Main scale of a vernier callipers reads 10mm in 10divisions.10divisions of verniers scale coincide with 9 divisions of main scale.When 2jaws of the callipers touch each other 5thdivision of vernier coincides with 9main scale divisions and zero of the vernier is to right of zero of main scale ,when a cylinder is tightly placed between two jaws,zero of the vernier scale lies slightly to left of 3.2cm and fourth vernier division coincides with a main scale division.Find diameter of cylinder

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$10 V S D = 9 M S D 1 V S D = \frac{9}{10} M S D = 0.9 M S D N o w l e a s t c o u n t = M S D - V S D = (1 - 0.9) m m = 0.1 m m f o r z e r o e r r o r = 5 \times l e a s t c o u n t = 5 \times 0.1 = 0.5 m m n o w d i m e t e r o f c y l i n d e r = {3.2 - 0.1} \times 10 (m m) + 4 \times 0.1 - 0.5} = + 30.9 m m R e g a r d s$

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Q. The main scale of a vernier callipers reads 10 mm in 10 divisions of Vernier scale coincide with 9 divisions of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly behind 3.2 cm and the fourth vernier division coincides with a main scale division. The diameter of the cylinder is :

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The vernier of a vernier scale is divided into 10 divisions which coincide with 9 divisions of the main scale, each main scale division being 0.5 mm. When the two jaws of the instrument are in contact with each other, the 4th division of the vernier scale coincides with a main scale division and the zero of the vernier lies to the right of the zero of the main scale. When a sphere is inserted between the jaws, the zero of vernier scale lies slightly to the right of 1.8 cm and the sixth vernier division coincides with a main division. The diameter of sphere will be

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Dear student 10 VSD = 9MSD1 VSD =910MSD =0.9MSDNow least count =MSD-VSD=(1-0.9)mm=0.1mmfor zero error =5×least count=5×0.1=0.5mmnow dimeter of cylinder={3.2-0.1}×10(mm)+4×0.1-0.5} =+30.9mmRegards