CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A screw gauge with  a pitch of 0.5mm and a circular scale with 50 divisions is used to measure the thickness of a thin 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 concides 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 0.5mm and the 25th division coincides with the main scale line?


A
0.75 mm
loader
B
0.80 mm
loader
C
0.70 mm
loader
D
0.50 mm
loader

Solution

The correct option is B 0.80 mm
$$LC=\dfrac { \text{pitch} }{ \text{circular scale divisions} } =\dfrac { 0.5 }{ 50 } mm=0.01mm$$

Negative zero error =  $$-5\times (0.01)mm=-0.05mm$$

Measured value = $$0.5mm+{ 25×0.01–(–0.05) }mm=0.8mm$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image