TS Intermediate Physics 1st Year Model Paper 2024

SECTION – A

Note: (i) Answer ALL Questions (ii) Each Question carries TWO marks (iii) All are very short answer type questions.

1. What is Physics?
Physics is the branch of science that deals with the study of matter, energy, and the interactions between them. It aims to understand the fundamental laws of nature that govern the behavior of the universe, from subatomic particles to large cosmic structures.

2. How can systematic errors be minimized or eliminated?
Systematic errors can be minimized or eliminated by:

Calibrating instruments before taking measurements.
Using more precise and accurate measuring devices.
Repeating measurements and taking averages.
Identifying and correcting any potential sources of error in the experiment setup or procedure.
Ensuring consistency in experimental conditions and methodology.

3. Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is the magnitude of their resultant?
The magnitude of the resultant force can be calculated using the law of cosines. The formula is:

$R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta}$

Where $F_1 = 3$

units, $F_2 = 5$

units, and $\theta = 60^\circ$

$R = \sqrt{3^2 + 5^2 + 2 \times 3 \times 5 \times \cos(60^\circ)}$

$R = \sqrt{9 + 25 + 30 \times \frac{1}{2}}$

$R = \sqrt{9 + 25 + 15}$

$R = \sqrt{49}$

$R = 7 \, \text{units}$

So, the magnitude of the resultant force is 7 units.

4. A horse has to pull harder during the start of the motion than later. Explain.
When a horse starts to move, it must overcome the inertia of the object (or cart) it is pulling. Inertia is the resistance of an object to changes in its state of motion, and it is greatest when the object is at rest. Once the object starts moving, the frictional forces are lower and the horse needs less force to maintain motion. Therefore, the horse has to pull harder at the beginning to overcome inertia and friction.

5. What is the principle behind the carburetor of an automobile?
The principle behind the carburetor is based on Bernoulli’s principle. The carburetor mixes air and fuel in the right proportions for combustion. As air flows through the carburetor, it passes through a narrow venturi, causing its velocity to increase and pressure to decrease. This drop in pressure draws fuel from the fuel reservoir, mixing with the air and creating a fuel-air mixture that is optimal for combustion in the engine.

6. Why are drops and bubbles spherical?
Drops and bubbles are spherical due to surface tension. Surface tension is the force that acts on the surface of a liquid, pulling the molecules together to minimize the surface area. A sphere has the smallest surface area for a given volume, so the liquid drop or bubble adopts a spherical shape to minimize the energy associated with surface tension.

7. Distinguish between heat and temperature.

Heat is a form of energy that flows from a hot object to a cooler one due to a temperature difference. It is measured in joules (J) or calories (cal).
Temperature is a measure of the average kinetic energy of the particles in a substance. It indicates how hot or cold an object is and is measured in degrees Celsius (°C), Fahrenheit (°F), or Kelvin (K).

8. Why are utensils coated black? Why the bottom of the utensils are made of copper?

Black coating: The black coating on utensils, typically made of black paint or blackened metal, increases the absorption of heat. Black surfaces absorb heat more efficiently than shiny or light-colored surfaces due to their high emissivity. This helps in faster heating of the utensils.
Copper bottom: Copper is a good conductor of heat, meaning it allows heat to flow quickly and evenly across the bottom of the utensil. This ensures uniform cooking and prevents hot spots. Copper is often used in cookware to improve heat distribution.

9. When does a real gas behave like an ideal gas?
A real gas behaves like an ideal gas under the following conditions:

High temperature: At high temperatures, the kinetic energy of gas molecules is high, which minimizes intermolecular forces, making the gas behave more ideally.
Low pressure: At low pressure, gas molecules are far apart, and intermolecular forces become negligible. This also helps the gas behave more like an ideal gas.
Low density: When the gas density is low, the interactions between gas molecules are weak, and the gas behaves more ideally. In these conditions, the behavior of a real gas closely matches the ideal gas law $PV = nRT$

10. State Dalton’s law of partial pressures.
Dalton’s Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures exerted by each gas in the mixture, when each gas occupies the same volume and temperature. Mathematically, it is expressed as:

$P_{\text{total}} = P_1 + P_2 + P_3 + \cdots + P_n$

$P_{total} = P_{1} + P_{2} + P_{3} + \dots + P_{n}$

Where:

$P_{\text{total}}$
$P_1, P_2, P_3, \dots, P_n$

Each partial pressure $P_i$

$P_{i}$ is the pressure that the gas would exert if it occupied the entire volume by itself at the same temperature.

SECTION – B

Note: (i) Answer ANY SIX questions. (ii) Each question carries FOUR marks. (iii) All are of short answer type questions.

11. State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.
Parallelogram Law of Vectors:
The parallelogram law of vectors states that if two vectors are represented by two adjacent sides of a parallelogram, then the diagonal of the parallelogram represents the resultant vector. The magnitude and direction of the resultant can be determined using this law.

Derivation for the magnitude and direction of the resultant vector:

Let two vectors A and B make an angle $\theta$

$θ$ with each other. The magnitude of the resultant vector $R$

$R$ can be obtained using the law of cosines:

$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$

$R = A^{2} + B^{2} + 2 A B cos θ$

Where:

$A$
$\theta$

The direction of the resultant vector $\alpha$

$α$ (the angle it makes with vector A) can be calculated using the law of sines:

$\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$

$tan α = A + B cos θ B sin θ$

Thus, the magnitude and direction of the resultant vector are determined.

12. Mention the methods used to decrease friction.
Some methods to decrease friction include:

Lubrication: Applying oils, greases, or other lubricants to reduce friction between surfaces.
Polishing or smoothing surfaces: Reducing the roughness of surfaces can lower friction.
Using ball bearings: Ball bearings reduce friction by replacing sliding motion with rolling motion.
Streamlining shapes: In fluids (like air or water), objects with streamlined shapes reduce friction and drag.
Using rollers or wheels: These reduce sliding friction by changing it into rolling friction, which is less.

13. Distinguish between center of mass and center of gravity.

Center of Mass: The center of mass of an object is the point where its mass can be considered to be concentrated for the purpose of analysis. It is a geometric property of the object and is independent of gravity.
Center of Gravity: The center of gravity of an object is the point where the total weight of the object acts. It depends on the gravitational field and is the point where the gravitational force can be assumed to act.

The center of gravity and center of mass coincide in a uniform gravitational field, but they may not coincide in non-uniform gravitational fields.

14. Define angular velocity (ω). Derive $v = r\omega$

$v = r ω$ for it.
Angular velocity (ω):
Angular velocity is defined as the rate at which an object rotates or revolves around an axis. It is the angle through which an object rotates per unit time. It is given by:

$\omega = \frac{\Delta \theta}{\Delta t}$

$ω = Δ t Δ θ$

Where:

$\omega$
$\Delta \theta$
$\Delta t$

Derivation of $v = r\omega$

$v = r ω$ :
Let an object move in a circular path with a radius $r$

$r$ . The tangential velocity $v$

$v$ is the distance traveled per unit time along the circumference. The distance traveled in one complete rotation is the circumference, $2\pi r$

$2 π r$ , and if the object completes one revolution in time $T$

$T$ , the angular velocity $\omega$

$ω$ is given by:

$\omega = \frac{2\pi}{T}$

$ω = T 2 π$

The tangential velocity $v$

$v$ is given by:

$v = \frac{\text{Distance traveled}}{\text{Time taken}} = \frac{2\pi r}{T}$

$v = Time taken Distance traveled = T 2 π r$

Now, using the relationship between angular velocity and time:

$v = r\omega$

$v = r ω$

This gives the relation between the tangential velocity and angular velocity.

15. What is orbital velocity? Obtain an expression for it.
Orbital velocity is the velocity required for an object to maintain a stable orbit around a celestial body (like a planet or a star). For an object to remain in orbit, the centripetal force must balance the gravitational force.

The gravitational force on an object of mass $m$

$m$ at a distance $r$

$r$ from the center of the celestial body is:

$F_{\text{gravity}} = \frac{GMm}{r^2}$

$F_{gravity} = r ^{2} GM m$

Where $G$

$G$ is the gravitational constant and $M$

$M$ is the mass of the celestial body.

The centripetal force required for circular motion is:

$F_{\text{centripetal}} = \frac{mv^2}{r}$

$F_{centripetal} = r m v ^{2}$

Equating the gravitational force and centripetal force for a stable orbit:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

$r ^{2} GM m = r m v ^{2}$

Simplifying and solving for the orbital velocity $v$

$v$ :

$v = \sqrt{\frac{GM}{r}}$

$v = r GM$

Thus, the orbital velocity is $v = \sqrt{\frac{GM}{r}}$

$v = r GM$ .

16. In what way is the anomalous behavior of water advantageous to aquatic animals?
Water exhibits anomalous behavior, such as its maximum density occurring at 4°C. When water cools below 4°C, it expands and becomes less dense, causing ice to float. This has several advantages for aquatic animals:

Insulation: Ice floating on the surface of water insulates the liquid water below, preventing it from freezing completely in cold environments.
Stable environment: Aquatic animals can survive in water below the ice layer, which remains liquid and at a relatively stable temperature, even when the surface water is freezing.

17. A man runs across the roof of a tall building and jumps horizontally onto the (lower) roof of an adjacent building. If his speed is 9 m/s and the horizontal distance between the buildings is 10 m and the height difference between the roofs is 9 m, will he be able to land on the next building? (Take $g = 10 \, \text{m/s}^2$

$g = 10 m/s^{2}$ )
We need to check if the man will be able to land on the next building. This is a projectile motion problem.

Horizontal velocity ( $u_x$
Horizontal distance ( $d$
Vertical displacement ( $h$
Gravitational acceleration ( $g$

Time to fall from the first building: Using the equation for vertical motion:

$h = \frac{1}{2} g t^2$

$h = 21 g t^{2}$ $9 = \frac{1}{2} \times 10 \times t^2$

$9 = 21 \times 10 \times t^{2}$ $t^2 = \frac{9}{5}$

$t^{2} = 59$ $t = \sqrt{\frac{9}{5}} \approx 1.34 \, \text{seconds}$

$t = 59 \approx 1.34 seconds$

Horizontal distance traveled: The horizontal distance traveled is given by:

$d = u_x \times t$

$d = u_{x} \times t$ $d = 9 \times 1.34 \approx 12.06 \, \text{m}$

$d = 9 \times 1.34 \approx 12.06 m$

Since the horizontal distance between the buildings is 10 m, and the man travels 12.06 m, he will overshoot and will not land on the adjacent roof.

18. Describe the behavior of a wire under gradually increasing load.
When a wire is subjected to a gradually increasing load:

Elastic region: Initially, the wire obeys Hooke’s Law, and the elongation is proportional to the load. The wire returns to its original length once the load is removed.
Plastic region: Beyond a certain load, the wire enters the plastic region. In this region, the wire experiences permanent deformation, and it does not return to its original length even when the load is removed.

Breaking point: If the load continues to increase, the wire will eventually break at a certain point known as the breaking or rupture point.

SECTION – C

Note: (i) Answer ANY TWO questions. (ii) Each question carries EIGHT marks. (iii) All are long answer type questions.

19. State the second law of thermodynamics. How is a heat engine different from a refrigerator?

Second Law of Thermodynamics:
The second law of thermodynamics states that the total entropy of an isolated system always increases over time. In simpler terms, energy spontaneously tends to disperse and spread out, and natural processes tend to increase disorder (entropy) in the system.

Mathematically:

$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$

$Δ S_{universe} = Δ S_{system} + Δ S_{surroundings} > 0$

Where $\Delta S$

$Δ S$ is the change in entropy.

Difference between a Heat Engine and a Refrigerator:

Heat Engine: A heat engine is a device that takes in energy by heat, converts part of it into work, and rejects the remaining energy to the surroundings. It operates in a cyclic process, absorbing heat from a hot reservoir and releasing heat to a cold reservoir.
Refrigerator: A refrigerator is a device that requires work to transfer heat from a colder body to a hotter body. It absorbs heat from the cold reservoir and expels heat to the warm reservoir, working against the natural flow of heat.

In short, a heat engine converts heat into work, while a refrigerator uses work to move heat.

20. Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is a seconds pendulum?

Motion of a Simple Pendulum as Simple Harmonic:
Consider a simple pendulum consisting of a mass $m$

$m$ suspended from a string of length $L$

$L$ and displaced by an angle $\theta$

$θ$ from the vertical. The restoring force is provided by the component of the gravitational force that acts along the arc of the pendulum’s motion.

The equation for the restoring force is:

$F = -mg \sin \theta$

$F = - m g sin θ$

For small angles (small angle approximation, $\sin \theta \approx \theta$

$sin θ \approx θ$ ), the force becomes:

$F = -mg\theta$

$F = - m g θ$

This force is proportional to the displacement $\theta$

$θ$ , and thus, the motion is simple harmonic.

Equation of Motion: The equation of motion is given by:

$I \alpha = -mg\theta$

$I α = - m g θ$

Where $I = mL^2$

$I = m L^{2}$ is the moment of inertia and $\alpha = \frac{d^2 \theta}{dt^2}$

$α = \frac{d ^{2} θ}{d t ^{2}}$ is the angular acceleration. Substituting, we get:

$mL^2 \frac{d^2 \theta}{dt^2} = -mg\theta$

$m L^{2} \frac{d ^{2} θ}{d t ^{2}} = - m g θ$ $\frac{d^2 \theta}{dt^2} = -\frac{g}{L} \theta$

$\frac{d ^{2} θ}{d t ^{2}} = - \frac{g}{L} θ$

This is the standard form of the equation for simple harmonic motion, where the angular frequency $\omega$

$ω$ is:

$\omega^2 = \frac{g}{L}$

$ω^{2} = \frac{g}{L}$

Thus, the angular frequency $\omega$

$ω$ is:

$\omega = \sqrt{\frac{g}{L}}$

$ω = \frac{g}{L}$

Time Period:
The time period $T$

$T$ of the pendulum is related to angular frequency by:

$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{g}}$

$T = \frac{2 π}{ω} = 2 π \frac{L}{g}$

Seconds Pendulum:
A seconds pendulum is a pendulum whose time period is exactly 2 seconds (1 second for a complete swing in each direction). For such a pendulum:

$T = 2 \, \text{seconds}$

$T = 2 seconds$

So, a seconds pendulum has a time period of 2 seconds.

21. State and prove the law of conservation of energy in the case of a freely falling body.

Law of Conservation of Energy:
The law of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another. The total energy in an isolated system remains constant.

In the case of a freely falling body:

Initially, the body has potential energy $PE = mgh$
As the body falls, the potential energy is converted into kinetic energy.

At any point during the fall, the total energy is the sum of the potential and kinetic energies:

$\text{Total Energy} = PE + KE = mgh + \frac{1}{2}mv^2$

$Total Energy = PE + K E = m g h + \frac{1}{2} m v^{2}$

As the body falls, the height $h$

$h$ decreases, and the potential energy is converted into kinetic energy $KE$

$K E$ . When the body reaches the ground, all of the potential energy is converted into kinetic energy, and the total energy remains constant.

At the point of reaching the ground:

$mgh = \frac{1}{2}mv^2$

$m g h = \frac{1}{2} m v^{2}$

Where $v$

$v$ is the final velocity of the body.

Thus, the law of conservation of energy holds because the total mechanical energy (sum of potential and kinetic energy) remains constant throughout the fall.

A machine gun fires 360 bullets per minute, and each bullet travels with a velocity of 600 m/s. If the mass of each bullet is 5 gm, find the power of the machine gun.

Given:

Number of bullets fired per minute = 360 bullets/minute = 6 bullets/second
Velocity of each bullet = 600 m/s
Mass of each bullet = 5 gm = 0.005 kg

The kinetic energy of one bullet is:

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.005 \times 600^2 = 0.0025 \times 360000 = 900 \, \text{J}$

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 0.005 \times 60 0^{2} = 0.0025 \times 360000 = 900 J$

Since 6 bullets are fired per second, the power (rate of energy transfer) of the machine gun is:

$P = \text{Energy per second} = 6 \times 900 = 5400 \, \text{J/s} = 5400 \, \text{W}$

$P = Energy per second = 6 \times 900 = 5400 J/s = 5400 W$

So, the power of the machine gun is 5400 W or 5.4 kW.