Asked in NUMS 2019 (cancelled sitting) 2019Moderate

A machine worker placed a cylinder with a diameter of 18 cm between the plates of a hydraulic press. If he applies a 4.25 x 10^5N force to the cylinder, the stress on the end of the cylinder due to the applied force is A x 107Pa. What is the value of A?

Correct answer: D. 1.67

  • A. 1
  • B. 0.5
  • C. 1.3
  • D. 1.67

Explanation

The area of the end of the cylinder is given by:A = (π/4)d2where d is the diameter of the cylinder.Substituting the given values, we get:A = (π/4)(0.18 m)2A = 0.02546 m2The stress on the end of the cylinder is given by:Stress = Force/AreaSubstituting the given values, we get:Stress = (4.25 x 105 N)/0.02546 m2Stress = 1.67 x 107 PaTherefore, the value of A is 1.67, which is option D.

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