JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions in circular scale and pitch of screw gauge is $0.1$ mm. When diameter of a sphere is measured, the reading of main scale is $5$ mm and $5 0^{th}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is _____mm.

A
$5.045$
B
$5.055$
C
$5.450$
D
$5.550$

Solution

Least count $= \frac{pitch}{circular divisions} = \frac{0.1}{100} = 0.001$ mm.

With the main-scale zero coinciding with the $5^{th}$ circular division at no load, the instrument has a positive zero error of $5 \times 0.001 = 0.005$ mm.

Observed reading: $5 mm + 50 \times 0.001 mm = 5.050$ mm.

Corrected diameter $= 5.050 - 0.005 = 5.045$ mm.

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