The diagrams show readings of a screw gauge. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The least count $(LC)$ of the screw gauge is calculated as: $LC = \frac{	ext{Pitch}}{	ext{Number of divisions on circular scale}} = \frac{0.5 \

Vedclass · Accepted Answer

$5.26$

Vedclass · Answer

(B) The least count $(LC)$ of the screw gauge is calculated as: $LC = \frac{	ext{Pitch}}{	ext{Number of divisions on circular scale}} = \frac{0.5 \ mm}{50} = 0.01 \ mm$.From figure $(i)$,the zero error is $+3$ divisions. Therefore,the zero error is $3 	imes 0.01 \ mm = +0.03 \ mm$.From figure $(ii)$,the main scale reading is $5.0 \ mm$ and the circular scale division coinciding with the reference line is $29$. The observed reading is $5.0 \ mm + (29 	imes 0.01 \ mm) = 5.29 \ mm$.The actual diameter is given by: $	ext{Diameter} = 	ext{Observed reading} - 	ext{Zero error} = 5.29 \ mm - 0.03 \ mm = 5.26 \ mm$.

Similar Questions

The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.

Which devices are used for the measurement of lengths of different orders?

$A$ travelling microscope has $20$ divisions per $cm$ on the main scale while its Vernier scale has total $50$ divisions and $25$ Vernier scale divisions are equal to $24$ main scale divisions. What is the least count of the travelling microscope in $cm$?

The main scale of a vernier calliper has $n$ divisions per $cm$. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. The least count of the vernier calliper is,

Vedclass Test Series

Exam Paper Generator

Online Exam Module