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Past Questions — 2010
The radius of a Hydrogen atom in its ground state is
A)
10⁻⁴ cm
B)
10⁻⁶cm
C)
10⁻⁸cm
D)
10⁻¹⁰cm
Bohr radius a0=0.529Å≈0.5×10^-8 cm, order 10^-8 cm.
For a free particle of momentum ħk moving in one dimension along the positive x-axis, the unnormalized wave function is
A)
sin kx
B)
cos kx
C)
eⁱᵏˣ
D)
e⁻ⁱᵏˣ
1. **Definition of a Free Particle**: A free particle is one where the potential energy V(x) = 0 everywhere. The time-independent Schrödinger equation for such a particle in one dimension is: - (ħ²/2m) d²ψ/dx² = Eψ.
2. **Wave Function Form**: The general solution to this differential equation is a plane wave of the form ψ(x) = Aeⁱᵏˣ + Be⁻ⁱᵏˣ, where k = √(2mE)/ħ.
3. **Momentum Operator**: In quantum mechanics, the momentum operator is given by p̂ = -iħ(d/dx). For a particle with a definite momentum p = ħk, the wave function must be an eigenfunction of the momentum operator: p̂ψ = pψ.
4. **Applying the Operator**:
- For ψ = eⁱᵏˣ: -iħ(d/dx)(eⁱᵏˣ) = -iħ(ik)eⁱᵏˣ = ħkeⁱᵏˣ. This corresponds to a particle moving in the positive x-direction with momentum +ħk.
- For ψ = e⁻ⁱᵏˣ: -iħ(d/dx)(e⁻ⁱᵏˣ) = -iħ(-ik)e⁻ⁱᵏˣ = -ħke⁻ⁱᵏˣ. This corresponds to a particle moving in the negative x-direction with momentum -ħk.
5. **Evaluating Options**:
- Option (a) sin(kx) and (b) cos(kx) are linear combinations of eⁱᵏˣ and e⁻ⁱᵏˣ; they represent standing waves (superpositions of particles moving in both directions) and do not have a definite momentum of ħk.
- Option (d) represents a particle moving in the negative x-direction.
- Option (c) eⁱᵏˣ correctly represents a particle moving in the positive x-direction with momentum ħk.
A sodium atom has 11 electrons around the nucleus. The eleventh electron as you go on filling up the shell from K shell is a
A)
s-electron
B)
p-electron
C)
d-electron
D)
f-electron
To determine the nature of the eleventh electron in a sodium (Na) atom, we follow the Aufbau principle, which dictates the order in which electrons fill atomic orbitals based on increasing energy levels.
1. **Atomic Number**: Sodium has an atomic number of 11, meaning it has 11 electrons.
2. **Filling Order**:
- **K shell (n=1)**: The first shell contains the 1s orbital. It can hold a maximum of 2 electrons. (Electrons 1-2 fill 1s).
- **L shell (n=2)**: The second shell contains the 2s and 2p orbitals.
- The 2s orbital holds 2 electrons (Electrons 3-4).
- The 2p orbital holds 6 electrons (Electrons 5-10).
- **M shell (n=3)**: After the L shell is full (10 electrons), the 11th electron enters the next lowest energy orbital.
3. **The 11th Electron**: According to the Aufbau sequence (1s → 2s → 2p → 3s), the 11th electron enters the **3s orbital**.
Since the electron resides in an 's' orbital, it is an s-electron.
- Option b (p-electron) is incorrect because the 3p orbital only begins filling after the 3s orbital is full.
- Option c (d-electron) and d (f-electron) are incorrect as these orbitals belong to higher energy levels or different shells entirely.
The output frequency of a full wave rectifier is
A)
the same as input frequency
B)
one half of the input frequency
C)
double of the input frequency
D)
independent of the input frequency
1. A rectifier converts alternating current (AC), which periodically reverses direction, into direct current (DC), which flows in one direction.
2. In a Half-Wave Rectifier, only one half-cycle of the input sine wave is passed to the output. Therefore, the output frequency is equal to the input frequency (f_out = f_in).
3. In a Full-Wave Rectifier (whether center-tapped or bridge), both the positive and negative half-cycles of the input AC signal are converted into positive pulses at the output.
4. Because the negative half-cycle is 'flipped' to become positive, the output waveform repeats itself twice for every one full cycle of the input waveform.
5. Mathematically, if the input frequency is f, the output ripple frequency is 2f. For example, if the input is 50 Hz, the output of a full-wave rectifier will have a ripple frequency of 100 Hz.
6. Therefore, the output frequency is double the input frequency.
Which of the following are not electromagnetic waves?
A)
Cosmic rays
B)
γ-rays
C)
β-rays
D)
X-rays
Beta rays are electrons (particles), not EM waves; gamma and X-rays are EM. Cosmic rays also particles but beta is textbook non-EM example.
The critical mass of a fissionable material can be reduced by
A)
heating it
B)
cooling it
C)
adding impurities to it
D)
surrounding it with a shield that will reflect neutrons
Critical mass is the minimum amount of fissionable material (like Uranium-235 or Plutonium-239) required to maintain a self-sustaining nuclear chain reaction. The primary factor that determines whether a mass is critical is the rate of neutron leakage from the surface of the material.
1. **Neutron Reflection (Option d):** When a fissionable mass is surrounded by a 'neutron reflector' (a shield made of materials like beryllium or graphite), neutrons that would otherwise escape the system are reflected back into the core. This increases the neutron flux within the material, allowing a smaller mass to achieve criticality. Therefore, a reflector reduces the required critical mass.
2. **Heating/Cooling (Options a & b):** While temperature can affect the cross-section of neutron absorption (Doppler broadening), it is not the primary method used to reduce the critical mass of a material in a fundamental sense; in many cases, heating can actually increase the critical mass due to thermal expansion (decreasing density).
3. **Adding Impurities (Option c):** Impurities generally act as 'neutron poisons' (materials with high neutron absorption cross-sections). Adding impurities increases the number of neutrons absorbed without causing fission, which increases the amount of material needed to sustain a reaction, thereby increasing the critical mass.
The dielectric between the conductors reduces the electric intensity
A)
to zero
B)
between them
C)
with no charge
D)
None of these
When a dielectric material is inserted between two charged conductors, the electric field (electric intensity) within the dielectric is reduced. This happens because the external electric field polarizes the molecules of the dielectric, creating an induced internal electric field that opposes the original external field. The net electric field (E_net) is given by E_net = E_0 / K, where E_0 is the original field and K is the dielectric constant. Therefore, the dielectric reduces the electric intensity specifically in the region 'between them' (the space occupied by the dielectric). Option (a) is incorrect because the field is reduced, not necessarily to zero (unless it is a perfect conductor). Option (c) is irrelevant to the mechanism of field reduction.
Among the configurations the lowest input impedance is obtained by
A)
CE
B)
CC
C)
CB
D)
both CB and CC
Based on Mizo customary knowledge / general exam pattern.
N-type semiconductor is obtained by doping germanium or silicon with an impurity like
A)
Aluminium
B)
Phosphorus
C)
Indium
D)
Gallium
To create an N-type (negative-type) semiconductor, a pure intrinsic semiconductor like Silicon (Si) or Germanium (Ge), which belong to Group IV of the periodic table (4 valence electrons), must be doped with a pentavalent impurity (Group V elements). Pentavalent impurities have 5 valence electrons; four form covalent bonds with the surrounding semiconductor atoms, leaving one extra electron free to move, thereby increasing the conductivity of the material via negative charge carriers.
1. Phosphorus (P) is a Group V element (pentavalent), making it the correct dopant for N-type semiconductors.
2. Aluminium (Al), Indium (In), and Gallium (Ga) are all Group III elements (trivalent). Doping with these elements creates 'holes' (absence of electrons), resulting in a P-type (positive-type) semiconductor.
During a negative beta decay
A)
an atomic electron is ejected
B)
an electron which is already present within the nucleus is ejected
C)
a neutron in the nucleus decays emitting an electron
D)
a part of the binding energy of the nucleus is converted into an electron
Negative beta decay ($\beta^-$ decay) occurs when a nucleus has too many neutrons to be stable. In this process, a neutron within the nucleus is transformed into a proton, an electron (the beta particle), and an electron antineutrino. The nuclear equation is: $n \rightarrow p + e^- + \bar{\nu}_e$.
Analysis of options:
- Option (a) is incorrect because the electron is not an 'atomic' electron (which orbits the nucleus) but is created inside the nucleus during the decay.
- Option (b) is incorrect because electrons are not 'present' within the nucleus prior to the decay process; they are leptons and do not exist as stable constituents of the nucleus.
- Option (c) is correct as it accurately describes the fundamental process of a neutron decaying into a proton and emitting an electron.
- Option (d) is incorrect because while mass-energy equivalence ($E=mc^2$) governs the process, the electron is a product of the weak nuclear interaction transforming a quark (down quark to up quark), not simply a conversion of binding energy into a particle.
∇²V = -4πρ represents
A)
Maxwell’s equation
B)
Laplace’s equation
C)
Poisson’s equation
D)
None of these
The given equation is ∇²V = -4πρ.
1. Analysis of the Equation:
- ∇² (the Laplacian operator) represents the divergence of the gradient of the scalar potential V.
- ρ represents the volume charge density.
- In Gaussian units (CGS), the equation is written as ∇²V = -4πρ. In SI units, it is written as ∇²V = -ρ/ε₀.
2. Identification:
- Poisson's Equation: This is the general form of the equation relating the electrostatic potential to the charge distribution in a region of space. It is expressed as ∇²V = -ρ/ε₀ (SI) or ∇²V = -4πρ (CGS).
3. Evaluating other options:
- Laplace's Equation: This is a special case of Poisson's equation where the charge density ρ = 0. It is written as ∇²V = 0. Since the given equation includes ρ, it is not Laplace's equation.
- Maxwell's Equations: These are a set of four fundamental equations that describe how electric and magnetic fields are generated and altered by each other and by charges and currents. While Poisson's equation is derived from one of Maxwell's equations (Gauss's Law), the specific mathematical form ∇²V = -4πρ is specifically known as Poisson's equation.
Therefore, the correct option is c.
Crystalline solids are
A)
anistropic
B)
isotropic
C)
amporphus
D)
None of these
Step-by-step reasoning: 1. Definition of Crystalline Solids: Crystalline solids are materials in which the constituent particles (atoms, ions, or molecules) are arranged in a highly ordered, repeating three-dimensional lattice structure. 2. Understanding Anisotropy: Anisotropy refers to the property of a material where its physical properties (such as electrical conductivity, refractive index, or thermal expansion) vary depending on the direction in which they are measured. Because the arrangement of atoms in a crystal lattice differs along different axes, the physical properties also differ. Therefore, crystalline solids are anisotropic. 3. Evaluating other options: (b) Isotropic: Isotropic materials have the same properties in all directions. This is a characteristic of amorphous solids (like glass), not crystalline solids. (c) Amorphous: This is a different category of solids characterized by a lack of long-range order; crystalline solids are the opposite of amorphous solids. (d) None of these: Incorrect, as option (a) is correct.
The photoelectric current in a photoelectric cell depends upon
A)
The nature of the metal used as the counter
B)
The wavelength of the incident light
C)
The intensity of the incident light
D)
All the above parameters
In the photoelectric effect, the photoelectric current is defined as the rate of flow of photoelectrons from the cathode to the anode. According to Einstein's photoelectric equation and experimental observations: 1. Photoelectric current is directly proportional to the intensity of the incident light (provided the frequency of the light is above the threshold frequency). This is because higher intensity means more photons striking the surface per unit time, which results in more electrons being ejected. 2. The wavelength (or frequency) of the incident light determines whether electrons are ejected at all (threshold frequency) and determines the maximum kinetic energy of the ejected electrons, but it does not determine the quantity of current once the threshold is exceeded. 3. The nature of the metal (work function) determines the threshold frequency, but not the current magnitude for a given intensity. Therefore, the current depends specifically on the intensity of the incident light.
Bohr magneton is the unit of moment of electron which has spin angular momentum equal to
A)
zero
B)
ħ/2
C)
3ħ/2
D)
5ħ/2
Bohr magneton corresponds to orbital moment with ħ; spin ħ/2 gives 1μB.
The difference between an amplifier and an oscillator is that an oscillator
A)
needs no external input signal
B)
needs no d.c. power supply
C)
delivers a reshaped version of the input signal
D)
delivers output signal without amplification of input signal
An amplifier is a device that increases the amplitude of an existing external input signal. It requires an input signal to function. In contrast, an oscillator is a circuit that generates its own periodic signal (such as a sine wave or square wave) without requiring an external AC input signal. It achieves this by using a feedback loop (positive feedback) and a frequency-determining network to convert DC power from a power supply into an AC signal.
Evaluating the options:
- Option (a) is correct because an oscillator is a self-starting signal generator.
- Option (b) is incorrect because both amplifiers and oscillators require a DC power supply (biasing) to operate the active components (transistors/op-amps).
- Option (c) is incorrect because reshaping a signal is typically the function of a wave-shaping circuit or a comparator, not the defining difference between an amplifier and an oscillator.
- Option (d) is incorrect because an oscillator actually involves amplification (the loop gain must be at least 1) to sustain oscillations; it does not 'deliver output without amplification'.
Suppose each state voltage gain is about 100, then the theoretical maximum voltage gain from a R. C coupled two stage transistor amplifier is
A)
100
B)
greater than 100 but less than 200
C)
200
D)
10,000
In a multi-stage amplifier, the total voltage gain (Av_total) is the product of the individual voltage gains of each stage. This is because the output of the first stage serves as the input to the second stage.
Given:
- Voltage gain of the first stage (Av1) = 100
- Voltage gain of the second stage (Av2) = 100
The theoretical maximum voltage gain for a two-stage RC coupled amplifier is calculated as:
Av_total = Av1 × Av2
Av_total = 100 × 100 = 10,000
Analysis of options:
- Option (a) 100: This is the gain of only one stage.
- Option (b) and (c): These assume additive gain (100 + 100), which is incorrect for cascaded amplifiers.
- Option (d) 10,000: This correctly applies the multiplicative property of cascaded gain.
The Schrödinger wave equation is
A)
the general differential equation of wave motion
B)
tatisfying the Broglie relation
C)
satisfying the uncertainty relation
D)
to obtain the energy state of the harmonic oscillator
The Schrödinger equation is the fundamental equation of quantum mechanics. It describes how the quantum state of a physical system (represented by the wave function Ψ) changes with time. Mathematically, it is a linear partial differential equation that serves as the 'general differential equation of wave motion' for matter waves, analogous to how the classical wave equation describes sound or light waves.
Evaluating the other options:
- Option (b): While the Schrödinger equation is consistent with the de Broglie relation (p = h/λ), it is not 'defined' as satisfying it; rather, the de Broglie relation is a prerequisite for the formulation of the equation.
- Option (c): The Heisenberg Uncertainty Principle is a consequence of the mathematical properties of the wave function (Fourier transforms), not the definition of the equation itself.
- Option (d): Obtaining the energy states of a harmonic oscillator is a specific application (a boundary value problem) of the Schrödinger equation, not the definition of the equation itself.
In nuclear reaction there is conservation of
A)
mass only
B)
energy only
C)
momentum only
D)
mass, energy and momentum
In nuclear reactions, several fundamental conservation laws must be obeyed:
1. **Conservation of Momentum**: Linear momentum is conserved in all nuclear reactions, as there are no external forces acting on the system.
2. **Conservation of Energy**: While 'rest mass' is not conserved (mass is converted to energy via E=mc²), the total energy (including rest mass energy and kinetic energy) is strictly conserved.
3. **Conservation of Mass (Mass Number)**: In the context of nuclear physics, the total number of nucleons (protons + neutrons) remains constant before and after the reaction. This is often referred to as the conservation of mass number.
Therefore, while mass-energy is interconvertible, the total quantity of mass-energy, the total nucleon count (mass number), and the momentum are all conserved. Option (d) is the most comprehensive and correct answer.
The work function is
A)
the same for all metal surfaces
B)
the same for all type of surfaces
C)
different for different metal surfaces
D)
dependent on the frequency of the incident light
Work function depends on material composition and surface conditions, varies between metals.
According to Bohr’s theory of the Hydrogen atom, the speed v of the electron in a stationary orbit n is related to the principal quantum number n as (C is a constant)
A)
v = C/n²
B)
v = C/n
C)
v = C x n
D)
v = C × n²
According to Bohr's model of the hydrogen atom, the electron moves in circular orbits where the angular momentum is quantized. The quantization condition is given by: mvr = nh/(2π), where m is the mass of the electron, v is the velocity, r is the radius of the orbit, and n is the principal quantum number (n = 1, 2, 3...).
Additionally, for a stable orbit, the centripetal force is provided by the electrostatic attraction between the nucleus and the electron:
(mv²/r) = (Ze²)/(4πε₀r²), where Z is the atomic number (for Hydrogen, Z=1).
By solving these two equations simultaneously for velocity (v), we find that:
v = (Ze²)/(2ε₀nh)
In this expression, Z, e, ε₀, n, and h are constants except for n. Therefore, we can group the constants as C = (Ze²)/(2ε₀h).
This simplifies the relationship to: v = C/n.
This means the velocity of the electron is inversely proportional to the principal quantum number n. As n increases (the electron moves to a higher energy level), the velocity of the electron decreases.
Comparing this to the options:
- Option (a) v = C/n² is incorrect (this describes the energy levels, where E ∝ 1/n²).
- Option (b) v = C/n is correct.
- Option (c) v = C x n is incorrect.
- Option (d) v = C x n² is incorrect.
Transformers are used in
A)
dc circuits alone
B)
ac circuits alone
C)
both ac and dc circuits
D)
integrated circuits
Transformers operate on the principle of Faraday's Law of Electromagnetic Induction. For a transformer to induce a voltage in the secondary coil, there must be a continuously changing magnetic flux in the core.
1. In AC (Alternating Current) circuits, the current changes direction and magnitude periodically, which creates a time-varying magnetic field. This changing flux induces an electromotive force (EMF) in the secondary winding, allowing the transformer to function.
2. In DC (Direct Current) circuits, the current is constant. Once the steady-state magnetic field is established, the magnetic flux does not change over time (dΦ/dt = 0). According to Faraday's Law, no voltage is induced in the secondary winding. Furthermore, since there is no back-EMF in a DC circuit, the primary winding would act as a simple low-resistance wire, potentially leading to overheating and burnout of the coil.
Therefore, transformers are exclusively used in AC circuits.
An electric dipole of moment p is placed in the position of stable equilibrium in a uniform electric field of intensity E. The torque required to rotate, when the dipole makes an angle q with the initial position is
A)
zero
B)
π/4
C)
π
D)
π/2
At stable equilibrium p∥E, θ=0, torque pE sinθ=0; maximum at 90°.
When transistor is operating in active region, collector junction is
A)
reverse biased of NPN transistor
B)
reverse biased for PNP transistor
C)
forward biased in both NPN and PNP transistor
D)
reverse biased in both NPN and PNP transistor
To operate a Bipolar Junction Transistor (BJT) in the 'Active Region' (which is the required region for linear amplification), specific biasing conditions must be met for its two junctions: the Emitter-Base (EB) junction and the Collector-Base (CB) junction.
1. Emitter-Base Junction: This junction must be Forward Biased to allow majority carriers to be injected from the emitter into the base.
2. Collector-Base Junction: This junction must be Reverse Biased. This creates a depletion region and a strong electric field that sweeps the carriers (electrons in NPN, holes in PNP) from the base into the collector.
This rule applies regardless of the transistor type:
- For an NPN transistor: The EB junction is forward biased (P-type base positive relative to N-type emitter) and the CB junction is reverse biased (N-type collector positive relative to P-type base).
- For a PNP transistor: The EB junction is forward biased (N-type base negative relative to P-type emitter) and the CB junction is reverse biased (P-type collector negative relative to N-type base).
Therefore, in the active region, the collector junction is reverse biased for both NPN and PNP transistors.
When the speed of electrons increases, the specific charge
A)
increases
B)
decreases
C)
remains unchanged
D)
increase and then decreases
The 'specific charge' of a particle is defined as the ratio of its electric charge (q) to its mass (m), expressed as q/m. For an electron, the charge (e ≈ 1.602 × 10⁻¹⁹ C) and the rest mass (m₀ ≈ 9.109 × 10⁻³¹ kg) are fundamental physical constants. In the context of classical physics and standard competitive exam questions of this level, the specific charge is treated as a constant property of the particle, regardless of its velocity. While relativistic mass increases as speed approaches the speed of light, the term 'specific charge' in general physics refers to the intrinsic property of the particle. Since the charge of the electron does not change with speed and the intrinsic mass remains constant, the specific charge remains unchanged. Therefore, option (c) is the correct answer.
A free neutron decays into a proton, an electron and
A)
a neutrino
B)
an antineutrino
C)
an alpha particle
D)
a beta particle
The process described is known as Beta-minus ($\beta^-$) decay. A free neutron is unstable and has a mean lifetime of about 880 seconds. It decays via the weak nuclear force according to the following equation: $n \rightarrow p + e^- + \bar{\nu}_e$.
1. **Conservation of Charge**: The neutron (charge 0) decays into a proton (+1) and an electron (-1), keeping the net charge 0.
2. **Conservation of Lepton Number**: The electron is a lepton (lepton number +1). To conserve the total lepton number of the universe (which was 0 before the decay), an antilepton must be produced. An electron antineutrino has a lepton number of -1, balancing the equation.
3. **Conservation of Energy and Momentum**: The discovery of the neutrino/antineutrino was originally proposed by Wolfgang Pauli to explain the continuous energy spectrum of the emitted electron, as a two-body decay would have resulted in a discrete energy peak.
Why other options are incorrect:
- (a) A neutrino is produced in $\beta^+$ decay (positron emission), not $\beta^-$ decay.
- (c) An alpha particle consists of two protons and two neutrons; it is far too massive to be a product of a single neutron decay.
- (d) A beta particle is simply another name for the electron (or positron) already mentioned in the question prompt.