Question 309: Units , Dimensions and Error Analysis

Question

A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the diameter of a wire of length 5.6 cm. The main scale reading is 1mm and 47
th
circular division coincides
with the main scale. Find the curved surface area of wire in cm
2
to appropriate significant figures. ( use $$\pi  = {{22} \over 7}$$ )

Accepted Answer

The distance moved on the linear scale when circular scale makes one complete rotation is p = 1 mm (pitch). The number of divisions on the circular scale is N = 100. Thus, one division on the circular scale is LC = $${p \over N} = {1 \over {100}}$$ = 0.01 mm. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR) is 47. Thus, the diameter of the wire is
d = LSR + CSR $$	imes$$ LC
= 1 + 47 $$	imes$$ 0.01 = 1.47 mm = 0.147 cm.
The curved surface area is S = 2$$\pi$$rL. That is,
$$S = 2\pi \left( {{d \over 2}} ight)L$$
$$S = \pi dL = \pi \left( {{{1.47} \over {10}}} ight)5.6 = $$ 2.5848 cm
2
= 2.6 cm
2
which is corrected to two significant digits.

Units , Dimensions and Error Analysis

Integer / Numeric Answer Type