Physics: CUET Mock Test - 1 - CUET MCQ

The zone near a charge where its attraction or repulsion force works, is known as the electric field of that charge. Theoretically, it is up to infinite but practically it has limitations. If a unit positive charge is kept in that field, it will undergo some force which is known as electric field intensity at that point.

Concept:

Electric field:

• ​It is defined as the force per unit of positive charge.
• Formula, Electric field, E = F/q where F = force, q = positive charge
• The SI unit of electric charge is N/C.

Electric field due to an infinitely long straight wire:

• Consider an infinitely long straight, uniformly charged wire. Let the linear charge density of this wire be λ. P is the point that is located at a perpendicular distance from the wire. The distance between point P and the wire is r.

​

• The electric field is due to an infinitely long straight wire, .

Explanation:
Assuming the wire is infinitely long, the electric field will be radial and will have a magnitude given by:

E =

where
λ is the linear charge density of the wire,
ε₀ is the permittivity of free space,
r is the distance from the wire.

• The direction of the electric field will be radial and pointing away from the wire if the charge on the wire is positive and towards the wire if the charge on the wire is negative.
• Note that the above formula assumes that the wire is thin compared to the distance from it, and that the electric field is being measured at a point far away from the ends of the wire. If these assumptions are not satisfied, the electric field may be different.

Calculation:

Let a charge "Q" is placed at a distance of "x" from (-q).

Let the force exerted by "-q" be F₁ and "4q" be F₂ on "Q".

To make the system in equilibrium, the net force acting on "Q" should be zero.

i.e., F₁ + F₂ = 0.-----(1)

F₁ = K(-q)Q/x²

F₂ = K(4q)Q/(r + x)²

Putting the value of F1 and F2 in equation (1), we get,

K(-q)Q/x2 + K(4q)Q/(r + x)2 = 0

(x + r)/x = 2

x = r

The charge "Q" is placed at a distance of "r" from charge "-q" on the extreme side of (-q).

The correct answer is option (1).

Concept:

• For a conducting spherical shell of radius R that carries a charge Q, the electric field inside the shell is zero.
• And outside the shell, the electric field behaves as if all the charge Q is concentrated at the center of the shell.
• Thus, the electric field outside the shell, at a distance r from the center, is given by Coulomb's law:

E = kQ/r²
where k is the Coulomb constant.

Therefore, the electric field outside the shell varies inversely with the square of the distance from the center of the shell.
That is, as the distance from the center of the shell increases, the electric field strength decreases rapidly.
So, the plot of electric field w.r.t the distance "r" would be,

The correct answer is option "2"

Concept:

We know that for a conducting sphere, the electric field outside the sphere is the same as the field due to a point charge located at the center of the sphere. So, we can use the formula for the electric field due to a point charge to find the net charge on the sphere.

The electric field due to a point charge q at a distance r from it is given by:

E = k*q/r²

where k is the Coulomb constant.

In this case, the electric field at a distance of 20 cm from the center of the sphere is given as 1.2 × 10³ NC^-1, and it points radially inwards.

This means that the point charge at the center of the sphere is negative.

So, we have:

E = k*q/r²

= q = ------(1)

Calculation:

Given that

E = 1.2 × 103 NC-1

r = 20 cm = 0.2 m.

k = 4.5 × 109

Substituting values of k, r, E in the equation we get,

q =

=

= 5.33 ×10-9 C

Since the point charge at the center of the sphere is negative.

q = - 5.3 ×10-9 C

The correct answer is option (4)

Concept:
The capacitance of a parallel plate capacitor is given by the formula:

C = εA/d
where
C is the capacitance,
ε is the permittivity of the medium between the plates,
A is the area of each plate, and d is the distance between the plates.

Calculation:
When a copper plate of thickness b is placed between the plates of the capacitor, the distance between the plates effectively reduces to d' = d - b, where b is the thickness of the copper plate. The permittivity of copper is very close to that of vacuum, so we can assume that the permittivity between the plates remains the same.

Therefore, the new capacitance of the capacitor is given by:
C' = εA/d'
C' = εA/(d - b)
C' =
The correct answer is option (2)

Semiconductors: A semiconductor is a material whose conductivity lies in between those of insulators and good conductors.

Examples: germanium and silicon.

In terms of energy band, semiconductors are materials that have almost an empty conduction band and an almost filled valence band with a very narrow energy gap (of the order of 1 eV) separating the two. At 0 K there are no electrons in the conduction band and the valence band is completely filled.

However, with the rise of temperature, some of the electrons from the valence band go to the conduction band overcoming the gap. So the conductivity of semiconductors increases with temperature.

Hence the correct answer is option 3.

Semiconductors: A semiconductor is a material whose conductivity lies in between those of insulators and good conductors.
Examples: germanium and silicon.
Hence the correct answer is option 3.

Insulators: Insulators are those materials in which the valence electrons are bonded very tightly to their parent atoms and there are no free electrons.
A large electric field is required to remove the electrons from the attraction of the nuclei.
In terms of energy bands, insulators (a) have a full valence band and (b) have an empty conductor band.
Hence the correct answer is option 2.

On the basis of the band theory of solids, the solids are classified as -
(i) conductors, (ii) insulators, and (iii) semiconductors.

(i) Conductors or metals: Conducting materials are those in which plenty of free electrons are available for electric conduction.
In terms of energy bands, it means that conductors are those which have overlapping valence and conduction bands.
Hence the correct answer is option 2.

On the basis of the band theory of solids, the solids are classified as -
(i) conductors, (ii) insulators, and (iii) semiconductors.
The superconductor is not included among these.
Hence the correct answer is option 4.

• Infrared radiation is a part of the electromagnetic spectrum. Infrared radiations are also known as thermal or heat waves.
• The range of wavelength is between 710nm to 1mm.
• These consist of heat inducing property. These rays have applications in the area of heat production. these rays keep earth hot.
• These are emitted by certain semiconductor light-emitting diodes.
• Water vapour traps heat radiated from Earth so as infra red rays.
• The thermal motion of the materials increases after they absorb infrared rays. 

So, Option-(i) and (iv) is are not correct.
Hence the correct answer is Option-1-Both (i) and (iv). 

The correct answer is (option 3) i.e. true - true - true
CONCEPT:
Coulomb’s Law talks about the magnitude of the attraction between the two charges. It says that the force is directly proportional to the product of the quantity of the two charges.
F ∝ q₁q₂, where (F is Force, q₁ and q₂ are charges).

• It is also inversely proportional to the square of the distance between them.

i.e., F ∝ 1/r² where (F is Force, r is the distance between the charges).

• Thus, the equation is  where k is the proportionality constant and is equal to 9 × 10⁹ Nm²/C².
• The Law was given by Charles-Augustin de Coulomb in the year 1784.

Coulomb’s Law was derived based on certain assumptions and can’t be used freely like other general formulas. The law is limited to the following points :

1. Coulomb’s Law is only applicable if the charges are static (in rest position)
2. Coulomb’s Law is easy to use while we are dealing with charges of regular shape and size, and it is very difficult to be applied to an irregular shape.
3. Coulomb’s Law is only valid when the distance between two charged particles is sufficiently larger than the size of both charges

The correct answer is Option 3.
Key Points

• Kirchhoff’s circuit laws lie at the heart of circuit analysis. 
• German physicist, Gustav Kirchhoff, developed a pair of laws that deal with the conservation of current and energy within electrical circuits.
• These two laws are commonly known as Kirchhoff’s Voltage and Current Law.
• These laws help calculate the electrical resistance of a complex network or impedance in the case of AC and the current flow in different network streams.
• Kirchhoff’s Current Law goes by several names as Kirchhoff’s First Law and Kirchhoff’s Junction Rule. Hence Statement b is correct.
  • According to the Junction rule, in a circuit, the total of the currents in a junction is equal to the sum of currents outside the junction.
  • The total current entering a junction or a node is equal to the charge leaving the node as no charge is lost.
  • The algebraic sum of the currents meeting at a junction in a circuit is zero. Hence statement a is incorrect.
• Kirchhoff’s Voltage Law goes by several names as Kirchhoff’s Second Law and Kirchhoff’s Loop Rule.
  • According to the loop rule, the sum of the voltages around the closed loop is equal to null.
  • The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero. Hence Statement c is correct.

Though water and ammonia are covalent molecules, they have a net dipole moment due to their distorted structure than the ideal one. Hydrochloric acid has a linear molecular structure but its net dipole moment is towards the Hydrogen atom. But methane has a symmetrical tetrahedral structure and its net dipole moment is zero hence it doesn’t behave as a permanent dipole. Methane solvents are therefore non-polar solvents.

Force on a unit point charge kept at a distance r from the charge = q/r². Therefore, work done to bring that point charge through a small distance dr =(q/(r²)) * (-dr). Therefore, the potential of that point is =

When the resistors are connected in series, and current is passed through them, the current passing through each of the resistor is the same. This is because, the resistors are connected end to the end and, therefore, there is only one path for the current to flow through.

According to Kirchhoff’s first law, the current flowing towards the junction is equal to the current leaving the junction. Mathematically, this law can be expressed as ∑ⁿ_{K – 1} I_K = 0 (where n is the number of branches carrying current towards or away from the junction).

If a unit positive charge is kept near a negative charge, the unit positive charge will be attracted towards the negative charge. That means the electric field is towards the negative charge. But in case of positive charge, the field is directed away from the charge.

The dipole moment is defined as the product of a charge and distance. The dimension of charge (current*time) is [I T] and the dimension of distance is [L]. Therefore the dimension of dipole moment is [L T I]. Its unit in the CGS and the SI system are esu*cm and C*m respectively.

In the SI system, electric potential due to a point charge at a distance r is Substituting the values, we get potential=  Though in practice, this huge value of electric potential is not present.

When three resistors are connected in series, then the equivalent resistance of this combination is R_s = R₁ + R₂ + R₃. So, if 3 resistors having resistances 10, 15, and 20 ohms, are connected in series, then equivalent resistance of this combination is R_s = 10 + 15 + 20 = 45 ohms.

The equation → ∑e = ∑IR is applicable to Kirchhoff’s second law. This law is also known as Kirchhoff’s loop rule. This expression tells us that in a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistance and currents flowing through them.

According to Gauss’s law, if there is no charge inside a closed surface, the field inside the closed surface will always be zero. We know the charge is distributed on the outer surface of a conducting hollow sphere because the charges want to maintain maximum distance among them due to repulsion. So there is no charge inside the sphere and hence no electric field.

If the distance of a point charge, kept on the axis of a dipole, is sufficiently large than the separation between the charges of a dipole, then it can be shown that electric field E at a distance r from the dipole varies as E α 1/(r³). Now if the distance becomes 2r, electric field intensity will decrease by a factor of 2³ = 8 times. Therefore the force F will now become F/8.

Work done = potential*charge by definition. We know that the potential of a point is the amount of work done to bring a unit charge from infinity to a certain point. Therefore, work done W = q * V = 4 * 10^-3 * 200J = 0.8J. The work done is positive in this case.

Domestic appliances in a house are connected in parallel combinations, and not in series combinations. This arrangement is done so that each of the appliances can switched on and off independently, which is essential in a house’s wiring.

According to sign convention while traversing a closed loop (in clockwise or anti-clockwise direction), if positive pole of the cell is encountered first then its emf is negative or else, it will be positive.

If the total charge of the sphere is q then the electric field at a distance of r is equal to  Therefore the electric field is proportional is (1/(r²)) (if r > radius of the sphere). But if r < radius of the sphere the electric field will be zero i.e. electric field inside a hollow sphere is always zero.

If the dipole moment of a dipole is p, then electric field intensity at a distance r from the dipole on its axis will be (2p/(r³)). But electric field intensity at a distance r from the dipole on its perpendicular bisector is (P/(r³)). Both the formulas are derived by assuming r >> length of the dipole. Thus E1= 2P/r³ and E2 = P/r³ and hence E1 = 2 E2.