Question

ICSE CLASS 9 PHYSICS
4 MARK MEASUREMENT AND UNITS QUESTION.



THERE ARE 100 DIVISIONS ONE THE CIRCULAR SCALE OF A SCREW GAUGE. IF THE HEAD OF THE SCREW IS GIVEN 5 REVOLUTIONS
, THE SPINDLE ADVANCES BY 5 MILLIMETRES. FIND ITS PITCH AND LEAST COUNT. WHEN WHEN READING, THE EDGE OF THE CIRCULAR SCALE IS FOUND TO BE BETWEEN 0.3 CM AND 0.4 CM AND THE 25TH DIVISION OF THE CIRCULAR SCALE COINCIDES COINCIDES WITH THE REFERENCE LINE. FIND THE READING.​

Answer 1

Here ya go mate

Least count = pitch/total no. of divisions in circular scale

=0.01cm/100 =1*10^-4cm

=0.01mm