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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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answer is 5.

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Detailed Solution

Let x = $l$ – 1.4Then, 1.4 + x + 5V = 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 (1S – 1V) = 5 C.

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