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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? |
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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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