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CBSE Class 12 Physics Case Study

Class 12th Physics Case Study

Case Study 1: Electric Charges and Fields

A large flat conducting sheet is placed horizontally on the ground and uniformly charged with surface charge density σ\sigmaσ. Above the sheet, a small spherical charged particle is suspended in equilibrium under the action of gravitational and electrostatic forces. The mass of the particle is mmm, and its charge is qqq. The electric field produced by the sheet exerts an upward force on the particle, balancing the downward gravitational force. The charged particle is observed to be stationary at a height hhh above the sheet.

This setup illustrates how electric fields can manipulate charged particles, and the role of surface charge density in producing uniform electric fields. It is assumed that air resistance is negligible and that the influence of any nearby charges is minimal.

Given the physical constants, g= \(\ {m/s}^{2} \) and ϵ0= \(\ 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^{-2}\) the relationship between the electric field and the gravitational force acting on the particle can be explored.

Questions:

  1. Calculate the magnitude of the electric field due to the infinite plane sheet of charge density \(\ \sigma \):
    • a) \(\frac{\sigma}{\epsilon_0}\)​
    • b) \(\frac{\sigma}{2\epsilon_0}\)​
    • c) \(\frac{2\sigma}{\epsilon_0}\)​
    • d) \(\frac{\epsilon_0}{\sigma}\)​​
  2. The force experienced by the charged particle due to the electric field is:
    • a) qE
    • b) \(\ E q^{2} \)
    • c) \(\frac{q}{E}\)​
    • d) \(\frac {E} {q} \)
  3. If the particle remains stationary, the relationship between gravitational force and electrostatic force is:
    • a) qE=mg
    • b) qE=2mg
    • c) mg=2qE
    • d) E=qmg
  4. What would happen to the force on the particle if the distance between the particle and the sheet were doubled?
    • a) The force would be halved
    • b) The force would remain the same
    • c) The force would double
    • d) The force would become zero

Answers:

  1. b) \(\frac{\sigma}{2\epsilon_0}\)​
  2. a) qE
  3. a) qE=mg
  4. b) The force would remain the same

Case Study 2: Current Electricity

In a laboratory, a simple electrical circuit is set up to study the heating effects of electric current. The circuit consists of a power supply of 12 V, a resistor of resistance 10 Ω , and an ammeter connected in series. A switch is provided to control the flow of current. When the switch is turned on, the ammeter reads a steady current value, and the resistor begins to heat up after a few seconds.

As time progresses, the temperature of the resistor rises steadily, indicating the conversion of electrical energy into heat energy. After a while, the experimenters measure the temperature change and note the power dissipated in the circuit. This experiment is an application of Joule's law of heating.

Questions:

  1. The current flowing through the circuit can be calculated using Ohm’s law. What is the value of the current?
    • a) 1.2 A
    • b) 0.83 A
    • c) 12 A
    • d) 120 A
  2. How much heat is produced by the resistor in 5 seconds? Use the formula H= \(\ I^{2} R \) (where H is the heat produced, I is the current, R is the resistance, and t is the time).
    • a) 60 J
    • b) 144 J
    • c) 120 J
    • d) 100 J
  3. If the resistance of the wire is doubled while keeping the voltage constant, what will happen to the current in the circuit?
    • a) The current will increase
    • b) The current will decrease
    • c) The current will remain constant
    • d) The current will become zero
  4. The power dissipated by the resistor is given by P=\(\ I^{2} R \) . What is the power dissipated in the circuit?
    • a) 14.4 W
    • b) 12 W
    • c) 1.44 W
    • d) 144 W

Answers:

  1. a) 1.2 A
  2. b) 144 J
  3. b) The current will decrease
  4. a) 14.4 W

Case Study 3: Electromagnetic Waves

A radio station broadcasts electromagnetic waves at a frequency of 100 MHz. The electromagnetic waves propagate through free space, carrying energy from the transmitter to radio receivers. The electric and magnetic fields in these waves oscillate perpendicular to each other and to the direction of wave propagation.

The speed of electromagnetic waves in free space is \(\ 3 \times 10^{8} m/s \) . The study of electromagnetic waves is essential in understanding the behavior of light, radio waves, and other forms of radiation. The wavelength, energy, and speed of these waves are all interrelated.

Questions:

  1. The wavelength of the electromagnetic wave is given by \(\ \lambda = \frac{c}{f} \)​, where ccc is the speed of light and fff is the frequency. Calculate the wavelength of the radio signal.
    • a) 3 meters
    • b) 0.3 meters
    • c) 30 meters
    • d) 300 meters
  2. The speed of electromagnetic waves in free space is:
    • a) \(\ 3 \times 10^{8} m/s \)
    • b) \(\ 1 \times 10^{8} m/s \)
    • c) \(\ 3 \times 10^{6} m/s \)
    • d) \(\ 1.5 \times 10^{8} m/s \)
  3. In an electromagnetic wave, the electric field and magnetic field are oriented:
    • a) Parallel to each other
    • b) Perpendicular to each other
    • c) Aligned in the same direction
    • d) Independent of each other
  4. Which type of wave listed below is not an electromagnetic wave?
    • a) Radio waves
    • b) X-rays
    • c) Sound waves
    • d) Microwaves

Answers:

  1. a) 3 meters
  2. a) \(\ 3 \times 10^{8} m/s \)
  3. b) Perpendicular to each other
  4. c) Sound waves

Case Study 4: Ray Optics

In an optics lab, a convex lens is used to focus the image of a distant object, such as a building, onto a screen. The lens has a focal length of 10 cm. The object is placed at a distance much greater than the focal length of the lens. The image produced on the screen is sharp, inverted, and real.

This experiment demonstrates the use of a convex lens in imaging distant objects, which forms the basis of camera lenses and telescopic systems. By adjusting the distance between the lens and the screen, the sharpness of the image can be controlled.

Questions:

  1. If the object is located very far from the lens, where will the image be formed?
    • a) At the focal point of the lens
    • b) At twice the focal length
    • c) At infinity
    • d) Between the optical center and the focus
  2. What is the nature of the image formed in this case?
    • a) Virtual and upright
    • b) Real and inverted
    • c) Virtual and inverted
    • d) Real and upright
  3. If the focal length of the lens is increased, how will it affect the position of the image?
    • a) The image will form closer to the lens
    • b) The image will form farther from the lens
    • c) The image position will not change
    • d) No image will be formed
  4. For distant objects, what can be said about the magnification of the image produced by the lens?
    • a) It is greater than 1
    • b) It is less than 1
    • c) It is equal to 1
    • d) It depends on the object size

Answers:

  1. a) At the focal point of the lens
  2. b) Real and inverted
  3. b) The image will form farther from the lens
  4. b) It is less than 1

Case Study 5: Electromagnetic Induction

In an experiment to study electromagnetic induction, a long solenoid with 1000 turns is connected to a galvanometer. A bar magnet is moved rapidly towards the solenoid, inducing an electromotive force (emf) in the coil. The direction and magnitude of the induced current are observed to depend on the speed of the magnet's movement and its polarity. The experiment demonstrates Faraday’s law of electromagnetic induction, which states that the emf induced in a circuit is directly proportional to the rate of change of magnetic flux through the circuit.

Questions:

  1. What happens to the induced emf if the speed of the magnet is doubled?
    • a) The induced emf remains the same
    • b) The induced emf is doubled
    • c) The induced emf is halved
    • d) The induced emf becomes zero
  2. The direction of the induced current is determined by:
    • a) Faraday's law only
    • b) Lenz's law
    • c) Ampere's law
    • d) Ohm's law
  3. If the north pole of the magnet is moved towards the solenoid, the direction of induced current is such that:
    • a) It attracts the magnet
    • b) It repels the magnet
    • c) There is no induced current
    • d) The current flows in a random direction
  4. What would happen if the bar magnet were stationary inside the solenoid?
    • a) A constant emf would be induced
    • b) The induced emf would increase
    • c) No emf would be induced
    • d) The current would alternate between positive and negative values

Answers:

  1. b) The induced emf is doubled
  2. b) Lenz's law
  3. b) It repels the magnet
  4. c) No emf would be induced

Case Study 6: Alternating Current

A household uses an alternating current (AC) supply with a voltage of 220 \(\ \text{V} \) at a frequency of 50 \(\ \text{Hz}\). Appliances such as refrigerators, fans, and lights are connected to the supply. The alternating current changes direction periodically, and the peak voltage is higher than the average voltage of 220 \(\ \text{V}\). The RMS (Root Mean Square) voltage is used to represent the effective voltage of the AC supply.

AC is preferred over DC (Direct Current) for long-distance power transmission due to lower energy losses. Transformers are used to step up or step down the voltage for different purposes.

Questions:

  1. What is the peak voltage of an AC supply with an RMS voltage of 220 \(\ \text{V}\) ?
    • a) 220 \(\ \text{V}\)
    • b) 311 \(\ \text{V}\)
    • c) 440 \(\ \text{V}\)
    • d) 156 \(\ \text{V}\)
  2. Which of the following appliances works on AC?
    • a) Battery-powered torch
    • b) Electric fan
    • c) Car engine
    • d) Solar panel
  3. The frequency of the AC supply in India is:
    • a) 60 \(\ \text{Hz}\)
    • b) 100 \(\ \text{Hz}\)
    • c) 50 \(\ \text{Hz}\)
    • d) 25 \(\ \text{Hz}\)
  4. Why is AC preferred over DC for long-distance power transmission?
    • a) AC can be easily transformed to higher or lower voltages
    • b) AC has lower transmission losses
    • c) AC is safer than DC
    • d) AC is cheaper to produce than DC

Answers:

  1. b) 311 \(\ \text{V}\)
  2. b) Electric fan
  3. c) 50 \(\ \text{Hz}\)
  4. a) AC can be easily transformed to higher or lower voltages

Case Study 7: Dual Nature of Radiation and Matter

In an experiment demonstrating the photoelectric effect, a metal plate is illuminated with light of varying frequencies. It is observed that electrons are emitted from the metal surface when the frequency of light exceeds a certain threshold. The kinetic energy of the emitted electrons depends on the frequency of the incident light, but not on its intensity. This experiment verifies Einstein's photoelectric equation and confirms the particle nature of light, where photons with energy E=\(\ \text {hf}\) transfer their energy to the electrons.

Questions:

  1. What is the relation between the kinetic energy of photoelectrons and the frequency of incident light?
    • a) Kinetic energy increases linearly with frequency
    • b) Kinetic energy decreases with frequency
    • c) Kinetic energy is independent of frequency
    • d) Kinetic energy depends only on the intensity of light
  2. Which of the following particles is emitted during the photoelectric effect?
    • a) Electrons
    • b) Protons
    • c) Neutrons
    • d) Photons
  3. The work function of a metal is defined as:
    • a) The minimum energy required to emit protons from the metal
    • b) The maximum energy required to emit neutrons from the metal
    • c) The minimum energy required to emit electrons from the metal
    • d) The energy of the incident photons
  4. If the intensity of incident light is increased while keeping the frequency constant, what happens to the number of photoelectrons emitted?
    • a) The number of photoelectrons increases
    • b) The number of photoelectrons decreases
    • c) The number of photoelectrons remains the same
    • d) The emission of photoelectrons stops

Answers:

  1. a) Kinetic energy increases linearly with frequency
  2. a) Electrons
  3. c) The minimum energy required to emit electrons from the metal
  4. a) The number of photoelectrons increases

Case Study 8: Wave Optics

In a Young’s double-slit experiment, coherent light of wavelength 600 \(\ \text{nm}\) passes through two narrow slits separated by a distance of 0.5 \(\ \text{mm}\). An interference pattern is observed on a screen placed at a distance of 2 \(\ \text{m}\) from the slits. The bright and dark fringes alternate, demonstrating the wave nature of light and the principle of superposition.

The distance between adjacent bright fringes (fringe width) is measured to be a few millimeters. By analyzing this pattern, the wavelength of light and the distance between the slits can be determined.

Questions:

  1. The condition for constructive interference (bright fringes) in the interference pattern is:
    • a)\(\ \Delta x = n \lambda\)
    • b) \(\ \Delta x = (2n+1)\lambda/2\)
    • c) \(\ \Delta x = \lambda/2\)
    • d) \(\ \Delta x = 0\)
  2. The fringe width β\betaβ in a double-slit experiment is given by:
    • a) \(\ \frac{D}{d}\)​
    • b) \(\ \frac{\lambda D}{d}\)​
    • c) \(\ \frac{d}{\lambda D}\)​
    • d) \(\ \lambda d D\)
  3. What happens to the fringe width if the distance between the slits is increased?
    • a) The fringe width increases
    • b) The fringe width decreases
    • c) The fringe width remains the same
    • d) The fringes disappear
  4. If the wavelength of light used is changed to 450 \(\ \text{nm}\), how will the fringe width change?
    • a) The fringe width will increase
    • b) The fringe width will decrease
    • c) The fringe width will remain constant
    • d) No change in the pattern

Answers:

  1. a) \(\ \Delta x = n \lambda\)
  2. b) \(\ \frac{\lambda D}{d}\)​
  3. b) The fringe width decreases
  4. b) The fringe width will decrease

Case Study 9: Semiconductor Electronics

A p-n junction diode is used in a circuit to convert alternating current (AC) to direct current (DC). When the diode is forward biased, current flows easily through the diode. However, in reverse bias, the current is blocked, except for a small leakage current. This property of diodes is used in rectifiers to convert AC to DC.

In the experiment, the output voltage of the rectifier is measured, and it is observed that the diode only allows current to flow during the positive half-cycle of the AC input. This behavior is fundamental to understanding the working of electronic circuits in devices such as mobile chargers.

Questions:

  1. In a p-n junction diode, when the p-side is connected to the positive terminal of the battery, the diode is said to be:
    • a) Reverse biased
    • b) Forward biased
    • c) Unbiased
    • d) Short-circuited
  2. What is the primary function of a p-n junction diode in a rectifier circuit?
    • a) Amplify the input signal
    • b) Convert AC to DC
    • c) Generate an AC signal
    • d) Block the input signal
  3. In reverse bias, the current flowing through a p-n junction diode is:
    • a) Very high
    • b) Zero
    • c) Very small
    • d) Infinite
  4. Which component is essential in a rectifier circuit to smooth out the DC output?
    • a) Resistor
    • b) Capacitor
    • c) Transformer
    • d) Inductor

Answers:

  1. b) Forward biased
  2. b) Convert AC to DC
  3. c) Very small
  4. b) Capacitor

Case Study 10: Electric Charges and Fields

Two identical conducting spheres are placed in proximity to each other. Initially, both spheres are uncharged. When a positively charged rod is brought near one of the spheres without touching it, the spheres are temporarily connected by a conductor. After the conductor is removed, it is observed that both spheres acquire equal but opposite charges. This experiment demonstrates the phenomenon of electrostatic induction, where charge separation occurs in the presence of a nearby charged body.

Questions:

  1. What type of charges do the spheres acquire after induction?
    • a) Both spheres acquire positive charge
    • b) Both spheres acquire negative charge
    • c) One sphere acquires positive and the other negative charge
    • d) No charges are acquired
  2. What role does the connecting conductor play in this process?
    • a) It neutralizes the charge
    • b) It allows charge to be transferred between the spheres
    • c) It prevents charge transfer
    • d) It enhances the charge on both spheres
  3. If the charged rod is removed after the process, what happens to the charges on the spheres?
    • a) The charges disappear
    • b) The charges remain the same
    • c) The spheres become neutral
    • d) The spheres gain more charge
  4. Which principle explains the charge acquired by the spheres?
    • a) Coulomb’s law
    • b) Law of conservation of energy
    • c) Law of conservation of charge
    • d) Faraday’s law

Answers:

  1. c) One sphere acquires positive and the other negative charge
  2. b) It allows charge to be transferred between the spheres
  3. b) The charges remain the same
  4. c) Law of conservation of charge

Case Study 11: Moving Charges and Magnetism

A straight conductor carrying a steady current I is placed in a magnetic field. The conductor experiences a force due to the interaction between the magnetic field and the current. The magnitude of the force is given by F=ILBsin⁡ F=ILBsinθ, where L is the length of the conductor, B is the magnetic field strength, and θ is the angle between the conductor and the magnetic field. The direction of the force is given by the right-hand rule.

Questions:

  1. What happens to the force if the angle between the conductor and the magnetic field is \(\ 90^\circ \)?
    • a) The force becomes zero
    • b) The force is maximum
    • c) The force is minimum
    • d) The force remains unchanged
  2. If the length of the conductor is doubled, how does the force change?
    • a) The force remains the same
    • b) The force is halved
    • c) The force is doubled
    • d) The force becomes zero
  3. In which direction is the force acting on the conductor if the current is flowing upwards and the magnetic field is directed to the right?
    • a) To the left
    • b) To the right
    • c) Into the page
    • d) Out of the page
  4. Which law governs the relationship between current, magnetic field, and force in this case?
    • a) Coulomb’s law
    • b) Ampere’s law
    • c) Faraday’s law
    • d) Lorentz force law

Answers:

  1. b) The force is maximum
  2. c) The force is doubled
  3. d) Out of the page
  4. d) Lorentz force law

Case Study 12: Electromagnetic Waves

The electromagnetic spectrum encompasses a wide range of wavelengths and frequencies, from radio waves to gamma rays. Different regions of the spectrum are used for various applications. For instance, microwaves are used in radar and cooking, infrared is used in remote controls and thermal imaging, visible light allows us to see, ultraviolet light is used in sterilization, X-rays are used in medical imaging, and gamma rays are employed in cancer treatment.

Questions:

  1. Which of the following waves has the longest wavelength?
    • a) X-rays
    • b) Microwaves
    • c) Ultraviolet
    • d) Gamma rays
  2. Which part of the electromagnetic spectrum is responsible for heating food in a microwave oven?
    • a) Infrared
    • b) Microwaves
    • c) Radio waves
    • d) Ultraviolet
  3. Gamma rays are used in cancer treatment because:
    • a) They have the longest wavelength
    • b) They have the highest frequency and energy
    • c) They are harmless to living tissues
    • d) They are easily absorbed by the skin
  4. What type of electromagnetic wave is used in remote controls?
    • a) Ultraviolet
    • b) Infrared
    • c) Visible light
    • d) X-rays

Answers:

  1. b) Microwaves
  2. b) Microwaves
  3. b) They have the highest frequency and energy
  4. b) Infrared

Case Study 13: Ray Optics and Optical Instruments

A converging (convex) lens is used to focus light from a distant object onto a screen. The focal length of the lens is 20 \(\ \text{cm}\). When the object is placed at infinity, the image is formed at the focus of the lens. As the object is moved closer to the lens, the image moves farther away from the lens and becomes magnified. This principle is used in optical instruments such as microscopes and telescopes to magnify small or distant objects.

Questions:

  1. When the object is at infinity, where is the image formed by the convex lens?
    • a) At the focal point
    • b) At twice the focal length
    • c) At the object position
    • d) At the center of curvature
  2. If the object is placed between the focal point and the lens, the image formed is:
    • a) Real and inverted
    • b) Virtual and magnified
    • c) Real and diminished
    • d) Virtual and diminished
  3. What is the nature of the image when the object is placed at twice the focal length from a convex lens?
    • a) Virtual and magnified
    • b) Real and inverted
    • c) Real and of the same size as the object
    • d) Virtual and diminished
  4. The magnification produced by a convex lens is positive when:
    • a) The image is real and inverted
    • b) The image is virtual and magnified
    • c) The image is real and of the same size as the object
    • d) The image is at infinity

Answers:

  1. a) At the focal point
  2. b) Virtual and magnified
  3. c) Real and of the same size as the object
  4. b) The image is virtual and magnified

Case Study 14: Nuclei

In a nuclear reactor, uranium-235 undergoes fission when bombarded by neutrons, releasing a large amount of energy. The reaction also produces additional neutrons, which can initiate further fission reactions in a chain reaction. To control this process, the reactor uses control rods made of materials that absorb neutrons, such as boron or cadmium. The energy released in the form of heat is used to produce steam, which drives turbines to generate electricity.

Questions:

  1. In a nuclear fission reaction, the nucleus of uranium-235 splits into:
    • a) Two smaller nuclei and no neutrons
    • b) Two smaller nuclei and two neutrons
    • c) Two smaller nuclei and three neutrons
    • d) Two smaller nuclei and one neutron
  2. What is the function of control rods in a nuclear reactor?
    • a) To absorb neutrons and control the chain reaction
    • b) To increase the number of neutrons
    • c) To generate electricity directly
    • d) To decrease the temperature
  3. Which material is commonly used in control rods?
    • a) Uranium
    • b) Boron
    • c) Plutonium
    • d) Thorium
  4. The energy released in nuclear fission is primarily in the form of:
    • a) Electrical energy
    • b) Heat energy
    • c) Light energy
    • d) Chemical energy

Answers:

  1. c) Two smaller nuclei and three neutrons
  2. a) To absorb neutrons and control the chain reaction
  3. b) Boron
  4. b) Heat energy

Case Study 15: Communication Systems

In modern communication systems, signals are transmitted from one point to another using different methods such as satellite communication, optical fibers, and wireless communication. A typical communication system consists of a transmitter, a channel, and a receiver. In wireless communication, the transmitter sends information using electromagnetic waves, while the receiver captures these waves and converts them back into a usable form. In optical fiber communication, light waves are used to transmit data over long distances with minimal loss. Satellites are often used for long-range communication, especially for broadcasting television signals and internet services.

Questions:

  1. Which component of a communication system converts electrical signals into electromagnetic waves?
    • a) Transmitter
    • b) Receiver
    • c) Antenna
    • d) Modulator
  2. What is the primary advantage of using optical fibers in communication?
    • a) Higher bandwidth and less signal loss
    • b) Low cost of installation
    • c) Ability to transmit sound signals
    • d) Simple setup and usage
  3. In satellite communication, what role does the satellite play?
    • a) It transmits signals directly to the receiver
    • b) It reflects the signals to distant locations
    • c) It amplifies the signals and sends them back to earth
    • d) It absorbs and stores signals
  4. Which type of wave is primarily used in wireless communication?
    • a) Radio waves
    • b) Infrared waves
    • c) X-rays
    • d) Gamma rays

Answers:

  1. a) Transmitter
  2. a) Higher bandwidth and less signal loss
  3. c) It amplifies the signals and sends them back to earth
  4. a) Radio waves