IIT JEE Main 2016 Physics Question No 22
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a time sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with
the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
|(1) 0.70 mm||(2) 0.50 mm||(3) 0.75 mm||(4) 0.80 mm|
We know that,
Least Count (LC) = pitch/total divisions
Here, LC = 0.5/50
=> LC = 0.01 mm
Here the error is negative and is given by
error = 5 × 0.01 = 0.05
Reading = 0.5 + 25 × 0.01 + 0.05
= 0.5 + 0.25 + 0.05
= 0.80 mm
Correct answer to given problem No 22 of IIT JEE Main Physics paper of 2016 is option number (4) that is 0.80 mm.