Question

The pitch of a screw gauge is 1 mm and there are 100 division on the circular scale. A students measures a diameter of a wire using this screw gauge and he gets main scale reading as 5 mm and circular divisions as 25. If the screw gauge has positive zero error $0.03mm$, the correct value of diameter will be

• A. $5.28$ mm
• B. $5.25$ mm
• C. $5.20$ mm
• D. $5.22$ mm

Answer 1

The correct option is D $5.22$ mm

The least count of the screw gauge $LC=$ pitch/ total number of circular divisions $=1/100=0.01mm$.
The final reading or diameter of the wire $D=MSR+(CSR\times LC)=5+(25\times 0.01)=5.25mm$
As the screw gauge has positive zero error so the zero error will be subtracted from final result.
Thus, the correct vale of diameter $=5.25-0.03=5.22mm$