Shortcut Methods and Tricks to Solve Numerical Problems

(1). Static Equilibrium Problems:

JEE Advanced Numerical:
- Resolve the force of 20 N into its horizontal (20 cos 30° = 17.32 N) and vertical (20 sin 30° = 10 N) components.
- The net force in the horizontal direction is 17.32 N - (0.2 x 10 x 9.8) N = 5.864 N.
- The acceleration of the block is therefore: a = F/m = 5.864 N / 10 kg = 0.5864 m/s².
CBSE Board Exam Numerical:
Resolve the force of 10 N into its horizontal (10 cos 60° = 5 N) and vertical (10 sin 60° = 8.66 N) components.
The maximum force that can be applied parallel to the plane without causing the block to slide down is: F = μs * N = 0.5 * 8.66 N = 4.33 N.

(2). Inclined Plane Problems:

JEE Advanced Numerical:
The acceleration of the body down the plane is given by: a = g * sin θ - μk * g * cos θ = 9.8 m/s² * sin 30° - 0.2 * 9.8 m/s² * cos 30° = 5.88 m/s².
CBSE Board Exam Numerical:
The maximum force that can be applied parallel to the plane without causing the block to slide down is: F = μs * m * g * cos θ = 0.4 * 5 kg * 9.8 m/s² * cos 45° = 17.15 N.

(3). Dynamics of Circular Motion Problems:

JEE Advanced Numerical:
- The centripetal force acting on the particle is given by: F = mv²/r = 2 kg * (10 m/s)² / 5 m = 40 N.
CBSE Board Exam Numerical:
- The centripetal force acting on the car is given by: F = mv²/r = 1000 kg * (20 m/s)² / 100 m = 40000 N.

(4). Fluid Pressure Problems:

JEE Advanced Numerical:
- The pressure at the bottom of the vessel is given by: P = ρgh = 1000 kg/m³ * 9.8 m/s² * 10 m = 98000 Pa.
CBSE Board Exam Numerical:
- The pressure at a point 1 m below the surface of water is given by: P = ρgh = 1000 kg/m³ * 9.8 m/s² * 1 m = 9800 Pa.

(5). Buoyant Force Problems:

JEE Advanced Numerical:
- The buoyant force acting on the block is given by: F = ρVg = 500 kg/m³ * 0.1 m³ * 9.8 m/s² = 490 N.
CBSE Board Exam Numerical:
- The buoyant force acting on the metal piece is given by: F = ρVg = 1000 kg/m³ * 0.2 m³ * 9.8 m/s² = 1960 N.