Q.

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 n5 th division of Vernier-Scale exactly coincides with a division of main-scale. Neglect any zero error.Now choose the correct option(s)

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a

The value of n is 40

b

The vernier constant of Vernier scale is 0.005 cm

c

The length of Pencil is 8.44 cm

d

The length of Pencil is 8.48 cm

answer is B.

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Detailed Solution

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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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 n5 th division of Vernier-Scale exactly coincides with a division of main-scale. Neglect any zero error.Now choose the correct option(s)