JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Units & Measurement

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The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between

5.10 c m

and

5.15 c m

of the main scale. The Vernier scale has 50 divisions equivalent to

2.45 c m

. The

2 4^{th}

division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

(A) $5.112 c m$
(B) $5.124 c m$ ✓ Correct
(C) $5.136 c m$
(D) $5.148 c m$

Answer: (B)

Solution

The least count of the Vernier callipers is calculated as follows:

1 MSD = 0.05 cm

1 VSD = 2.45/50 = 0.049 cm

Least count = 1 MSD - 1 VSD = 0.05 - 0.049 = 0.001 cm

The reading is: Main scale reading + (Vernier coinciding division × Least count)

Diameter = 5.10 + (24 × 0.001) = 5.10 + 0.024 = 5.124 cm

Hence the answer is (B).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Units & Measurement

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