JEE PYQ: Mathematics In Physics Question 6
Question 6 - 2021 (25 Feb Shift 1)
The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lines 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while $72^{\text{nd}}$ division on circular scale coincides with the reference line. The radius of the wire is:
(1) 1.64 mm
(2) 1.80 mm
(3) 0.82 mm
(4) 0.90 mm
Show Answer
Answer: (3)
Solution
Least count $= \frac{\text{pitch}}{\text{no. of div.}} = \frac{1 \text{ mm}}{100} = 0.01$ mm +ve error $= 8 \times$ L.C. $= +0.08$ mm measured reading $= 1$ mm $+ 72 \times$ L.C. $= 1$ mm $+ 0.72$ mm $= 1.72$ mm True reading $= 1.72 - 0.08$ $= 1.64$ mm Radius $= \frac{1.64}{2} = 0.82$ mm
BETA
