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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° 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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