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Past Questions — 2010
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.
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.
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.
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.
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.
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.
The electron emitted in beta radiation originates from
A)
inner orbits of atoms
B)
free electrons existing in nuclei
C)
decay of a neutron in a nucleus
D)
photon escaping from the nucleus
Beta radiation (specifically beta-minus decay) occurs when a nucleus has too many neutrons to be stable. In this process, a neutron inside the nucleus transforms into a proton, an electron (the beta particle), and an electron antineutrino. The reaction is represented as: n → p + e⁻ + νₑ. The electron is created at the moment of decay; it does not previously exist as a 'free electron' inside the nucleus, nor does it come from the atomic orbits.
Analysis of other options:
- Option (a) refers to processes like X-ray emission or Auger electrons, not beta decay.
- Option (b) is incorrect because electrons cannot exist freely inside a nucleus due to their charge and the laws of quantum mechanics (Heisenberg uncertainty principle/confinement).
- Option (d) describes Gamma radiation, which consists of high-energy photons, not electrons.
The continuous X-ray spectrum at constant voltage will have
A)
a minimum wavelength
B)
a maximum wavelength
C)
a minimum frequency
D)
a maximum frequency
In the production of X-rays (Bremsstrahlung), electrons are accelerated by a constant voltage (V) and strike a target. The maximum energy an electron can transfer to a photon occurs when the electron loses all its kinetic energy in a single collision. This energy is given by E = hν = eV, where h is Planck's constant, ν is frequency, e is the charge of an electron, and V is the accelerating voltage. Since wavelength (λ) is inversely proportional to frequency (λ = c/ν), the maximum energy corresponds to the shortest possible wavelength. This is known as the Duane-Hunt law, expressed as λ_min = hc/eV. Therefore, for a constant voltage, the continuous X-ray spectrum is characterized by a distinct 'cutoff' or minimum wavelength (λ_min), below which no X-rays are emitted. There is no theoretical maximum wavelength or maximum frequency for the continuous spectrum in this context, as electrons can lose any fraction of their energy, leading to longer wavelengths.
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.
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.
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.
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.
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.
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.
∇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.
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.
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°.
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.
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 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.
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.
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.
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.
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.