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$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1 )$ divisions of the vernier scale. If each division of main scale is $a$ units , then the least count of the instrument is
$a$
$\frac {a}{N}$
$\frac{N}{{N + 1}} \times a$
$\frac{a}{{N + 1}}$
Solution
$\begin{array}{l}
No\,of\,divisions\,on\,main\,scale\, = N\\
No\,of\,divisions\,on\,vernier\,scale\, = N + 1\\
Size\,of\,main\,scale\,divisions\, = a\\
Let\,size\,of\,vernier\,scale\,division\,be\,b\,then\,we\,have\\
aN = b\,\left( {N + 1} \right) \Rightarrow b = \frac{{aN}}{{N + 1}}\\
Least\,count\,is\,a – b = a – \frac{{aN}}{{N + 1}}\\
= a\left[ {\frac{{N + 1 – N}}{{N + 1}}} \right] = \frac{a}{{N + 1}}
\end{array}$
Similar Questions
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
| Measurement condition | Main scale reading | Circular scale reading |
| Two arms of gauge touching each other without wire | $0$ division | $4$ division |
| Attempt-$1$: With wire | $4$ division | $20$ division |
| Attempt-$2$: With wire | $4$ division | $16$ division |
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
