Topic: Strength of Materials
Three metal balls are suspended by three wires (a, b & c) of equal length arranged in sequence as shown in figure. The masses of the balls, starting at the top are 2 kg, 4kg and 3 kg. The wire diameters in the same order are 2 mm, 1.2 mm & 1 mm respectively. What is the correct order of increasing stresses in the wires?
- A. a, c, b
- B. b, c, a
- C. a, b, c
- D. All are having same stress
Answer: A. a, c, b
Explanation: Stress = Force / Area. Area is proportional to diameter squared (d²). For wire a: Force supports all three masses = 9kg, stress is proportional to 9 / (2^2) = 2.25. For wire b: Force supports two bottom masses = 7kg, stress is proportional to 7 / (1.2^2) = 4.86. For wire c: Force supports bottom mass = 3kg, stress is proportional to 3 / (1^2) = 3. Order of increasing stress is 2.25 (a) < 3.0 (c) < 4.86 (b).