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EEJ MAIN Physics QUESTION #7034
Question 1
A student records the length of a rod as $3.50 \text{ cm}$. Identify the instrument used for this measurement based on its precision.
  • A screw gauge with a pitch of $1 \text{ mm}$ and $100$ circular scale divisions
  • A screw gauge with a pitch of $1 \text{ mm}$ and $50$ circular scale divisions
  • A standard metre scale
  • A vernier calliper where $10$ divisions of the vernier scale coincide with $9$ main scale divisions (main scale has $10 \text{ divisions per cm}$)✔️
Correct Answer Explanation

The recorded value $3.50 \text{ cm}$ has a precision of $0.01 \text{ cm}$. We must find an instrument with a least count (LC) of $0.01 \text{ cm}$.

  • Option A: $LC = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} = 0.001 \text{ cm}$
  • Option B: $LC = \frac{1 \text{ mm}}{50} = 0.02 \text{ mm} = 0.002 \text{ cm}$
  • Option C: $LC = 0.1 \text{ cm}$
  • Option D: $1 \text{ MSD} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm}$. Since $10 \text{ VSD} = 9 \text{ MSD}$, $1 \text{ VSD} = 0.9 \text{ MSD}$. $LC = 1 \text{ MSD} - 1 \text{ VSD} = 0.1 \text{ MSD} = 0.01 \text{ cm}$.

Thus, Option D matches the required precision.