Legal Metrology • 2010 • Physics Paper-I
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.
Legal Metrology • 2010 • Physics Paper-I
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.
Legal Metrology • 2010 • Physics Paper-I
Water is flowing through a horizontal pipe line having a constriction. Then
A)
pressure will be the same throughout the length of the pipe
B)
pressure will be greater at the constriction
C)
pressure will be greater in the wider portion
D)
None of these
This problem is solved using two fundamental principles of fluid dynamics: the Continuity Equation and Bernoulli's Principle.
1. **Continuity Equation**: For an incompressible fluid flowing through a pipe, the flow rate must remain constant. The equation is $A_1v_1 = A_2v_2$, where $A$ is the cross-sectional area and $v$ is the velocity. This means that in a constriction (where the area $A$ decreases), the velocity $v$ of the water must increase to maintain the flow rate.
2. **Bernoulli's Principle**: For a horizontal pipe (where height $h$ is constant), the energy conservation equation is $P + \frac{1}{2}\rho v^2 = \text{constant}$. This indicates an inverse relationship between the pressure ($P$) and the square of the velocity ($v$).
3. **Application**:
- In the **wider portion**, the area is large, so the velocity is low. According to Bernoulli's equation, a lower velocity results in a higher pressure.
- In the **constriction**, the area is small, so the velocity is high. A higher velocity results in a lower pressure.
Therefore, the pressure is greater in the wider portion of the pipe and lower at the constriction (this is known as the Venturi Effect).
Legal Metrology • 2010 • Physics Paper-I
When a compact disc is illuminated by a source of white light, coloured lines are observed. This is due to
A)
dispersion
B)
diffraction
C)
interference
D)
polarization
A Compact Disc (CD) contains a spiral track of tiny pits and lands that act as a reflection grating. When white light hits the surface, the closely spaced tracks act as a diffraction grating. Diffraction occurs when light waves encounter these periodic structures and bend around them, causing the light to spread and overlap. Because white light consists of different wavelengths (colors), each wavelength is diffracted at a slightly different angle (according to the grating equation d sin θ = nλ). This separates the white light into its constituent colors, creating the observed rainbow-like colored lines.
Why other options are incorrect:
- a. Dispersion: While dispersion also separates white light into colors, it occurs due to the variation of the refractive index of a medium with wavelength (e.g., light passing through a glass prism). A CD is a reflective surface, not a refractive medium.
- c. Interference: While diffraction is a form of interference (the superposition of waves from different parts of the wavefront), the specific mechanism of a grating creating a spectrum is categorized as diffraction.
- d. Polarization: Polarization refers to the orientation of the oscillations of the light wave and does not cause the separation of white light into colors.
Legal Metrology • 2010 • Physics Paper-I
When a beam of light is used to determine the position of an object, the maximum accuracy is achieved if the light is
A)
polarized
B)
of longer wavelength
C)
of shorter wavelength
D)
of high intensity
The accuracy with which the position of an object can be determined is limited by the phenomenon of diffraction. According to the Rayleigh criterion, the minimum resolvable distance (the limit of resolution) is proportional to the wavelength of the light used. The formula for the limit of resolution (d) is given by d = 1.22 * (λ / D), where λ is the wavelength of the light and D is the aperture diameter. To achieve maximum accuracy (which means minimizing the limit of resolution 'd'), the wavelength (λ) must be as small as possible. Therefore, light of a shorter wavelength (such as blue or ultraviolet light) provides higher resolution and better accuracy in determining the position of an object compared to longer wavelengths (such as red or infrared light). Polarization and intensity do not fundamentally change the diffraction limit of the optical system.
Legal Metrology • 2010 • Physics Paper-I
Which one of the following statements is true?
A)
Both light and sound waves can travel in vacuum
B)
Both light and sound waves in air are transverse
C)
The sound waves in air are longitudinal, while the light waves are transverse
D)
Both light and sound waves in air are longitudinal
To determine the correct statement, we must analyze the fundamental properties of sound and light waves:
1. **Nature of Sound Waves**: Sound is a mechanical wave that requires a medium (solid, liquid, or gas) to propagate. In air, sound travels as a series of compressions and rarefactions where the particles of the medium vibrate parallel to the direction of wave propagation. This is the definition of a **longitudinal wave**. Therefore, sound cannot travel in a vacuum.
2. **Nature of Light Waves**: Light is an electromagnetic wave consisting of oscillating electric and magnetic fields. These fields oscillate perpendicular to the direction of wave propagation, which is the definition of a **transverse wave**. Unlike sound, electromagnetic waves do not require a medium and can travel through a vacuum.
Evaluating the options:
- **Option a**: Incorrect. Sound waves cannot travel in a vacuum.
- **Option b**: Incorrect. Sound waves in air are longitudinal, not transverse.
- **Option c**: Correct. It accurately identifies sound waves in air as longitudinal and light waves as transverse.
- **Option d**: Incorrect. Light waves are transverse, not longitudinal.
Legal Metrology • 2010 • Physics Paper-I
Suppose temperature of the sun goes down by a factor of two, then the total power emitted by the sun will go down by a factor of
A)
2
B)
4
C)
8
D)
16
The total power emitted by a blackbody (like the sun) is governed by the Stefan-Boltzmann Law. According to this law, the total power (P) emitted per unit surface area of a blackbody is proportional to the fourth power of its absolute temperature (T). The formula is expressed as: P = σAT⁴, where σ is the Stefan-Boltzmann constant and A is the surface area.
1. Let the initial temperature be T₁ and the initial power be P₁ = σAT₁⁴.
2. The problem states the temperature goes down by a factor of two, so the new temperature T₂ = T₁ / 2.
3. The new power P₂ = σA(T₂⁴) = σA(T₁ / 2)⁴.
4. Simplifying the expression: P₂ = σA(T₁⁴ / 2⁴) = σA(T₁⁴ / 16).
5. Comparing P₂ to P₁: P₂ = P₁ / 16.
Therefore, the total power emitted by the sun will go down by a factor of 16. Options a, b, and c are incorrect because they assume linear, squared, or cubed relationships, whereas the physical law dictates a fourth-power relationship.
Legal Metrology • 2010 • Physics Paper-I
According to Huygen’s principle, light is a form of
A)
particle
B)
rays
C)
wave
D)
None of these
Huygens' Principle was proposed by Christiaan Huygens in 1678 to explain the propagation of light. The core of this principle is the assertion that every point on a wavefront acts as a source of secondary spherical wavelets that spread out in the forward direction at the speed of light. The new wavefront is the surface tangent to these secondary wavelets. This theoretical framework explicitly treats light as a wave rather than a particle (Newton's corpuscular theory) or simple geometric rays. While modern physics recognizes the wave-particle duality of light, Huygens' principle specifically defines and describes light as a wave to explain phenomena such as reflection, refraction, and diffraction.
Legal Metrology • 2010 • Physics Paper-I
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.
Legal Metrology • 2010 • Physics Paper-I
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.
Legal Metrology • 2010 • Physics Paper-I
A body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If R is the radius of the earth, the maximum height attained by the body is
A)
R/6
B)
R/3
C)
2R/3
D)
R
Energy conservation: v=ve/2 ⇒ 1/(R+h)=3/(4R) ⇒ h=R/3.
Legal Metrology • 2010 • Physics Paper-I
At what temperature do the Celsius and Fehrenheit scales give the same reading?
A)
80º F
B)
-40º C
C)
0º C
D)
40º R
To find the temperature where the Celsius (C) and Fahrenheit (F) scales are equal, we use the standard conversion formula: F = (9/5)C + 32.
Step-by-step calculation:
1. Set F = C (since we are looking for the point where the readings are the same). Let this common value be 'x'.
2. Substitute x into the formula: x = (9/5)x + 32.
3. Subtract (9/5)x from both sides: x - (9/5)x = 32.
4. Combine the x terms: (5/5)x - (9/5)x = 32, which simplifies to (-4/5)x = 32.
5. Solve for x: x = 32 * (-5/4).
6. x = 8 * (-5) = -40.
Therefore, -40º C is equal to -40º F.
Evaluating the options:
- Option (a) 80º F: Not equal to 80º C.
- Option (b) -40º C: Correct, as -40º C = -40º F.
- Option (c) 0º C: Equal to 32º F.
- Option (d) 40º R: The Rankine scale is an absolute scale; 40º R is extremely cold and does not correspond to the intersection of C and F.
Legal Metrology • 2010 • Physics Paper-I
The unit of Stefan’s constant in S I are
A)
J s-1 m-2 K-4
B)
W s-1 m-2K
C)
J m-2K-4
D)
W m-2 K-4
Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a black body per unit time (emissive power, E) is directly proportional to the fourth power of its absolute temperature (T). The formula is given by: E = σT⁴, where σ is the Stefan-Boltzmann constant.
1. Analyze the units of the variables:
- E (Emissive Power) = Energy per unit area per unit time = (Joules) / (square meters × seconds) = J s⁻¹ m⁻².
- Since 1 Watt (W) = 1 Joule per second (J s⁻¹), the unit of E is W m⁻².
- T (Absolute Temperature) is measured in Kelvin (K).
2. Rearrange the formula to solve for σ:
σ = E / T⁴
3. Substitute the units:
Unit of σ = (W m⁻²) / (K⁴) = W m⁻² K⁻⁴.
Comparing this to the options:
- Option (a) J s-1 m-2 K-4 is equivalent to W m-2 K-4, but standard notation for the constant usually uses Watts or Joules/sec. However, looking at the options provided, (d) is the most standard representation of the unit.
- Option (b) is incorrect as it includes s-1 and K instead of K-4.
- Option (c) is incorrect as it lacks the time component (seconds/Watts).
- Option (d) W m-2 K-4 matches the derived unit perfectly.
Legal Metrology • 2010 • Physics Paper-I
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).
Legal Metrology • 2010 • Physics Paper-I
Which of the following quantity is always positive in a S.H.M.?
A)
F.a
B)
v.r
C)
a.r
D)
F.r
In Simple Harmonic Motion (S.H.M.), the restoring force (F) is directly proportional to the displacement (r) from the equilibrium position and acts in the opposite direction. This is expressed by the equation: F = -kr, where k is the force constant. According to Newton's Second Law, F = ma, where 'm' is the mass and 'a' is the acceleration. Therefore, F and a are always in the same direction (since mass m is always positive). The dot product of two vectors in the same direction is the product of their magnitudes: F · a = |F||a| cos(0°) = |F||a|. Since magnitudes are always non-negative and the vectors are collinear and codirectional, F · a is always positive (or zero at the equilibrium point).
Evaluating other options:
- b. v · r: Velocity (v) and displacement (r) are in the same direction when moving away from equilibrium and opposite when moving toward it. Thus, v · r can be positive or negative.
- c. a · r: Since a = - (k/m)r, acceleration and displacement are always in opposite directions. Therefore, a · r = |a||r| cos(180°) = -|a||r|, which is always negative.
- d. F · r: Since F = -kr, force and displacement are always in opposite directions. Therefore, F · r = |F||r| cos(180°) = -|F||r|, which is always negative.
Legal Metrology • 2010 • Physics Paper-I
A gale blows over a house. The force due to gale on the roof is
A)
in the downward direction
B)
in the upward direction
C)
in the horizontal direction
D)
zero
This phenomenon is explained by Bernoulli's Principle, which states that for an incompressible, non-viscous fluid (like air), an increase in the speed of the fluid occurs simultaneously with a decrease in static pressure.
1. When a gale (strong wind) blows over a house, the air velocity above the roof is significantly higher than the air velocity inside the house (where the air is relatively stagnant).
2. According to Bernoulli's equation (P + 1/2ρv² = constant), the high velocity of the wind over the roof results in a region of low pressure.
3. The air inside the house remains at atmospheric pressure, which is higher than the pressure on the exterior surface of the roof.
4. This pressure difference (higher pressure inside, lower pressure outside) creates a net upward force, often referred to as 'lift'.
Therefore, the force exerted by the gale on the roof is directed upwards, which is why roofs are often lifted off houses during severe storms or hurricanes.
Legal Metrology • 2010 • Physics Paper-I
When applied to solar radiation, Planck’s law reduces to Wien’s law in the
A)
ultraviolet region
B)
microwave region
C)
infrared region
D)
visible region
Planck's Law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T. The formula is given by: B(λ, T) = (2hc²/λ⁵) / (exp(hc/λkT) - 1).
Wien's Approximation (Wien's Distribution Law) is a simplification of Planck's Law. It is valid when the term (hc/λkT) is much larger than 1 (hc/λkT >> 1). In this limit, the '-1' in the denominator of Planck's equation becomes negligible, and the expression reduces to: B(λ, T) ≈ (2hc²/λ⁵) exp(-hc/λkT).
For solar radiation (where the surface temperature of the sun is approximately 5778 K), the condition hc/λkT >> 1 is satisfied when the wavelength λ is very small. Small wavelengths correspond to the high-frequency, high-energy end of the spectrum, which is the ultraviolet (UV) region.
Conversely, for very long wavelengths (infrared or microwave regions), the Rayleigh-Jeans law is the appropriate approximation of Planck's law, as the term (hc/λkT) becomes very small.
Legal Metrology • 2010 • Physics Paper-I
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.
Legal Metrology • 2010 • Physics Paper-I
Golden view of sea shell is due to
A)
diffraction
B)
dispersion
C)
polarization
D)
reflection
The iridescent or 'golden' colors seen on the surface of sea shells (such as nacre or mother-of-pearl) are caused by the phenomenon of diffraction and interference. The inner layer of these shells consists of microscopic, parallel plates of aragonite (calcium carbonate) separated by organic layers. When light hits these periodic structures, it is diffracted and interferes constructively or destructively depending on the wavelength and the angle of observation. This results in the shifting, rainbow-like colors known as iridescence.
Why other options are incorrect:
- b. Dispersion: This is the splitting of white light into its constituent colors through a medium with a varying refractive index (like a prism). While it produces colors, it does not create the structural iridescence seen in shells.
- c. Polarization: This refers to the orientation of light wave oscillations. While it can affect how light is reflected, it is not the cause of the multi-colored appearance of shells.
- d. Reflection: Simple specular reflection only reflects the color of the light source; it does not split light into a spectrum of colors unless combined with interference or diffraction.
Legal Metrology • 2010 • Physics Paper-I
The internal energy of a perfect gas is
A)
partly kinetic and partly potential
B)
wholly potential
C)
wholly kinetic
D)
depends on the ratio of two specific heats
1. Definition of a Perfect Gas: A perfect gas (often used interchangeably with an ideal gas in this context) is a theoretical gas composed of point masses that move randomly and collide elastically.
2. Intermolecular Forces: The fundamental assumption of a perfect/ideal gas is that there are no intermolecular forces of attraction or repulsion between the gas molecules.
3. Potential Energy: In thermodynamics, internal potential energy arises from the forces of attraction or repulsion between molecules. Since these forces are assumed to be zero for a perfect gas, the internal potential energy is zero.
4. Kinetic Energy: The internal energy of a gas consists of the sum of the kinetic energies of its molecules (translational, rotational, and vibrational). For a perfect gas, the internal energy is solely a function of temperature and consists entirely of this kinetic energy.
5. Conclusion: Because there is no potential energy component, the internal energy of a perfect gas is wholly kinetic.
Comparison with other options:
- Option (a) describes a real gas, where intermolecular forces exist.
- Option (b) is incorrect as gases possess kinetic energy due to molecular motion.
- Option (d) is incorrect because while the specific heat ratio (gamma) relates to the degrees of freedom, the nature of the energy itself remains kinetic.
Legal Metrology • 2010 • Physics Paper-I
An ideal black body at room temperature is thrown into a furnace. It is observed that
A)
initially it is the darkest body and at later times the brightest
B)
it is the darkest body at all times
C)
it cannot be distinguished at all times
D)
initially it is the darkest body and at later times it cannot be distinguished
1. Definition of a Black Body: An ideal black body is defined as a body that absorbs all incident electromagnetic radiation regardless of frequency or angle of incidence. It reflects no light.
2. Initial State: When the black body at room temperature is first thrown into the furnace, the furnace walls are at a much higher temperature and are emitting intense radiation. The black body absorbs all this incident radiation and reflects none. Since it emits very little radiation (due to its low initial temperature) and reflects nothing, it appears as the darkest object in the furnace.
3. Heating Process: As the black body remains in the furnace, it absorbs energy and its temperature increases.
4. Final State: According to the Stefan-Boltzmann Law, the power radiated by a body is proportional to the fourth power of its absolute temperature (P = σAT⁴). A black body is the most efficient emitter of radiation at a given temperature. Once the black body reaches the same temperature as the furnace walls, it will emit the maximum possible amount of radiation for that temperature. Since the furnace walls are likely not perfect black bodies (they reflect some light), the ideal black body will emit more radiation than the walls, making it appear as the brightest object in the furnace.
5. Conclusion: The body transitions from being the darkest (maximum absorption, minimum emission) to the brightest (maximum emission at equilibrium temperature).
Legal Metrology • 2010 • Physics Paper-I
In Rayleigh scattering, the amount of scattered light is proportional to
A)
1/λ
B)
1/λ^2
C)
1/λ^3
D)
1/λ^4
Rayleigh scattering occurs when light is scattered by particles that are much smaller than the wavelength of the light (such as gas molecules in the atmosphere). According to the Rayleigh scattering law, the intensity of the scattered light (I) is inversely proportional to the fourth power of the wavelength (λ). Mathematically, this is expressed as I ∝ 1/λ⁴. This explains why shorter wavelengths (blue light) are scattered much more efficiently than longer wavelengths (red light), which is why the sky appears blue during the day and the sun appears red at sunset. Options a, b, and c are incorrect because they do not follow the derived physical law for scattering by small particles.
Legal Metrology • 2010 • Physics Paper-I
Deviation produced in a grating is independent of
A)
wavelength of light
B)
grating constant or element
C)
refractive index of grating material
D)
None of these
The deviation (diffraction angle) produced by a diffraction grating is governed by the grating equation: d sin(θ) = nλ, where 'd' is the grating element (the distance between two adjacent slits), 'θ' is the angle of diffraction (deviation), 'n' is the order of the spectrum, and 'λ' is the wavelength of the incident light.
1. Wavelength (λ): The equation shows that sin(θ) is directly proportional to λ. Therefore, deviation depends on the wavelength of light (Option a is incorrect).
2. Grating constant/element (d): The equation shows that sin(θ) is inversely proportional to d. Therefore, deviation depends on the grating constant (Option b is incorrect).
3. Refractive Index: A diffraction grating works on the principle of interference of light waves originating from multiple slits. Unlike a prism, where deviation is caused by refraction (and thus depends on the refractive index of the material), a grating's deviation is a result of diffraction and interference. The material's refractive index does not enter the grating equation. Thus, the deviation is independent of the refractive index of the grating material (Option c is correct).
Legal Metrology • 2010 • Physics Paper-I
A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth; R being the radius of the earth. What will be the time period of another satellite at a height 2.5R from the surface of the earth?
A)
6^2 hours
B)
6^2.5 hours
C)
6^3 hours
D)
12 hours
Step-by-step solution:
1. **Kepler's Third Law**: The square of the orbital period (T) is proportional to the cube of the semi-major axis (r) of the orbit: T² ∝ r³.
2. **Identify Orbital Radii**:
- The orbital radius is the sum of the Earth's radius (R) and the height above the surface (h).
- For the first satellite (geostationary): r₁ = R + 6R = 7R.
- For the second satellite: r₂ = R + 2.5R = 3.5R.
3. **Set up the Ratio**:
(T₂ / T₁)² = (r₂ / r₁)³
(T₂ / T₁)² = (3.5R / 7R)³
(T₂ / T₁)² = (1/2)³ = 1/8
4. **Solve for T₂**:
T₂² = T₁² / 8
T₂ = T₁ / √8
5. **Analyze the Geostationary Period (T₁)**:
A geostationary satellite has a period of 24 hours.
T₂ = 24 / √8 ≈ 24 / 2.828 ≈ 8.48 hours.
6. **Evaluating the Options**:
- Option (a): 6² = 36 hours (Incorrect based on calculation, but let's re-examine the question's logic).
- Option (b): 6^2.5 ≈ 88 hours.
- Option (c): 6³ = 216 hours.
- Option (d): 12 hours.
Wait, let's re-read the prompt's specific values. If the first satellite is at 6R *above the surface*, r₁ = 7R. If the first satellite is at 6R *from the center*, r₁ = 6R.
If r₁ = 6R and r₂ = 3.5R, T₂ = 24 * (3.5/6)^1.5 ≈ 11.3 hours.
However, looking at the options provided in this specific exam paper (MPSC 2010), there is a common typo in the question's phrasing or options. If we assume the ratio of radii was intended to be such that T₂ = T₁ * (r₂/r₁)^1.5, and we test the options:
If T₂ = 36 hours (6²), it would be slower than the geostationary one, which is impossible since it is closer to Earth (2.5R < 6R).
Correction on Calculation: If the question meant the first satellite is at height h₁ and the second at h₂, and we use the ratio:
(T₂/24)² = (3.5/7)³ = 1/8
T₂² = 576 / 8 = 72
T₂ = √72 ≈ 8.48 hours.
Given the options provided in the official key for this specific legacy paper, option (a) '6^2' is often listed, but mathematically, for a satellite closer to earth, the period must be LESS than 24 hours. There is a high probability of a typo in the original exam's options or the height values. However, based on the standard provided answer key for this specific MPSC 2010 paper, 'a' is the marked answer, likely due to a calculation error in the paper's design where they might have inverted the ratio or used different constants. Mathematically, none of the options perfectly match 8.48h, but 'a' is the designated key.
Legal Metrology • 2010 • Physics Paper-I
Phase difference between two successive Fresnel’s Half Period Zones (H.P.Z.) is
A)
Zero
B)
π/2
C)
π
D)
2π
Fresnel's Half Period Zones (H.P.Z.) are defined as concentric rings of a wavefront such that the distance from the source to the boundary of each zone increases by exactly half a wavelength (λ/2) compared to the previous zone.
1. Path Difference: By definition, the path difference between the boundaries of two successive H.P.Z.s is Δx = λ/2.
2. Phase Difference Formula: The relationship between path difference (Δx) and phase difference (φ) is given by the formula: φ = (2π / λ) * Δx.
3. Calculation: Substituting Δx = λ/2 into the formula: φ = (2π / λ) * (λ / 2) = π radians.
Because the phase difference is π (or 180 degrees), the waves originating from successive zones interfere destructively. This is why the total amplitude at a point is the alternating sum of the amplitudes of the individual zones.
- Option (a) Zero: This would imply constructive interference and identical phase.
- Option (b) π/2: This represents a quarter-cycle shift.
- Option (d) 2π: This represents a full cycle, which would mean the zones are in phase, contradicting the definition of 'Half Period' zones.