A Student measured the diameter of a small steel ball using a screw. g...
Solution:
Given data:
Main scale reading = 5 mm = 0.5 cm
Circular scale reading = 25 divisions
Least count of screw gauge = 0.001 cm
Zero error of screw gauge = -0.004 cm
We need to find the correct diameter of the steel ball.
Calculation:
Total reading on screw gauge = Main scale reading + Circular scale reading x Least count
= 0.5 cm + 25 x 0.001 cm
= 0.525 cm
Corrected reading = Total reading - Zero error
= 0.525 cm - (-0.004 cm)
= 0.529 cm
Therefore, the correct diameter of the steel ball is 0.529 cm.
Answer: (iv) 0.529 cm
Explanation:
The screw gauge is a measuring instrument used for measuring the diameter of small objects. It consists of a U-shaped frame, a screw, a thimble, and a spindle. The thimble is graduated with a circular scale, and the spindle has a measuring face. The main scale is engraved on the U-shaped frame, and it is used to determine the integer part of the measurement. The circular scale is used to determine the decimal part of the measurement.
In this problem, the main scale reading is 5 mm, and the circular scale reading is 25 divisions. The least count of the screw gauge is 0.001 cm. Therefore, the total reading on the screw gauge is calculated as 0.5 cm + 25 x 0.001 cm = 0.525 cm. However, the screw gauge has a zero error of -0.004 cm, which means that the actual measurement is 0.004 cm less than the measurement shown on the screw gauge. Therefore, the corrected reading is calculated as 0.525 cm - (-0.004 cm) = 0.529 cm. Hence, the correct diameter of the steel ball is 0.529 cm.
A Student measured the diameter of a small steel ball using a screw. g...
Reading = main scale reading + L.C× cirular scale reading + or - (error)... ( if - ve error then add, if +ve error then subtract) . reading = 0.5cm+ 0.001×25 + 0.004 = 0.529cm