25*60
= 15 Volts.
The correct answer is option (3)

Calculation:

In the given potentiometer experiment, the balancing length l is 1.6 m, and the total length of the wire AB is 2 m. The resistance per unit length of the wire AB is R = 10/2 = 5 Ω/m.

Let V be the emf of the standard cell to be determined. At the balancing point, the potential difference across the wire AB is equal to the emf of the standard cell, i.e., V = IR, where I is the current flowing through the wire AB.

The total resistance in the circuit is the sum of the resistance of the standard cell (0.2 Ω), the internal resistance of the cell in the potentiometer (2 Ω), and the resistance of the wire AB up to the balancing point (5 Ω/m x 1.6 m = 8 Ω).

The current I in the circuit can be calculated using the potential difference across the circuit, which is the sum of the emf of the cell in the potentiometer and the potential drop across the wire AB up to the balancing point. Thus, I = (2.4 V)/(2 Ω + 5 Ω/m x 1.6 m) = 0.24 A.

Substituting the value of I and R into the equation V = IR gives V = 0.24 A x (5 x 1.6 + 0.2) = 1.968 V.

The correct answer is option (2) i.e., nearly equal to 1.9 V.

Concept:

• The drift velocity of electrons refers to the average velocity at which free electrons drift in a conductor under the influence of an electric field.
• When a voltage is applied to a conductor, the electric field generated accelerates free electrons and they start to move in the direction of the electric field.
• However, due to the collisions with atoms and other electrons present in the conductor, the free electrons do not move with a constant velocity, but rather move with a small, average velocity in the direction of the electric field. This average velocity is called the drift velocity.
• The drift velocity of electrons in a conductor depends on various factors such as the applied voltage, the type of material of the conductor, the density of free electrons, and the temperature.
• However, in a typical metallic conductor at room temperature, the drift velocity of electrons is typically on the order of a few millimeters per second, which is quite slow compared to the speed of light or even the speed of sound.

The drift velocity of electrons is directly proportional to the voltage applied across a conductor.

The other factors mentioned in the options, such as temperature, length of the conductor, and area of cross section of conductor, can affect the resistance of the conductor, but they do not directly affect the drift velocity of electrons.

So, the correct answer to the question is (2)

Additional Information

Drift velocity of electrons is directly proportional to the electric field applied to the conductor.

This means that as the strength of the electric field increases, the drift velocity of the electrons in the conductor also increases. Mathematically, the relationship between the drift velocity and the electric field is described by the following equation:

vd = μE

where vd is the drift velocity,

E is the electric field strength, and

μ is the electron mobility, which is a material property that describes how easily electrons can move through a material.

The higher the electron mobility, the higher the drift velocity for a given electric field strength.

Let the distance between any two charges is d. Therefore the net potential energy of the system is  But the total energy of the system is zero. So, -qQ - qQ + q²/2 = 0 hat means q = 4Q, i.e. q/Q = 4.

Electric field intensity is defined as the force on a unit positive charge kept in an electric field. Hence we can simply consider its dimension as (the dimension of force/the dimension of charge). The dimension of force is [MLT^-2] and the dimension of charge is [IT]. Therefore the dimension of field intensity is [M L T^-3 I^-1].

Torque = NIBA sinθ.
Torque = 50 × 20 × 0.2 × 200 × 10^-4 × sin (90)
Torque = 4Nm.

dB = 

dB = 1.4 × 10^-9 T.

We can consider the relation that includes power and resistance, i.e. Power = voltage²/resistance . Since, the resistors are connected in parallel, the voltage across them will be the same. From this relation, power and resistance are inversely proportional to each other.
Thus,  
So, the power dissipated is in the ratio is 1:3.

Kirchhoff’s Law is used to analyze circuits. This law is important because they represent connections of a circuit. Kirchhoff’s Law provide the constraints that let us find the current flowing and voltage across every circuit element.

The electric field at a distance of r from a charge q is equal to  Let the electric field intensity will be zero at a distance of x cm from +4q charge, so the fields due to the two charges will balance each other at that point. Therefore  Solving this we get x=20cm. Therefore the point will be 20cm away from the +4q charge.

A non-polar substance consists of a huge number of dipoles in it, but they are oriented randomly and hence the net dipole moment of the substance becomes 0 and it acts as a non-polar substance. But if it is placed in an electric field, the dipoles present in it will orient themselves in the direction of the field and hence a net dipole moment will be observed, known as induced dipole moment. No current flow or oscillation of the substance will be observed.

Net electrostatic field is zero in the interior of a conductor. When a conductor is placed in an electric field, its free electrons begin to move in the opposite direction. Negative charges are induced on the left end and positive charges on the right end of the conductor. The process continues until the electric field set up by induced charges becomes equal and opposite the external field.

The equivalent overall resistance of the parallel combination is:

R1 is in series with R2; So, R3 = R1 + R2 = 10 + 20 = 30 ohms.
Now, we can employ the method of cross-multiplication:
For 20 ohms resistor → 30 W power consumed
For 30 ohms resistor combination → x
20x = 30 × 30
x = (30 × 30)/20
x = 45
Therefore, the power consumed by the parallel combination is 45 ohms.

Kirchhoff’s 2^nd year is applied in a closed loop. Kirchhoff’s 2^nd law supports the law of conservation of energy. This means that energy is neither created nor destroyed in the closed loop. Whatever energy enters the loop, same amount leaves the loop.

Ampere’s circuital law states that the line integral of magnetic field around any closed path in vacuum is equal to μ₀ times the total current passing the closed path. It is given by: ∮ B .dl = μ₀I
Ampere’s circuital law is analogous to Gauss’s law in electrostatics.

The magnetic flux through a surface is the surface integral of the normal component of the magnetic field flux density passing through that surface. The SI unit of magnetic flux is weber (Wb). One weber is the flux produced when a uniform magnetic field of one tesla acts normally over an area of 1 cm².

E=-(dV/dx) where E is the field intensity, V is potential and x is distance. Therefore unit of electric field intensity will be (unit of potential/unit of distance) = V/m. Electric flux has unit V * m, V is the unit of electric potential whereas charge has a unit of Coulomb or esu.

Right-hand thumb rule can be used to find the direction of the magnetic field at a point near a current-carrying conductor. Right hand rule states that, if the thumb of the right hand is in the direction of the current flow then, the curl fingers show the direction of the magnetic field.

Torque (Max) = NIBA.
Τ_max = 400 × 10³ × 0.2 × 320 × 10^-4
T_max = 2560 Nm.

Magnetic flux = Magnetic field × Area.
Flux = 
Flux = 
Flux = [M L² A^-1 T^-2].

Torque on a planar current loop depends upon current, the strength of the magnetic field, area of the loop and the orientation of the loop in the magnetic field. It is independent of the shape of the loop.

1 Wb = 1 T × 1 m²
Wb = 10⁴ G × 10⁴ cm².
1 Wb = 10⁸ Maxwell.

The principle of superposition applies to both fields. This is because the magnetic field is linearly related to its source, namely, the current element and the electrostatic field is related linearly to its source, the electric charge.

In classical electrodynamics, the magnetic field given by a current loop and the electric field caused by the corresponding dipoles in sheets are very similar, as far as we are far away from the loop, which enables us to deduce Ampere’s magnetic circuital law from the Biot-Savart law easily.

The SI unit of magnetic pole strength is ampere meter (Am).
The strength of a magnetic pole is said to be one-ampere meter if it repels an equal and similar pole with a force of 10^-7 N when placed in vacuum at a distance of one meter from it.

A normal to a plane can be drawn from either side. If the normal drawn to a plane points out in the direction of the field, then θ = 0^o and the flux are taken as positive. If the flux is positive and increasing the voltage will be negative.

The potential difference between any two points on an equipotential surface is zero.
Work done=Test charge x potential difference(0)
=Zero .