I am a vernier calipers. I have a task to measure the length of a pencil as shown in the figure. The vernier constant of my vernier scale is 0.05 times the value of one division of main-scale.

My Vernier Scale has n-divisions and coincides with (n – 4) divisions of main scale. While measuring the length of pencil, the $\frac{n}{5}$ th division of Vernier-Scale exactly coincides with a division of main-scale. Neglect any zero error.
Now choose the correct option(s)

Question

I am a vernier calipers. I have a task to measure the length of a pencil as shown in the figure. The vernier constant of my vernier scale is 0.05 times the value of one division of main-scale.

My Vernier Scale has n-divisions and coincides with (n – 4) divisions of main scale. While measuring the length of pencil, the $\frac{n}{5}$ th division of Vernier-Scale exactly coincides with a division of main-scale. Neglect any zero error.
Now choose the correct option(s)

Infinity Learn · Accepted Answer

L.C.$$=0.05$$ times $$1MSD=0.05$$×$$0.1cm=0.005$$  cm​$$=0.05mmLet$$ $$1$$  V.S.D.$$=x$$​∴L.C.$$=1M$$.S.D.−$$1$$  V.S.D.⇒    $$0.05mm=1mm$$−x⇒$$x=0.95$$  mm​Given : (n$$−$$4)×$$1mm=n$$×$$0.95$$​mm⇒   $$n=80$$ Length of pencil $$=8.4cm+$$[$$0.005cm$$×(805)]$$=8.48$$ cm

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