In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brough in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible and 46^th division the circular scale coincide with the reference line. The diameter of the wire is $\times 10^{- 2} m m$ .

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

2.2

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$L e a s t c o u n t = \frac{P i t c h}{N o . o f c i r c u l a r d i v i s i o n s}$

$= \frac{0.5 m m}{100}$

Least count $= 5 \times 10^{- 3} m m$

Positive error = MSR + CSR (LC)

$= O + 6 (5 \times 10^{- 3})$

$= 30 \times 10^{- 3} m m$

Diameter = $(4 \times 0.5) + (46 \times 5 \times 10^{- 3}) -$ $(6 \times 5 \times 10^{- 3})$

$= 2.2 m m$

Watch 3-min video & get full concept clarity