General Physics Ques 23
- A screw gauge with a pitch of $0.5 \mathrm{~mm}$ and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45 th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet, if the main scale reading is $0.5 \mathrm{~mm}$ and the 25 th division coincides with the main scale line?
(2016 Main)
(a) $0.75 \mathrm{~mm}$
(b) $0.80 \mathrm{~mm}$
(c) $0.70 \mathrm{~mm}$
(d) $0.50 \mathrm{~mm}$
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Answer:
Correct Answer: 23.( b )
Solution:
- Least count
$ \quad=\frac{\text { pitch }}{\text { number of divisions on circular scale }}=\frac{0.5 \mathrm{~mm}}{50} $
$ \therefore \quad L C =0.01 $
$ \text { Negative zero error }=-5 \times L C=-0.005 \mathrm{~mm} $
$ \text { Measured value }=\text { main scale reading }+ \text { screw gauge reading } $
$\text { zero error }$
$ \quad=0.5 \mathrm{~mm}+{25 \times 0.01-(-0.05)} \mathrm{mm}$
= 0.8mm
BETA
