First slide

Measurement

Question

In a screw gauge the zero of main scale coincides with fifth division of circular scale, the circular scale contains 50divisions Now a ball placed between screws, it takes four 0.5mm on main scale in one rotation and it takes four rotatory. To measure the ball then find diameter of the ball is (Index line coincides with 25line on circular scale)

Difficult

A

1.1 mm
B

4.4 mm
C

2.2 mm
D

3.3 mm

Solution

Error is $+ 5 = c o r r e c t i o n = - 5$
Least court $(L . C) = \frac{0.5}{50} m m = 0.01 m m$
Diameter of the ball $= M . S . R \times n o . r o t a t r y + C H S R \times L . C$
$= 4 \times 0.5 + (25 - 5) \times 0.01 = 2 + 20 \times 0.01 = 2.2 m m$

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