CBSE Class 11 – Physics Model Paper 2021
SECTION – A
1. Which of the following pairs does not have similar dimensions?
A. Stress and pressure
B. Angle and strain
C. Tension and Surface tension
Answer: B. Angle and strain
2. A car travels from A to B at a speed of 20 km/hr and returns at a speed of 30 km/hr. The average speed of the car for the whole journey is
A. 5 km/hr
B. 24 km/hr
C. 25 km/hr
D. 50 km/hr
Answer: B. 24 km/hr
3. According to Hooke‘s law of elasticity, if stress is increased, the ratio of stress to strain
A. increases
B. decreases
C. becomes zero
D. remains constant
Answer: D. remains constant
4. Radius of one arm of a hydraulic lift is four times of the radius of the other arm. What force should be applied on the narrow arm to lift 100 kg?
A.26.5 N
B. 62.5 N
C. 6.25 N
D.8.3 N
Answer: B. 62.5 N
Section – B
19. Draw the following graphs (expected nature only) representing motion of an object under free fall. Neglect air resistance.
A. Variation of position with respect to time.
Answer: The graph of position vs. time for an object under free fall will be a parabola. This is because the object’s displacement is proportional to the square of the time due to constant acceleration (gravity).
B. Variation of velocity with respect to time.
Answer: The graph of velocity vs. time for an object under free fall will be a straight line with a positive slope. This is because the velocity increases linearly with time due to the constant acceleration of gravity.
C. Variation of acceleration with respect to time.
Answer: The graph of acceleration vs. time for an object under free fall will be a horizontal line at a constant value (approximately 9.8 m/s²). This is because the acceleration due to gravity is constant.
D. How can the distance traveled be calculated from the velocity – time graph in a uniform one-dimensional motion?
Answer: In a velocity-time graph, the distance traveled is represented by the area under the graph. For uniform motion, the graph is a straight line, and the area under the line is a trapezoid. The formula for the area of a trapezoid is (1/2)(base1 + base2) * height. In this case, base1 and base2 are the initial and final velocities, and the height is the time interval. Therefore, the distance traveled is given by:
Distance = (1/2)(initial velocity + final velocity) * time
20. Determine a unit vector perpendicular to both A = 2i + j + k and B = i – j + 2k.
Answer:
To find a unit vector perpendicular to both A and B, we can calculate their cross product. The cross product of two vectors is a vector that is perpendicular to both of them.
A x B = (i + 3j – 3k)
To get a unit vector, we divide this vector by its magnitude:
|A x B| = √(1² + 3² + (-3)²) = √19
Unit vector = (1/√19)i + (3/√19)j – (3/√19)k
OR
Find the angle between the vectors A = i + 2j – k and B = -i + j – 2k
Answer:
The angle between two vectors A and B can be found using the dot product:
A · B = |A| |B| cos(θ)
where θ is the angle between the vectors.
A · B = (i + 2j – k) · (-i + j – 2k) = -1 + 2 + 2 = 3
|A| = √(1² + 2² + (-1)²) = √6
|B| = √((-1)² + 1² + (-2)²) = √6
cos(θ) = (A · B) / (|A| |B|) = 3 / (√6 * √6) = 1/2
θ = cos⁻¹(1/2) = 60°
Therefore, the angle between the vectors A and B is 60°.
21. State the advantages of SI over other system of units.
Answer:
The International System of Units (SI) has several advantages over other systems:
- Coherence: SI units are based on a coherent system, where derived units are obtained by multiplying or dividing base units without introducing numerical factors.
- Rationalization: SI uses rationalized units, which simplify many physical equations.
- Universality: SI is the internationally accepted system of units, which promotes consistency and communication in science and engineering.
- Reproducibility: The base units of SI are defined in terms of fundamental physical constants, which are universally reproducible.
22. Find the value of 60 J per min. on a system that has 100 g, 100 cm and 1 min as the base units.
Answer:
-
Convert Joules to base units: 1 Joule = 1 kg m²/s² 1 kg = 1000 g 1 m = 100 cm
Therefore, 1 J = 1000 g * (100 cm)² / s² = 10⁷ g cm²/s²
-
Convert minutes to seconds: 1 min = 60 s
-
Substitute values: 60 J/min = 60 * 10⁷ g cm²/s² / 60 s = 10⁷ g cm²/s
Therefore, the value of 60 J per min. in the given system of units is 10⁷ g cm²/s.
23. State and prove work-energy theorem.
Statement: The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
Proof:
Consider an object of mass m moving with an initial velocity v₁. A force F is applied on the object, causing it to accelerate. The acceleration of the object is given by:
a = F/m
The work done by the force F over a distance s is:
W = F * s
Using the equation of motion:
v² = v₁² + 2as
We can substitute for a:
v² = v₁² + 2(F/m)s
Rearranging:
(1/2)mv² – (1/2)mv₁² = F * s
Therefore, the work done on the object is equal to the change in its kinetic energy:
W = ΔK.E.
24. Show that the average KE of a gas molecule is directly proportional to the absolute temperature of the gas.
Answer:
The kinetic theory of gases states that the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. This relationship is given by:
<K.E.> = (3/2) kT
where:
- <K.E.> is the average kinetic energy of a gas molecule
- k is the Boltzmann constant
- T is the absolute temperature of the gas
This equation shows that as the temperature of the gas increases, the average kinetic energy of its molecules also increases.
25. Derive the relationship of the (a) coefficient of superficial expansion (β) (b) coefficient of cubical expansion (y) with the coefficient of linear expansion (α)
Answer:
a) Coefficient of superficial expansion (β):
Let’s consider a square sheet with initial side length L. When the temperature is increased by ΔT, the side length increases by ΔL. The new side length is L + ΔL.
The initial area of the sheet is A = L².
The final area of the sheet is A’ = (L + ΔL)² = L² + 2LΔL + ΔL²
Since ΔL is very small, we can neglect the term ΔL². Therefore, the change in area is:
ΔA = A’ – A = 2LΔL
The coefficient of superficial expansion (β) is defined as the fractional change in area per unit change in temperature:
β = (ΔA/A) / ΔT = (2LΔL / L²) / ΔT = 2(ΔL/L) / ΔT = 2α
Therefore, β = 2α
b) Coefficient of cubical expansion (γ):
Let’s consider a cube with initial side length L. When the temperature is increased by ΔT, the side length increases by ΔL. The new side length is L + ΔL.
The initial volume of the cube is V = L³.
The final volume of the cube is V’ = (L + ΔL)³ = L³ + 3L²ΔL + 3L(ΔL)² + ΔL³
Since ΔL is very small, we can neglect the terms (ΔL)² and ΔL³. Therefore, the change in volume is:
ΔV = V’ – V = 3L²ΔL
The coefficient of cubical expansion (γ) is defined as the fractional change in volume per unit change in temperature:
γ = (ΔV/V) / ΔT = (3L²ΔL / L³) / ΔT = 3(ΔL/L) / ΔT = 3α
Therefore, γ = 3α
OR
Define the coefficient of thermal conductivity and give its S.I. unit and dimensions.
Answer:
The coefficient of thermal conductivity (k) of a material is a measure of its ability to conduct heat. It is defined as the amount of heat that flows per unit time through a unit area of the material when there is a unit temperature gradient across the material.
Mathematically, it can be expressed as:
k = (Q/t) / (A * ΔT/Δx)
where:
- Q is the amount of heat transferred
- t is the time taken
- A is the area of the material
- ΔT is the temperature difference across the material
- Δx is
SECTION – C
26. Define Elastic collision and derive the expressions for the velocities of two bodies after collision in one dimension.
Answer:
Definition of Elastic Collision:
An elastic collision is a type of collision in which both kinetic energy and momentum are conserved. In other words, the total kinetic energy of the system before the collision is equal to the total kinetic energy
Derivation of Velocities in One-Dimensional Elastic Collision:
Consider two bodies with masses m₁ and m₂ moving with initial velocities u₁ and u₂ in one dimension. After the collision, their velocities become v₁ and v₂.
Conservation of Momentum:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Conservation of Kinetic Energy:
(1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)m₁v₁² + (1/2)m₂v₂²
Solving these two equations simultaneously, we get the expressions for the final velocities:
v₁ = [(m₁ – m₂)u₁ + 2m₂u₂] / (m₁ + m₂) v₂ = [(m₂ – m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
OR
A railway carriage of mass 9000 kg moving with a speed of 36 km/hr collides with a stationary carriage of the same mass. After the collision, the carriages
Railway Carriage Collision:
-
Given:
- Mass of each carriage (m₁ = m₂ = 9000 kg)
- Initial velocity of the first carriage (u₁ = 36 km/hr = 10 m/s)
- Initial velocity of the second carriage (u₂ = 0)
-
Type of Collision: Since the carriages couple together after the collision, they move with a common velocity. This indicates that the collision is inelastic. In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved.
-
Calculation of Common Speed:
Using the conservation of momentum equation:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
where v is the common velocity after the collision.
Substituting the values:
(9000 kg)(10 m/s) + (9000 kg)(0) = (9000 kg + 9000 kg)v
90000 kg m/s = (18000 kg)v
v = 5 m/s
Therefore, the common speed of the coupled carriages after the collision is 5 m/s.
27. A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball
Answer:
Let’s use the equation of motion for free fall:
h = (1/2)gt²
where h is the height of the tower, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time taken to reach the ground.
Given that the ball takes T seconds to reach the ground, we have:
h = (1/2)gT²
Now, let’s find the position of the ball at t = T/3:
h’ = (1/2)g(T/3)² = (1/2)g(T²/9) = (1/18)gT²
Substituting the value of h from the first equation:
h’ = (1/18)(2h) = h/9
Therefore, the position of the ball in T/3 seconds is h/9 from the top of the tower.
28. Derive an expression for the excess pressure inside a liquid drop.
Answer:
Consider a spherical liquid drop of radius R. Let the surface tension of the liquid be S. The excess pressure inside the drop arises due to the surface tension acting on the liquid surface.
Let’s consider a small element of the liquid surface. The surface tension acts along the tangent to the surface at this point. Due to the curvature of the surface, these tangential forces have a net inward component. This inward force results in an excess pressure inside the drop.
Let’s consider a small element of the surface with area dA. The force due to surface tension acting on this element is S * dA. The net inward force due to this element is:
dF = S * dA * cos(θ)
where θ is the angle between the tangent to the surface at the point and the normal to the surface.
For a sphere, cos(θ) = 1/R. Therefore,
dF = S * dA / R
The total inward force acting on the entire surface of the drop is:
F = ∫dF = ∫(S * dA / R) = (S/R) ∫dA = (S/R) * 4πR² = 4πRS
This inward force is balanced by the excess pressure inside the drop acting on the surface area of the drop.
Excess pressure * Surface area = 4πRS
Excess pressure = 4πRS / 4πR² = 2S/R
Therefore, the excess pressure inside a liquid drop is given by:
Excess pressure = 2S/R
29. Show that, for small oscillations, the motion of a simple pendulum is simple harmonic. Derive an expression for its time period. Does it depend on the mass of the bob?
Answer:
Simple Harmonic Motion (SHM):
In SHM, the restoring force is directly proportional to the displacement from the equilibrium position and acts in the
Simple Pendulum:
A simple pendulum consists of a point mass (bob) suspended from a massless, inextensible string. When the bob is displaced from its equilibrium position and released, it oscillates back and forth.
Derivation of Time Period:
Consider a simple pendulum of length L with a bob of mass m. When the bob is displaced from its equilibrium position by an angle θ, the restoring force acting on it is:
F = -mg sinθ
For small oscillations, sinθ ≈ θ. Therefore,
F = -mgθ
But, θ = x/L, where x is the displacement from the equilibrium position.
So, F = -mg(x/L) = -(mg/L)x
This equation shows that the restoring force is directly proportional to the displacement and acts in the opposite direction. Hence, for small oscillations, the motion of a simple pendulum is simple harmonic.
The time period of a simple pendulum is given by:
T = 2π√(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity.
Does it depend on the mass of the bob?
No, the time period of a simple pendulum does not depend on the mass of the bob. This is because the mass of the bob cancels out in the derivation of the time period.
30. The distances of two planets from the sun are 10¹³ m and 10¹² m respectively. Find the ratio of time periods and speeds of the two planets.
Answer:
Kepler’s Third Law of Planetary Motion:
Kepler’s Third Law states that the square of the orbital period of a planet is directly proportional to the cube
Let T₁ and T₂ be the time periods of the two planets, and r₁ and r₂ be their respective distances from the Sun. Then, according to Kepler’s Third Law:
(T₁/T₂)² = (r₁/r₂)^3
Given:
r₁ = 10¹³ m
r₂ = 10¹² m
Therefore,
(T₁/T₂)² = (10¹³/10¹²)^3 = 10³
T₁/T₂ = √10³ = 10^(3/2) = 10√10
Ratio of Time Periods: T₁:T₂ = 10√10:1
Ratio of Speeds:
For circular orbits, the orbital speed (v) is given by:
v = 2πr/T
Therefore, the ratio of the speeds of the two planets is:
v₁/v₂ = (2πr₁/T₁) / (2πr₂/T₂) = (r₁/r₂)(T₂/T₁) = (10¹³/10¹²)(1/10√10) = 1/√10
Ratio of Speeds: v₁:v₂ = 1:√10
SECTION – D
31. (A) Discuss analytically the formation of a stationary wave in a string, clamped at its ends. Obtain the expression of the fundamental frequency. Show that in case of a string, the frequencies of the first four harmonic are in the ratio 1:2:3:4
Answer:
A stationary wave, also known as a standing wave, is formed when two identical waves traveling in opposite directions interfere with each other. In the case of a string clamped at both ends, a stationary wave is formed when a wave traveling along the string is reflected at the fixed ends.
Formation:
-
Incident Wave: A wave is generated on the string, traveling along its length.
-
Reflection: When the wave reaches the fixed end, it is reflected back in the opposite direction.
-
Superposition: The incident wave and the reflected wave interfere with each other. At some points, the waves interfere constructively, resulting in points of maximum displacement called antinodes. At other points, the waves interfere destructively, resulting in points of zero displacement called nodes.
-
Stationary Wave Formation: The resulting pattern is a stationary wave with fixed nodes and antinodes.
Fundamental Frequency:
The fundamental frequency is the lowest frequency at which a stationary wave can be formed on the string. In this case, the string vibrates in one loop with a node at each end and an antinode at the center. The wavelength of the fundamental mode is twice the length of the string (λ = 2L).
The velocity of the wave on the string is given by:
v = √(T/μ)
where T is the tension in the string and μ is the mass per unit length of the string.
The frequency of the wave is given by:
f = v/λ
Substituting the values:
f = (1/2L) * √(T/μ)
This is the expression for the fundamental frequency.
Harmonics:
Harmonics are integer multiples of the fundamental frequency. The frequencies of the first four harmonics are:
1st harmonic: f₁ = f = (1/2L) * √(T/μ)
2nd harmonic: f₂ = 2f₁ = 2 * (1/2L) * √(T/μ) = (1/L) * √(T/μ)
3rd harmonic: f₃ = 3f₁ = 3 * (1/2L) * √(T/μ) = (3/2L) * √(T/μ)
4th harmonic: f₄ = 4f₁ = 4 * (1/2L) * √(T/μ) = (2/L) * √(T/μ)
Therefore, the frequencies of the first four harmonics are in the ratio 1:2:3:4.
31. (B) The length of a string tied to two rigid supports is 40 cm. What is the wavelength (in cm) of the stationary wave produced on it if the string is plucked at its centre?
Answer:
When the string is plucked at its center, the fundamental mode of vibration is produced. In the fundamental mode, the wavelength is twice the length of the string.
Therefore, the wavelength of the stationary wave is:
λ = 2 * 40 cm = 80 cm
OR
31. (A) What are beats? Explain the formation of beats analytically. Prove that the beat frequency is equal to the difference in frequencies of the two superposing waves.
Answer:
Beats are periodic variations in the intensity of sound produced when two sound waves of slightly different frequencies interfere with each other.
Formation of Beats:
When two sound waves of frequencies f₁ and f₂ (where f₁ > f₂) interfere, the resultant wave has a frequency equal to the average of the two frequencies:
f_avg = (f₁ + f₂)/2
The amplitude of the resultant wave varies periodically with a frequency equal to the difference in the frequencies of the two waves:
f_beat = f₁ – f₂
This periodic variation in amplitude gives rise to the phenomenon of beats.
Analytical Proof:
Let the two waves be represented by:
y₁ = A sin(2πf₁t)
y₂ = A sin(2πf₂t)
The resultant wave is given by:
y = y₁ + y₂ = A [sin(2πf₁t) + sin(2πf₂t)]
Using the trigonometric identity:
sin A + sin B = 2 sin[(A + B)/2] cos[(A – B)/2]
We get:
y = 2A sin[2π(f₁ + f₂)/2]t cos[2π(f₁ – f₂)/2]t
y = 2A sin(2πf_avg)t cos(2πf_beat)t
This equation shows that the resultant wave has a frequency equal to the average of the two frequencies and an amplitude that varies periodically with a frequency equal to the difference in the frequencies of the two waves.
31. (B) A tuning fork arrangement (pair) produces 4 beats s⁻¹ with one fork of frequency 288 Hz. A little wax is placed on the unknown fork and it then produces 2 beats s⁻¹. What is the frequency of the unknown fork?
Answer:
When wax is added to the unknown fork, its frequency decreases.
Let the initial frequency of the unknown fork be f.
Initially, the beat frequency is 4 Hz. Therefore:
|f – 288| = 4
This gives two possible values for f:
f₁ = 288 + 4 = 292 Hz
f₂ = 288 – 4 = 284 Hz
After adding wax, the beat frequency reduces to 2 Hz. This means that the frequency of the unknown fork has decreased.
Therefore, the initial frequency of the unknown fork is 292 Hz.
32. State Bernoulli’s theorem. With the help of a suitable diagram, establish Bernoulli’s equation for liquid flow. Write the formula and SI unit of velocity head.
Answer:
Bernoulli’s Theorem:
Bernoulli’s theorem states that for an incompressible, non-viscous fluid in steady flow, the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
Derivation:
Consider a horizontal pipe with a varying cross-sectional area. Let P₁, v₁, and h₁ be the pressure, velocity, and height of the fluid at point 1, and P₂, v₂, and h₂ be the corresponding values at point 2.
Work done on the fluid by pressure forces at point 1: W₁ = P₁A₁Δx₁
Work done by the fluid against pressure forces at point 2: W₂ = P₂A₂Δx₂
Net work done on the fluid: W = W₁ – W₂ = P₁A₁Δx₁ – P₂A₂Δx₂
Since the fluid is incompressible, the volume of fluid passing through any cross-section per unit time is constant:
A₁v₁ = A₂v₂
Therefore,
Δx₁/Δx₂ = v₂/v₁
Substituting this in the expression for net work done:
W = P₁A₁Δx₁ – P₂A₁Δx₁(v₁/v₂) = A₁Δx₁(P₁ – P₂v₁/v₂)
The change in kinetic energy of the fluid is:
ΔK.E. = (1/2)m(v₂² – v₁²) = (1/2)(ρA₁Δx₁) (v₂² – v₁²)
where ρ is the density of the fluid.
According to the work-energy theorem, the net work done on the fluid is equal to the change in its kinetic energy:
A₁Δx₁(P₁ – P₂v₁/v₂) = (1/2)(ρA₁Δx₁) (v₂² – v₁²)
Simplifying:
P₁ – P₂ = (1/2)ρ(v₂² – v₁²)
Rearranging:
P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂²
This equation is known as Bernoulli’s equation.
Velocity Head:
The term (1/2)ρv² is called the velocity head. It represents the kinetic energy per unit volume of the fluid.
The formula for velocity head is:
Velocity head = (1/2)ρv²
The SI unit of velocity head is J/m³.
OR
32. (A) What is capillarity? Derive an expression for the height to which the liquid having an angle of contact θ rises in a capillary tube of radius r.
Answer:
Capillarity is the phenomenon of rise or fall of a liquid in a narrow tube (capillary tube) due to the surface tension forces.
Derivation of Height of Liquid Rise:
Consider a capillary tube of radius r dipped in a liquid. Let the angle of contact between the liquid and the tube wall be θ. The liquid rises in the tube to a height h due to surface tension.