Practice <strong>NEET UG Physics</strong> MCQ questions covering Mechanics, Thermodynamics, Waves & Optics, Modern Physics, Electricity & Magnetism, and Electromagnetic Induction. Aligned with NCERT Class 11 & 12 Physics syllabus.
A ball is dropped from a height of 80 m. If air resistance is negligible, what is the velocity of the ball just before it hits the ground? (g = 10 m/s²)
Answer: B
Using v² = u² + 2as, where u = 0, a = 10 m/s², s = 80 m. v² = 0 + 2(10)(80) = 1600, so v = 40 m/s
Q.22Easy
Which of the following is NOT a consequence of Newton's first law of motion?
Answer: C
Option C is the statement of Newton's second law (F = ma), not the first law. Newton's first law deals with inertia and absence of acceleration.
Q.23Easy
A vehicle increases its speed from 18 km/h to 54 km/h in 8 seconds. What is the acceleration of the vehicle?
Answer: A
Initial velocity u = 18 km/h = 5 m/s. Final velocity v = 54 km/h = 15 m/s. Time t = 8 s. Using a = (v - u)/t = (15 - 5)/8 = 810 = 1.25 m/s²
Q.24Medium
A man pulls a rope attached to a 50 kg box with a force at 30° above the horizontal. If the coefficient of kinetic friction is 0.1 and the pulling force is 200 N, what is the acceleration of the box? (g = 10 m/s²)
Answer: C
Horizontal component: Fₓ = 200 cos(30°) = 200 × (√23) ≈ 173.2 N. Vertical component: Fᵧ = 200 sin(30°) = 100 N. Normal force: N = mg - Fᵧ = 500 - 100 = 400 N. Friction: f = μN = 0.1 × 400 = 40 N. Net force: F_net = 173.2 - 40 = 133.2 N. Acceleration: a = 133.502 ≈ 2.8 m/s²
Q.25Easy
A particle moves along a circular path of radius 5 m with a constant speed of 20 m/s. What is the centripetal acceleration?
An elevator with a mass of 1000 kg is moving upward with acceleration 2 m/s². What is the tension in the cable? (g = 10 m/s²)
Answer: C
Using Newton's second law: T - mg = ma. T = m(g + a) = 1000(10 + 2) = 1000 × 12 = 12000 N
Q.27Medium
A block of mass 4 kg is placed on an inclined plane at angle 37° with the horizontal. The coefficient of static friction is 0.8. Is the block in equilibrium?
Answer: A
Component of weight along plane = mg sin(37°) = 4 × 10 × 0.6 = 24 N. Normal force N = mg cos(37°) = 4 × 10 × 0.8 = 32 N. Maximum static friction = μₛN = 0.8 × 32 = 25.6 N. Since 24 N < 25.6 N, the block is in equilibrium.
Q.28Medium
Two forces of magnitude 10 N and 15 N are acting at an angle of 90° to each other. What is the magnitude of the resultant force?
Answer: B
When two perpendicular forces act, resultant R = √(F₁² + F₂²) = √(10² + 15²) = √(100 + 225) = √325 ≈ 18.03 N
Q.29Medium
The equation of motion for a body is s = 4t + 3t². What is the acceleration of the body?
Answer: B
Given s = 4t + 3t². Velocity v = ds/dt = 4 + 6t. Acceleration a = dv/dt = 6 m/s² (constant)
Q.30Medium
A 1500 kg car traveling at 30 m/s collides with a wall and comes to rest in 0.5 seconds. What is the average force exerted by the wall on the car?
Answer: B
Using F = ma, first find acceleration: a = (v - u)/t = (0 - 30)/0.5 = -60 m/s². Force F = ma = 1500 × 60 = 90000 N (magnitude)
Q.31Medium
A body of mass m is projected vertically upward with velocity v₀. At what height will its kinetic energy equal its potential energy (taking ground as reference)?
Answer: C
At height h: KE = ½m(v₀² - 2gh) and PE = mgh. When KE = PE: ½m(v₀² - 2gh) = mgh. ½v₀² - gh = gh. ½v₀² = 2gh. h = v₀²/4g
Q.32Medium
A cyclist moving at 10 m/s on a circular track of radius 50 m leans at an angle θ to the vertical. What is the angle of lean for friction-free motion?
Answer: A
For friction-free circular motion: tan θ = v²/(rg) = (10)²/(50 × 10) = 500100 = 0.2. Therefore θ = tan⁻¹(0.2)
Q.33Easy
A body slides down a frictionless incline of angle 30°. What is the acceleration along the plane?
Answer: C
On a frictionless incline: a = g sin θ = 10 × sin(30°) = 10 × 0.5 = 5 m/s²
Q.34Easy
A particle experiences two perpendicular accelerations: aₓ = 3 m/s² and aᵧ = 4 m/s². What is the magnitude of the net acceleration?
A block of mass 5 kg is pushed on a horizontal surface by a 30 N force at 20° below the horizontal. The coefficient of kinetic friction is 0.3. Find the acceleration. (g = 10 m/s²)
Answer: C
Horizontal component: Fₓ = 30 cos(20°) ≈ 28.2 N. Vertical component: Fᵧ = -30 sin(20°) ≈ -10.3 N (downward). Normal force: N = mg + 10.3 = 50 + 10.3 = 60.3 N. Friction: f = 0.3 × 60.3 ≈ 18.1 N. Net force: F_net = 28.2 - 18.1 ≈ 10.1 N. Acceleration: a ≈ 10.51 ≈ 2 m/s². Closest to 1.8 m/s².
Q.36Easy
A particle moving in a circle of radius r = 2 m completes 5 revolutions in 10 seconds. What is its angular velocity?
