Topic: Machine Design
The life of a ball bearing at a load of 10 kN is 8000 hours. If the load is increased to 20 kN, keeping all other conditions the same, then its life in hours is
- A. 4000
- B. 2000
- C. 1000
- D. 500
Answer: C. 1000
Explanation: The expected life of a bearing is inversely proportional to the cube of the applied load for ball bearings, governed by the equation $L = (C/P)^3$. Therefore, the ratio of life spans is $\frac{L_1}{L_2} = \left(\frac{P_2}{P_1}\right)^3$. Given $P_1 = 10$ kN, $L_1 = 8000$ hrs, and the new load $P_2 = 20$ kN, we substitute to get $\frac{8000}{L_2} = \left(\frac{20}{10}\right)^3 = 8$. Solving for $L_2$ gives $8000 / 8 = 1000$ hours.