Answer:
The value of minimum possible radius of the specimen without fracture
= 4.23 mm
Step-by-step explanation:
Given data
Applied load F = 500 N
Flexural strength = 105 M Pa
Separation between load points = 50 mm
The minimum possible radius of the specimen without fracture is given by
![r_} = \sqrt[3]{((500) (50))/((105) (3.14) ) }](https://img.qammunity.org/2021/formulas/engineering/college/5xpctqgtwepqift8nj218yzrodw4blt5qs.png)
= 4.23 mm
This is the value of minimum possible radius of the specimen without fracture.