The Vernier of a Vernier scale is divided into 10 divisions which coincide with 9 divisions of the main scale, each main scale division being 0.5 mm. When the two jaws of the instrument are in contact with each other, the 4th division of the Vernier scale coincides with a main scale division and the zero of the Vernier lies to the right of the zero of the main scale. When a sphere is inserted between the jaws, the zero of Vernier scale lies slightly to the right of 1.8 cm and the sixth Vernier division coincides with a main division. The diameter of sphere will be (in cm).

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answer is 1.81.

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Detailed Solution

Least court $(L C) = 1 M S D - 1 V S D = 0.05 m m - \frac{9}{10} (0.05) m m = 0.005 c m$
$Reading of scale=Diameter=MSR+n (L e a s t c o u n t) - z e r o e r r o r$
$= [1.8 + 6 \times (0.005) - 4 \times (0.005)] c m$
$D = 1.81 c m$

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