One main scale division of a vernier callipers is $a$ $cm$ and $n ^{\text {th }}$ division of the vernier scale coincide with $( n -1)^{\text {th }}$ division of the main scale. The least count of the callipers in $mm$ is
$\frac{10 na }{( n -1)}$
$\frac{10 a }{( n -1)}$
$\left(\frac{ n -1}{10 n }\right) a$
$\frac{10 a }{ n }$
Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
A specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale, there are $10$ equal divisions and they match with $11$ main scale divisions. Then, the least count of the Vernier calliper is ........... $mm$
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5 \ mm$ and there are $50$ divisions on the circular scale. The reading on the main scale is $2.5 \ mm$ and that on the circular scale is $20$ divisions. If the measured mass of the ball has a relative error of $2 \%$, the relative percentage error in the density is
Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0. 1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as
S.No. | $MS\;(cm)$ | $VS$ divisions |
$(1)$ | $0.5$ | $8$ |
$(2)$ | $0.5$ | $4$ |
$(3)$ | $0.5$ | $6$ |
If the zero error is $- 0.03\,cm,$ then mean corrected diameter is ........... $cm$