Thermodynamics, Waves, and Mechanics: Physics Q&A
Thermodynamics, Waves, and Mechanics: Physics Q&A
Q.1 Zeroth’s Law of Thermodynamics
Answer: The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third one, then they are in thermal equilibrium with each other.
Q.2 Sign Convention in Thermodynamics
Answer: The sign convention used in the measurement of heat, work, and internal energy is:
- Heat added to the system is positive.
- Work done by the system is positive.
- Change in internal energy is positive if the temperature increases.
Q.3 First Law of Thermodynamics
Answer: The first law of thermodynamics is a statement of the law of conservation of energy. It states that energy can neither be created nor destroyed, but it can be transferred from one form to another. This means that the total amount of energy in the universe is constant. The first law of thermodynamics can be expressed mathematically as: ΔU = Q – W
Q.5 Work Done in Isothermal and Adiabatic Processes
Answer:
Isothermal Process:
- An isothermal process is a thermodynamic process in which the temperature of the system remains constant.
- The work done by an ideal gas during an isothermal process is given by: W = nRT ln(V2/V1)
Adiabatic Process:
- An adiabatic process is a thermodynamic process in which no heat is transferred to or from the system.
- The work done by an ideal gas during an adiabatic process is given by: W = (P2V2 – P1V1)/(1 – γ)
Q.6 Carnot Engine and Carnot Cycle
Answer: A Carnot engine is a theoretical heat engine that operates in a cyclic process called the Carnot cycle. It provides the maximum possible efficiency for converting heat into work. It consists of:
- Hot Reservoir (Th) – Provides heat at constant high temperature.
- Cold Reservoir (Tc) – Absorbs rejected heat at constant low temperature.
- Working Substance – Usually an ideal gas that undergoes expansion and compression.
- Cylinder & Piston – The gas expands and compresses in a thermally insulated cylinder.
The Carnot cycle consists of four reversible processes:
- Isothermal Expansion (A → B)
The gas absorbs heat from the hot reservoir at constant Th. It expands, doing work on the surroundings. - Adiabatic Expansion (B → C)
The gas expands further without heat exchange. Its temperature decreases from Th to Tc. - Isothermal Compression (C → D)
The gas releases heat to the cold reservoir at constant Tc. The gas compresses, reducing its volume. - Adiabatic Compression (D → A)
The gas is compressed without heat exchange. Its temperature rises back to Th, completing the cycle.
Q.7 Efficiency of Carnot’s Engine
Answer:
Carnot Engine Efficiency
The efficiency of a Carnot engine is given by:
η = 1 – (Tc/Th)
where:
Th = Temperature of the hot reservoir (Kelvin)
Tc = Temperature of the cold reservoir (Kelvin)
This formula shows that efficiency depends only on temperatures, not the working substance.
Q.8 P-V Diagram for Carnot Cycle
Answer:
P-V Diagram for a Carnot Cycle
A Carnot cycle is represented on a P-V diagram as a rectangle with curved sides.
- Isothermal Expansion (A to B):
The gas expands isothermally (constant temperature) while absorbing heat from the hot reservoir. This is represented by the top, horizontal line of the rectangle. - Adiabatic Expansion (B to C): The gas continues to expand, but now adiabatically (no heat exchange). This is represented by the curved line sloping downwards to the right.
- Isothermal Compression (C to D):
The gas is compressed isothermally, rejecting heat to the cold reservoir. This is represented by the bottom, horizontal line of the rectangle. - Adiabatic Compression (D to A):
The gas is further compressed adiabatically, returning to its initial state. This is represented by the curved line sloping upwards to the left.
Q.9 Working of a Heat Engine
Answer: A heat engine is a system that converts heat or thermal energy into mechanical energy, which can then be used to do mechanical work. It does this by bringing a working substance from a high temperature state to a low temperature state. A heat source generates thermal energy that brings the working substance to the high temperature state. The working substance generates work in the working body of the engine while transferring the heat to the cold sink until the working substance reaches a low temperature state. During this process, some of the thermal energy is converted into work by exploiting the physical properties of the working substance.
Q.10 Carnot Engine Efficiency: Ice and Steam Points
Answer: The efficiency of a Carnot engine is determined by the temperatures of the hot and cold reservoirs between which it operates. In this case:
Ice point: 273 K (cold reservoir)
Steam point: 373 K (hot reservoir)
The formula for Carnot efficiency is: η = 1 – (Tc/Th)
η = 1 – (273 K / 373 K)
η = 0.268
So, the efficiency of a Carnot engine operating between the ice point and steam point is approximately 0.268, or 26.8%.
Q.11 Maximum Possible Engine Efficiency
Answer: The maximum possible efficiency of a heat engine operating between two temperatures is given by the Carnot efficiency:
η = 1 – (Tc/Th)
Tc = 200 K
Th = 700 K
Therefore: η = 1 – (200 K / 700 K)
η ≈ 0.714
So, the maximum possible efficiency of this engine is approximately 71.4%.
Q.12 Progressive vs. Transverse Waves
Answer:
Progressive Waves:
These waves transfer energy through a medium. The wave profile goes forward with the wave’s speed. Energy is transferred.
Transverse Waves:
These are a type of progressive wave. The particles of the medium vibrate perpendicular to the direction of wave propagation. They have crests and troughs.
Q.13 Periodicity of sin(ωt) and cos(ωt)
Answer: A function is periodic if it repeats after a time T
Sin(ωt+T) = sin(ωt)
Cos(ωt+T) = cos(ωt)
Q.14 Simple Harmonic Motion (SHM)
Answer:
- Proportionality: The acceleration is directly proportional to the displacement.
- Direction: The acceleration is always directed towards the equilibrium position.
- Periodic Motion: SHM is a repetitive or oscillatory motion.
- SHM can be represented by sin or cos functions.
Q.15 SHM: Displacement, Velocity, Acceleration, Time Period
Answer:
- The displacement of a particle in SHM is a sinusoidal function of time. It can be represented as: x(t) = A cos(ωt + φ)
- Velocity is the first derivative of displacement with respect to time: v(t) = dx/dt = -Aω sin(ωt + φ)
- Acceleration is the first derivative of velocity with respect to time: a(t) = dv/dt = -Aω² cos(ωt + φ)
- The time period is the time taken for one complete oscillation. It’s related to the angular frequency by: T = 2π/ω
Q.16 What is a Beat?
Answer: Beats are the periodic variations in amplitude that occur when two waves of slightly different frequencies interfere with each other. This interference results in alternating loud and soft sounds. Essentially, it is the periodic variation of sound intensity, due to the superposition of two sound waves of slightly differing frequencies.
Q.17 Total Energy of a Particle in SHM
Answer: The total energy of a particle executing SHM is the sum of its kinetic energy and potential energy.
The kinetic energy of a particle is given by: KE = 1/2 mv^2
The potential energy of a particle is given by: PE = 1/2 kx^2
The total energy of the particle is then given by: E = KE + PE = 1/2 mv^2 + 1/2 kx^2
Substituting the expressions for velocity and displacement, we obtain:
E = 1/2 m(-Aω sin(ωt + φ))^2 + 1/2 k(A cos(ωt + φ))^2
Simplifying, we obtain:
E = 1/2 mA^2ω^2 sin^2(ωt + φ) + 1/2 kA^2 cos^2(ωt + φ)
Using the identity sin^2(θ) + cos^2(θ) = 1, we obtain:
E = 1/2 mA^2ω^2 (sin^2(ωt + φ) + cos^2(ωt + φ))
Therefore, the total energy of a particle executing SHM is constant and is proportional to the square of the amplitude.
Q.18 Time Period and Frequency of Oscillations
Answer:
- The time period of a horizontal oscillation of a massless loaded spring is given by: T = 2π√(m/k)
- The time period of a vertical oscillation of a massless loaded spring is given by: T = 2π√(m/k)
Q.19 Root Mean Square (RMS) Speed
Answer: The formula for RMS speed (vrms) is: vrms = √(3RT/M)
The formula shows that RMS speed is:
- Directly proportional to the square root of the temperature. Higher temperatures mean faster molecules.
- Inversely proportional to the square root of the molar mass. Lighter molecules move faster than heavier ones at the same temperature.
Q.20 Pressure Exerted by a Gas
Answer: The pressure exerted by a gas in a cubical vessel is given by: P = (1/3)ρv^2
The pressure is proportional to the density of the gas and the square of the RMS speed. This means that the pressure increases as the density of the gas increases or as the temperature increases.
Q.21 Law of Equipartition of Energy
Answer: The law of equipartition of energy states that, in thermal equilibrium, the average energy of a molecule is equally distributed among all its degrees of freedom, and the average energy associated with each degree of freedom is 1/2 kT, where k is the Boltzmann constant and T is the absolute temperature.
Derivation
- Degrees of Freedom:
A degree of freedom is an independent way a molecule can possess energy (e.g., translational, rotational, vibrational). - Quadratic Energy Terms:
The equipartition theorem applies to degrees of freedom where the energy is a quadratic function of a variable. - Average Energy:
According to statistical mechanics, the average energy associated with each quadratic degree of freedom is given by 1/2 kT. - Total Average Energy:
If a molecule has ‘f’ degrees of freedom, the total average energy of the molecule is: E = (f/2) kT
Q.22 Uses of Avogadro’s Number
Answer: Avogadro’s number is used for:
- Calculating the number of atoms or molecules in a given mass of a substance.
- Calculating the molar mass of a substance.
- Relating macroscopic measurements to microscopic quantities of atoms and molecules.
Q.23 Mean Free Path
Answer: The mean free path is the average distance that a molecule travels between collisions. It is given by: λ = 1/(nσ)
The mean free path is inversely proportional to the number density of the gas and the collision cross-section of the molecules. This means that the mean free path decreases as the number density of the gas increases or as the collision cross-section of the molecules increases.
Q.24 Pascal’s Law of Fluid Pressure
Answer: Pascal’s Law states: A pressure change at any point in a confined incompressible fluid is transmitted equally to all points throughout the fluid.
To prove Pascal’s Law, consider a small, right-angled triangular prism of fluid within a larger container, where every point on the prism is at the same depth, meaning the pressure acting on each face is essentially the same; by analyzing the forces acting on this prism and setting them equal to zero, we can demonstrate that the pressure is transmitted equally in all directions throughout the fluid.
Q.25 Experimental Verification of Pascal’s Law
Answer: Experimentally verifying Pascal’s Law involves demonstrating that a pressure change at one point in a confined fluid is transmitted equally throughout the fluid. Here’s a common experimental setup:
Materials:
- A spherical glass bulb with multiple narrow, vertical tubes extending from it at various points.
- A piston or plunger that fits tightly into an opening on the bulb.
- Water or another incompressible fluid.
- A ruler or measuring scale.
Procedure:
- Fill the Bulb: Fill the spherical glass bulb completely with the chosen fluid, ensuring no air bubbles are trapped. The fluid should rise to a certain level in each of the vertical tubes.
- Establish Equilibrium: Allow the fluid to settle so that the levels in all the vertical tubes are stable. This represents the initial, uniform pressure state.
- Apply Pressure: Gently push the piston or plunger into the opening of the bulb. This will increase the pressure within the fluid.
- Observe the Levels: Carefully observe the change in fluid levels in each of the vertical tubes. You should see that the fluid level rises in all the tubes.
- Measure the Change: Use the ruler or measuring scale to measure the change in fluid level in each tube.
- Compare the Changes: Compare the changes in fluid level in all the tubes. If Pascal’s Law holds, the changes in fluid level should be approximately equal, indicating that the pressure increase was transmitted equally throughout the fluid.
Explanation:
- The rise in fluid level in each tube is proportional to the increase in pressure at that point.
- If the pressure increase is transmitted equally, the fluid levels in all tubes will rise by the same amount.
- The use of the spherical bulb is to show that the pressure is transmitted in all directions.
Precautions:
- Ensure there are no air bubbles in the fluid, as they can compress and affect the results.
- Apply pressure gently and steadily to avoid sudden changes that could cause inaccuracies.
- Use a fluid that is truly incompressible for best results.
Q.26 Applications of Pascal’s Law
Answer: Applications include hydraulic jacks, hydraulic brakes, and hydraulic pumps. It’s also used in the human heart and in some construction machines.
Q.27 Pressure Exerted by a Liquid Column
Answer: (Derivation required here)
Q.28 What is Gauge Pressure?
Answer: Gauge pressure is the pressure of a system relative to atmospheric pressure. It’s the difference between the absolute pressure and the atmospheric pressure.
Formula: Gauge pressure = Absolute pressure – Atmospheric pressure
Q.29 Coefficient of Viscosity
Answer: The coefficient of viscosity (η) measures a fluid’s resistance to flow, or its internal friction. A high coefficient indicates a thick fluid. A low coefficient indicates a thin fluid.
Dimensions: [η] = ML⁻¹T⁻¹
Units:
- SI Unit: Pascal-second (Pa·s) or kilogram per meter-second (kg/(m·s)).
- CGS Unit: Poise (P). (1 Pa·s = 10 Poise).
Q.31 Terminal Velocity
Answer: (Derivation required here)
Q.32 Streamline, Turbulent, and Laminar Flow
Answer:
- Laminar Flow:
Smooth, parallel layers of fluid. Particles move in orderly, predictable paths. Occurs at lower velocities. - Streamline Flow:
Often synonymous with laminar. Focuses on individual particle paths that don’t cross. Represents smooth, ordered flow. - Turbulent Flow:
Chaotic, irregular fluid motion. Eddies and vortices cause significant mixing. Occurs at higher velocities.
Q.33 What is an Ideal Fluid?
Answer: An ideal fluid is a theoretical concept in fluid dynamics. It’s a fluid that’s assumed to have these properties:
- Incompressible: Its density stays constant, no matter the pressure.
- Non-viscous: It has no internal friction, so it flows without any resistance.
- Irrotational: The flow is smooth, with no swirling or eddies.
- Steady flow: The velocity of the fluid at any point does not change over time.
Q.34 Equation of Continuity
Answer: The equation of continuity for an incompressible, non-viscous fluid with steady flow is: A₁v₁ = A₂v₂
Q.35 Bernoulli’s Principle
Answer: (Statement, proof, and applications required here)
Q.36 Surface Energy and Surface Tension
Answer:
Surface Energy:
This is the potential energy stored at the surface of a liquid due to the attractive forces between its molecules. It represents the extra energy needed to create a new surface area.
Surface Tension:
This is the force acting along the edge of a liquid surface that tries to minimize the surface area. It is measured in Newtons per meter (N/m).
Proof:
Imagine a rectangular frame with one movable side on a liquid surface.
Work done by surface tension: When the movable side is pulled by a small distance dx, the force exerted by surface tension is γL. The work done is then γL × dx.
Area increase: The increase in surface area due to this movement is L × dx.
Surface energy per unit area: By definition, surface energy per unit area is the work done per unit area increase. Therefore, surface energy per unit area = (γL * dx) / (L * dx) = γ.
Q.37 Pressure Difference Across a Curved Liquid Surface
Answer: (Diagram and explanation required here)
Q.38 Excess Pressure Inside Liquid Drop and Soap Bubble
Answer:
Liquid Drop:
- Fs (tension) = 2πRT
- Fp (pressure) = ΔPπR²
- Fs = Fp → 2πRT = ΔPπR²
- ΔP = 2T/R
Soap Bubble:
- Fs (tension) = 4πRT (two surfaces)
- Fp (pressure) = ΔPπR²
- Fs = Fp → 4πRT = ΔPπR²
- ΔP = 4T/R
Q.39 Angle of Contact
Answer: The angle of contact is the angle formed between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid.
Factors it depends on:
- Nature of the liquid and solid:
The adhesive forces between the liquid and solid molecules, and the cohesive forces within the liquid, play a significant role.
Q.40 Ascent Formula
Answer:
Ascent Formula Derivation:
- Upward Force (Tension): Fup = 2πrTcosθ
- Downward Force (Weight): Fdown = πr²hρg
- Equilibrium: Fup = Fdown
- 2πrTcosθ = πr²hρg
- h = 2Tcosθ / rρg
Q.41 Bulk Modulus Calculation
Answer: Bulk Modulus: B = – (ΔP) / (ΔV/V)
Initial pressure (P1) = 1.00 x 10⁵ Pa
Final pressure (P2) = 1.165 x 10⁵ Pa
ΔV/V = 0.1
Change in Pressure (ΔP):
- ΔP = P2 – P1
- ΔP = (1.165 x 10⁵ Pa) – (1.00 x 10⁵ Pa)
- ΔP = 0.165 x 10⁵ Pa = 1.65 x 10⁴ Pa
Calculating Bulk Modulus (B):
- B = – (ΔP) / (ΔV/V)
- B = – (1.65 x 10⁴ Pa) / (0.1)
- B = -1.65 x 10⁵ Pa
Considering the absolute value
- Bulk modulus is always a positive value, therefore the negative sign is dropped.
- B = 1.65 x 10⁵ Pa
Therefore, the bulk modulus of the medium is 1.65 x 10⁵ Pascals.
Q.42 Definitions: Elasticity
Answer:
- Elastic Limit: The maximum stress a material can endure without permanent deformation, ensuring it returns to its original shape upon stress removal.
- Perfect Elastic Body: A theoretical material that instantaneously and completely reverts to its initial form after any deforming force is removed.
- Malleability: A material’s capacity to be shaped into thin sheets by hammering or rolling, demonstrating its ability to withstand compressive forces.
- Ductility: A material’s ability to be stretched into wires under tensile stress, indicating its capacity for plastic deformation without fracturing.
- Modulus of Elasticity: A material’s stiffness, quantifying its resistance to elastic deformation as the ratio of stress to strain within the elastic region.
Q.43 Types of Stress and Strain
Answer:
Stress Types:
- Tensile: Stretching or pulling forces.
- Compressive: Squeezing or pushing forces.
- Shear: Tangential forces causing sliding.
- Hydraulic/Bulk: Uniform pressure from all sides.
Strain Types:
- Tensile: Elongation or increase in length.
- Compressive: Shortening or decrease in length.
- Shear: Angular distortion or deformation.
- Volume/Bulk: Change in volume relative to original volume.
Q.44 Types of Moduli of Elasticity
Answer:
- Young’s Modulus: Stretch/compression stiffness.
- Shear Modulus: Twist/slide stiffness.
- Bulk Modulus: Volume compression resistance.
Q.45 Young’s Modulus Calculation
Answer:
Calculating Volume (V):
- V = mass / density
- V = 0.05 kg / 8000 kg/m³
- V = 6.25 x 10⁻⁶ m³
Calculating Area (A):
- A = Volume / Length
- A = 6.25 x 10⁻⁶ m³ / 2.5 m
- A = 2.5 x 10⁻⁶ m²
Calculating Stress (σ):
- σ = Force / Area
- σ = 98 N / 2.5 x 10⁻⁶ m²
- σ = 3.92 x 10⁷ Pa
Calculating Strain (ε):
- ε = Change in Length / Original Length
- ε = 0.002 m / 2.5 m
- ε = 0.0008
Calculating Young’s Modulus (E):
- E = Stress / Strain
- E = 3.92 x 10⁷ Pa / 0.0008
- E = 4.9 x 10¹⁰ Pa
Q.46 Hooke’s Law
Answer: Hooke’s Law states that within the elastic limit of a material, the stress is directly proportional to the strain.
(Load vs. Extension graph required here)
Q.47 Poisson’s Ratio
Answer: Poisson’s ratio is a measure of a material’s tendency to deform in directions perpendicular to the direction of an applied force. Specifically, it’s the ratio of lateral strain to longitudinal strain when a material is stretched or compressed.
Expression:
ν = – (Lateral Strain) / (Longitudinal Strain)
Or, in terms of changes in dimensions:
ν = – (Δd/d) / (ΔL/L)
Significance of the Negative Sign:
The negative sign in the expression is crucial because it accounts for the opposite nature of lateral and longitudinal strains in most materials.
- When a material is stretched, its width or diameter decreases. The negative sign ensures that Poisson’s ratio is a positive value for typical materials.
- Conversely, when a material is compressed, its width or diameter increases. Again, the negative sign maintains a positive ratio.
Therefore, the negative sign is there to ensure that the poissons ratio is a positive value, since the lateral and longitudinal strains are opposite in nature.
Q.48 Lateral Strain Calculation
Answer:
Poisson’s Ratio Formula:
Poisson’s ratio (ν) = – (Lateral Strain) / (Longitudinal Strain)
Calculating Lateral Strain:
- Lateral Strain = – 0.5 * (2 x 10⁻³)
- Lateral Strain = – 1 x 10⁻³
Therefore, the lateral strain is -1 x 10⁻³. The negative sign indicates that the lateral deformation is opposite to the longitudinal deformation.
Q.49 Fundamental vs. Derived Physical Quantities
Answer:
Fundamental Physical Quantities:
These are the base units of measurement, independent and irreducible, forming the foundation for all other physical quantities in science.
Derived Physical Quantities:
Quantities formed by combining fundamental units through mathematical operations, providing measurements of more complex physical phenomena in the universe.
Q.50 Applications of Dimensional Analysis
Answer: Dimensional analysis is a fundamental aspect of measurement and is applied in real-life physics. We make use of dimensional analysis for three prominent reasons:
- To check the consistency of a dimensional equation.
- To derive the relation between physical quantities in physical phenomena.
- To change units from one system to another.
Q.51 Force Expression Derivation
Answer: F = ma.
Q.52 Significant Figures
Answer: 1- 1 ,2- 4 ,3- 4 ,4- 4,5- 4.
Q.53 Significant Figures in Calculations
Answer: 1- 2.535 ,2- 30.0 g, 3- 4800 ,4- 4 x 10⁻⁴
Q.54 Acceleration Calculation
Answer:
- Given Velocity Equation: v = 9t² + 3t + 1
- Differentiate Velocity with Respect to Time (dv/dt):
- dv/dt = d(9t²)/dt + d(3t)/dt + d(1)/dt
- dv/dt = (2 * 9t) + (1 * 3) + 0
- dv/dt = 18t + 3
- Acceleration (a) is the Derivative of Velocity:
- a = dv/dt
- a = 18t + 3
- Substitute t = 1 Second:
- a = 18(1) + 3
- a = 18 + 3
- a = 21
Therefore, the acceleration at t = 1 second is 21 units/second².
Q.55 Velocity Calculation
Answer:
To find the velocity, you need to differentiate the displacement equation with respect to time.
Given:
- Displacement (x) = 7t² + 8t + 3
- Time (t) = 2 seconds
Differentiate Displacement to Find Velocity:
- Velocity (v) = dx/dt
- v = d(7t² + 8t + 3)/dt
- v = 14t + 8
Substitute t = 2 Seconds:
- v = 14(2) + 8
- v = 28 + 8
- v = 36
The velocity of the particle along the x-axis at t = 2 seconds is 36 units per second.
Q.56 Distance Traveled from Graph
Answer:
Region 1 (0-5 seconds): Trapezoid
* Area = (1/2) * (sum of parallel sides) * height
* Area = (1/2) * (0 m/s + 50 m/s) * 5 s
* Area = (1/2) * 50 m/s * 5 s
* Area = 125 meters
Region 2 (5-7 seconds): Rectangle
* Area = base * height
* Area = 2 s * 50 m/s
* Area = 100 meters
Region 3 (7-10 seconds): Triangle
* Area = (1/2) * base * height
* Area = (1/2) * 3 s * 50 m/s
* Area = 75 meters
Total Distance
* Add the areas of all three regions:
* Total Distance = 125 m + 100 m + 75 m
* Total Distance = 300 meters
Therefore, the total distance traveled between 0 and 10 seconds, based on the given graph, is 300 meters.
Q.57 Define the following:
i) Equal vector ⅱ) Modulus of a vector iii) unit vector.
Answer: * Equal Vectors: Two vectors are equal if they have the same magnitude and direction.
* Modulus of Vector:The magnitude or length of a vector, representing its size.
* Unit Vector: A vector with a magnitude of 1, used to indicate direction.
Q.58 A body is Simultaneously given 2 velocity 30m/s towards east and other 40m/s towards north. Find the resultant velocity.
Answer: 1. Given Velocities
* v₁ = 30 m/s (east)
* v₂ = 40 m/s (north)
2. Magnitude of Resultant Velocity (v)
* Using Pythagorean theorem:
v² = v₁² + v₂²
v² = (30 m/s)² + (40 m/s)²
v² = 900 m²/s² + 1600 m²/s²
v² = 2500 m²/s²
v = √(2500 m²/s²)
v = 50 m/s
3. Direction of Resultant Velocity (θ)
* Using the tangent function: tan θ = v₂ / v₁
tan θ = (40 m/s) / (30 m/s)
tan θ = 4/3
θ = tan⁻¹(4/3)
θ ≈ 53.1°
Therefore, the resultant velocity is 50 m/s at 53.1° north of east.
Q.59 Two vectors both equal in magnitude, have their resultant equal in magnitude of origin.Find the angle between two vectors.
Answer: 1. Visualizing the Vectors
Imagine two vectors, A and B, with the same magnitude. When these vectors are added, they form a resultant vector R with the same magnitude as the original vectors. Since the magnitude of the resultant is equal to the magnitudes of the original vectors, the three vectors form an equilateral triangle.
2. Equilateral Triangle
In an equilateral triangle, all sides are equal, and all angles are 60°.
3. Angle Between Vectors
Therefore, the angle between the two vectors A and B is 60°.
Q.63 Derive the trajectory of a projectile , time of maximum height, Time of flight, horizontal flight, Range for a projectile fired with a velocity and an angle theta with the horizontal.
Answer: * Trajectory: A projectile follows a curved, parabolic path due to gravity, determined by initial velocity and launch angle, resulting in a predictable arc.
* Time of Maximum Height: The time taken to reach peak altitude, calculated using initial vertical velocity and gravity, representing the ascent duration before descent.
* Time of Flight: The total airborne duration, twice the time to maximum height, influenced by initial vertical velocity and gravity, indicating the complete flight time.
* Horizontal Range: The total horizontal distance traveled, determined by initial velocity, launch angle, and gravity, representing the projectile’s landing point distance.