Step-by-step explanation:
(a) The given data is as follows.
Load applied (P) = 1000 kg
Indentation produced (d) = 2.50 mm
BHI diameter (D) = 10 mm
Expression for Brinell Hardness is as follows.
HB =
Now, putting the given values into the above formula as follows.
HB =
=
=
= 200
Therefore, the Brinell HArdness is 200.
(b) The given data is as follows.
Brinell Hardness = 300
Load (P) = 500 kg
BHI diameter (D) = 10 mm
Indentation produced (d) = ?
d =
![\sqrt{(D^(2) - [D - (2P)/(HB) \pi D]^(2))}](https://img.qammunity.org/2020/formulas/chemistry/college/af9knu86cnsia4isth5jxutrzpwn8h6877.png)
=
![\sqrt{(10 mm)^(2) - [10 mm - (2 * 500 kg)/(300 * 3.14 * 10 mm)]^(2)}](https://img.qammunity.org/2020/formulas/chemistry/college/fzzxchez0nh382uqe5h9xy1soa8w1twigy.png)
= 4.46 mm
Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.