JEE Advance - Physics (2004 - No. 1)
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 47th circular division coincides
with the main scale. Find the curved surface area of wire in cm2 to appropriate significant figures. ( use $$\pi = {{22} \over 7}$$ )
2.5 cm2
2.58 cm2
2.5848 cm2
2.6 cm2
2.7 cm2
Explanation
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 $$\times$$ LC
= 1 + 47 $$\times$$ 0.01 = 1.47 mm = 0.147 cm.
The curved surface area is S = 2$$\pi$$rL. That is,
$$S = 2\pi \left( {{d \over 2}} \right)L$$
$$S = \pi dL = \pi \left( {{{1.47} \over {10}}} \right)5.6 = $$ 2.5848 cm2 = 2.6 cm2
which is corrected to two significant digits.
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,
$$S = 2\pi \left( {{d \over 2}} \right)L$$
$$S = \pi dL = \pi \left( {{{1.47} \over {10}}} \right)5.6 = $$ 2.5848 cm2 = 2.6 cm2
which is corrected to two significant digits.
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