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Past Questions — 2010
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.
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.
Even though we can get X-ray by the bombardment of any charged particles on a target, what is the reason for using electrons to produce X-rays?
A)
Electrons have smallest possible charge
B)
Electrons are common constituents of all matter
C)
Electron mass is very small
D)
Electrons can be produced very easily
To produce X-rays (specifically Bremsstrahlung or 'braking radiation'), a charged particle must be decelerated rapidly upon hitting a target. According to classical electromagnetism, the acceleration (or deceleration) of a charged particle is inversely proportional to its mass (F = ma, so a = F/m). Because electrons have an extremely small mass compared to other charged particles like protons or alpha particles, they undergo much more violent acceleration/deceleration when interacting with the electric fields of the target nuclei. This high rate of deceleration results in the emission of high-energy photons in the X-ray spectrum. While electrons are common (option b) and easy to produce (option d), these are practical conveniences rather than the physical reason why they are the most efficient particles for generating X-rays. The fundamental physical reason is their low mass, which allows for the necessary high acceleration to emit X-ray photons.
A CRT generally has
A)
electrostatic focusing
B)
magnetic focusing
C)
electromagnetic focusing
D)
optical focusing
A Cathode Ray Tube (CRT) requires a method to focus the electron beam into a small, sharp spot on the phosphor screen to ensure image clarity. There are two primary methods used in CRTs: Electrostatic focusing (using electric fields via anodes) and Electromagnetic focusing (using a magnetic lens created by a coil). While basic or low-cost CRTs use electrostatic focusing, the standard 'general' high-quality CRT (especially those used in oscilloscopes and older television sets) utilizes electromagnetic focusing because it allows for a more precise, adjustable focus and is more effective for beam deflection. In the context of technical examinations for electronics, 'electromagnetic focusing' is typically the preferred answer as it represents the more sophisticated and widely applicable technology for professional CRT applications. Option (a) is a method but less versatile; options (b) and (d) are incorrect because 'magnetic focusing' is a subset of electromagnetic focusing, and 'optical focusing' applies to light, not electron beams.
In 1935, Yukawa suggested that nuclear forces arise as a result of interchange of certain particles between nucleons. These particles are
A)
protons
B)
mesons
C)
photons
D)
positrons
In 1935, Japanese physicist Hideki Yukawa proposed a theory to explain the strong nuclear force that binds protons and neutrons (nucleons) together in the nucleus. He hypothesized that this force is mediated by the exchange of a particle with a mass intermediate between that of an electron and a proton. He named these particles 'mesons' (from the Greek word 'mesos', meaning middle). Specifically, the exchange of pi-mesons (pions) is responsible for the long-range part of the nuclear force.
Why other options are incorrect:
- a. Protons: These are the nucleons themselves, not the exchange particles that mediate the force between them.
- c. Photons: These are the gauge bosons for the electromagnetic force, not the strong nuclear force.
- d. Positrons: These are the antiparticles of electrons (leptons) and are not involved in the mediation of the strong nuclear force.
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.
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.
The moderators are required
A)
for thermalization of fast neutrons
B)
to increase the energy of the neutrons
C)
to increase the number of neutrons
D)
to decrease the number of delayed neutrons
In a nuclear reactor, the fission of Uranium-235 produces 'fast neutrons' which have very high kinetic energy. However, the probability (cross-section) of these fast neutrons inducing further fission in U-235 is very low. To sustain a chain reaction, these fast neutrons must be slowed down to 'thermal energies' (energies comparable to the ambient temperature of the medium). This process of slowing down neutrons is called 'thermalization'. A moderator (such as graphite, heavy water, or light water) is a material composed of light nuclei that slows down neutrons through successive elastic collisions.
Analysis of other options:
- Option (b) is incorrect because moderators decrease the kinetic energy of neutrons, they do not increase it.
- Option (c) is incorrect because moderators do not create new neutrons; that is the role of the fission process itself.
- Option (d) is incorrect because delayed neutrons are a result of radioactive decay of fission fragments and are not managed or decreased by the moderator.
The half life of a substance is 20 minutes. What is the time interval between 33% decay and 67% decay?
A)
40 minutes
B)
20 minutes
C)
30 minutes
D)
25 minutes
Based on Mizo customary knowledge / general exam pattern.
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.
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.
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.
Positive rays
A)
are coming from anode of the discharge tube
B)
are canal rays
C)
consists of positively charged positrons
D)
consists of a particles
Based on Mizo customary knowledge / general exam pattern.
The circuits used to generate their own sinusoidal signals are called
A)
generators
B)
mixers
C)
oscillators
D)
modulators
An oscillator is an electronic circuit that produces a periodic, oscillating electronic signal, often a sine wave or a square wave, without requiring an external input signal. It converts direct current (DC) from a power supply into an alternating current (AC) signal. This is typically achieved using a feedback loop (positive feedback) and an amplifier.
Why other options are incorrect:
- a. Generators: While electrical generators produce AC, in the context of electronics circuits, the specific term for a signal-generating circuit is an oscillator. 'Generator' is a broader term often referring to mechanical-to-electrical energy conversion.
- b. Mixers: A mixer is a circuit that takes two different frequencies and produces a new frequency (the sum or difference of the two), rather than generating its own signal from scratch.
- d. Modulators: A modulator is a circuit that varies a property of a high-frequency carrier signal (like amplitude or frequency) based on a modulating signal to transmit information.
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°.
U238 is bombarded with slow neutrons. The results in the
A)
formation of an atom with atomic number 93
B)
formation of an isotope of Uranium with mass number 239
C)
splitting up of uranium nucleus into two nearly equal fragments
D)
splitting up of Uranium nucleus into two exactly equal fragments
Step-by-step reasoning:
1. The process described is neutron capture. When a Uranium-238 (U-238) nucleus absorbs a slow (thermal) neutron, it becomes Uranium-239 (U-239).
2. The nuclear equation for this step is: ^{238}_{92}U + ^{1}_{0}n \rightarrow ^{239}_{92}U.
3. U-239 is unstable and undergoes beta-minus decay to become Neptunium-239 (atomic number 93), which then decays into Plutonium-239. However, the immediate result of the bombardment is the formation of U-239.
4. Evaluating options:
- Option (a) refers to the subsequent decay product (Neptunium), not the immediate result of bombardment.
- Option (b) correctly identifies the immediate product: an isotope of Uranium with mass number 239.
- Options (c) and (d) describe nuclear fission. While U-235 is fissile (splits easily with slow neutrons), U-238 is 'fertile' and generally requires fast neutrons to undergo fission. Slow neutrons primarily lead to capture in U-238, not splitting.
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.
Consider an α, β and γ-particles each having an energy 1 MeV. Arrange these particles in order of increasing distances that they travel in air
A)
α, β, γ
B)
γ, β, α
C)
γ, α, β
D)
α, γ, β
The distance a particle travels in air (penetrating power) depends on its mass, charge, and nature of interaction with matter. Given that all three particles have the same energy (1 MeV), we compare their properties:
1. Alpha (α) particles: These are helium nuclei consisting of 2 protons and 2 neutrons. They have a large mass and a high positive charge (+2e). Due to their size and charge, they interact strongly with air molecules via Coulombic forces, losing energy rapidly. Consequently, they have the shortest range in air (typically a few centimeters).
2. Beta (β) particles: These are high-energy electrons (or positrons). They have a much smaller mass than alpha particles and a lower charge (-1e or +1e). Because they are smaller and less charged, they interact less frequently with matter than alpha particles, allowing them to travel further (typically a few meters).
3. Gamma (γ) rays: These are high-energy electromagnetic photons. They have zero mass and zero electric charge. They interact with matter much less frequently than charged particles (via photoelectric effect, Compton scattering, or pair production), giving them the highest penetrating power and the longest range in air (potentially hundreds of meters).
Therefore, the order of increasing distance traveled in air is Alpha < Beta < Gamma.
Characteristic or line X-ray spectra are not likely to be excited by the following
A)
Electron bombardment
B)
Irradiation by short X-rays
C)
Proton bombardment
D)
Neutron absorption
Characteristic X-rays are produced when an inner-shell electron (usually from the K or L shell) is ejected from an atom, creating a vacancy. When an electron from a higher energy outer shell drops down to fill this vacancy, a photon with energy equal to the difference between the two shells is emitted. This energy is unique (characteristic) to the element.
- Option (a) Electron bombardment: High-energy electrons can knock out inner-shell electrons via collisions, triggering characteristic X-ray emission.
- Option (b) Irradiation by short X-rays: High-energy (short wavelength) X-ray photons can ionize inner shells via the photoelectric effect, triggering characteristic X-ray emission.
- Option (c) Proton bombardment: High-energy protons can interact with inner-shell electrons through Coulombic forces, ejecting them and triggering characteristic X-ray emission.
- Option (d) Neutron absorption: Neutrons are electrically neutral. They interact primarily with the nucleus, not the electrons. While neutron capture can lead to gamma-ray emission or internal conversion (which might indirectly lead to X-rays), the process of 'neutron absorption' itself does not directly excite the electronic transitions required for characteristic X-ray spectra in the same way that charged particles or high-energy photons do. Therefore, it is the least likely method among the choices to excite characteristic X-ray spectra.
After the completion of the average life of a radioactive substance, the activity is
A)
reduced to zero
B)
increased to infinity
C)
reduced to a small value
D)
maintained as a constant
1. Understanding Radioactive Decay: Radioactive decay follows an exponential decay law. The activity (A) of a radioactive substance at time (t) is given by the formula A = A₀e^(-λt), where A₀ is the initial activity and λ is the decay constant.
2. Average Life (τ): The average life (or mean life) is the reciprocal of the decay constant (τ = 1/λ). It represents the average time a nucleus survives before decaying.
3. Activity after Average Life: To find the activity after one average life, we substitute t = τ = 1/λ into the decay equation:
A = A₀e^(-λ * (1/λ)) = A₀e^(-1)
4. Calculation: The value of e^(-1) is approximately 1/2.718 ≈ 0.368. This means that after one average life, the activity is reduced to approximately 36.8% of its original value.
5. Evaluating Options:
- (a) Reduced to zero: Incorrect. Exponential decay asymptotically approaches zero but never reaches it in a finite time.
- (b) Increased to infinity: Incorrect. Activity decreases over time.
- (c) Reduced to a small value: Correct. 36.8% is a significant reduction from the original activity.
- (d) Maintained as a constant: Incorrect. Activity decreases continuously as the number of radioactive nuclei decreases.
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.