Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once.
Find: (i) pitch of the screw gauge,
(ii) Least count of the screw gauge, and
(iii) The diameter of the wire.

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Solution

(i) Pitch = distance covered in one revolution = 1mm
(ii) Least count $=$ Pitch /total number of divisioins on circular head
$= 1 m m / 50 = 0.02 m m$
(iii) Diameter of the wire $= m a i n s c a l e r e a d i n g + c i r c u l a r s c a l e r e a d i n g$
$= 4.94 m m$

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Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once.Find: (i) pitch of the screw gauge,(ii) Least count of the screw gauge, and(iii) The diameter of the wire.

(i) Pitch = distance covered in one revolution = 1mm(ii) Least count =Pitch /total number of divisioins on circular head =1mm/50=0.02mm(iii) Diameter of the wire =mainscalereading+circularscalereading =4.94mm

Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once.
Find: (i) pitch of the screw gauge,
(ii) Least count of the screw gauge, and
(iii) The diameter of the wire.

(i) Pitch = distance covered in one revolution = 1mm
(ii) Least count $=$ Pitch /total number of divisioins on circular head
$= 1 m m / 50 = 0.02 m m$
(iii) Diameter of the wire $= m a i n s c a l e r e a d i n g + c i r c u l a r s c a l e r e a d i n g$
$= 4.94 m m$