In a screw gauge, when the circular scale is given five complete rotations, it moves linearly by $2.5 \text{ mm}$. If the circular scale has $100$ divisions, the least count of the screw gauge is . . . . . . $\text{mm}$.
- A$1 \times 10^{-2}$
- B$1 \times 10^{-3}$
- C$5 \times 10^{-2}$
- D$5 \times 10^{-3}$
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Similar Questions
$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :
$A$ Vernier calipers has $1 \,mm$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is (in $\,mm$)
In a vernier callipers,each $cm$ on the main scale is divided into $20$ equal parts. If the $10^{th}$ vernier scale division coincides with the $9^{th}$ main scale division,then the value of the vernier constant will be $\dots \; \times 10^{-2} \; mm$.
If in a Vernier callipers $10 \,VSD$ coincides with $8 \,MSD$,then the least count of the Vernier calliper is ............ $m$ [given $1 \,MSD = 1 \,mm$].
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View SolutionThe least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.
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