Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
- (A)If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
- (B)If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.✓ Correct
- (C)If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.✓ Correct
- (D)If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
Answer: (B), (C)
Solution
For Vernier callipers: 1 MSD = 1/8 cm = 1.25 mm. Since 5 VSD = 4 MSD, 1 VSD = 0.8 MSD. Least count = 1 MSD - 1 VSD = 0.2 MSD = 0.25 mm.
For screw gauge: pitch = 2 divisions on linear scale. Least count = pitch/100.
(A) If pitch = 2 × 0.25 = 0.5 mm, then LC = 0.5/100 = 0.005 mm. So (A) is wrong, (B) is correct.
(C) If LC of linear scale = 2 × 0.25 = 0.5 mm, and pitch = 2 × (LC of linear scale) = 1 mm, then LC of screw = 1/100 = 0.01 mm. So (C) is correct.
(D) With same conditions as (C), LC = 0.01 mm, not 0.005 mm. So (D) is wrong.
Correct options: (B) and (C).
For screw gauge: pitch = 2 divisions on linear scale. Least count = pitch/100.
(A) If pitch = 2 × 0.25 = 0.5 mm, then LC = 0.5/100 = 0.005 mm. So (A) is wrong, (B) is correct.
(C) If LC of linear scale = 2 × 0.25 = 0.5 mm, and pitch = 2 × (LC of linear scale) = 1 mm, then LC of screw = 1/100 = 0.01 mm. So (C) is correct.
(D) With same conditions as (C), LC = 0.01 mm, not 0.005 mm. So (D) is wrong.
Correct options: (B) and (C).
JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options