JEE Advanced 2022 · Paper 2 · Physics Q17

← JEE Advanced 2022 · Paper 2

PhysicsSingle correct+3 / -1Question 35 of 54

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0.5 mm. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

Measurement condition	Main scale reading	Circular scale reading
Two arms of gauge touching each other without wire	0 division	4 divisions
Attempt-1: With wire	4 divisions	20 divisions
Attempt-2: With wire	4 divisions	16 divisions

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

2.22 \pm 0.02\ \text{mm},\ \pi(1.23 \pm 0.02)\ \text{mm}^2

2.22 \pm 0.01\ \text{mm},\ \pi(1.23 \pm 0.01)\ \text{mm}^2

2.14 \pm 0.02\ \text{mm},\ \pi(1.14 \pm 0.02)\ \text{mm}^2

2.14 \pm 0.01\ \text{mm},\ \pi(1.14 \pm 0.01)\ \text{mm}^2

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped (marks to all) in the official key. Below is the standard working.

Least count. One full turn moves the main scale by two divisions, i.e. $2\times0.5=1.0$ mm. So the least count is $\frac{1.0}{100}=0.01$ mm.

Zero error. $+4$ divisions $=+0.04$ mm.

Readings.

Attempt 1: $4(0.5)+20(0.01)-0.04=2.16$ mm.
Attempt 2: $4(0.5)+16(0.01)-0.04=2.12$ mm.

Result. The mean is $d=2.14$ mm. Half the spread is $\pm0.02$ mm, so $d=2.14\pm0.02$ mm. The area is $\pi\left(\frac d2\right)^2=\pi(1.07)^2\approx\pi(1.14)\ \text{mm}^2$ , which is option (C).

Why it was likely dropped. Strictly, the relative error in the area is twice that in the diameter, giving $\pi(1.14\pm0.04)$ . No option has this.

Topics:Vernier callipers and screw gauge Errors in measurement

← Previous question Next question →

JEE Advanced 2022 · Paper 2, Physics question 17