← Legal Metrology
Past Questions
When n-p-n transistor is used as an amplifier, then
A)
holes move from emitter to base
B)
electrons move from base to collector
C)
holes move from base to emitter
D)
electrons move from collector to base
In an n-p-n transistor operating as an amplifier (Active Region), the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.
1. **Emitter to Base**: The emitter is heavily doped n-type material. When forward-biased, it injects a large number of majority carriers (electrons) into the p-type base.
2. **Base Action**: The base is very thin and lightly doped. Only a small fraction of these electrons recombine with holes in the base (creating a small base current).
3. **Base to Collector**: The vast majority of the electrons injected from the emitter diffuse across the thin base and are swept across the reverse-biased collector-base junction by the strong electric field into the collector.
Therefore, the primary current flow in an n-p-n transistor consists of electrons moving from the emitter, through the base, and into the collector.
Evaluating the options:
- (a) Incorrect: In n-p-n, electrons (not holes) move from emitter to base.
- (b) Correct: Electrons that survive the base region move from the base to the collector.
- (c) Incorrect: Holes are minority carriers in the emitter; they do not drive the amplification process in this direction.
- (d) Incorrect: Electrons move from base to collector, not collector to base.
In Quantum mechanical tunneling process energy
A)
remains conserved
B)
remains non-conserved
C)
varies with the advance penetrating wave
D)
None of the above
Quantum tunneling is a phenomenon where a particle penetrates through a potential energy barrier that it classically would not have enough energy to surmount. In this process, the total energy (E) of the particle remains constant throughout the entire region—before, during, and after the tunneling event.
1. **Conservation of Energy**: In the Schrödinger equation for a stationary state, the energy E is a constant of motion. The particle does not 'gain' energy to jump over the barrier, nor does it 'lose' energy to pass through it.
2. **The Mechanism**: Inside the barrier (where the potential V > E), the wave function becomes exponentially decaying rather than oscillating. However, the energy eigenvalue associated with that wave function remains the same.
3. **Analysis of Options**:
- Option (a) is correct because the total energy of the system is conserved.
- Option (b) is incorrect because energy conservation is a fundamental law of physics and is maintained in quantum tunneling.
- Option (c) is incorrect because the 'penetrating wave' describes the probability amplitude of the particle's position, not a change in the particle's energy level.
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.
The small value of binding energy of nucleus in the low mass number is due to
A)
surface effects
B)
low mass
C)
repulsive force which arises from protons
D)
more potential energy of the nucleons
The binding energy per nucleon (BE/A) is a measure of nuclear stability. According to the Semi-Empirical Mass Formula (Liquid Drop Model), the total binding energy is influenced by several terms. For nuclei with low mass numbers (small A), the 'surface term' plays a dominant role. Nucleons on the surface of the nucleus have fewer neighbors to interact with via the strong nuclear force compared to nucleons in the interior. In small nuclei, a larger fraction of the total nucleons reside on the surface relative to the volume. This results in a significant reduction in the total binding energy per nucleon, leading to a smaller value of binding energy for low mass numbers.
Regarding other options:
- (b) Low mass itself is a characteristic, not the physical cause of the reduced binding energy.
- (c) Repulsive forces from protons (Coulomb term) actually become more significant in HEAVY nuclei (high mass number), which is why BE/A decreases for very large nuclei.
- (d) Potential energy is a component of the binding energy calculation, but the specific reason for the dip at low mass numbers is the surface-to-volume ratio.
Two charged spheres separated at a distance d exerts a force F on each other. If they are immersed in a liquid of dielectric constant 2, then the force (if all conditions are same) is
A)
F/2
B)
F
C)
2F
D)
4F
1. According to Coulomb's Law, the electrostatic force (F) between two point charges (q1 and q2) separated by a distance (d) in a vacuum or air is given by: F = (1 / 4πε₀) * (q1q2 / d²).
2. When the charges are immersed in a medium (liquid) with a dielectric constant (K), the force (F') between them is reduced. The formula becomes: F' = (1 / 4πε₀K) * (q1q2 / d²).
3. This can be rewritten in terms of the original force F as: F' = F / K.
4. In this specific problem, the dielectric constant (K) of the liquid is given as 2.
5. Substituting K = 2 into the formula: F' = F / 2.
6. Therefore, the force is halved when the spheres are immersed in the liquid. Option (a) is correct, while options (b), (c), and (d) are incorrect as they suggest the force remains the same or increases, which contradicts the physics of dielectric media.
Transistor is a
A)
voltage operated device
B)
current operated device
C)
both current and voltage operated device
D)
None of the above
A Bipolar Junction Transistor (BJT) is fundamentally a current-controlled device. This is because the output current (collector current, Ic) is controlled by the input current (base current, Ib). The relationship is defined by the equation Ic = β * Ib, where β (beta) is the current gain of the transistor. A small change in the base current results in a large change in the collector current, making it a current-operated device. In contrast, Field Effect Transistors (FETs) are voltage-operated devices because their output current is controlled by the voltage applied to the gate terminal. Since the term 'Transistor' in general competitive exams of this nature typically refers to the BJT unless specified otherwise, it is classified as a current operated device.
The displacement current arise due to
A)
positive charges only
B)
negative charges only
C)
both positive and negative charges
D)
time varying electric field
The concept of displacement current was introduced by James Clerk Maxwell to resolve an inconsistency in Ampere's Law when applied to a charging capacitor. According to Maxwell, a changing electric field produces a magnetic field, just as a flow of electric charge (conduction current) does. Mathematically, the displacement current density (Jd) is defined as Jd = ε₀(∂E/∂t), where ε₀ is the permittivity of free space and ∂E/∂t is the rate of change of the electric field with respect to time. Therefore, displacement current does not involve the actual movement of charges (positive or negative) through a medium, but is instead a result of a time-varying electric field. Options a, b, and c are incorrect because they describe conduction currents (the physical movement of charges), whereas displacement current exists even in a vacuum where no free charges are present.
A non conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre
A)
zero
B)
decreases as r increases, for 0 < r < ∞
C)
decreases as r increases, for R < r < ∞
D)
is discontinuous at r = R
To find the electric field (E) of a uniformly charged non-conducting solid sphere of radius R, we use Gauss's Law.
1. Inside the sphere (0 ≤ r ≤ R): The charge enclosed by a Gaussian surface of radius r is proportional to the volume, so q_enclosed = Q * (r³/R³). Applying Gauss's Law: E(4πr²) = (Q r³/R³) / ε₀, which simplifies to E = (Q / 4πε₀R³) * r. In this region, the electric field increases linearly with r.
2. Outside the sphere (r > R): The entire charge Q is enclosed. Applying Gauss's Law: E(4πr²) = Q / ε₀, which simplifies to E = Q / (4πε₀r²). In this region, the electric field follows the inverse square law and decreases as r increases.
Evaluating the options:
- (a) Zero: Incorrect. The field is only zero at the exact center (r=0).
- (b) Decreases as r increases for 0 < r < ∞: Incorrect. It increases for 0 < r < R.
- (c) Decreases as r increases for R < r < ∞: Correct. As derived above, E ∝ 1/r² for r > R.
- (d) Is discontinuous at r = R: Incorrect. At r = R, both formulas yield E = Q / (4πε₀R²), meaning the electric field is continuous at the boundary.
Which of the interaction cause the non-conservation of orbital angular momentum of the electrons in an atom?
A)
Spin-orbit interaction
B)
Spin-spin interaction
C)
Electrostatic interaction between electrons
D)
Electrostatic interaction between electrons and nucleus
In an atom, the orbital angular momentum (L) and the spin angular momentum (S) are separate quantities. However, the spin-orbit interaction describes the coupling between the electron's spin and its orbital motion. This interaction arises because, from the electron's frame of reference, the nucleus orbits the electron, creating a magnetic field that interacts with the electron's intrinsic spin magnetic moment. This coupling results in a torque that causes L and S to precess around the total angular momentum vector J (where J = L + S). Because of this exchange of angular momentum between the spin and orbital components, neither L nor S is independently conserved; only the total angular momentum J remains a constant of motion.
Regarding the other options:
- (b) Spin-spin interaction: While it exists, it is much weaker and primarily affects the relative orientation of spins, not the orbital angular momentum.
- (c) Electrostatic interaction between electrons: This can cause changes in L, but in the context of standard atomic physics problems regarding the 'fine structure' and the fundamental cause of non-conservation of L in a single-electron or central-field approximation, spin-orbit is the primary driver.
- (d) Electrostatic interaction between electrons and nucleus: This is a central force interaction. In a purely central potential, orbital angular momentum L is strictly conserved.
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.
Minority carriers produced by thermal agitation or vibration in a semi-conductor are
A)
holes in N-type and free electrons in P-type
B)
holes in N- and P-type
C)
free electrons in N-type and hole in P-type
D)
free electrons in N-type and P-type
In semiconductor physics, charge carriers are divided into majority and minority carriers based on the type of doping.
1. **N-type Semiconductor**: Doped with pentavalent impurities. Electrons are the majority carriers. However, thermal agitation (thermal energy) occasionally breaks covalent bonds between silicon/germanium atoms, creating electron-hole pairs. In N-type material, these thermally generated holes are the minority carriers because they are far fewer in number compared to the electrons provided by the dopants.
2. **P-type Semiconductor**: Doped with trivalent impurities. Holes are the majority carriers. Similarly, thermal agitation creates electron-hole pairs. In P-type material, these thermally generated free electrons are the minority carriers because they are far fewer in number compared to the holes provided by the dopants.
Therefore, minority carriers are holes in N-type and free electrons in P-type semiconductors.
Analysis of other options:
- Option b is incorrect because holes are majority carriers in P-type.
- Option c describes the majority carriers for both types.
- Option d is incorrect because electrons are majority carriers in N-type.
The minimum energy required to ionize hydrogen atom from its ground state is about
A)
13.6 eV
B)
1.36 eV
C)
0.136 eV
D)
136 eV
1. Definition: Ionization energy is the minimum energy required to remove an electron completely from an atom in its ground state (bringing it to an infinite distance where its potential energy is zero).
2. Bohr Model Formula: The energy of an electron in the n-th orbit of a hydrogen atom is given by the formula: E_n = -13.6 / n^2 eV.
3. Ground State: For the ground state, n = 1. Therefore, E_1 = -13.6 / (1)^2 = -13.6 eV. The negative sign indicates that the electron is bound to the nucleus.
4. Ionization Energy Calculation: To ionize the atom, the electron must be moved from n = 1 to n = ∞. The energy at infinity (E_∞) is 0 eV.
Ionization Energy = E_∞ - E_1 = 0 - (-13.6 eV) = 13.6 eV.
5. Analysis of Options:
- Option (a) 13.6 eV is the standard accepted value for the ionization energy of hydrogen.
- Options (b), (c), and (d) are incorrect as they differ by powers of 10 from the physically derived constant.
An LC-circuit cannot produce oscillations when
A)
capacitance is large
B)
inductance is large
C)
resistance is large
D)
All the given above
In a real-world LC circuit (which is actually an RLC circuit), the ability to produce oscillations depends on the damping factor. The condition for oscillations to occur is that the circuit must be underdamped. The damping is caused by the resistance (R) in the circuit, which dissipates energy as heat. Mathematically, for a series RLC circuit, oscillations occur if R < 2√(L/C). If the resistance (R) is too large (specifically, if the circuit becomes critically damped or overdamped), the energy is dissipated faster than it can be transferred between the inductor (L) and the capacitor (C), and the system will return to equilibrium without oscillating. Large values of inductance (L) or capacitance (C) merely change the frequency of the oscillations (f = 1 / (2π√LC)) but do not inherently prevent oscillations from occurring; in fact, increasing L or C while keeping R constant can actually make the circuit more likely to oscillate by reducing the damping ratio.
In Raman spectra Q branch is
A)
absent
B)
present
C)
presence or appearance depends on the state of polarization of the molecule
D)
None of the above
In Raman spectroscopy of diatomic molecules, the selection rules govern which transitions are allowed. The Raman transitions are categorized into three branches: the S-branch (ΔJ = +2), the O-branch (ΔJ = -2), and the Q-branch (ΔJ = 0). For a linear diatomic molecule, the selection rule for rotational Raman transitions is ΔJ = 0, ±2. However, for the specific case of the pure rotational Raman spectrum of a diatomic molecule, the Q-branch (ΔJ = 0) corresponds to the Rayleigh line (the elastic scattering peak) and is not considered part of the Raman shifted spectrum. More importantly, in the context of vibrational-rotational Raman spectra of simple diatomic molecules, the Q-branch (ΔJ = 0) is forbidden by selection rules for molecules with no electronic angular momentum along the internuclear axis (like H2, N2, O2). Therefore, in the standard Raman spectra of these molecules, the Q-branch is absent, leaving only the O and S branches. While some polyatomic molecules or specific electronic states can show a Q-branch, in the general context of basic molecular physics exams referring to diatomic Raman spectra, the Q-branch is identified as absent.
The radius of a nucleus is
A)
proportional to its mass number
B)
inversely proportional to its mass number
C)
proportional to the cube root of its mass number
D)
not related to its mass number
The radius of an atomic nucleus can be modeled by assuming that the nucleus is a sphere with a constant density. The volume (V) of a sphere is given by V = (4/3)πR³, where R is the radius. Since the volume of the nucleus is proportional to the number of nucleons (mass number, A), we have V ∝ A. Therefore, (4/3)πR³ ∝ A, which simplifies to R³ ∝ A. Taking the cube root of both sides, we get R ∝ A^(1/3). This is expressed by the empirical formula R = R₀A^(1/3), where R₀ is a constant (approximately 1.2 to 1.5 fm).
Analysis of options:
- Option (a) is incorrect because the radius does not grow linearly with mass; if it did, the nucleus would become less dense as it grew.
- Option (b) is incorrect because the radius increases as more nucleons are added, it does not decrease.
- Option (c) is correct as it follows the derived relationship R ∝ A^(1/3).
- Option (d) is incorrect because the mass number directly determines the volume and thus the radius of the nucleus.
Energy is not carried by
A)
transverse progressive wave
B)
longitudinal progressive wave
C)
stationary wave
D)
electromagnetic wave
To determine which wave does not carry energy, we must distinguish between progressive waves and stationary waves:
1. **Progressive Waves (Options a, b, and d):** A progressive wave (whether transverse, longitudinal, or electromagnetic) is a wave that travels through a medium or space, transporting energy from one point to another. For example, sound waves (longitudinal) and light waves (electromagnetic) carry energy from the source to the receiver.
2. **Stationary Waves (Option c):** Also known as standing waves, these are formed by the superposition of two identical progressive waves traveling in opposite directions. In a stationary wave, the energy is 'trapped' between the nodes. While the particles of the medium vibrate, there is no net transfer of energy from one position to another along the wave. The energy oscillates between kinetic and potential forms within the antinodes, but the average energy flux across any plane is zero.
Therefore, energy is not carried (transported) by a stationary wave.
For a rigid sphere of radius a the scattering cross-section is given by
A)
2 pa2
B)
3 pa2
C)
4 pa2
D)
pa2
In the context of atomic and nuclear physics, the scattering of particles by a rigid sphere (hard sphere scattering) is a fundamental problem. For a hard sphere of radius 'a', the total scattering cross-section (σ) is calculated based on the impact parameter. In the classical limit, any particle with an impact parameter b < 2a will strike the sphere and be scattered. However, the standard result for the total cross-section of a hard sphere in the low-energy (s-wave) limit of quantum mechanics, or the effective area presented by the sphere for scattering, is given by σ = 4πa². This is exactly four times the geometric cross-section (πa²) of the sphere. This phenomenon occurs because the wave nature of the particle leads to diffraction-like scattering, effectively doubling the radius of the interaction in the cross-section calculation (σ = π(2a)² = 4πa²). Note: In the options provided, 'p' is used as a typographical representation of the mathematical constant π (pi).
A rocket is fired from the earth to the moon. The distance between the earth and the moon is r and the mass of the earth is 81 times the mass of the moon. The gravitational force on the rocket will be zero, when its distance from the moon is
A)
r/20
B)
r/15
C)
r/10
D)
r/5
To find the point where the gravitational force on the rocket is zero, the gravitational pull from the Earth must be equal and opposite to the gravitational pull from the Moon.
1. Let $M_e$ be the mass of the Earth and $M_m$ be the mass of the Moon.
2. Given: $M_e = 81 M_m$.
3. Let $r$ be the total distance between the Earth and the Moon.
4. Let $x$ be the distance of the rocket from the Moon. Therefore, the distance of the rocket from the Earth is $(r - x)$.
5. At the point where the net force is zero:
$F_{earth} = F_{moon}$
$rac{G M_e m}{(r - x)^2} = rac{G M_m m}{x^2}$
6. Simplifying the equation by canceling $G$ and $m$ (mass of the rocket):
$rac{M_e}{(r - x)^2} = rac{M_m}{x^2}$
7. Substitute $M_e = 81 M_m$:
$rac{81 M_m}{(r - x)^2} = rac{M_m}{x^2}$
$rac{81}{(r - x)^2} = rac{1}{x^2}$
8. Taking the square root of both sides:
$rac{9}{r - x} = rac{1}{x}$
9. Solving for $x$:
$9x = r - x$
$10x = r$
$x = r/10$
Thus, the gravitational force on the rocket will be zero when its distance from the moon is $r/10$.
At 0K which of the following properties of a gas will be zero?
A)
Kinetic energy
B)
Potential energy
C)
Mass
D)
Density
1. Definition of Absolute Zero: 0K (Kelvin) is known as absolute zero, the lowest possible temperature theoretically achievable.
2. Kinetic Theory of Gases: According to the kinetic molecular theory, the temperature of a gas is a direct measure of the average kinetic energy of its molecules. The relationship is expressed as KE = (3/2)kT, where k is the Boltzmann constant and T is the absolute temperature.
3. Analysis at 0K: When T = 0K, the mathematical product (3/2)k(0) equals zero. This implies that at absolute zero, the translational motion of the molecules ceases, and the kinetic energy becomes zero (classically).
4. Evaluating other options:
- Potential energy (b) does not become zero; in fact, intermolecular forces (van der Waals forces) still exist, and potential energy depends on the position of molecules relative to each other.
- Mass (c) is an intrinsic property of matter and remains constant regardless of temperature.
- Density (d) is mass per unit volume. While a gas may liquefy or solidify at 0K (increasing density), the density itself does not become zero; it actually increases as the volume decreases.
As a consequence of Brewster’s law for polarization, it follows that the angle of polarization depends upon
A)
the wavelength of light employed
B)
plane of polarization’s orientation
C)
the amplitude of light employed
D)
None of these
Brewster's Law states that the tangent of the polarizing angle (Brewster's angle, θp) is equal to the ratio of the refractive indices of the two media involved. Mathematically, it is expressed as: tan(θp) = n2 / n1, where n2 is the refractive index of the denser medium and n1 is the refractive index of the rarer medium (usually air, where n1 ≈ 1).
Analyzing the options:
1. Wavelength of light (a): While the refractive index (n) of a material can vary slightly with wavelength (a phenomenon called dispersion), Brewster's Law itself defines the angle based on the refractive index. In the context of standard physics problems regarding the 'consequence' of the law, the angle is a function of the material properties (refractive index), not the wavelength of the light source itself.
2. Plane of polarization's orientation (b): The plane of polarization is a result of the light hitting the surface at the Brewster angle; the angle does not depend on the orientation of the plane.
3. Amplitude of light (c): The refractive index and the resulting angle of polarization are independent of the intensity or amplitude of the incident light.
Since the angle of polarization depends solely on the refractive indices of the two media, and none of the provided options (wavelength, orientation, or amplitude) are the primary determinants defined by Brewster's Law, the correct option is 'None of these'.
Theory of relativity shows that Newtonian mechanics is true
A)
approximately for all velocities
B)
approximately for low velocities
C)
approximately for high velocities
D)
wholly
The Special Theory of Relativity, proposed by Albert Einstein, generalizes the laws of physics to include objects moving at speeds close to the speed of light (c). According to relativity, the Lorentz factor γ = 1 / sqrt(1 - v²/c²) determines the magnitude of relativistic effects such as time dilation and length contraction. When the velocity (v) is very small compared to the speed of light (v << c), the term v²/c² becomes negligible (approximately 0), and γ becomes approximately 1. In this limit, the relativistic equations reduce exactly to the classical equations of Newtonian mechanics. Therefore, Newtonian mechanics is not 'wholly' true (as it fails at high speeds) nor is it true for high velocities; it serves as a highly accurate approximation for the 'low velocity' regime encountered in everyday human experience.
A telescope is on the moon. If a tiny drop of ink falls on its objective, then
A)
moon will appear black
B)
field of view will be dark
C)
there will be a dark spot in the field of view
D)
brightness of the image will be slightly reduced
To understand this, we must consider how a telescope forms an image. The objective lens (or mirror) collects light from a distant object. Every point on the object sends a cone of light rays that hits the entire surface of the objective lens. All these rays are then refracted/reflected to converge at a single point in the image plane.
1. If a tiny drop of ink falls on the objective, it blocks only a small fraction of the light rays coming from every point of the observed object.
2. Since the image of any single point is formed by the integration of light from the entire aperture of the objective, blocking a small part of that aperture does not 'block' the image of a specific part of the object. Instead, it slightly reduces the total amount of light reaching the focal point for every part of the image.
3. Consequently, the entire image becomes slightly dimmer (reduced brightness), but no specific 'dark spot' appears in the field of view because the remaining clear parts of the lens still collect light from the area behind the ink drop.
Why other options are incorrect:
- Option (a): The moon is where the telescope is located; the telescope is looking at other objects, not the moon itself.
- Option (b): The field of view remains visible; it does not go completely dark unless the entire lens is covered.
- Option (c): A dark spot would only appear if the ink were placed at the focal plane (the eyepiece or the sensor), not on the objective lens. On the objective, it acts as a small stop, affecting intensity rather than geometry.
A particle is moving in a circle with uniform speed. Its motion is
A)
periodic and simple harmonic
B)
periodic but not simple harmonic
C)
aperiodic
D)
None of the above
1. **Analysis of Periodicity**: A particle moving in a circle with uniform speed returns to its starting position after a fixed time interval (the time period T = 2πr/v). Since the motion repeats itself at regular intervals of time, it is by definition 'periodic'.
2. **Analysis of Simple Harmonic Motion (SHM)**: For a motion to be Simple Harmonic, it must satisfy two conditions: (i) the motion must be oscillatory (back and forth about a mean position), and (ii) the restoring force (or acceleration) must be directly proportional to the displacement from the mean position and directed towards it (F = -kx).
3. **Comparison**: In uniform circular motion, the particle does not oscillate back and forth; it moves continuously in one direction along the circumference. While the projection of this motion onto any diameter of the circle is SHM, the actual 2D motion of the particle itself is not. The acceleration (centripetal acceleration) is always directed toward the center, but the displacement is measured along the arc or as a vector from the center, which does not satisfy the linear restoring force requirement of SHM.
4. **Conclusion**: The motion is periodic because it repeats, but it is not simple harmonic because it is not an oscillation satisfying the SHM differential equation.
Which one of the following relationships between the acceleration a and displacement y of a particle represents simple harmonic motion?
A)
a = 0.5y
B)
a = 400y^2
C)
a = -20y
D)
a = -3y^2
Simple Harmonic Motion (SHM) is defined as a type of periodic motion where the restoring force (and consequently the acceleration) is directly proportional to the displacement from the equilibrium position and is always directed opposite to the displacement. Mathematically, the condition for SHM is expressed as: a ∝ -y, or a = -ω²y, where 'a' is acceleration, 'y' is displacement, and 'ω' is the angular frequency (a positive constant).
Analyzing the options:
- Option (a) a = 0.5y: The acceleration is proportional to displacement but has a positive sign. This represents an unstable equilibrium where the particle would accelerate away from the center, not oscillate.
- Option (b) a = 400y²: The acceleration is proportional to the square of the displacement. This is non-linear motion and does not satisfy the SHM condition.
- Option (c) a = -20y: This fits the form a = -ω²y (where ω² = 20). The acceleration is proportional to the displacement and acts in the opposite direction (restoring force), which is the defining characteristic of SHM.
- Option (d) a = -3y²: While the sign is negative, the acceleration depends on the square of the displacement, making it non-linear and therefore not SHM.
If the temperature of the source is increased, the efficiency of the Carnot’s engine
A)
increases
B)
decreases
C)
remains constant
D)
first increases and then remains constant
Carnot efficiency η=1-Tc/Th, increasing Th raises η.