Two full turns of the circular scale of screw gauge cover a distance of $1\,mm$ on scale. The total number of divisions on circular scale is $50$. Further, it is found that screw gauge has a zero error of $+0.03\,mm$. While measuring the diameter of a thin wire a student notes the main scale reading of $3\,mm$ and the number of circular scale division in line, with the main scale is $35$. The diameter of the wire is .......... $mm$
$3.32$
$3.73$
$3.67$
$3.38$
The main scale of a vernier calliper has $n$ divisions/ $\mathrm{cm}$. $n$ divisions of the vernler scale coincide with $(\mathrm{n}-1)$ divisions of maln scale. The least count of the vernler calliper is,
If in a Vernier callipers $10 \,VSD$ coincides with $8 \,MSD$, then the least count of Vernier calliper is ............ $m$ [given $1 \,MSD =1 \,mm ]$
The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5\, mm$, calculate the minimum inaccuracy in the measurement of distance.
Answer the following :
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
$(b)$ A screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only ?