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# The pitch of the screw gauge is 1mm and

Question:

The pitch of the screw gauge is $1 \mathrm{~mm}$ and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while $72^{\text {nd }}$ division on circular scale coincides with the reference line. The radius of the wire is

1. $1.64 \mathrm{~mm}$

2. $0.82 \mathrm{~mm}$

3. $1.80 \mathrm{~mm}$

4. $0.90 \mathrm{~mm}$

Correct Option: , 2

Solution:

Least count $=\frac{1 \mathrm{~mm}}{100}=0.01 \mathrm{~mm}$

zero error $=+8 \times \mathrm{LC}=+0.08 \mathrm{~mm}$

$=(1 \mathrm{~mm}+72 \times \mathrm{LC})-($ Zero error $)$

$=(1 \mathrm{~mm}+72 \times 0.01 \mathrm{~mm})-0.08 \mathrm{~mm}$

$=1.72 \mathrm{~mm}-0.08 \mathrm{~mm}$

$=1.64 \mathrm{~mm}$

therefore, radius $=\frac{1.64}{2}=0.82 \mathrm{~mm}$.