The number of circular divisions on the shown screw gauge is 50. It moves 2 mm on the main scale for two complete rotations. The main scale reading is 5. The diameter of the ball is :
PLEASE ANSWER WITH A COMPLETE SOLUTION.
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Answered by
1
Answer:
As shown in first figure the zero error of the screw gauge is:
ZE=
50
5×0.5
=0.05mm
Least count of circular scale =
50
0.5
=0.01mm
Circular scale reading =25×0.01=0.25mm
Main scale reading =2×0.5=1mm
Actual measurement is =MSR+CSR − ZE =1+0.25−0.05=1.20mm
Answered by
0
Answer:
As shown in first figure the zero error of the screw gauge is:
ZE= 5×0.5/50 = 0.05mm
Least count of circular scale = 0.5 / 50 = 0.01mm
Circular scale reading =25×0.01=0.25mm
Main scale reading =2×0.5=1mm
Actual measurement is =MSR+CSR − ZE =1+0.25−0.05=1.20mm
so , 1.20 is the correct ans
Explanation:
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