JEE Main 2022 · 29 July · morning · Single correct · +4 / −1
In an experiment to find out the diameter of wire using screw gauge, the following observations were noted :
(A) Screw moves 0.5 mm on main scale in one complete rotation
(B) Total divisions on circular scale =50
(C) Main scale reading is 2.5 mm
(D) 45ᵗʰ division of circular scale is in the pitch line
(E) Instrument has 0.03 mm negative error
Then the diameter of wire is :
- (a) 2.92 mm
- (b) 2.54 mm
- (c) 2.98 mm
- (d) 3.45 mm
Answer and solution
Answer (c) 2.98 mm
Solution
Physics sir · 4 steps
Read the gauge first, then remember that a negative zero error is always added back to the reading.
- Least count. One full turn moves the screw by the pitch, 0.5 mm, and the circular scale has 50 divisions. So the least count is 0.5/50 = 0.01 mm.
- Observed reading. Main scale reading is 2.5 mm, and the 45ᵗʰ circular division is on the pitch line. Observed reading = 2.5 + 45 × 0.01 = 2.95 mm.
True reading = Observed reading - (zero error)
- Correct the error. A negative zero error means the zero error is -0.03 mm. So the true reading is 2.95 - (-0.03) = 2.95 + 0.03 = 2.98 mm.
- So. The diameter of the wire is 2.98 mm, option (c).
The idea to keep. Always subtract the zero error with its sign. A negative zero error makes the gauge read too low, so the correction adds it back. A positive one gets subtracted.
Answer from the NTA final answer key. Solution by PrepWiser.