QJEEAdv15Ph84a0: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph84a0 Solution: Consider a Vernier callipers in which each 1 cm on the main…

In given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = ${1 \over 8}$ = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.8 $\times$ 0.125 = 0.1 cm. The least count of the Vernier callipers is given by

LC = 1 MSD $-$ 1 VSD = 0.125 $-$ 0.1 = 0.025 cm.

In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation. Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get p = 2l. The least count of the screw gauge is defined as ratio of the pitch to the number of divisions on the circular scale (n) i.e.,

$LC' = {p \over n} = {{2l} \over {100}} = {l \over {50}}$ ....... (1)

If p = 2 LC = 2(0.025) = 0.05 cm, then $l = {p \over 2} = 0.025$ cm. Substitute l in equation (1) to get the least count of the screw gauge

$LC' = {{0.025} \over {50}} = 5 \times {10^{ - 4}}$ cm = 0.005 mm.

If l = 2 LC = 2(0.025) = 0.05 cm then equation (1) gives

$LC' = {{0.05} \over {50}} = 1 \times {10^{ - 3}}$ cm = 0.01 mm.