Test pitch of screw gauge is 1mm and surplus scale
as 200 div. While measuring diameter of a wire
the main scale reads 2mm and 45th mark on circular
the main scale reads 35 div and 4th div of vernies coincide
with main scale. Find least count and radius of cylinder
Answers
Thus the least count = 0.01 mm and the surface area is 2.6 cm^2
Explanation:
The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm2 ) of the wire in appropriate number of significant figures.
Solution:
L.C. of the screw gauge = Pitch / Number of divisions on circular scale
L.C = 1 / 100 = 0.01 mm
Given: Main scale reading = 1 mm
Circular scale reading = 0.01 × 47 = 1.47 mm = 0.147 cm
Diameter of wire =Diameter of wire = Total reading =1.47 mm=0.147 cm
Curved surface area = π × dl = π × (0.29) (0.147) = 2.586 ≈ 2.6 cm^2
Thus the least count = 0.01 mm and the surface area is 2.6 cm^2