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 verniercalliper where the 10 divisions in vernier scale matches with 9 division in main scale &main scale has 10 divisions in 1 cm.
Correct Answer: D
Solution :
Least count = 0.01 cmi.e. 0.1 mm For verniercalliper of10 v.s.d. = 9 M.s. d. 1 v.s.d. = 0.9 M.s.d. L.C. = 1 M.s.d = (0.1) M.s.d. 1 v.s.d. = 0.1 cm \[\Rightarrow \]L.c = 0.01 cmYou need to login to perform this action.
You will be redirected in
3 sec