Identify the instrument having minimum least count out of the following options.

Question

Identify the instrument having minimum least count out of the following options. (A) A vernier calipers with 20 divisions on the vernier scale coinciding with 19 main scale divisions. (B) A screw gauge of pitch 1 mm and 100 divisions on the circular scale. (C) A spherometer of pitch 0.1 mm and 100 divisions on the circular scale. (D) An optical instrument that can measure length to within a wavelength to light.

Tardigrade Admin · Accepted Answer

(a) Least count of vernier calipers
= 1MSD – 1 VSD =1 MSD -

\frac{19}{20}

MSD
=

\frac{1}{20} MS D = \frac{1}{20} mm = \frac{1}{200} c m = 0.005 c m

(∵ 1 MS D = 1 mm)

(b) Least count of screw gauge=

\frac{P i t c h}{N o . o f d i v i s i o n s o n c i rc u l a r sc a l e}

=

\frac{1}{100} mm = \frac{1}{1000} c m = 0.001 c m

(c) Least count of spherometer =

\frac{P i t c h}{N o . o f d i v i s i o n s o n c i r u l a r sc a l e}

=

\frac{0.1 mm}{100} = \frac{1}{1000} mm = 0.0001 c m

(d) Wavelength of light,

λ \approx 1 0^{- 5} c m = 0.00001 c m

Clearly the optical instrument is the most precise.

Q. Identify the instrument having minimum least count out of the following options.

A vernier calipers with 20 divisions on the vernier scale coinciding with 19 main scale divisions.

A screw gauge of pitch 1 mm and 100 divisions on the circular scale.

A spherometer of pitch 0.1 mm and 100 divisions on the circular scale.

An optical instrument that can measure length to within a wavelength to light.