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Kerala Plus Two Physics Exam 2025: Check Paper Analysis, Question Paper and Answer Key Details

Last Updated: Mar 5, 2025, 19:36 IST

The Kerala Plus Two Physics Exam 2025 was conducted successfully, by following the latest syllabus and exam pattern. Students found the paper well-balanced, with a mix of conceptual and numerical questions. The difficulty level ranged from moderate to challenging, with some sections requiring critical thinking. The question paper, answer key, and expert analysis are now available to help students assess their performance.

Kerala Plus Two Physics Exam 2025: Check Paper Analysis, Question Paper and Answer Key Details
Kerala Plus Two Physics Exam 2025: Check Paper Analysis, Question Paper and Answer Key Details

The Kerala Plus Two Physics Exam 2025 was held as per the schedule set by the Directorate of Higher Secondary Education (DHSE), Kerala. This exam is a crucial milestone for Class 12 students, as it plays a significant role in shaping their higher education prospects. The paper tested students' understanding of core physics concepts, including mechanics, electromagnetism, modern physics, and optics, while also evaluating their problem-solving skills through numerical and application-based questions.

A thorough analysis of the question paper reveals a well-structured exam that adhered to the prescribed syllabus and blueprint. Students who had a strong conceptual grasp and practiced numericals regularly found the paper manageable. With the release of the official question paper and answer key, students can now review their responses and estimate their scores, aiding in better preparation for future competitive exams.

Kerala Plus Two Physics Exam 2025 Question Paper

Students can check the Kerala Plus Two Physics Exam 2025 Question Paper.

Kerala Plus Two Physics Exam 2025 Question Paper 2025 

Kerala Plus Two Physics Exam 2025 Answer Key 2025

Now that the paper is over, students might be waiting for the answer key. The answer key helps the students to estimate their total score. Students can check the link to the answer key. 

1.The electrostatic force per unit charge is known as _______.
Answer: (c) Electric field

2.A circle of radius r is drawn with a charge +q placed at the centre. The work done in moving a point charge once around the circumference of the circle is _______.
Answer: Zero

3.Which of the following gives the polarity of the induced emf?
(i) Biot-Savart Law
(ii) Lenz's Law
(iii) Ampere's Circuital Law
(iv) Fleming's Right-Hand Rule
Answer: (ii) Lenz's Law

4.The frequencies of gamma rays, ultraviolet rays, and X-rays are v₁, v₂, and v₃ respectively. Then
(i) v₁ = v₂ = v₃
(ii) v₁ > v₃ > v₂
(iii) v₁ > v₂ > v₃
(iv) v₃ > v₂ > v₁
Answer: (iii) v₁ > v₂ > v₃

5.If the radius of the first electron orbit of hydrogen is r₀, the radius of the second electron orbit of hydrogen is _______.
Answer: 4r₀

6.Two thin lenses of power +4 D and -2 D are in contact. The focal length of the combination is _______.
Answer: 50 cm

7._______ process is responsible for the production of energy in the sun. (Nuclear fission / Nuclear fusion)
Answer: Nuclear fusion

Answer any 5 questions from 8 to 14. Each carries 2 scores. 

8.Show that the resistance of a conductor can be expressed by:

Where symbols have their usual meanings.

Answer: Resistance R of a conductor is defined as R=V/I, ……….(i)

where 

V = voltage across the conductor and 

I = current flowing through it.

Using Ohm's law, we can express current I as I=neAvd, …………(ii)

where 

n = number density of charge carriers, 

e = charge of an electron, 

A = cross-sectional area of the conductor, and 

vd = drift velocity of the charge carriers.

vd can be expressed in terms of the mean free time τ and the mean free path l as 

vd = l/τ, Substituting this in (ii) we get:

I=neAl/τ

Now substituting I in (i), we have 

R=V/I = Vτ/neAl

Rearranging gives us 

R=ml/ne2τA, where m is the mass of the charge carriers.

9. Relative permeability of a material μr < 1, identify the magnetic material. Write the relation between relative permeability and magnetic susceptibility.

Answer: The material is diamagnetic.

The relation between magnetic susceptibility χm and relative permeability μr is

μr = 1 + χm 

Q10. Instantaneous current and voltage in an AC circuit

Given:
i=10sin⁡(314t)i = 10 \sin(314t)i=10sin(314t)
v=50sin⁡(314t+π/2)v = 50 \sin(314t + \pi/2)v=50sin(314t+π/2)

(a) Phase difference between voltage and current:
The voltage leads the current by π/2 (90 degrees).

(b) Power dissipation in the circuit:
Since the phase difference is π/2, the power dissipation is:

Answer:  (a)Microwaves.

(b) Current produced between plates of a capacitor:
Answer: (iii) 

Q12. 

(a) Definition of Wavefront:
A wavefront is the locus of points having the same phase of oscillation in a wave.

(b) Refracted wavefront for a convex lens:
The plane wavefront bends and converges after passing through the convex lens, forming a curved wavefront.

Q13. What is meant by ionisation energy ? Write its value for hydrogen atom.

Ionization energy: The minimum energy required to remove an electron from an isolated neutral atom.

For a hydrogen atom, ionization energy = 13.6 eV.

Kerala Plus Two Physics Exam 2025 Answer Key 2025 (Link Active Soon)

Anisha Mishra
Anisha Mishra

Executive - Editorial

Anisha Mishra is journalist with over 3 years of experience in covering the Indian education sector. She has worked extensively in the K12 domain, with focus on the state board as well as central board examinations, policy structure of seconday and higher secondary education as well as the entrance examinations like JEE, NEET, CLAT, etc. Her extensive experience in the domain has helped her provide students with concise accurate information in all aspectes of school life and education. Her key interest lies in decoding the changes in the curriculum, NEP implementation and changing education ecosystem in the country. Besides working, she enjoys traveling, exploring new places and cultures, and painting.

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First Published: Mar 5, 2025, 16:05 IST

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