A screw gauge has a negative error of 8 divisions. While measuring the diameter of a wire, the reading on the main scale is 3 divisions and the 24th circular scale division coincides with the base line. If the number of divisions on the main scale are 20 to a centimeter and the circular scale has 50 divisions, then the correct diameter of the wire is

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Solution

Given $1 M . S . D = \frac{1}{20} c m = 0.05 c m & 50 C . S . D = 1 M . S . D \Rightarrow 1 C . S . D = \frac{1}{50} M . S . D = 0.001 c m$
Reading = 3 M.S.D + (24+8) C.S.D $= (3 \times 0.05 + 32 \times 0.001) c m = 0.182 c m$

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Given 1 M.S.D =120 cm = 0.05 cm& 50 C.S.D = 1 M.S.D⇒1 C.S.D = 150 M.S.D = 0.001 cm Reading = 3 M.S.D + (24+8) C.S.D =(3×0.05 + 32×0.001) cm=0.182 cm

Given $1 M . S . D = \frac{1}{20} c m = 0.05 c m & 50 C . S . D = 1 M . S . D \Rightarrow 1 C . S . D = \frac{1}{50} M . S . D = 0.001 c m$
Reading = 3 M.S.D + (24+8) C.S.D $= (3 \times 0.05 + 32 \times 0.001) c m = 0.182 c m$