#### Answer

$d = 1.45 mm$

#### Work Step by Step

Given:
10-mm diameter Brinell hardness indicator produced an indentation of 2.50 mm in diameter in steel alloy when load of 1000 kg was used
Required:
diameter of indentation to yield hardness of 300 HB when 500-kg load is used
Solution:
Using equation for HB in Table 6.5:
$d = \sqrt D^{2} - [D - \frac{2P}{(HB)πD}]^{2} = \sqrt (10 mm)^{2} - [(10 mm) - \frac{2 (500 kg)}{(300)π(10 mm)}]^{2} = 1.45 mm$