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°), then the least count of the instrument is

The correct answer is one minute.

To find the least count of the instrument, we can use the standard relation for a vernier scale:
$\text{Least Count (LC)}=1\text{ MSD}-1\text{ VSD}$
where MSD represents Main Scale Division and VSD represents Vernier Scale Division.

From the given problem, the smallest division of the main scale (1 MSD) is half-a-degree:
$1\text{ MSD}={0.5}^{\circ }$
Since $1^{\circ }=60\text{ minutes (}{60}^{\prime }\text{)}$, we can convert this to minutes:
$1\text{ MSD}=30\text{ minutes (}{30}^{\prime }\text{)}$

We are also given that 29 divisions of the main scale exactly coincide with 30 divisions of the vernier scale:
$30\text{ VSD}=29\text{ MSD}$
Solving for 1 VSD gives:
$1\text{ VSD}=\frac{29}{30}\text{ MSD}$

Substituting this relation into the least count formula:
$\text{LC}=1\text{ MSD}-\frac{29}{30}\text{ MSD}$

$\text{LC}=\left(1-\frac{29}{30}\right)\text{ MSD}$

$\text{LC}=\frac{1}{30}\text{ MSD}$

Now, substituting the value of 1 MSD in minutes (${30}^{\prime }$):
$\text{LC}=\frac{1}{30}\times {30}^{\prime }=1^{\prime }\text{ (one minute)}$

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