QJEEAdv04Ph6586: JEE Advanced 2004 Physics - Class 11 JEEAdv04Ph6586 Solution: A screw gauge having 100 equal divisions and a pitch of length…

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 = $$\frac{p}{N}=\frac{1}{100}$$ = 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 $$\times$$ LC

= 1 + 47 $$\times$$ 0.01 = 1.47 mm = 0.147 cm.

The curved surface area is S = 2$$\pi$$rL. That is,

$$S=2\pi \left(\frac{d}{2}\right)L$$

$$S=\pi dL=\pi \left(\frac{1.47}{10}\right)5.6=$$ 2.5848 cm² = 2.6 cm²

which is corrected to two significant digits.