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

Question

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 (A) 1.64 mm (B) 0.82 mm (C) 1.80 mm (D) 0.90 mm

Tardigrade Admin · Accepted Answer

Least count =(1 mm /100)=0.01 mm zero error =+8 × LC =+0.08 mm True reading (Diameter) =(1 mm +72 × LC )- text (Zero error) =(1 mm +72 × 0.01 mm ) -0.08 mm =1.72 mm -0.08 mm therefore radius =(1.64/2) =0.82 mm

$1.64 mm$

$0.82 mm$

$1.80 mm$

$0.90 mm$

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

1.64mm

0.82mm

1.80mm

0.90mm

Least count =1001mm​=0.01mm zero error =+8×LC=+0.08mm True reading (Diameter) =(1mm+72×LC)− (Zero error) =(1mm+72×0.01mm)−0.08mm =1.72mm−0.08mm therefore radius =21.64​ =0.82mm

$1.64 mm$

$0.82 mm$

$1.80 mm$

$0.90 mm$

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