Practice Section

Timed. Tiered. Real exam-style. Easy (Boards) → NEET → JEE Main → JEE Advanced. Start the timer, attempt, reveal answers. No cheating allowed — yourself.

20 Questions 4 Levels Timed Mode Answer Analysis

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🎯 Target

These should be solved in under 60 seconds each. If you're slower, revise core concepts. All are direct formula application.

A car accelerates from rest at 3 m/s² for 10 seconds. What is its final velocity?

Easy Boards Concept: v = u + at

✅ Solution

v = u + at = 0 + 3 × 10 = 30 m/s
Use v = u + at. u = 0 (from rest), a = 3 m/s², t = 10 s.

A man walks 8 m east, then 6 m west. What is his displacement from start?

EasyBoards

✅ Solution

Displacement = 8 − 6 = 2 m east. Net = algebraic sum with direction. Distance = 14 m (total path).

A body dropped from rest falls under gravity (g = 10 m/s²). Distance fallen in 3 seconds is:

EasyBoards

✅ Solution

h = ½gt² = ½ × 10 × 9 = 45 m. u = 0, a = g = 10 m/s², t = 3 s. Use s = ut + ½at².

In a v-t graph, what does the area under the curve represent?

EasyBoards

✅ Solution

Displacement (with sign). Area above t-axis = positive displacement. Area below = negative displacement. Slope of v-t graph = acceleration.

Which kinematic equation does NOT involve time?

EasyBoards

✅ Solution

v² = u² + 2as — contains v, u, a, s but no t. Use this whenever time is absent from the problem.

🎯 Target

1.5 to 2 minutes per question. Focus on correct equation selection and sign convention. These closely mirror actual NEET questions.

A ball is thrown upward with velocity 20 m/s. Time to reach maximum height is: (g = 10 m/s²)

MediumNEET

✅ Solution

At max height, v = 0. Use v = u − gt → 0 = 20 − 10t → t = 2 s

A particle starts from rest and moves with uniform acceleration. Ratio of distances in 1st, 2nd, and 3rd second is:

MediumNEET

✅ Solution

Using sₙ = u + a(n−½) with u=0: s₁ = a/2, s₂ = 3a/2, s₃ = 5a/2. Ratio = 1 : 3 : 5 (odd number series). Classic NEET trap.

A car moving at 72 km/h brakes to rest over 40 m. The retardation is:

MediumNEET

✅ Solution

72 km/h = 20 m/s. v²=u²+2as → 0 = 400+2(a)(40) → a = −400/80 = −5 m/s² → retardation = 5 m/s²

A particle has velocity −5 m/s and acceleration +3 m/s². Which statement is correct?

MediumNEET Conceptual

✅ Solution

v = −5 (negative direction), a = +3 (positive direction). v and a are in opposite directions → speed decreases. The particle is decelerating even though a is positive.

Two trains A and B are approaching each other. A at 60 m/s, B at 40 m/s. They are 200 m apart. Time to meet?

MediumNEET Relative

✅ Solution

Relative velocity = 60 + 40 = 100 m/s (approaching). t = d/v_rel = 200/100 = 2 s

🎯 Target

2 minutes per question. Multi-step thinking. Careful with signs. Draw a quick diagram before solving.

A particle starts from rest and has acceleration a = 2t m/s². Velocity at t = 3 s is:

Medium-HardJEE Main

✅ Solution

a = 2t → variable. Integrate: v = ∫2t dt = t² + C. At t=0, v=0 → C=0. v(3) = 9 m/s. Cannot use v=u+at as a is not constant!

A stone is thrown vertically up from the top of a 25 m tower with initial velocity 20 m/s. How long does it take to reach the ground? (g = 10 m/s²)

Medium-HardJEE Main

✅ Solution

Up = +ve, origin at top. s = −25 m (ground below). s = ut + ½at² → −25 = 20t − 5t² → 5t² − 20t − 25 = 0 → t² − 4t − 5 = 0 → (t−5)(t+1) = 0 → t = 5 s

Distance covered by a particle in the 5th second of motion starting from rest with a = 4 m/s² is:

MediumJEE Main

✅ Solution

sₙ = u + a(n − ½) = 0 + 4(5 − ½) = 4 × 4.5 = 18 m. This is distance in the 5th second specifically, not total 5 seconds.

🎯 Target

3-4 minutes per question. These require calculus, careful setup, or multi-concept application. Don't guess. Either solve it or skip it — there's negative marking.

Velocity of a particle is v = 3x² + 2 where x is position. Find acceleration at x = 1 m.

HardJEE Advanced

✅ Solution

a = v·dv/dx. v = 3x²+2, dv/dx = 6x. At x=1: v = 5, dv/dx = 6. a = 5 × 6 = 30 m/s²

A particle starts from origin with v₀ = 4 m/s. Retardation = kv. k = 2 s⁻¹. Find position when v = v₀/e.

Very HardJEE Advanced

✅ Solution

a = −kv. x = ∫v dv/a = ∫v dv/(−kv) = −(1/k)∫dv = −(v−v₀)/k. When v = v₀/e: x = (v₀ − v₀/e)/k = v₀(1−1/e)/k = 4(1−1/e)/2 = 2(1−1/e) ≈ 1.26 m. Exact: 2(1−1/e) m

The x-t graph of a particle is a straight line with negative slope. This means the particle: [Select ALL correct options]

MediumMultiple Correct

✅ Solution

Straight line → constant slope → constant velocity (A correct). Zero curvature → zero acceleration (C correct). Negative slope = negative velocity, which still has positive speed magnitude (D wrong). Not decelerating since a = 0 (B wrong). Correct: A and C

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