Case 1: A spherometer has a least count of $0.005 m m$ and its head scale is divided into $200$ equal divisions. Let the distance between consecutive threads on the spherometer screw be $d_{1}$ mm.
Case 2: A vernier calipers has each division on its main scale to be $1 m m$ . On vernier scale, $99 m m$ is divided into $100$ division. Let the vernier constant be $d_{2}$ mm.
Case 3: The pitch of a screw gauge is $1 m m$ and there are $100$ divisions on its circular scale. When no article is placed in between its jaws, the zero of the circular scale coincides with the reference line. When a steel wire is placed between the jaws, two main scale division are clearly visible and $67$ divisions on the circular scale are observed. Let the diameter of the wire be $d_{3}$ mm.
The values of $d_{1}$ , $d_{2}$ and $d_{3}$ are:

d_{1} = 1, d_{2} = 1.99, d_{3} = 0.01

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d_{1} = 0.01, d_{2} = 9.9, d_{3} = 1

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d_{1} = 1, d_{2} = 0.01, d_{3} = 2.67

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d_{1} = 0.01, d_{2} = 2.67, d_{3} = 0.01

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Solution

The correct option is D $d_{1} = 1, d_{2} = 0.01, d_{3} = 2.67$
For $d_{1}$ :
$d_{1}$ $=$ Pitch $=$ Distance between consecutive threads
$=$ (least count) $\times$ (total number of division on head scale)
$=$ (0.005 mm) (200)
$=$ 1 mm
For $d_{2}$ :
$d_{2} =$ Vernier constant
$= 1 m m - (\frac{99}{100}) m m$
$=$ 0.01 mm
For $d_{3}$ :
$p = 1 m m$
$N = 100$
Least count $= \frac{p}{N} = \frac{1 m m}{100} = 0.01 m m$
Main scale reading $= M S R = 2 \times (1 m m) = 2 m m$
Circular scale reading $= C S R = 67 \times (0.01) = 0.67 m m$
$d_{3} = M S R + C S R$
$= 2 + 0.67$
$= 2.67 m m$

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