Question

A hole of $25\text{H}_7$ is to be checked. IT7 tolerance = $0.021\text{ mm}$. Hole limits are:

• A. 25.000 -- 25.021 mm
• B. 24.979 -- 25.000 mm
• C. 25.021 -- 25.042 mm
• D. 24.979 -- 25.021 mm

Hint:

Uppercase letters refer to holes, while lowercase letters refer to shafts.
For any 'H' hole, the lower deviation is always $0$, while for any 'h' shaft, the upper deviation is always $0$. This simplifies limits calculations instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This metrology question requires us to determine the upper and lower limits of a hole designated as $25\text{H}_7$ using the standard ISO system of limits and fits.

Step 2: Key Formula or Approach:
For a hole with fundamental deviation letter 'H':
- The fundamental deviation (lower deviation, $EI$) is exactly zero.
- This means the lower limit of the hole is equal to the basic size:
\[ \text{Lower Limit} = \text{Basic Size} \]
The upper limit of the hole is calculated as:
\[ \text{Upper Limit} = \text{Basic Size} + \text{Tolerance} \]

Step 3: Detailed Explanation:

• Identify the given parameters:
- Basic size of the hole = $25.000 \text{ mm}$
- Tolerance grade IT7 value = $0.021 \text{ mm}$

• Apply the rules for the 'H' letter designation:
- Since it is an 'H' hole, the fundamental deviation is zero.
- Therefore, the minimum limit of the hole is:
\[ \text{Lower Limit} = 25.000 \text{ mm} \]

• Compute the maximum limit of the hole by adding the tolerance:
\[ \text{Upper Limit} = 25.000 \text{ mm} + 0.021 \text{ mm} = 25.021 \text{ mm} \]

• The limits of size for the hole are therefore $25.000 \text{ mm}$ (lower limit) and $25.021 \text{ mm}$ (upper limit).

Step 4: Final Answer:
The hole limits are 25.000 -- 25.021 mm.