1.

Thedivisionson themainscaleof a screw gaugeare1 mmapartand thescrewof thespindleadvances by 5mainscaledivisions when thespindleis given5 completerotations. Howmanydivisionsare tobeprovided on thecircular scalefor theleastcountof theinstrument is tobe 1 mu m ? Whatchanges arerequied of thenumberif divisionscan beonly500for thesame leastcount?

Answer»

Solution : Leastcount = `(1 MSD)/(N)`
`=(0.5 mm)/(10) =0.05 mm`
Diameter =x +1Y
9.75 mm =(x+0.05y) mm
=x+ 0.05 y
Sincey `lt ` 10 and0.05y `lt`0.5. 9.75can be
written as 9.5 + 0.25
(9.5 mm) +(0.25 mm) = (xxmm )+(0.05 y mm)
SINCE 1 MSD=0.5mm
The MSR`=(9.5)/(0.5) =10 " and " 0.05 y0.25`
(b)9.75 =(x+ly) -0.35
=9.75 +0.35
=10.10
=10+ 0.1
x=10 =MSR`=(10)/(0.5) =20`
0.05 =0.1
`y= (0.1)/(0.05) =2 (=VCD)`
MSR =20 ,VCD =2


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