A student uses an error-free screw gauge to measure the thickness of a uniform glass plate. The pitch of the screw gauge is 0.5 mm and there are 50 divisions on its circular scale. During the measurement, the linear scale reads 2.5 mm and the 34^ th division of the circular scale aligns with the reference line. The thickness of the glass plate is:

Step-by-step solution

Least count of the screw gauge is given by: LC = Pitch Number of divisions on circular scale LC = 0.5 mm 50 = 0.01 mm The thickness of the glass plate is calculated as: Thickness = Linear scale reading + ( Circular scale reading LC ) Thickness = 2.5 mm + (34 0.01 mm ) Thickness = 2.5 mm + 0.34 mm = 2.84 mm Converting to centimeters: 2.84 mm = 0.284 cm Answer: 0.284 cm