A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5 \mathrm{~mm}$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively:
Correct Option: 2
(2) Given : No. of division on circular scale of screw gauge
$=50$
Pitch $=0.5 \mathrm{~mm}$
Least count of screw gauge
$=\frac{\text { Pitch }}{\text { No. of division on circular scale }}$
$=\frac{0.5}{50} \mathrm{~mm}=1 \times 10^{5} \mathrm{~m}=10 \mu \mathrm{m}$
And nature of zero error is positive.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.