in an experiment the angles are required to be using an instrument, `29` divisions of the main scale exactly coincide with the `30` divisions of the vernier scale. If the sallest division of the main scale is half- a degree `(= 0.5^(@)`, then the least count of the instrument is :

Options

A. one minute
B. half minute
C. one degree
D. half degree

Answer

To find the least count of the instrument, we can follow these steps:

### Step 1: Understand the relationship between main scale divisions (MSD) and vernier scale divisions (VSD).
According to the problem, 29 divisions of the main scale coincide with 30 divisions of the vernier scale. This can be expressed mathematically as:
\[ 30 \, \text{VSD} = 29 \, \text{MSD} \]

### Step 2: Calculate the value of one vernier scale division (VSD).
From the above relationship, we can find the value of one VSD:
\[ 1 \, \text{VSD} = \frac{29}{30} \, \text{MSD} \]

### Step 3: Calculate the least count (LC).
The least count is defined as:
\[ \text{Least Count} = 1 \, \text{MSD} - 1 \, \text{VSD} \]
Substituting the value of VSD from Step 2:
\[ \text{Least Count} = 1 \, \text{MSD} - \left(\frac{29}{30} \, \text{MSD}\right) \]
\[ \text{Least Count} = 1 \, \text{MSD} - \frac{29}{30} \, \text{MSD} = \left(1 - \frac{29}{30}\right) \, \text{MSD} = \frac{1}{30} \, \text{MSD} \]

### Step 4: Substitute the value of 1 MSD.
The problem states that the smallest division of the main scale is half a degree (0.5°):
\[ 1 \, \text{MSD} = 0.5° \]
Now substituting this value into the least count formula:
\[ \text{Least Count} = \frac{1}{30} \times 0.5° = \frac{0.5°}{30} = \frac{0.5}{30}° = \frac{1}{60}° \]

### Final Answer:
Thus, the least count of the instrument is:
\[ \text{Least Count} = \frac{1}{60}° \text{ or } 1 \text{ minute} \]

---