A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickenss 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 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?
A. 0.80 mm
B. 0.70 mmC. 0.50 mm
D. 0.75 mm
Answers
Final Answer : A) 0.80 mm
Error and Measurement :D
Screw Gauge (Micrometer)
Logic Used:
1) Least count of a screw gauge :
2) Concept of Zero error
Steps:
1) Pitch of screw : 0.5mm
No. of circular divisions: 50
=>Least count :0.5/50=0.01mm
2)This is given in Question: "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."
Above statement means that :
Thickness of sheet will be recorded lesser than what is its actual thickness due to some material stucked between the jaws.
OR We can say Spindle (Rotation of screw ) has to cover (50-45) =5 more divisions to get to the zero of main -scale.
Hence,Reading will 5 divisions lesser than thickness of Sheet.
Mathematically speaking, Zero Error: (-5*Least count) = -0.05mm
2) Thickness of Sheet :
Main Scale Reading+ Circular scale Reading*Least Count -Zero Error
=> 0.5mm + 25*0.01mm -(-0.05)
=>0.5mm +0.25mm+0.05mm
=> 0.80mm
Hence Thickness of Sheet is 0.80mm
Answer:0.80mm
Explanation:
Actual reading=main scale reading+vernier scale reading×least count -zero error
So reading=0.50+25×0.01-(-0.05)=0.80mm
Thank you!