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The CBSE Class 12th board exams are a crucial milestone in students’ academic journey, shaping their future career paths. The marks students score in this class help them become eligible for various competitive exams and colleges. Each subject of class 12 has their importance; in this blog, we will discuss the Most Challenging Class 12th Physics Problems and tips to solve them.
Physics is often considered one of the most difficult subjects due to its abstract concepts and complex problem-solving. Many students, whether they have opted for Maths or Biology streams, face difficulty in solving complex problems in Physics.
To excel in Physics, it’s essential to identify the most challenging topics. In this blog post, we’ll delve into the areas that students commonly find difficult, provide a breakdown of the types of questions asked, and offer effective tips to help you overcome these hurdles. Whether you’re struggling with electrostatics, magnetism, optics, or modern physics, this blog will equip you with the tips and strategies to conquer these topics and achieve your academic goals.
Electrostatics is an important chapter heavily weighted in the CBSE Class 12 board exams. The chapter covers electrostatics and current electricity, and students are asked a total of 16 mark questions. Sometimes, students find it difficult to understand the concepts of electric potential and potential energy, especially in complex configurations.
Numerical problems involving electrostatic potential, equipotential surface, potential due to electric dipole, capacitance, and energy stored in capacitors. Conceptual questions on electric field lines, dipole behaviour, and Gauss’s Law applications.
The parallel plate capacitor has plates with an area of 90 cm² each, separated by a distance of 2.5 mm. It is charged by connecting it to a 400 V power supply.
The area of the plates of a parallel plate capacitor is given as,
A = 90 cm² = 90*10¯⁴ m² = 0.009 m²
The distance between the plates d = 2.5 mm = 2.5 * 10¯³
The potential difference across plates- V = 400 V
Capacitance of the capacitor is given: C = Aɛ₀/d
Here, ɛ₀ = Permittivity of free space = 8.85 * 10¯¹² F/m
C = (8.85 * 10¯¹²) * (0.009) / (2.5*10¯³)
C = 3.186 * 10¯¹¹ F
The relation for electrostatic energy stored in the capacitor is given by U = ½ CV²
U = 1/ 2 * (3.186 * 10¯¹¹) * (400) ²
U = 2.55 * 10¯⁶
The electrostatic energy stored by the capacitor = 2.55 * 10¯⁶ J.
Volume of the given capacitor,
V = A*d = 90*10¯⁴ * 2.5*10¯³ = 2.25*10¯⁵ m³
u = U/V = 2.55 * 10¯⁶ * 2.25*10¯⁵
u = 0.113 J/ m³
Relationship between u and E
E = V/d
E = 400 / 2.5*10¯³ = 160000 V/m
The energy density u in the electric field is given by:
u = ½ ɛ₀E²
Another important yet challenging chapter of class 12 Physics. Students often struggle in analysing complex circuits involving multiple resistors, capacitors, and batteries, incorporating concepts like Kirchhoff’s Laws, Potentiometer, and Meter Bridge.
The chapter includes both numerical and conceptual-based questions. Numerical problems involving circuit analysis, potential difference calculations & resistance determination. Conceptual questions on Kirchhoff’s Laws, potentiometer principle, and meter bridge applications.
Using Krichhoff’s law, determine the current I₁, I₂, I₃ for the below network:
In the above figure, applying the junction rule at point F
I₁ = I₂ + I₃ —– (1)
Loop rule for BAFCB
2I₁ + 6I₂ – 24 + 27 = 0
2I₁ + 6I₂ + 3 = 0 —– (2)
Loop rule for FCDEF
27 + 6I₂ – 4I₃ = 0 —– (3)
Substituting I₁ in equation (2)
(2) => 2(I₂ + I₃) + 6I₂ + 3 = 0
(2) => 2I₃ + 8I₂ + 3 = 0
2 * (2) => 2 (2I₃ + 8I₂ + 3) = 0
2 * (2) => 4I₃ + 16I₂ + 6 = 0 ——(4)
(4) + (3) => 27 + 6 + 22 I₂ = 0
=> I₂ = -33/ 22 = -3/ 2
=> I₂ = -1.5 A
Substituting I₂ in equation (2)
2I₁ + 6 (-1.5) + 3 = 0
=> 2I₁ = 6
=> I₁ = 3 A
Substitute I₁, I₂ in equation (1)
=> 3 = -1.5 + I₃
=> I₃ = 4.5 A
The currents are:- I₁ = 3 A, I₂ = -1.5 A and I₃ = 4.5 A
Both topics are important for the CBSE Class 12 board, JEE and NEET exams. These topics have a higher weight in the exam, and students find it challenging to understand the concept, including EMF and its applications, especially in AC circuits. AC generators, transformers and LC oscillations are important topics under these.
Numerical problems involving induced emf, power factor, and resonance frequency. Conceptual questions on AC circuits, transformers, and LC oscillations.
An infinitesimal bar magnet of dipole moment M is pointing and moving with speed v in the direction x. A closed circular conducting loop of radius a and negligible self-inductance lies in the y – z plane with its centre at x=0 & its axis coinciding with the x-axis. Determine the force that opposes the motion of the magnet, assuming that the resistance of the loop is R. Also, consider that the distance x of the magnet from the centre of the loop is much greater than the radius a.
Magnetic field intensity due to a bar magnet at a distance can be written as:
B = (μ₀/ 4π) (2M/x²)
Magnetic flux linked with the loop can be written as:
Φ = BA = (μ₀/ 4π) (2M/x²) πa²
Emf induced in the loop can be written as:
ɛ = – dΦ/dt = (μ₀/ 4π) (6 πa²M/ x⁴) dx/dt
ɛ = (μ₀/ 4π) (6 πa²M/ x⁴) v
The induced current can be written as:
i = ɛ/R = (μ₀/ 4π) (6 πa²M/ Rx²) v
Let F be the force opposing the motion of the magnet. Using energy conservation, we can write
Power invested due to opposing force = Rate of heat dissipation in the coil
Fv = i²R
Fv = [(μ₀/ 4π) (6 πa²M/ Rx²) v] ² R
Fv = [(μ₀/ 2) (3a²M/ Rx⁴) v] ² R
Fv = 9/4 [(μ₀²M² a⁴ v/ Rx⁸]
Electromagnetic waves play a crucial role in understanding how energy and information are transmitted through space. They are essential for comprehending the properties of light, including how it propagates through various materials, such as tissue, and how it scatters and is absorbed. Students often encounter difficulties in grasping these abstract concepts and their applications, which can lead to misidentifying different types of waves. Additionally, students face many other challenges in this chapter.
Numerical problems involving electromagnetic wave properties and polarisation. Conceptual questions on the nature of electromagnetic waves and their applications.
The expression of electric field for a light beam travelling in the X-direction is E = 250 sin w (t – x/c) v/m
A proton travels along the y-direction with a speed of 2.2 * 10⁷ m/s. Calculate the maximum electric force and maximum magnetic force on the electron.
Maximum magnitude of electric field
E₀ = 250 V/m
Maximum magnitude of magnetic field
B₀ = E₀ / c = 250/ 3*10⁸
= 8.3 * 10¯⁷ t, along the z-direction
Maximum electric force on the proton is
Fр = qE₀
= 1.6 * 10¯¹⁹ * 250
= 4.0 * 10¯¹⁷ N
Maximum magnetic force on the proton is
Fm = qvB₀
= 1.6 * 10¯¹⁹ * 2.2 * 10¯⁷ * 8.3 * 10¯⁷
= 2.92 * 10¯¹⁸ N
Class 12th Important to Read: |
Optics is one of the vast chapters in class 12 Physics; the chapters cover topics from wave and ray optics. Sometimes, students face challenges while studying this chapter, which consists of interpreting the ray diagrams, understanding polarisation, understanding different derivations and remembering them, linking theory to real-world scenarios, understanding the concept of interference and diffraction, and so many others.
Different numerical and conceptual questions asked from this chapter include numerical problems involving interference, diffraction, and lens formula and conceptual questions on the wave nature of light, interference patterns, and optical instruments.
State Brewster Law. Using this law prove that, at the polarising angle of incidence, reflected and transmitted rays are perpendicular to each other.
According to the Brewster Law, the longest of angle of polarization for a transparent medium is equal to the refractive index of the medium.
µ = tan iр
using Snell’s law
when i = iр, µ = sin iр / sin rр —-(1)
Also, we have, tan iр = sin iр / cos iр —–(2)
From equation (1) & (2)
sin iр / sin rр = sin iр / cos iр
=> sin rp = cos ip
=> sin rp = sin (90° – ip)
Therefore, rp + ip = 90°
Therefore, at the polarizing angle of incidence, the reflected and transmitted rays are found to be perpendicular to each other.
This unit includes chapters on the dual nature of radiation and matter, atoms, nuclei, and semiconductor electronics. The challenges that students face in these chapters are understanding the concepts of quantum mechanics, nuclear physics, spectra of the hydrogen atoms, Bhohr’s model of atoms, fission and fusion, transistor amplifiers and oscillators and many others.
The chapters contain a variety of numerical and conceptual questions. The numerical problems focus on topics such as energy levels, spectral lines, ionisation energy, radioactive decay, nuclear binding energy, nuclear reactions, Zener diode characteristics, transistor biasing, and amplifier gain. Additionally, there are conceptual questions related to atomic structure, spectral series, nuclear structure, nuclear forces, nuclear energy, and semiconductor devices, along with their applications.
Draw a labelled circuit diagram of a common emitter transistor amplifier. Draw the input & output waveforms and also state the relation between the input & output signal.
Below given is the circuit diagram of a common emitter transistor amplifier:
Input Waveform:
Output Waveform:
There is 180° phase difference between the output and input waveform. Also, the output waveform is amplified compared to the input waveform.
Remember, persistence and consistent practice are key to master any challenging topics from any subject.
All the best for your board exams!
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