A self-aligning ball bearing has a basic dynamic load rating (C10, for 106 revolutions) of 35 kN. If the equivalent radial load on the bearing is 45 kN, the expected life (in 10⁶ revolutions) is

The correct option is below 0.5.

Step-by-step Explanation:

To determine the expected life of a ball bearing, we use the standard load-life relationship equation:

$$L={\left(\frac{C}{P}\right)}^p$$

where:
- $L$ is the expected rating life in million revolutions (${10}^6$ revolutions).
- $C$ is the basic dynamic load rating.
- $P$ is the equivalent dynamic radial load.
- $p$ is the load-life exponent, which is equal to 3 for ball bearings (and 10/3 for roller bearings).

Given Data:
- Basic dynamic load rating, $C=35\text{kN}$
- Equivalent radial load, $P=45\text{kN}$
- For a self-aligning ball bearing, the exponent is $p=3$.

Calculation:
Substitute the given values into the load-life formula:

$$L={\left(\frac{35}{45}\right)}^3$$

Simplify the fraction inside the parentheses:

$$\frac{35}{45}=\frac{7}{9}\approx 0.7778$$

Now, raise this ratio to the power of 3:

$$L\approx {\left(0.7778\right)}^3\approx 0.47$$

The calculated expected life is approximately $0.47\times {10}^6$ revolutions. Since 0.47 is less than 0.5, this value falls in the range of below 0.5.