VITEEE Physics Test - 9 - JEE MCQ

The problem involves calculating the current flowing through an inductor.

To determine the current (I) through an inductor connected to an alternating current (a.c.) source, we can use the following steps:

• The formula for the inductive reactance (X_L) is given by: X_L = 2πfL, where:
• f = frequency (in Hz)
• L = inductance (in H)
• Substituting the values:
• f = 50 Hz
• L = 1 H
• Calculating X_L: X_L = 2π × 50 × 1 = 100π ≈ 314.16 Ω
• Using Ohm's law (V = IR), we rearrange to find the current:
• I = V / X_L
• Substituting the voltage (V) of 200 V:
• I = 200 V / 314.16 Ω ≈ 0.64 A

Thus, the current flowing through the inductor is approximately 0.64 A.

The charge on a 4μF capacitor can be calculated using the formula:

• Q = C × V where:
• Q is the charge in coulombs,
• C is the capacitance in farads, and
• V is the voltage across the capacitor in volts.

Given that the capacitance is 4μF (or 4 × 10^-6 F), you need to know the voltage applied to find the charge.

If the voltage across the capacitor is V volts, then:

• The charge Q will be 4 × 10^-6 F × V.

To determine the exact charge, you should substitute the value of V into the equation.

To find the increase in mass of the copper pole:

• The mass decrease of the zinc pole is 0.13 g in 30 minutes.
• The electrochemical equivalent for Zn is 32.5 g/C and for Cu is 31.5 g/C.
• Using the formula for mass change:
  • Mass of Cu deposited = (Mass of Zn lost) × (Electrochemical equivalent of Cu / Electrochemical equivalent of Zn).
• Calculating the increase in mass of Cu:
  • Increase in mass = 0.13 g × (31.5 / 32.5).
  • Increase in mass ≈ 0.126 g.

The increase in mass of the positive Cu pole is approximately 0.126 g.

According to max. power transfer theorem, power transferred to the load will be maximum when load resistance is equal to internal resistance
i.e., R = r/2

To move the null point from the 7th wire to the 9th wire in a potentiometer, consider the following approaches:

• Attach a resistor in series with the battery to increase the overall resistance.
• Increase the resistance in the main circuit to reduce the current flow.
• Decrease the resistance in the main circuit to allow more current.
• Reduce the applied emf to lower the voltage across the potentiometer.

Choose the option that best aligns with your needs to achieve the desired null point adjustment.

If two identical heaters rated at 220 V and 1000 W are connected in parallel across a 220 V supply, the total power output can be calculated as follows:

• Each heater operates at 1000 W.
• In a parallel configuration, the total power is the sum of the individual powers.
• Thus, the combined power is:
• 1000 W (first heater) + 1000 W (second heater) = 2000 W.
• Therefore, the total power of the heaters is 2000 W.

No. of turns across primary N_p = 50
Number of turns across secondary N_s = 1500
Magnetic flux linked with primary, Φ = Φ₀ + 4t
∴ Voltage across the primary,

Heat radiation propagates at the speed of light, which is the fastest known speed in the universe.

• Heat radiation is a form of energy transfer that occurs through electromagnetic waves.
• It includes infrared radiation, which is felt as heat.
• This type of radiation travels through a vacuum, unlike sound waves, which require a medium.
• In summary, heat radiations move at the same speed as light waves.

The potential at the centre of a hollow metal sphere is the same as the potential on its surface.

This is because:

• The electric field inside a hollow conductor is zero.
• Since the electric field is zero, there is no change in potential within the sphere.
• As a result, the potential at any point inside, including the centre, is equal to the potential at the surface.

Therefore, if the potential on the surface is 80 V, the potential at the centre of the sphere is also 80 V.

When an electron is brought closer to another electron, the electrostatic potential energy of the system experiences the following changes:

• The potential energy increases as they approach each other.
• This is due to the repulsive force between the two negatively charged electrons.
• As they get closer, work must be done against this repulsion, leading to a rise in energy.

In summary, bringing two electrons closer results in an increase in their electrostatic potential energy due to their like charges repelling each other.

To determine the capacitance of the capacitor when the distance between the plates is halved, we can use the formula for the capacitance of a parallel plate capacitor:

• The formula is given by: C = ε₀(A/d), where:
• C is the capacitance,
• ε₀ is the permittivity of free space,
• A is the area of the plates,
• d is the distance between the plates.
• Initially, the capacitance is 10 μF with a distance d of 8 cm.
• When the distance is reduced to 4 cm, the new capacitance can be calculated as:
• Since capacitance is inversely proportional to d, when d is halved, C will double.
• Thus, the new capacitance is 10 μF × 2 = 20 μF.

To determine the new potential difference after charging the shell:

• The initial potential difference between the solid sphere and the hollow shell is denoted as V.
• The solid conducting sphere has a charge of Q.
• The hollow shell initially is uncharged.
• When the hollow shell is given a charge of -3Q, it will influence the potential.
• The sphere induces a charge of +Q on the inner surface of the shell.
• This results in a net charge of -4Q on the outer surface of the shell.
• The potential of the outer surface of the shell will change due to this additional charge.
• Consequently, the new potential difference between the surfaces can be expressed as 2V.

To find the electrostatic potential at point P, located at a distance R/2 from the centre of the shell, consider the following:

• The spherical conducting shell has a total charge of q.
• Charge Q is situated at the centre of the shell.
• Inside a conductor, the electric field is zero, meaning that the potential is constant throughout the conductor.
• Thus, the potential at any point within the shell, including point P, is determined solely by the charges inside and on the shell.
• The potential at point P can be calculated using the formula: V = k(Q + q)/r, where k is the electrostatic constant, and r is the distance from the charge to point P.
• As point P is at R/2, the potential becomes: V = (1/4πε₀) * (Q + q) / (R/2).
• Simplifying gives: V = 2(Q + q)/4πε₀R.

The electrostatic potential at point P is therefore: 2(q + Q)/4πε₀R.

To calculate the force on a straight wire in a magnetic field, use the formula:

• F = BIL

Where:

• F is the force on the wire (in Newtons)
• B is the magnetic field strength (in Teslas)
• I is the current (in Amperes)
• L is the length of the wire (in metres)

Given the values:

• Length of the wire (L) = 0.5 m
• Current (I) = 1.2 A
• Magnetic field (B) = 2 T

Now, substituting the values into the formula:

• F = 2 T × 1.2 A × 0.5 m
• F = 1.2 N

Thus, the force on the wire is 1.2 N.

In the process of radioactive decay, the emitted β-particles, which are negatively charged, originate from specific interactions within the nucleus.

• These β-particles are essentially electrons produced during the transformation of neutrons.
• When a neutron decays, it converts into a proton, releasing an electron in the process.
• This emitted electron is what we refer to as a β-particle.

Packing fraction = (M - A)/A , where M is the atomic mass and A is the mass number.

The energy released from the fission of a U²³⁵ nucleus is approximately:

• 200 MeV per fission event.

This energy release is significant and is a key factor in the functioning of nuclear reactors and weapons.

The maximum kinetic energy (K.E.) of photoelectrons can be calculated using the formula:

• K.E. = E - Φ

Where:

• E is the energy of the incident photon (3.4 eV).
• Φ is the work function of the metal (2 eV).

Now, substituting the values:

• K.E. = 3.4 eV - 2 eV
• K.E. = 1.4 eV

Thus, the maximum kinetic energy of the photoelectrons is 1.4 eV.

Work function W₀ = hc/λ

In a prism made of glass, the refractive index varies with light colour:

• The refractive index for red light is 1.520.
• The refractive index for blue light is 1.525.

As the refractive index increases, the angle of minimum deviation also changes. The relationship can be summarised as follows:

• A lower refractive index typically results in a smaller angle of minimum deviation.
• Thus, for the given values:
• D₁ (red light) is likely to be less than D₂ (blue light).

In conclusion, it can be inferred that:

• D₁ is less than D₂.

The condition for achieving dispersion with zero deviation when combining two prisms is derived from their dispersive powers and mean angular deviations.

• For two prisms, the dispersive powers are represented as ω and ω'.
• The mean angular deviations are denoted as d and d'.
• The condition for zero deviation with dispersion can be mathematically expressed.
• It involves the relationship between the dispersive powers and the mean deviations of both prisms.
• The correct expression reflects the balance needed to achieve this optical condition.

The Fermi level in an intrinsic semiconductor is positioned:

• Midway between the conduction band and valence band.
• This positioning indicates that the chances of finding electrons in the conduction band are equal to those in the valence band.
• It reflects the equilibrium state of the semiconductor, where no doping alters the balance of charge carriers.

In an n-p-n transistor, the following movements occur:

• Holes move from the emitter to the base.
• Holes then migrate from the base to the collector.
• Negative charge flows from the emitter to the base.
• Negative charge moves from the collector to the base.

In an n-p-n transistor, the p-type material serves as the base component.

This is a key part of the transistor's operation, as it plays a crucial role in controlling current flow. The following points explain its significance:

• Base: The p-type material is the base, which is sandwiched between two n-type materials.
• Control: The base allows the transistor to control the flow of current between the emitter and collector.
• Amplification: It enables the transistor to amplify signals, making it essential for various electronic applications.

The valency of an impurity added to germanium crystal to create a P-type semiconductor is 3.

P-type semiconductors are formed by introducing impurities that have fewer valence electrons than the semiconductor material itself.

• The base material, germanium, has a valency of 4.
• To achieve P-type characteristics, an impurity with a valency of 3, such as boron, is used.
• This results in the creation of "holes," which are positive charge carriers.

Thus, the correct valency of the impurity is 3.

When the P end of a P-N junction is connected to the negative terminal of the battery and the N end to the positive terminal of the battery, the P-N junction behaves like:

• The P-N junction becomes forward biased.
• Current can flow easily through the junction.
• This state allows the junction to function as a conductor.
• In this condition, it does not act as an insulator or a super-conductor.
• It effectively demonstrates the properties of a semiconductor.