JEE Main 2022 · 29 Jul · Shift 1 · Physics Q50

PhysicsSingle correct+4 / -1Question 50 of 90

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

(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

Answer and worked solution

Official answer: C

Least count $\frac{0.5}{50}=0.01$ mm. Reading $2.5+45\times0.01=2.95$ mm.

Negative error of 0.03 mm is added back: 2.98 mm.

(C)

Topics:Vernier callipers and screw gauge

JEE Main 2022 · 29 Jul · Shift 1, Physics question 50