KCET Physics All Formulas PDF

KCET Physics All Formulas PDF - KCET 2025 Physics is one of the most important sections. As per the previous year's analysis, the KCET Physics section consists of formula-based questions. As per the information brochure, the KCET Physics Syllabus is divided into two parts - First and Second PUC syllabi as directed by the Department of Pre-University Education of Karnataka State. To score 160+ in KCET, candidates should focus on the KCET formulas to improve their preparation strategy. Before the KCET exam , candidates are advised to check the KCET Physics All Formulas PDF from this article to ace their KCET 2025 preparation.

Check Also: How to Score 50+ Marks in KCET Chemistry 2025

KCET Physics All Formulas PDF

Source - TestPrep Cart

KCET Physics Syllabus 2025

The table below highlights the syllabus of KCET Physics 2025.

KCET 2025 Physics - Topic Wise Weightage

Some of the topics holding the highest weight in KCET Physics are Gravitation, Magnetism Matter, etc. Candidates can check the topic-wise weightage of KCET 2025 Physics.

KCET 2025 First PUC Important Topics

The table below shows the KCET Physics important topics from the first PUC.

KCET 2025 Second PUC Important Topics

The table below shows the KCET Physics important topics from the second PUC.

KCET Physics - Most Asked Questions from Previous Year Question Papers

Candidates can check the KCET physics questions below.

1. A moving coil galvanometer is converted into an ammeter of range 0 to 5mA. The galvanometer resistance is 90Ω and the shunt resistance has a value of 10Ω. If there are 50 divisions in the galvanometer-turned ammeter on either side of zero, its current sensitivity is -
2. 1 × 105 A/div
3. 2 × 104 A/div
4. 1 × 105 div/A
5. 2 × 104 div/A

B. The energy gap of an LED is 2.4 eV. When the LED is switched ‘ON’, the momentum of the emitted photons is

1. 2.56 × 10-27 kg.m.s-1
2. 1.28 × 10-11 kg.m.s-1
3. 0.64 × 10-27 kg.m.s-1
4. 1.28 × 10-27 kg.m.s-1

C. If the length of a conductor is increased by 20% and the cross-sectional area is decreased by 4%, then calculate the percentage change in the resistance of the conductor.

Answer- 25%

D. A cell of emf 90V is connected across a series combination of two resistors each of 100Ω resistance A voltmeter of resistance 400Ω is used to measure the potential difference across each resistor The reading of the voltmeter will be :

1. 45V
2. 40V
3. 80V
4. 90V

E. A hollow cylindrical conductor has a length of 314m, while its inner and outer diameters are 4mm and 8mm respectively The resistance of the conductor is n × 10 −3 Ω n × 10 −3 Ω. If the resistivity of the material is 24 × 10 − 8 Ω 24 × 10 −8 Ω m The value of n is ___.

SOLUTION -

R=ρ A ℓ, the cross-sectional area is π(b2−a2)

R=ρ π (b2−a2) ℓ​=3.14×(42−22)×10−62.4×10−8×3.14​

=2×10−3Ω

∴ n=2

F. A metallic rod of length 1 m held along the east-west direction is allowed to fall freely. Given the horizontal component of earth’s magnetic field BH = 3 × 10-5 T. The emf induced in the rod at an instant t = 2s after it is released is (Take g = 10 ms-2 )

1. 3 × 10-3 V
2. 3 × 10-4 V
3. 6 × 10-3 V
4. 6 × 10-4 V

G. The energy gap of an LED is 2.4 eV. When the LED is switched ‘ON’, the momentum of the emitted photons is -

1. 2.56 × 10-27 kg.m.s-1
2. 1.28 × 10-11 kg.m.s-1
3. 0.64 × 10-27 kg.m.s-1
4. 1.28 × 10-27 kg.m.s-1

H. In Young’s double-slit experiment, the intensity of light passing through each of the two double slits is 2 × 10-2 Wm-2. The screen-slit distance is very large in comparison with the slit-slit distance. The fringe width is β. The distance between the central maximum and a point P on the screen is 33 β. Then the total light intensity at that point is -

1. 4 × 10-2 Wm-2
2. 2 × 10-2 Wm-2
3. 16 × 10-2 Wm-2
4. 8 × 10-2 Wm-2

I. If the area of cross-section is halved and the length of a wire having Young's modulus Y is doubled, then its new Young's modulus will be -

1. Y
2. 4Y
3. Y/2
4. Y/4

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