General Physics Ques 12
- The pitch and the number of divisions, on the circular scale for a given screw gauge are $0.5 \mathrm{~mm}$ and 100 , respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale for a thin sheet are $5.5 \mathrm{~mm}$ and 48 respectively, the thickness of this sheet is
(2019 Main, 9 Jan II)
(a) $5.950 \mathrm{~mm}$
(b) $5.725 \mathrm{~mm}$
(c) $5.755 \mathrm{~mm}$
(d) $5.740 \mathrm{~mm}$
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Answer:
Correct Answer: 12.( c )
Solution:
- For a measuring device, the least count is the smallest value that can be measured by measuring instrument.
Least count $=\frac{\text { Minimum reading on main-scale }}{\text { Total divisions on the scale }}$
Here, screw gauge is used for measurement therefore,
$ \begin{aligned} \mathrm{LC} & =\frac{\text { Pitch }}{\text { number of division }} \\ \mathrm{LC} & =\frac{0.5}{100} \mathrm{~mm} \\ \mathrm{LC} & =5 \times 10^{-3} \mathrm{~mm} \end{aligned} $
According to question, the zero line of its circular scale lies 3 division below the mean line and the readings of main scale $=5.5 \mathrm{~mm}$
The reading of circular scale $=48$
then the actual value is given by actual value of thickness $(t)$ $=($ main scale reading $)+$ (circular scale reading + number of division below mean line) $\times$ LC
$ \begin{array}{ll} \Rightarrow & t=5.5 \mathrm{~mm}+(48+3) \times 5 \times 10^{-3} \mathrm{~mm} \\ \Rightarrow & t=5.755 \mathrm{~mm} \end{array} $
BETA
