KCET Practice Test 2026

About the KCET exam

The Karnataka Common Entrance Test (KCET) is a engineering examination administered in India. It is designed to assess candidates' knowledge and proficiency across 4 key subject areas including Physics, Chemistry, Mathematics, Biology. This exam is widely recognized as an essential benchmark for aspiring engineers.

Use short KCET practice sets to identify topics to revisit. Choose a subject and difficulty, answer questions, then review the explanations. New accounts get 30 question credits; further new questions use paid credits. Retaking previously seen questions is free.

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8 free KCET practice questions (2026)

Answer these first — no signup, no card. Every answer is explained. For more practice, sign up for 30 free question credits and choose a subject above.

1. The dimensions of electrical permittivity of free space (ε₀) are:

   • [M⁻¹ L⁻³ T⁴ A²]
   • [M L⁻³ T⁴ A²]
   • [M⁻¹ L³ T⁻⁴ A⁻²]
   • [M⁻¹ L⁻³ T⁻² A²]
   Show answer & explanation

   Answer: A. [M⁻¹ L⁻³ T⁴ A²]

   From Coulomb's law, F = (1 / 4πε₀) * (q₁q₂ / r²). Therefore, ε₀ = [q²] / [F * r²]. Since [q] = [AT] and [F] = [MLT⁻²], we get [ε₀] = [A²T²] / ([MLT⁻²] * [L²]) = [M⁻¹ L⁻³ T⁴ A²].

   Topic: Physics · medium

2. A stone dropped from the top of a tower of height h reaches the ground in t seconds. What is the time taken by the stone to reach half the height of the tower?

   • t / 2
   • t / √2
   • t / 2√2
   • t / 4
   Show answer & explanation

   Answer: B. t / √2

   Using s = (1/2)gt²: for full height h, h = (1/2)gt². For half height h/2, h/2 = (1/2)g(t')². Dividing the two equations gives (t')² = t² / 2, which yields t' = t / √2.

   Topic: Physics · medium

3. If the magnitude of the sum of two vectors is equal to the magnitude of their difference, then the angle between the two vectors is:

   • 0 degrees
   • 45 degrees
   • 90 degrees
   • 180 degrees
   Show answer & explanation

   Answer: C. 90 degrees

   |A + B| = |A - B| implies A² + B² + 2AB cos(θ) = A² + B² - 2AB cos(θ). This gives 4AB cos(θ) = 0, so cos(θ) = 0, which means θ = 90 degrees.

   Topic: Physics · easy

4. Two projectiles are thrown with the same initial speed at angles θ and (90 - θ) with the horizontal. The ratio of their horizontal ranges is:

   • 1 : 1
   • 1 : 2
   • 2 : 1
   • tan(θ) : 1
   Show answer & explanation

   Answer: A. 1 : 1

   The horizontal range is given by R = u² sin(2θ) / g. For complementary angles θ and 90 - θ, sin(2(90 - θ)) = sin(180 - 2θ) = sin(2θ). Thus, the ranges are equal, and the ratio is 1 : 1.

   Topic: Physics · easy

5. A man of mass m is standing in a lift which is moving downwards with a uniform acceleration 'a' (where a < g). The apparent weight of the man is:

   • m(g + a)
   • m(g - a)
   • mg
   • zero
   Show answer & explanation

   Answer: B. m(g - a)

   When a lift accelerates downwards with acceleration 'a', the effective acceleration due to gravity is (g - a). Therefore, the normal reaction (apparent weight) is R = m(g - a).

   Topic: Physics · easy

6. A body of mass m is accelerated from rest by a constant engine power P. The velocity of the body as a function of time t is proportional to:

   • t
   • t²
   • t^(1/2)
   • t^(3/2)
   Show answer & explanation

   Answer: C. t^(1/2)

   Power P = F * v = m * (dv/dt) * v. Integrating both sides, P dt = m v dv, which gives P * t = (1/2) * m * v². Therefore, v = √(2Pt/m), meaning v is proportional to t^(1/2).

   Topic: Physics · medium

7. What is the ratio of the radii of gyration of a circular ring and a circular disc of the same mass and radius about their respective geometrical axes?

   • √2 : 1
   • 1 : √2
   • 2 : 1
   • 1 : 2
   Show answer & explanation

   Answer: A. √2 : 1

   For a ring, I_ring = MR² = M k₁², so k₁ = R. For a disc, I_disc = (1/2)MR² = M k‒², so k‒ = R / √2. The ratio k₁ / k‒ is equal to R / (R / √2) = √2 : 1.

   Topic: Physics · medium

8. The value of acceleration due to gravity 'g' at a height h = R (where R is the radius of the Earth) above the surface of the Earth is:

   • g / 2
   • g / 3
   • g / 4
   • g / 9
   Show answer & explanation

   Answer: C. g / 4

   The acceleration due to gravity at height h is given by g' = g * [R / (R + h)]². Substituting h = R gives g' = g * [R / 2R]² = g / 4.

   Topic: Physics · medium

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KCET Exam Subjects

Available practice subjects: Physics, Chemistry, Mathematics, Biology. Select a subject for focused preparation. Coverage is not a guarantee of the complete official syllabus.

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