3 Chegg 6 Complessd em 01:29:07 Lo Ls os - Question 4 © E tL Asecments ome 30 2 1s regi ch are. Cause oe ASTM gram sae aero mam 2 — : B 5 » a. oe FE EB 3 3 i
Question:
3 Chegg 6 Complessd em 01:29:07 Lo
Ls os
- Question 4 © E
tL Asecments ome 30 2 1s regi ch are. Cause oe ASTM gram sae aero mam
2 —
: B
5 » a.
oe FE
EB 3
3
i
This is AI generated Response.
Asked by: Veera manikanta
Created at: 2025-08-23 12:56:13
Asked by: Veera manikanta
Created at: 2025-08-23 12:56:13
Final Answer:
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Let's break down the problem:
**Question:**
A specimen is observed under a microscope with a magnification of 300. Over a 3 in² region, the specimen has about 60 grains. Calculate the ASTM grain size number for the material.
---
### ASTM Grain Size Number Formula
The ASTM grain size number (n) is given by:
\[
n = 1 + \log_2 N
\]
Where \(N\) is the number of grains per square inch at 100× magnification.
Given:
- Area = 3 in²
- Number of grains in that area = 60
- Magnification = 300×
First, find the number of grains per square inch at 300×:
\[
N_{300} = \frac{60}{3} = 20 \text{ grains/in}^2 \text{ at 300×}
\]
Now convert this to the equivalent number at 100×:
\[
N_{100} = N_{300} \left(\frac{300}{100}\right)^2 = 20 \times 9 = 180 \text{ grains/in}^2 \text{ at 100×}
\]
Now, plug into the ASTM formula:
\[
n = 1 + \log_2(180)
\]
\[
\log_2(180) = \frac{\log_{10}(180)}{\log_{10}(2)} \approx \frac{2.2553}{0.3010} \approx 7.495
\]
\[
n = 1 + 7.495 = 8.495
\]
---
### Closest Answer
From the options, **7.5** is the closest and most reasonable match.
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**Final Answer:**
**7.5**
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