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A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° - 90° side on a horizontal table. A cubical block of mass m rests on 60° - 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)
  • a)
    g
  • b)
    g/√2
  • c)
    g/√3
  • d)
    g/√5
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A triangular block of mass M with angles 30°, 60° and 90° ...

Use result, a = gtan θ, a = gtan 30° = g/√3
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A triangular block of mass M with angles 30°, 60° and 90° ...
Understanding the Problem
To analyze the situation, we have a triangular block (mass M) inclined at 60 degrees resting on a horizontal surface with a cubical block (mass m) on its incline. The goal is to find the acceleration of the triangular block (M) that keeps the cubical block (m) stationary relative to it.
Forces Acting on the Cubical Block
- Weight Force (mg): Acts vertically downward.
- Normal Force (N): Acts perpendicular to the surface of the triangular block.
Since the triangular block is accelerating, the cubical block experiences pseudo forces due to this acceleration.
Components of Forces
1. Normal Force (N): Acts perpendicular to the inclined plane.
2. Gravitational Force Resolution:
- Along the incline: mg * sin(60)
- Perpendicular to the incline: mg * cos(60)
Setting Up the Equation
For the cubical block to remain stationary relative to the triangular block, the net force acting along the incline must equal the force due to the acceleration of block M:
- Let 'a' be the acceleration of block M.
- The effective force along the incline is given by the component of gravitational force minus the pseudo force due to acceleration.
Using the equilibrium condition:
0 = mg * sin(60) - m * a * cos(60)
Solving for Acceleration (a)
After simplifying the above equation and substituting values for sin(60) and cos(60):
- a = (g * sin(60)) / cos(60)
- This simplifies to a = g * (√3 / 2) / (1/2) = g / √3.
Thus, the required acceleration to keep the cubical block stationary relative to the triangular block is:
Final Answer: g / √3 (Option C)
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A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° - 90° side on a horizontal table. A cubical block of mass m rests on 60° - 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g/√2c)g/√3d)g/√5Correct answer is option 'C'. Can you explain this answer?
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A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° - 90° side on a horizontal table. A cubical block of mass m rests on 60° - 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g/√2c)g/√3d)g/√5Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° - 90° side on a horizontal table. A cubical block of mass m rests on 60° - 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g/√2c)g/√3d)g/√5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° - 90° side on a horizontal table. A cubical block of mass m rests on 60° - 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g/√2c)g/√3d)g/√5Correct answer is option 'C'. Can you explain this answer?.
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