Find the least count of a vernier caliper whose 50 vernier scale divisions match with 49 main scale divisions?

Given: least count of main scale is 1 mm.

0.1 mm

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0.02 mm

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0.05 mm

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0.001 mm

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Solution

The correct option is B
0.02 mm

Since 50 vernier scale divisions match with 49 main scale divisions, vernier scale division is:
$1 V S D = \frac{49 \times 1 m m}{50}$

Using the formula for least count,
$L C = 1 M S D - 1 V S D$
$L C = 1 m m - \frac{49}{50} m m$
$L C = 0.02 m m$

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Find the least count of a vernier caliper whose 50 vernier scale divisions match with 49 main scale divisions?

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A Vernier caliper has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which matches with 16 main scale divisions. For this Vernier caliper, the least count is:

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Find the least count of a vernier caliper whose 50 vernier scale divisions match with 49 main scale divisions? Given: least count of main scale is 1 mm.

The correct option is B 0.02 mm Since 50 vernier scale divisions match with 49 main scale divisions, vernier scale division is: 1 VSD=49×1 mm50 Using the formula for least count, LC=1 MSD−1 VSD LC=1 mm−4950 mm LC=0.02 mm

Find the least count of a vernier caliper whose 50 vernier scale divisions match with 49 main scale divisions?

Given: least count of main scale is 1 mm.

The correct option is B
0.02 mm

Since 50 vernier scale divisions match with 49 main scale divisions, vernier scale division is:
$1 V S D = \frac{49 \times 1 m m}{50}$

Using the formula for least count,
$L C = 1 M S D - 1 V S D$
$L C = 1 m m - \frac{49}{50} m m$
$L C = 0.02 m m$