Question

A theodolite circle is divided into degrees and half degrees.If the length of 59 main scale divisions is taken as length of vernier and divided into 60 parts the least count of vernier,in seconds is

Answer 1

Answer:

The least count of the theodolite is 30"

Explanation:

Since 59 main scale divisions are equal to the 60 divisions on vernier scale

Therefore, the vernier is a direct vernier

Here $n=60$

Value of one smallest division on the main scale $=0.5^\circ$

Therefore, the least count of vernier is given by

$\text{LC}=\frac{s}{n}$

$\implies \text{LC}=\frac{0.5^\circ}{60}$

$\implies \text{LC}=\frac{0.5\times 60'\times 60"}{60}$

$\implies \text{LC}=30"$

Therefore, the least count of the theodolite is 30"

Hope this answer is helpful.