JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 19)
In an experiment to find out the diameter of wire using screw gauge, the following observations were noted :
_29th_July_Morning_Shift_en_19_1.png)
(A) Screw moves $$0.5 \mathrm{~mm}$$ on main scale in one complete rotation
(B) Total divisions on circular scale $$=50$$
(C) Main scale reading is $$2.5 \mathrm{~mm}$$
(D) $$45^{\text {th }}$$ division of circular scale is in the pitch line
(E) Instrument has 0.03 mm negative error
Then the diameter of wire is :
2.92 mm
2.54 mm
2.98 mm
3.45 mm
Explanation
L.C. = $${{0.5} \over {50}}$$ mm = 0.01 mm
$$d = (2.5 + 45 \times 0.01 + 0.03)$$ mm
= 2.98
Comments (0)
