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Past Questions — 2010
The particle which most easily penetrates through the nucleus of an atom is
A)
neutron
B)
electron
C)
proton
D)
alpha particle
To determine which particle penetrates the nucleus most easily, we must consider the forces acting between the particle and the nucleus. The nucleus of an atom consists of protons and neutrons and is positively charged.
1. **Coulombic Repulsion**: Protons, alpha particles (which consist of two protons and two neutrons), and electrons are all charged particles. Protons and alpha particles are positively charged, meaning they experience strong electrostatic repulsion (the Coulomb barrier) as they approach the positively charged nucleus. This repulsion prevents them from entering the nucleus unless they possess extremely high kinetic energy.
2. **Neutrons**: A neutron is electrically neutral. Because it carries no charge, it does not experience the Coulombic repulsion from the nucleus. This allows it to approach and penetrate the nucleus much more easily than charged particles, regardless of its energy level.
3. **Electrons**: While electrons are negatively charged and are attracted to the nucleus, they are very light and typically exist in orbitals around the nucleus. In the context of 'penetrating' the nucleus from the outside (as in bombardment experiments), the neutral charge of the neutron makes it the most effective probe for nuclear reactions.
Therefore, the neutron is the particle that most easily penetrates the nucleus.
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.
What is the average binding energy of a nucleon in the nucleus of an atom?
A)
7.8 eV
B)
7.8 KeV
C)
7.8 MeV
D)
78 eV
Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons (nucleons). The 'average binding energy per nucleon' is a key characteristic of nuclear stability. For most stable nuclei (excluding the very lightest ones like Hydrogen or Deuterium), the binding energy per nucleon typically ranges from approximately 7 to 8.8 MeV. Specifically, for a wide range of medium-to-heavy nuclei, the average value is often cited around 7.8 to 8 MeV.
Evaluating the options:
- Option (a) 7.8 eV: This is far too low; electron binding energies in atoms are in this range, not nuclear binding energies.
- Option (b) 7.8 KeV: This is still too low by a factor of 1,000.
- Option (c) 7.8 MeV: This aligns with the established physical constant for nuclear binding energy (Mega-electronvolts).
- Option (d) 78 eV: This is far too low.
Therefore, the only physically plausible answer in the context of nuclear physics is 7.8 MeV.
An electric dipole placed in a non-uniform electric field experiences
A)
both a torque and a net force
B)
only a force but no torque
C)
only a torque but no net force
D)
no torque and no net force
An electric dipole consists of two equal and opposite charges (+q and -q) separated by a small distance d.
1. **Net Force**: In a uniform electric field, the force on +q (+qE) and -q (-qE) cancel each other out, resulting in a net force of zero. However, in a **non-uniform electric field**, the magnitude of the electric field is different at the positions of the two charges (E₁ ≠ E₂). Therefore, the forces acting on the two charges do not cancel out, resulting in a non-zero net force (F_net = q(E₂ - E₁)).
2. **Torque**: Torque (τ = p × E) occurs whenever the dipole moment (p) is not aligned with the electric field. In a non-uniform field, as long as the dipole is not perfectly aligned with the local field lines, it will experience a torque that attempts to rotate it into alignment.
Since a non-uniform field creates a difference in force magnitude between the two poles and generally maintains a directional component relative to the dipole's orientation, the dipole experiences both a translational force and a rotational torque.
Electric Intensity of a point varies as r⁻¹ for
A)
a point charge
B)
spherically symmetric charge distribution
C)
a plane infinite sheet of charge
D)
a line charge of infinite length
To determine how the electric intensity (Electric Field, E) varies with distance (r), we analyze the standard formulas for different charge distributions:
1. **Point Charge (Option a):** According to Coulomb's Law, the electric field is given by E = (1 / 4πε₀) * (q / r²). Thus, E ∝ r⁻², which is an inverse-square law.
2. **Spherically Symmetric Charge Distribution (Option b):** For a point outside a sphere of charge, the shell theorem states it behaves like a point charge at the center. Thus, E ∝ r⁻².
3. **Infinite Plane Sheet of Charge (Option c):** Using Gauss's Law, the electric field is E = σ / (2ε₀), where σ is the surface charge density. This value is constant and does not depend on the distance r (E ∝ r⁰).
4. **Infinite Line Charge (Option d):** Using Gauss's Law for a cylindrical surface around a line charge with linear charge density λ, the electric field is given by E = λ / (2πε₀r). In this case, E is inversely proportional to the distance r, meaning E ∝ r⁻¹.
Therefore, the electric intensity varies as r⁻¹ only for an infinite line charge.
Full wave rectifier is preferred over half wave rectifier because of the difference in
A)
the output level
B)
frequency
C)
ripples
D)
construction
A rectifier converts alternating current (AC) to direct current (DC). The 'ripple' refers to the residual AC component remaining in the DC output.
1. **Half-Wave Rectifier**: Only one half-cycle of the AC input is used, while the other is blocked. This results in large gaps between pulses, leading to a high ripple factor (1.21). This makes the output very 'un-smooth' and difficult to filter.
2. **Full-Wave Rectifier**: Both half-cycles of the AC input are converted to DC. This results in a continuous series of pulses with no gaps, significantly reducing the ripple factor (0.482).
Because the ripple is much lower in a full-wave rectifier, it is far easier to smooth the output using a filter capacitor to obtain a steady DC voltage. While full-wave rectifiers have a higher output level (average voltage) and higher ripple frequency, the primary technical reason they are 'preferred' in power supply design is the significant reduction in the ripple factor, which improves efficiency and output quality.
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.
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 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.
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.
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.
Beta rays emitted by a radioactive material are
A)
electromagnetic radiation
B)
the electrons orbiting around the nucleus
C)
charged particles emitted by the nucleus
D)
neutral particles
Beta decay is a type of radioactive decay in which a nucleus emits a beta particle. There are two types of beta decay: Beta-minus (β⁻), where a neutron turns into a proton and emits an electron, and Beta-plus (β⁺), where a proton turns into a neutron and emits a positron. Both electrons and positrons are charged particles.
Analysis of options:
- Option (a) is incorrect because electromagnetic radiation refers to Gamma rays (photons), not Beta rays.
- Option (b) is incorrect because Beta particles are created inside the nucleus during the decay process; they are not the existing electrons orbiting the nucleus.
- Option (c) is correct because Beta rays consist of high-energy electrons or positrons, which are charged particles originating from the nucleus.
- Option (d) is incorrect because neutral particles emitted by the nucleus are called neutrons (or gamma rays, which are photons), whereas beta particles carry a charge of -1 or +1.
Diode is used as a/an
A)
oscillator
B)
amplifier
C)
rectifier
D)
modulator
A diode is a semiconductor device that allows current to flow in only one direction (forward bias) and blocks it in the opposite direction (reverse bias). This unidirectional property is the fundamental requirement for rectification, which is the process of converting Alternating Current (AC), which periodically reverses direction, into Direct Current (DC), which flows in one direction.
Why other options are incorrect:
- a. Oscillator: An oscillator requires an active gain element (like a transistor or Op-Amp) and a feedback network to generate a periodic waveform. A diode alone cannot sustain oscillations.
- b. Amplifier: An amplifier increases the power or amplitude of a signal. This requires a device with gain (like a Bipolar Junction Transistor or FET). A diode is a passive device in terms of power gain and cannot amplify a signal.
- d. Modulator: Modulation is the process of varying a carrier signal's properties. While diodes can be used in specific types of detectors (demodulators), the primary and most fundamental use of a diode among the given choices is as a rectifier. Modulation itself typically requires more complex circuitry involving active components.
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 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.
If M is the mass of a nucleus and A its atomic mass, then the packing fraction is
A)
(M - A) / M
B)
(M - A) / A
C)
(M - A) / (M + A)
D)
(M - A) / M
In nuclear physics, the packing fraction is a measure of the stability of a nucleus. It is defined as the difference between the actual mass of the nucleus (M) and its mass number (A, which represents the mass of the nucleus if it were simply the sum of its constituent nucleons), divided by the mass number (A).
Mathematically, the formula is:
Packing Fraction = (Actual Mass of Nucleus - Mass Number) / Mass Number
Packing Fraction = (M - A) / A
- Option (a) and (d) are identical and use M in the denominator, which is incorrect as the normalization is done relative to the mass number A.
- Option (c) uses the sum of M and A in the denominator, which does not correspond to any standard physical definition of packing fraction.
- Option (b) correctly represents the ratio of the mass defect (relative to the mass number) to the mass number itself.
The dimensions of the coefficient of thermal conductivity are
A)
ML-1 T-2K-1
B)
ML-2 T-3 K-1
C)
ML-1 T-1 K-1
D)
ML T-3 K-1
k from Fourier law: [W/mK] → M L T^-3 K^-1.
The hidden heat which vaporises water at its boiling point is called
A)
latent heat
B)
thermal heat
C)
heat of vaporisation
D)
quantity of heat
Step-by-step reasoning: 1. The question asks for the specific term for 'hidden heat' used to vaporize water at its boiling point. 2. 'Hidden heat' is a general term for 'Latent Heat' (from Latin 'latere', meaning to lie hidden), which is heat absorbed or released during a phase change without a change in temperature. 3. Latent heat is categorized based on the phase change: 'Latent heat of fusion' (solid to liquid) and 'Latent heat of vaporization' (liquid to gas). 4. Since the question specifically mentions the process of vaporizing water at its boiling point, the most precise and technically correct term is the 'heat of vaporisation'. 5. While 'latent heat' (Option a) is a correct general category, 'heat of vaporisation' (Option c) is the specific answer for the process described. In competitive exams, the most specific correct answer is preferred over a general one. 6. 'Thermal heat' and 'quantity of heat' are generic terms and do not describe the specific physical phenomenon of phase change at a constant temperature.
As the diameter of the objective lens of a telescope increases, the resolution of the telescope
A)
decreases
B)
increases
C)
remains the same
D)
depends on the focal length of the lens
The resolution of a telescope refers to its ability to distinguish between two closely spaced objects. This is limited by the phenomenon of diffraction. According to Rayleigh's criterion, the minimum angular separation (θ) that a telescope can resolve is given by the formula: θ = 1.22 * (λ / D), where λ is the wavelength of light and D is the diameter of the objective lens (aperture).
1. Inverse Relationship: The formula shows that the angular resolution θ is inversely proportional to the diameter D.
2. Understanding 'Resolution': In optics, a 'smaller' value of θ means the telescope can resolve finer details (better resolution).
3. Effect of Increasing Diameter: As the diameter (D) of the objective lens increases, the value of θ decreases.
4. Conclusion: A smaller θ corresponds to a higher resolving power. Therefore, as the diameter of the objective lens increases, the resolution of the telescope increases.
Why other options are incorrect:
- (a) decreases: This contradicts the physics of diffraction; a larger aperture reduces diffraction blur.
- (c) remains the same: Resolution is physically dependent on the aperture size.
- (d) depends on the focal length: While focal length affects magnification, the fundamental limit of resolution (the ability to see detail) is determined by the aperture diameter, not the focal length.
Which of the following is not Maxwell’s thermodynamic relation?
A)
(∂S/∂V)_T = (∂P/∂T)_V
B)
(∂S/∂P)_T = -(∂V/∂T)_P
C)
(∂T/∂P)_S = (∂V/∂S)_P
D)
(∂P/∂V)_T = (∂S/∂T)_V
Maxwell's relations are derived from the exact differentials of the four thermodynamic potentials: Internal Energy (U), Enthalpy (H), Helmholtz Free Energy (A), and Gibbs Free Energy (G).
1. From Internal Energy (dU = TdS - PdV):
(∂S/∂V)_T = (∂P/∂T)_V => This matches Option (a). This is a valid Maxwell relation.
2. From Helmholtz Free Energy (dA = -SdT - PdV):
(∂S/∂P)_T = -(∂V/∂T)_P => This matches Option (b). This is a valid Maxwell relation.
3. From Gibbs Free Energy (dG = VdP - SdT):
(∂V/∂T)_P = -(∂S/∂P)_T (which is the same as option b) and (∂S/∂P)_T = -(∂V/∂T)_P.
4. From Enthalpy (dH = TdS + VdP):
(∂S/∂P)_T = -(∂V/∂T)_P (again) and (∂T/∂P)_S = (∂V/∂S)_P => This matches Option (c). This is a valid Maxwell relation.
Evaluating Option (d): (∂P/∂V)_T = (∂S/∂T)_V.
According to the first Maxwell relation, (∂S/∂V)_T = (∂P/∂T)_V. Option (d) incorrectly swaps the variables and the derivatives. There is no thermodynamic identity that equates the isothermal pressure-volume gradient to the isochoric entropy-temperature gradient in this manner. Therefore, option (d) is not a Maxwell relation.
The mean free path of a molecule of a gas depends
A)
Only on its diameter (d)
B)
Only on the number density (n) of the molecules
C)
On both d and n
D)
Neither on d nor on n
Mean free path λ=1/(√2 π d² n) depends on molecular diameter d and number density n.
Which of the following produces a plane wave front?
A)
Point source
B)
Line source
C)
Extended source
D)
None of the above
A wavefront is a surface over which the phase of the wave is constant. Different types of sources produce different types of wavefronts:
1. **Point Source**: A point source emits waves equally in all directions. The resulting wavefronts are concentric spheres, known as **Spherical Wavefronts**. (Option a is incorrect).
2. **Line Source**: A linear source (like a long thin glowing wire) produces wavefronts that are concentric cylinders, known as **Cylindrical Wavefronts**. (Option b is incorrect).
3. **Extended Source**: An extended source is essentially a collection of many point sources. The resulting wavefront is a superposition of many spherical wavefronts, which generally does not result in a plane wavefront unless specific conditions (like distance) are met. (Option c is incorrect).
**What produces a Plane Wavefront?**
A plane wavefront is produced when a spherical wavefront from a distant source (at infinity) reaches a point. As the radius of the sphere becomes infinitely large, the curvature becomes negligible, and the wavefront appears as a plane. Therefore, none of the sources listed (point, line, or extended) inherently produce a plane wavefront at their origin; a plane wavefront is an approximation of a spherical wavefront from a source at an infinite distance.
The Maxwell velocity distribution is valid for a classical gas
A)
in equilibrium irrespective of the nature of particle interaction
B)
strictly under contact interaction amongst particles
C)
under a steady flow of particles
D)
only in the absence of interparticle interaction
The Maxwell-Boltzmann velocity distribution describes the speeds of particles in an ideal gas. The fundamental assumption of an ideal gas, which is required for the Maxwell distribution to be strictly valid, is that there are no intermolecular forces (interparticle interactions) except during brief, elastic collisions.
1. Why 'd' is correct: The derivation of the Maxwell distribution relies on the assumption that particles move independently between collisions and that their potential energy is negligible compared to their kinetic energy. This is only true in the absence of long-range interparticle interactions (like electrostatic or van der Waals forces).
2. Why 'a' is incorrect: If there are strong interactions (like in a real gas or liquid), the distribution deviates from the Maxwellian form because the potential energy of the particles becomes significant.
3. Why 'b' is incorrect: 'Contact interaction' (hard-sphere collisions) is used to redistribute energy to reach equilibrium, but the distribution itself describes the state where no persistent attractive or repulsive forces exist between particles.
4. Why 'c' is incorrect: The Maxwell distribution is derived for a system in thermal equilibrium, not for a system under a steady flow (which would imply a non-equilibrium state or a bulk drift velocity).
In an inelastic collision of two bodies, which of the following do not change after the collision?
A)
Total kinetic energy
B)
Total linear momentum
C)
Total energy
D)
Both (b) and (c)
To determine what remains unchanged in an inelastic collision, we must analyze the fundamental laws of physics:
1. **Total Linear Momentum (Option b):** According to the Law of Conservation of Momentum, the total linear momentum of an isolated system remains constant regardless of whether the collision is elastic or inelastic, provided no external forces act on the system. Therefore, total linear momentum does not change.
2. **Total Energy (Option c):** According to the Law of Conservation of Energy, energy cannot be created or destroyed, only transformed. In an inelastic collision, some kinetic energy is converted into other forms (such as heat, sound, or internal deformation energy), but the *total* energy of the universe/system remains constant. Therefore, total energy does not change.
3. **Total Kinetic Energy (Option a):** By definition, an inelastic collision is one in which kinetic energy is NOT conserved. A portion of the initial kinetic energy is dissipated into other forms of energy. Therefore, total kinetic energy changes.
Since both total linear momentum and total energy remain unchanged, the correct choice is option (d).
Light appears to travel in a straight line, because
A)
its velocity is very large
B)
it is not absorbed by surrounding
C)
its wavelength is very small
D)
it is not reflected by surrounding
Light is an electromagnetic wave. All waves exhibit a phenomenon called diffraction, which is the bending of waves around the corners of an obstacle or through an aperture. The degree of diffraction depends on the ratio of the wavelength (λ) to the size of the obstacle or opening (d). Specifically, diffraction is significant only when the wavelength is comparable to the size of the obstacle (λ ≈ d). Because the wavelength of visible light is extremely small (approximately 400 nm to 700 nm), it does not noticeably bend around everyday macroscopic objects. Consequently, in most practical scenarios, light appears to travel in straight lines (rectilinear propagation).
Analysis of other options:
- Option (a): Velocity does not determine the path of a wave; waves of all speeds can travel in straight lines or bend.
- Option (b): Absorption relates to the intensity of light, not its direction of travel.
- Option (d): Reflection is a change in direction upon hitting a surface, but it does not explain why light travels in a straight line between two points.