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Standard IX

Physics

Vernier Caliper

A vernier cal...

Question

A vernier calliper measures the diameter of rod as $1.24 c m$ . The vernier callipers has $10$ divisions which slide over the main scale of $1 M S D = 0.5 m m .$ Find the main scale reading and vernier coincidence.

V C D = 8

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V C D = 3

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V C D = 2

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V C D = 4

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Solution

The correct option is A $V C D = 8$
Diameter of rod $= 1.24 c m = 12.4 m m$
$N = 10$
$1 M S D = 0.5 m m$
$∴ L C = \frac{1 M S D}{N}$
$= \frac{0.5 m m}{10}$
$= 0.05 m m$
Diameter $= 12.4 m m$
$= 12 m m + 0.4 m m$
Diameter $= 12 m m + (8 \times 0.05 m m)$
But Diameter $= M S R + (V C D \times L C)$
$\Rightarrow M S R = 12 m m o r 1.2 c m$
$V C D = 8$

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A vernier calliper measures the diameter of rod as 1.24cm. The vernier callipers has 10 divisions which slide over the main scale of 1MSD=0.5mm. Find the main scale reading and vernier coincidence.

The correct option is A VCD=8Diameter of rod =1.24cm=12.4mmN=101MSD=0.5mm∴LC=1MSDN=0.5mm10=0.05mmDiameter =12.4mm=12mm+0.4mmDiameter =12mm+(8×0.05mm)But Diameter =MSR+(VCD×LC)⇒MSR=12mmor1.2cmVCD=8

A vernier calliper measures the diameter of rod as $1.24 c m$ . The vernier callipers has $10$ divisions which slide over the main scale of $1 M S D = 0.5 m m .$ Find the main scale reading and vernier coincidence.