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Past Questions — 2010
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 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 kinetic energy of a 300 K thermal neutron is
A)
300 eV
B)
300 MeV
C)
0.026 eV
D)
0.026 MeV
A 'thermal neutron' is a neutron that is in thermal equilibrium with its surrounding medium. According to the kinetic theory of gases, the average kinetic energy (E) of a particle at a temperature (T) is given by the formula: E = (3/2)kT, where k is the Boltzmann constant (1.38 × 10⁻²³ J/K).
1. Calculation in Joules:
E = (3/2) * (1.38 × 10⁻²³ J/K) * 300 K
E = 6.21 × 10⁻²¹ Joules
2. Conversion to electron-volts (eV):
Since 1 eV = 1.602 × 10⁻¹⁹ Joules,
E = (6.21 × 10⁻²¹ J) / (1.602 × 10⁻¹⁹ J/eV)
E ≈ 0.0387 eV
However, in nuclear physics, the 'most probable energy' or the characteristic energy often cited for thermal neutrons at room temperature (approx 293-300 K) is based on the formula E = kT (the characteristic thermal energy).
Using E = kT:
E = (1.38 × 10⁻²³ J/K * 300 K) / (1.602 × 10⁻¹⁹ J/eV)
E ≈ 0.0258 eV ≈ 0.026 eV.
Comparing this to the options:
- Option (a) 300 eV is far too high.
- Option (b) 300 MeV is the energy scale of fast neutrons/nuclear reactions, not thermal neutrons.
- Option (c) 0.026 eV matches the calculated characteristic thermal energy.
- Option (d) 0.026 MeV is 26,000 eV, which is too high.
If the oscillator is in its normal state, then the probability of finding the particle outside the classical limit is approximately
A)
16%
B)
24%
C)
48%
D)
84%
1. **Understanding the 'Normal State'**: In the context of a quantum harmonic oscillator, the 'normal state' refers to the ground state (n=0). The wave function for the ground state is a Gaussian: ψ₀(x) = (α/π)^{1/4} exp(-αx²/2), where α = mω/ħ.
2. **Defining the Classical Limit**: The classical limit (classical turning points) occurs where the total energy equals the potential energy. For the ground state, E₀ = ½ħω and V(x) = ½mω²x². Setting E₀ = V(x) gives: ½ħω = ½mω²x² → x² = ħ/(mω). Let the classical turning point be ±x₀, where x₀ = √(ħ/mω).
3. **Calculating Probability**: The probability of finding the particle outside the classical limit is the integral of the probability density |ψ₀(x)|² from -∞ to -x₀ and from x₀ to +∞. Due to symmetry, this is 2 * ∫[x₀ to ∞] (α/π)^{1/2} exp(-αx²) dx.
4. **Substitution**: Let u = √(α)x. Then dx = du/√α. When x = x₀ = √(ħ/mω), u = √(mω/ħ) * √(ħ/mω) = 1.
5. **Evaluating the Integral**: The probability P = 2 * (1/√π) * ∫[1 to ∞] exp(-u²) du. This integral is related to the complementary error function: P = erfc(1).
6. **Numerical Value**: The value of erfc(1) is approximately 0.1573, which is roughly 15.7%.
7. **Conclusion**: Rounding 15.7% to the nearest option provided, we get 16%. Therefore, option (a) is the correct answer.
Pauli’s Exclusion Principle states that
A)
two electrons can have all the quantum numbers same
B)
no two electrons can have all the quantum numbers same
C)
particles with integer and half integer spin cannot exist in the same state
D)
None of the above
Pauli's Exclusion Principle, formulated by Wolfgang Pauli in 1925, is a fundamental principle in quantum mechanics. It states that no two fermions (particles with half-integer spin, such as electrons) in an atom can occupy the same quantum state simultaneously. In terms of quantum numbers, this means that no two electrons in a single atom can have the exact same set of four quantum numbers: the principal quantum number (n), the azimuthal quantum number (l), the magnetic quantum number (m), and the spin quantum number (s).
Analysis of options:
- Option (a) is incorrect because it contradicts the principle; if two electrons had the same quantum numbers, they would be in the same state, which is forbidden.
- Option (b) is correct as it accurately describes the principle.
- Option (c) is incorrect because it misrepresents the nature of particles. Particles with integer spin (bosons) and half-integer spin (fermions) are different classes of particles; the exclusion principle specifically applies to fermions, not to the coexistence of different types of particles in a system.
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.
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 cause of a potential barrier in a P-N diode is the
A)
concentration of positive charge near the junction
B)
depletion of negative charges near the junction
C)
concentration of positive and negative charges near the junction
D)
depletion of positive charges near the junction
When a P-type semiconductor (rich in holes/positive charges) and an N-type semiconductor (rich in electrons/negative charges) are joined to form a P-N junction, a process called diffusion occurs. Electrons from the N-side diffuse into the P-side, and holes from the P-side diffuse into the N-side. Near the junction, these mobile charge carriers recombine and neutralize each other. This leaves behind immobile ionized impurity atoms: positive donor ions on the N-side and negative acceptor ions on the P-side. The resulting 'concentration' of these fixed positive and negative charges creates an internal electric field and a region devoid of mobile carriers known as the depletion region. This electric field acts as a potential barrier that prevents further diffusion of majority carriers. Therefore, the potential barrier is caused by the concentration of these fixed positive and negative ions near the junction. Options a, b, and d are incorrect because they only describe one side of the charge distribution, whereas the barrier is a result of the dipole-like arrangement of both positive and negative ions across the junction.
The magnitude of the induced emf produced in a coil when a magnet is inserted into it does not depend upon the
A)
number of turns in the coil
B)
resistance of the coil
C)
magnetic moment of the magnet
D)
speed of approach of the magnet
According to Faraday's Law of Electromagnetic Induction, the magnitude of the induced electromotive force (emf) is given by the formula: ε = -N (dΦ/dt), where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux.
1. Number of turns (Option a): The emf is directly proportional to N. Increasing the turns increases the induced emf.
2. Magnetic moment of the magnet (Option c): The magnetic flux Φ depends on the strength of the magnet (its magnetic moment). A stronger magnet creates a larger change in flux, increasing the induced emf.
3. Speed of approach (Option d): The term 'dt' in the formula represents time. A faster speed of approach means the flux changes over a shorter period of time (higher dΦ/dt), which increases the induced emf.
4. Resistance of the coil (Option b): The induced emf (voltage) is a property of the change in magnetic field and the geometry of the coil. While the induced CURRENT (I = ε/R) depends on the resistance according to Ohm's Law, the induced EMF itself is independent of the resistance of the circuit.
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.
∇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 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.
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.
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 output of OR gate is 1
A)
if both inputs are zero
B)
if either or both inputs are 1
C)
only if both inputs are 1
D)
if either input is zero
An OR gate is a basic digital logic gate that implements logical addition. The output of an OR gate is HIGH (1) if at least one of its inputs is HIGH (1).
Truth Table for a 2-input OR gate:
- Input A=0, Input B=0 -> Output = 0
- Input A=0, Input B=1 -> Output = 1
- Input A=1, Input B=0 -> Output = 1
- Input A=1, Input B=1 -> Output = 1
Analysis of options:
- Option (a) is incorrect because if both inputs are 0, the output is 0.
- Option (b) is correct because it accurately describes the condition where the output is 1 (at least one input is 1).
- Option (c) is incorrect because this describes the behavior of an AND gate.
- Option (d) is incorrect because if one input is 0, the output depends on the other input; it is not a sufficient condition for the output to be 1.
The phenomenon of nuclear fission can be understood on the basis of
A)
liquid drop model of the nucleus
B)
shell model of the nucleus
C)
independent particle model of the nucleus
D)
meson theory of nuclear forces
The phenomenon of nuclear fission is best explained by the Liquid Drop Model, proposed by George Gamow and further developed by Niels Bohr and John Wheeler. In this model, the nucleus is treated as a drop of incompressible nuclear fluid held together by the strong nuclear force (analogous to surface tension). When a heavy nucleus (like Uranium-235) absorbs a neutron, it becomes unstable and begins to oscillate. If the oscillation is strong enough, the 'drop' becomes elongated, forming a 'dumbbell' shape. Once the electrostatic repulsion between the protons in the two ends overcomes the surface tension (nuclear force), the nucleus splits into two smaller fragments, releasing a large amount of energy.
Why other options are incorrect:
- b. Shell Model: This model explains the 'magic numbers' of nucleons and the ground-state properties (spin, parity) of nuclei, rather than the collective deformation required for fission.
- c. Independent Particle Model: This is essentially the basis for the shell model, focusing on individual nucleons moving in a mean potential field.
- d. Meson Theory: Proposed by Hideki Yukawa, this explains the mechanism of the nuclear force (the exchange of pions) but does not provide a macroscopic model to explain the splitting of a nucleus.
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.
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.
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°.
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.
The average number of neutrons released by the fission of one Uranium atom is
A)
1
B)
2
C)
2.5
D)
3
In the process of nuclear fission of Uranium-235 (the most common fissile isotope), a single slow neutron triggers the nucleus to split into two smaller daughter nuclei (fission fragments). During this process, several neutrons are released. While the number of neutrons released per single fission event can vary (typically between 2 and 4), the statistical average for U-235 is approximately 2.4 to 2.5 neutrons. This value is critical for maintaining a chain reaction; since one neutron is required to trigger the next fission, an average of 2.5 means there is a net gain of 1.5 neutrons per event, allowing the reaction to be sustained or accelerated. Option (a) 1 would result in a steady state or decay; Option (b) 2 is too low for the average; Option (d) 3 is higher than the observed average for standard U-235 fission.
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.
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 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'.