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JEE Main 202229 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Evening Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2022 · 29 June · Evening Shift · Physics · Experimental Physics

The Vernier constant of Vernier callipers is 0 . 1mm and it has...

The Vernier constant of Vernier callipers is 0.1mmandithaszeroerrorof(−0.05cm)0 . 1mm and it has zero error of \left(- 0 . 05cm\right). While measuring diameter of a sphere, the main scale reading is 1.7cm1 . 7cm and coinciding vernier division is 55 . The corrected diameter will be _____×10−2cm\times {10}^{- 2}cm.

Correct answer

0

Step-by-step solution

Vernier constant is the least count of the vernier V.C=0.1mm=0.01cmV . C = 0 . 1mm = 0 . 01cm

zero error =−0.05cm= - 0 . 05cm

∴∴correction=0.05cm= 0 . 05cm

M.S.R.=1.7cmM . S . R . = 1 . 7cm

V.S.R.=5V.C.=5×0.01=0.05cmV . S . R . = 5V . C . = 5 \times 0 . 01 = 0 . 05cm

The diameter of the sphere will be,

diameter=MSR+VSR+= MSR + VSR +correction

1.7+0.05+0.05=1.8cm1 . 7 + 0 . 05 + 0 . 05 = 1 . 8cm

=180×10−2cm= 180 \times {10}^{- 2}cm

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