JEE Main Units and Measurements Previous year Question

16.(JEE Main 2016 (Offline) )

A screw gauge with a pitch of 0.5 mm 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 45^th 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 25^th division coincides with the main scale line?

(A) 0.70 mm

(B) 0.50 mm

(C) 0.75 mm

(D) 0.80 mm

Correct answer is (D)

Least count = $\frac{0.5}{50}$ = 0.01 mm

Zero error = (45 - 50) $\times$ 0.01 mm = - 0.05 mm

Thickness of sheet = (0.5 + 25 $\times$ 0.01) - (-0.05)

= 0.50 + 0.30 = 0.80 mm

17.(JEE Main 2014 (Offline) )

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

(A) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

(B) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

(C) A meter scale.

(D) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

Correct answer is (D)

Measured length of rod = 3.50 cm

That means least count of the measuring instrument should be 0.01 cm = 0.1 mm

For vernier scale 1 main scale division = 1 mm

And 9 MSD = 10 VSD

Least count = 1 MSD - 1 VSD

= 1 - 0.9 = 0.1 mm

18.(AIEEE 2008 )

Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

(A) 3.32 mm

(B) 3.73 mm

(C) 3.67 mm

(D) 3.38 mm

Correct answer is (D)

Least count of screw gauge = $\frac{0.5}{50} m m$ = 0.01mm

Main scale reading = 3 mm

Vernier scale reading = 35

$∴$ Reading = [Main scale reading + circular scale reading $\times$ L.C] - (zero error)

= [3 + 35 $\times$ 0.01] - (-0.03) = 3.38 mm