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

Units and Measurements PYQ — JEE Main 2009 · Shift 0 · Q76 with Solution

View written solutionFree

Correct answer: A

Given data
- $29$ divisions of the main scale coincide with $30$ divisions of the vernier scale.
- Value of $1$ main scale division (MSD) $= 0.5^\circ$ .
Find the value of 1 vernier scale division (VSD)

Since $30\,\text{VSD} = 29\,\text{MSD}$

therefore, $1\,\text{VSD} = \frac{29}{30}\,\text{MSD}$
Least count of a vernier instrument

For a direct vernier, $\text{Least Count} = 1\,\text{MSD} - 1\,\text{VSD}$

Substitute $1\,\text{VSD} = \frac{29}{30}\,\text{MSD}$ : $\text{LC} = 1\,\text{MSD} - \frac{29}{30}\,\text{MSD}$ $\text{LC} = \frac{1}{30}\,\text{MSD}$
Now use $1\,\text{MSD} = 0.5^\circ$

$\text{LC} = \frac{1}{30} \times 0.5^\circ = \frac{0.5}{30}^\circ = \frac{1}{60}^\circ$
Convert degree into minutes

We know, $1^\circ = 60'$

Hence, $\frac{1}{60}^\circ = 1'$
Evaluate options
- A: one minute $\rightarrow$ correct
- B: half minute $\rightarrow$ incorrect
- C: one degree $\rightarrow$ incorrect
- D: half degree $\rightarrow$ incorrect

Therefore, the least count of the instrument is one minute.