While measuring length of an object, it was observed that the zero of the Vernier lies between $1.4$ and $1.5$ of the main scale and the fifth Vernier division coincides with a main scale division. If the length of the object measured is $l$ , then the value of $(l - 1.4)$ in terms of the least count $C$ of the instrument is

C

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1.45 C

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4 C

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5 C

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Solution

The correct option is D $5 C$
Let $x = l - 1.4$
$l = x + 1.4$
Then as zero of the Vernier lies between $1.4$ and $1.5$ of the scale and the fifth Vernier division coincides with a main scale division
$1.4 + x + 5 V = 1.4 + 5 S,$
where $S$ denotes the length of one main scale division and
$V$ denotes the length of one vernier scale division.
Hence, $x = 5 S - 5 V = 5 (1 S - I V) = 5 C$ . $(∵ C = 1 S - 1 V)$

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