Errors in measurementMedium
Question
In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm . When diameter of a sphere is measured, the reading of main scale is 5 mm and $50^{\text{th}\text{~}}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is $\_\_\_\_$ mm.
Options
A.5.045
B.5.055
C.5.450
D.5.550
Solution
Sol. [Key Concept: Reading $=$ MSD $+$ CSD$\times$LC $-$ zero error.]
Least count : $\frac{0.1}{100}\text{ }mm = 0.001\text{ }mm$
Reading $= MSD + (CSD) \times LC -$ zero error
$$= 5\text{ }mm + 50 \times 0.001 - 5 \times 0.001 = 5.045\text{ }mm $$
Create a free account to view solution
View Solution FreeMore Errors in measurement Questions
In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm . If the circular ...If the percentage error in the measurement of radius of a sphere is 2%, then the maximum percentage errors in the measur...A student measured the diameter of a wire using a screw gauge with least count 0.001 cm and listed the measurements. the...Significant figure in 3400 are -...One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10 . ...