Question

While measuring the diameter of a lead-shot using a screw gauge , the reading on the pitch scale is found to be 7.5 mm and that on the head scale is 48 . If the least count is 0.01 mm and zero error is +0 l. 05 mm , find the diameter of the lead-shot .​

Answer 1

Answer:

Here is your answer

P. S. R = 7.5 mm

H. S. R = 48

L. C = 0.01 mm

Error = +0.05 mm , correction = -0. 05 mm

Diameter of lead-shot = P. S. R + H. S. R × L. C + Correction

=> 7.5 mm + 48 × 0.01 mm - 0.05 mm

=> 7.98 mm - 0.05 mm

=> 7.93 mm

The diameter of the lead shot is 7.93 mm.

Answer 2

Answer:

P. S. R = 7.5 mm

H. S. R = 48

L. C = 0.01 mm

Error = +0.05 mm

correction = -0. 05 mm

Diameter of lead-shot = P. S. R + H. S. R × L. C + Correction

→ 7.5 mm + 48 × 0.01 mm - 0.05 mm

→ 7.98 mm - 0.05 mm

→ 7.93 mm

Explanation:

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