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  4. The pitch of the screw gauge is 1 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

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Q. The pitch of the screw gauge is $1 mm$ 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

JEE MainJEE Main 2021Physical World, Units and Measurements

Solution:

Least count $=\frac{1 \,mm }{100}=0.01 \,mm$
zero error $=+8 \times LC =+0.08 \,mm$
True reading (Diameter)
$=(1\, mm +72 \times LC )-\text { (Zero error) } $
$=(1 mm +72 \times 0.01 mm ) -0.08 \,mm $
$=1.72 \,mm -0.08 \,mm $
therefore radius $=\frac{1.64}{2}$
$=0.82\,mm$