Question

The distance moved by the screw of a screw gauge is $2mm$ in four rotations and there are $50$ divisions on its cap. When nothing is put between its jaws, ${30}^{th}$ division of circular scale coincides with reference line, with zero of circular scale lying above the reference line. When a plate is placed between the jaws, main scale reads $2$ divisions and circular scale reads $20$ divisions. Thickness of plate is:

• A. $0.9mm$
• B. $1.2mm$
• C. $1.4mm$
• D. $1.5mm$

Answer 1

Answer: (A)

$0.9mm$

In one rotation, the scale moves $0.5mm$. Hence in two rotations, $0.5\times 2=1mm$.

Reading on circular scale is $\frac{0.5}{50}\times 20=0.2mm$.

Now, error is $-\frac{0.5}{50}\times 30=-0.3$.

Hence, correct reading is $1+0.2-0.3=0.9mm$