JEE Physics 2023 Question with Solution

Given: Least count is $0.1 \, \text{mm}$. When the jaws are closed, the vernier zero lies to the right of the main scale zero and the **$6$**th vernier division coincides, so the zero error is positive. While measuring the bob, the vernier zero lies between $3 \, \text{cm}$ and $3.3 \, \text{cm}$, and the **$4$**th vernier division coincides.

Find: The diameter of the spherical bob.

From the given working,

$$\text{Zero error} = 6 \times 0.1 \, \text{mm} = 0.6 \, \text{mm}$$

This is a positive zero error.

The observed diameter is

$$\text{Diameter} = \text{Main Scale Reading} + \text{Vernier Scale Reading} \times \text{Least Count}$$ $$\text{Diameter} = 3.2 \, \text{cm} + 4 \times 0.1 \, \text{mm} = 3.2 \, \text{cm} + 0.4 \, \text{mm} = 3.24 \, \text{cm}$$

Now subtract the positive zero error:

$$\text{Actual diameter} = 3.24 \, \text{cm} - 0.06 \, \text{cm} = 3.18 \, \text{cm}$$

Therefore, the diameter of the bob is $3.18 \, \text{cm}$. The solution working gives $3.18 \, \text{cm}$, which corresponds to option C, although the provided solution incorrectly labels the correct option as D.