A screw gauge of pitch $$0.5 \text{ mm}$$ is used to measure the diameter of uniform wire of length $$6.8 \text{ cm}$$, the main scale reading is $$1.5 \text{ mm}$$ and circular scale reading is $$7$$. The calculated curved surface area of wire to appropriate significant figures is [Screw gauge has $$50$$ divisions on the circular scale]

We are given a screw gauge with pitch = 0.5 mm and 50 divisions on the circular scale.

Step 1: Find the Least Count (LC)

$$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions}} = \dfrac{0.5}{50} = 0.01 \text{ mm}$$

Step 2: Find the diameter of the wire

$$d = \text{MSR} + \text{CSR} \times \text{LC} = 1.5 + 7 \times 0.01 = 1.57 \text{ mm}$$

Step 3: Calculate the curved surface area

The curved surface area of a cylinder is:

$$A = \pi d L$$

where $$d = 1.57 \text{ mm} = 0.157 \text{ cm}$$ and $$L = 6.8 \text{ cm}$$.

$$A = \pi \times 0.157 \times 6.8 = \pi \times 1.0676 = 3.3543 \text{ cm}^2$$

Step 4: Apply significant figures

The diameter (1.57 mm) has 3 significant figures, and the length (6.8 cm) has 2 significant figures. The result should be rounded to 2 significant figures:

$$A \approx 3.4 \text{ cm}^2$$

Therefore, the correct answer is Option B.