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Question 51

Bihar Police ASI Operations 2026• Test

Which of the following thermodynamic law gives the concept of entropy?

  1. AOption (a)

    First law of thermodynamics

  2. BOption (b)

    Second law of thermodynamics

  3. COption (c)

    Third law of thermodynamics

  4. DOption (d)

    Zeroth law of thermodynamics

View answer & explanation

Correct answer

Answer: (b)B.

Second law of thermodynamics

Why this answer is correct

The concept of entropy is given by the Second Law of Thermodynamics.

Key Points Entropy is a fundamental concept in thermodynamics that measures the degree of disorder or randomness in a system. It is introduced by the Second Law of Thermodynamics, which states:

Second Law of Thermodynamics: In any irreversible process, the entropy of a closed system always increases, and for reversible processes, the entropy remains constant.

Entropy: A thermodynamic property denoted by S, representing the unavailability of a system's energy to do work.

SI Unit: Joule per Kelvin (J/K)

Formula: Change in entropy, ΔS = Qrev / T, where:

Qrev: Heat transferred during a reversible process

T: Temperature in Kelvin

Question 52

Bihar Police ASI Operations 2026• Test

Minimum energy required to free the electron from the ground state of hydrogen atom is -

  1. AOption (a)

    \( 10.2 \text{ eV} \)

  2. BOption (b)

    \( 11.3 \text{ eV} \)

  3. COption (c)

    \( 13.6 \text{ eV} \)

  4. DOption (d)

    \( 14.6 \text{ eV} \)

View answer & explanation

Correct answer

Answer: (c)C.

\( 13.6 \text{ eV} \)

Why this answer is correct

Concept

  • The ground state of a hydrogen atom corresponds to the lowest energy level, where the principal quantum number \(n = 1\).
  • To free an electron means to provide enough energy to move it from its bound state to an infinitely far distance from the nucleus (where its potential energy is zero).
  • The minimum energy required to remove an electron from an atom in its ground state is called the ionization energy.

Formula Used

The energy of an electron in the \(n^{th}\) orbit of a hydrogen-like atom is given by:

\(E_n = -\frac{13.6 \times Z^2}{n^2} \text{ eV}\)

Where:

  • \(Z\) is the atomic number (for Hydrogen, \(Z = 1\)).
  • \(n\) is the principal quantum number.

Calculation

Given:

For Hydrogen atom, \(Z = 1\).

For ground state, \(n = 1\).

Energy in the ground state:

\(E_1 = -\frac{13.6 \times (1)^2}{(1)^2} \text{ eV}\)

\(E_1 = -13.6 \text{ eV}\)

Energy required to free the electron (to reach \(n = \infty\), where \(E_\infty = 0\)):

\(\text{Required Energy} = E_\infty - E_1\)

\(\Rightarrow \text{Required Energy} = 0 - (-13.6 \text{ eV})\)

\(\Rightarrow \text{Required Energy} = 13.6 \text{ eV}\)

Hence, the minimum energy required to free the electron from the ground state of a hydrogen atom is \(13.6 \text{ eV}\).

Question 53

Bihar Police ASI Operations 2026• Test

An electron in the ground state of a hydrogen atom absorbs 12.09 eV energy. The angular momentum of the electron increases by

  1. AOption (a)

    \((\mathrm{h} / 2 \pi)\)

  2. BOption (b)

    \(2(h / 2 \pi)\)

  3. COption (c)

    \(3(\mathrm{~h} / 2 \pi)\)

  4. DOption (d)

    \(4(h / 2 \pi)\)

View answer & explanation

Correct answer

Answer: (b)B.

\(2(h / 2 \pi)\)

Why this answer is correct

Concept

  • According to Bohr's model of the hydrogen atom, energy levels are quantized, and the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2}$ eV.
  • When an electron absorbs energy, it transitions from a lower energy level (ground state, $n=1$) to a higher energy level ($n$).
  • Bohr's second postulate states that the angular momentum ($L$) of an electron is quantized and is given by $L = \frac{nh}{2\pi}$.

Formula Used

Energy of an electron in $n^{th}$ orbit:

\(E_n = -\frac{13.6}{n^2}\text{ eV}\)

Angular momentum of an electron:

\(L = \frac{nh}{2\pi}\)

Change in angular momentum:

\(\Delta L = L_{final} - L_{initial}\)

Calculation

Given:

Initial state (ground state), \(n_1 = 1\)

Energy in ground state, \(E_1 = -13.6\text{ eV}\)

Energy absorbed, \(\Delta E = 12.09\text{ eV}\)

The energy of the electron in the excited state ($E_n$) will be:

\(E_n = E_1 + \Delta E\)

\(E_n = -13.6 + 12.09 = -1.51\text{ eV}\)

Now, find the principal quantum number ($n$) for this energy level:

\(-1.51 = -\frac{13.6}{n^2}\)

\(n^2 = \frac{13.6}{1.51} \approx 9\)

\(n = 3\)

The electron transitions from \(n = 1\) to \(n = 3\).

Initial angular momentum (\(n=1\)):

\(L_1 = 1 \cdot \frac{h}{2\pi}\)

Final angular momentum (\(n=3\)):

\(L_3 = 3 \cdot \frac{h}{2\pi}\)

Increase in angular momentum:

\(\Delta L = L_3 - L_1\)

\(\Delta L = \frac{3h}{2\pi} - \frac{h}{2\pi}\)

\(\Delta L = \frac{2h}{2\pi} = 2\left(\frac{h}{2\pi}\right)\)

Hence, the angular momentum of the electron increases by \(2(h / 2 \pi)\).

Question 54

Bihar Police ASI Operations 2026• Test

What is the de Broglie wavelength associated with a ball having kinetic energy equal to 67.5 J? The mass of the ball is 150 g. (Planck's constant = \(6.63 \times 10 ^ {- 34} \)Js )

  1. AOption (a)

    1.04 x 10-34 m

  2. BOption (b)

    3.27 x 10-35 m

  3. COption (c)

    1.47 x 10-34 m

  4. DOption (d)

    4.66 x 10-35 m

View answer & explanation

Correct answer

Answer: (c)C.

1.47 x 10-34 m

Why this answer is correct

Calculation:

Given:

Kinetic energy, KE = 67.5 J

Mass of the ball, m = 150 g = 0.150 kg

Planck’s constant, h = 6.63 × 10-34 J s

The de Broglie wavelength λ is given by

λ = h / p

where p is the momentum of the ball. The momentum in terms of kinetic energy is

p = √(2 × m × KE)

Substituting the values:

p = √(2 × 0.150 × 67.5)

⇒ p = √(20.25)

⇒ p = 4.5 kg m s-1

Now, use the de Broglie relation:

λ = h / p

⇒ λ = (6.63 × 10-34) / 4.5

⇒ λ ≈ 1.47 × 10-34 m

Therefore, the correct answer is 1.47 × 10-34 m.

Question 55

Bihar Police ASI Operations 2026• Test

The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?

  1. AOption (a)

    Brackett series

  2. BOption (b)

    Paschen series

  3. COption (c)

    Lyman series

  4. DOption (d)

    Balmer series

View answer & explanation

Correct answer

Answer: (d)D.

Balmer series

Why this answer is correct

Explanation:

The Balmer series corresponds to the set of hydrogen spectral lines that appear in the visible region of the electromagnetic spectrum. These transitions occur when an electron falls to the n = 2 energy level.

The Lyman series lies in the ultraviolet region and is produced when electrons transition to the n = 1 level, resulting in higher-energy photon emissions.

The Paschen, Brackett, Pfund, and Humphrey series all lie in the infrared region. These series arise from electronic transitions to the n = 3, n = 4, n = 5, and n = 6 levels respectively, leading to lower-energy infrared radiation.

Question 56

Bihar Police ASI Operations 2026• Test

Given below are two statements :

Statement I : Stopping potential in photoelectric effect does not depend on the power of the light source.

Statement II : For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.

In the light of above statements, choose the most appropriate answer from the options given below.

  1. AOption (a)

    Statement I is incorrect but statement II is correct

  2. BOption (b)

    Both Statement I and Statement II are incorrect

  3. COption (c)

    Statement I is correct but statement II is incorrect 

  4. DOption (d)

    Both statement I and statement II are correct 

View answer & explanation

Correct answer

Answer: (d)D.

Both statement I and statement II are correct 

Why this answer is correct

Explanation:

The stopping potential (VS) is given by the equation:

VS = (K Emax) / e

Where K is a constant, Emax is the maximum kinetic energy of the emitted electrons, and e is the charge of an electron.

Alternatively, it can be expressed as:

VS = (hC / λ - φ) / e

Where h is Planck’s constant, C is the speed of light, λ is the wavelength of the incident light, and φ is the work function of the material.

           

From this expression, we can see that the stopping potential depends on the frequency (which is related to the wavelength λ) of the incident light, but it does not depend on the intensity or power of the light used.

Question 57

Bihar Police ASI Operations 2026• Test

A square loop of side a carries current I and has area vector along x̂. It is placed in a uniform magnetic field B = B ŷ. The torque on the loop is

  1. AOption (a)

    Ia2B ẑ

  2. BOption (b)

    -Ia2B ẑ

  3. COption (c)

    IaB x̂

  4. DOption (d)

    0

View answer & explanation

Correct answer

Answer: (a)A.

Ia2B ẑ

Why this answer is correct

Solution:

Concept:

A current loop behaves like a magnetic dipole. The torque on a magnetic dipole in a magnetic field is τ = μ × B.

Calculation:

The magnetic moment of the loop is

μ = IA n̂

Here,

A = a2

and the area vector is along x̂.

Therefore,

μ = Ia2x̂

The field is

B = B ŷ

Hence,

τ = μ × B = Ia2B(x̂ × ŷ)

Since x̂ × ŷ = ẑ,

τ = Ia2B ẑ

Question 58

Bihar Police ASI Operations 2026• Test

The force on a charged particle moving in a magnetic field is zero when:

  1. AOption (a)

    The velocity of the particle is perpendicular to the magnetic field

  2. BOption (b)

    The particle is at rest

  3. COption (c)

    The particle is moving in a circular path

  4. DOption (d)

    The velocity of the particle is parallel to the magnetic field

View answer & explanation

Correct answer

Answer: (d)D.

The velocity of the particle is parallel to the magnetic field

Why this answer is correct

The correct answer is The velocity of the particle is parallel to the magnetic field.

Key Points

  • The magnetic force acting on a charged particle moving through a magnetic field is governed by the Lorentz Force equation, which is expressed as F = q(v × B).
  • In terms of magnitude, the force is calculated using the formula F = qvB sin θ, where q is the charge, v is the velocity, B is the magnetic field induction, and θ is the angle between the velocity vector and the magnetic field vector.
  • When a particle moves parallel to the magnetic field, the angle θ between the velocity and the field is 0°.
  • Since the trigonometric value of sin 0° is exactly 0, the entire product qvB sin 0° becomes zero, resulting in no magnetic force being exerted on the particle.
  • Similarly, if the particle moves anti-parallel (opposite direction) to the field, the angle θ is 180°, and since sin 180° = 0, the force remains zero.

 Additional Information

  • Perpendicular Motion:
  • If the velocity is perpendicular to the magnetic field (θ = 90°), the force is at its maximum value (F = qvB) because sin 90° = 1.
  • This maximum force acts as a centripetal force, forcing the charged particle into a circular trajectory.
  • Particle at Rest:
  • If a charged particle is stationary (v = 0), the magnetic force is zero. This distinguishes magnetic fields from electric fields, as the latter can exert force on charges even at rest.
  • Work and Energy:
  • The magnetic force is always orthogonal to the direction of motion (velocity). Consequently, the work done by the magnetic field on the particle is zero, meaning the speed and kinetic energy of the particle remain constant.

Question 59

Bihar Police ASI Operations 2026• Test

In the following circuit diagram, when \(3\Omega\)  resistor is removed, then equivalent resistance of the network

  1. AOption (a)

    increases

  2. BOption (b)

    decreases

  3. COption (c)

    Cremains same

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

Cremains same

Why this answer is correct

The equivalent Wheatstone's bridge network of the given circuit is shown in figure. Here,

The points B and D are at the same potential as the bridge is balanced. So, the\(3\Omega\)  resistance in BD  arm is ineffective and can be omitted from the circuit.

Hence, equivalent resistance remains same when \(3\Omega\)   resistor is removed.

Question 60

Bihar Police ASI Operations 2026• Test

The Kirchoff's first law and second law are respectively based on -

  1. AOption (a)

    conservation of charge, conservation of energy

  2. BOption (b)

    conservation of charge, conservation of momentum

  3. COption (c)

    conservation of energy, conservation of charge

  4. DOption (d)

    conservation of momentum, conservation of charge

View answer & explanation

Correct answer

Answer: (a)A.

conservation of charge, conservation of energy

Why this answer is correct

Concept

  • Kirchhoff's First Law (Junction Rule): It states that the algebraic sum of currents meeting at any junction in a circuit is zero.
  • Kirchhoff's Second Law (Loop Rule): It states that the algebraic sum of the potential differences (including electromotive forces and resistive voltage drops) around any closed loop in a circuit is zero.

Explanation

1. Kirchhoff's First Law (Current Law):

According to this law, \(\sum I = 0\). This means the total current entering a junction must equal the total current leaving it. Since current is the rate of flow of electric charge (\(I = \frac{dq}{dt}\)), this law implies that charge cannot be created or destroyed at a junction. Thus, it is based on the conservation of charge.

2. Kirchhoff's Second Law (Voltage Law):

According to this law, \(\sum V = 0\) for a closed loop. This means that if a unit charge starts at a point and travels around a complete loop to return to the same point, the total work done on it must be zero. Since potential difference is defined as work done (energy) per unit charge, this law is a direct consequence of the conservation of energy.

Conclusion

The first law is based on the conservation of charge, and the second law is based on the conservation of energy.

Hence, Kirchhoff's first law and second law are respectively based on conservation of charge and conservation of energy.