Answer:
Option B,C
Explanation:
For Vernier Callipers
$1MSD=\frac{1}{8}cm$
5 VSD=4 MSD
$\therefore$ $1VSD=\frac{4}{5}MSD$
= $\frac{4}{5}\times\frac{1}{8}=\frac{1}{10}cm$
Least count of Vernier callipers
= 1MSD-1VSD
= $\frac{1}{8}cm-\frac{1}{10}cm=0.025 cm$
(a) and (b)
Pitch of screw gauge
= $2\times0.025 =0.05 cm$
Least count of screw gauge
$=\frac{0.05}{100}cm=0.005mm$
(c) and (d)
Least count of linear scale of screw gauge=0.05
Pitch= $0.05\times 2=0.1cm$
Least count of screw gauge
$=\frac{0.1}{100}cm=0.01 mm$