A spring is placed between the jaws of screw gauge such that the spring is not all compressed. The main scale reads 2 division and circular scale reads 28 divisions. Now we turn the circular scale by `18^(@)` such that the spring is compessed. The circular scale has 200 divisions and the least count of the main scale is `1mm`. What is the force exerted by the spring on the jaws if the spring constnat is `100N//m`

Options

A. `4mN`
B. `3mN`
C. `5mN`
D. `7mN`

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the Given Information
- Main scale reading (MSR) = 2 divisions
- Circular scale reading (CSR) = 28 divisions
- Circular scale has 200 divisions
- Least count of the main scale = 1 mm
- Spring constant (k) = 100 N/m
- The circular scale is turned by 18 degrees.

### Step 2: Calculate the Least Count
The least count (LC) of the screw gauge can be calculated using the formula:
\[ \text{Least Count} = \frac{\text{Smallest division on the main scale}}{\text{Total number of divisions on the circular scale}} \]
Here, the smallest division on the main scale is 1 mm, and the total number of divisions on the circular scale is 200.

\[ \text{Least Count} = \frac{1 \text{ mm}}{200} = 0.005 \text{ mm} = 5 \times 10^{-3} \text{ mm} \]

### Step 3: Convert the Circular Scale Reading to mm
When the circular scale is turned by 18 degrees, the amount of compression (Δx) can be calculated as follows:
\[ \Delta x = \text{Least Count} \times \left(\frac{\text{Angle turned}}{360^\circ}\right) \]
Substituting the values:
\[ \Delta x = 0.005 \text{ mm} \times \left(\frac{18}{360}\right) = 0.005 \text{ mm} \times \frac{1}{20} = 0.00025 \text{ mm} \]
Now, convert this to meters:
\[ \Delta x = 0.00025 \text{ mm} \times \frac{1 \text{ m}}{1000 \text{ mm}} = 0.00000025 \text{ m} = 2.5 \times 10^{-4} \text{ m} \]

### Step 4: Calculate the Force Exerted by the Spring
The force exerted by the spring can be calculated using Hooke's Law:
\[ F = k \cdot \Delta x \]
Substituting the values:
\[ F = 100 \text{ N/m} \times 2.5 \times 10^{-4} \text{ m} = 0.025 \text{ N} \]

### Final Answer
The force exerted by the spring on the jaws is **0.025 N**.

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