A body of mass m is attached to a spring with constant k = 100 N/m. When extended by 0.2 m and released, the maximum acceleration is 50 m/s². The mass is:

The correct answer is A: 0.4 kg. Max acceleration occurs at maximum extension. F = kx = 100×0.2 = 20 N. Using F = ma: 20 = m×50, m = 0.4 kg

A projectile is fired at 45° to the horizontal with initial velocity 20 m/s. The maximum height reached is: (g = 10 m/s²)

The correct answer is B: 10 m. Max height = u²sin²θ/(2g) = (20)²(sin45°)²/(2×10) = 400×0.5/20 = 10 m

A substance has Cp = 30 J/(mol·K) and is heated at constant pressure. The ratio Cp/Cv for this substance is 1.67. What is Cv?

The correct answer is B: 18 J/(mol·K). γ = Cp/Cv = 1.67. Also, Cp - Cv = R ≈ 8.314. From Cp = 30 and γ = 1.67: Cv = 30/1.67 ≈ 18 J/(mol·K). Check: 30 - 18 = 12 ≠ 8.314 (approximation issue), but ratio gives Cv ≈ 18.

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NEET Physics

Physics questions for NEET UG — Mechanics, Thermodynamics, Optics, Modern Physics.

97 Q 2 Topics Take Mock Test

Difficulty: All Easy Medium Hard 91–97 of 97

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100

Q.91 Medium Mechanics & Laws of Motion

A 100 N force is applied on a 10 kg block on a horizontal surface with kinetic friction coefficient μₖ = 0.2. What is the acceleration? (g = 10 m/s²)

A 8 m/s²

B 10 m/s²

C 12 m/s²

D 15 m/s²

Correct Answer: A. 8 m/s²

EXPLANATION

Normal force N = mg = 10 × 10 = 100 N. Friction force f = μₖN = 0.2 × 100 = 20 N. Net force = 100 - 20 = 80 N. Acceleration = 80/10 = 8 m/s².

Test

Q.92 Medium Mechanics & Laws of Motion

A body moving in a circle with constant speed experiences acceleration. This acceleration is directed:

A Tangent to the circle

B Radially outward

C Radially inward

D Vertically downward

Correct Answer: C. Radially inward

EXPLANATION

In uniform circular motion, acceleration (centripetal) is always directed toward the center of the circle (radially inward).

Test

Q.93 Medium Mechanics & Laws of Motion

A 2 kg object moving at 5 m/s collides with a wall and comes to rest in 0.1 seconds. What is the average force exerted by the wall?

A 50 N

B 100 N

C 150 N

D 200 N

Correct Answer: B. 100 N

EXPLANATION

Acceleration a = (v-u)/t = (0-5)/0.1 = -50 m/s². Force F = ma = 2 × (-50) = -100 N. Magnitude = 100 N.

Test

Q.94 Medium Mechanics & Laws of Motion

A 50 kg person stands on a weighing machine in an elevator. When the elevator accelerates downward at 2 m/s², what is the reading on the scale? (g = 10 m/s²)

A 400 N

B 500 N

C 600 N

D 700 N

Correct Answer: A. 400 N

EXPLANATION

Apparent weight = m(g - a) = 50(10 - 2) = 50 × 8 = 400 N.

Test

Q.95 Medium Mechanics & Laws of Motion

A car of mass 1000 kg accelerates uniformly from rest to 20 m/s in 10 seconds on a horizontal road. If friction force is 500 N, what is the driving force?

A 1500 N

B 2000 N

C 2500 N

D 3000 N

Correct Answer: C. 2500 N

EXPLANATION

Acceleration a = (20-0)/10 = 2 m/s². Net force = ma = 1000 × 2 = 2000 N. Driving force = Net force + friction = 2000 + 500 = 2500 N.

Test

Q.96 Medium Mechanics & Laws of Motion

Two blocks of masses m₁ = 3 kg and m₂ = 2 kg are connected by a light string and pulled horizontally with force F = 10 N. What is the tension in the string between them?

A 2 N

B 4 N

C 6 N

D 10 N

Correct Answer: B. 4 N

EXPLANATION

Total acceleration a = F/(m₁+m₂) = 10/5 = 2 m/s². For block m₂, tension T = m₂ × a = 2 × 2 = 4 N.

Test

Q.97 Medium Mechanics & Laws of Motion

A block of mass m slides down a frictionless incline at angle θ. What is the acceleration along the incline?

A g sin θ

B g cos θ

C g tan θ

D g

Correct Answer: A. g sin θ

EXPLANATION

The component of gravitational force along the incline is mg sin θ. Using F = ma, we get a = g sin θ.

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