Experimental Physics

JEE Main level, numerical answer, medium

32 of 70

Question 32 of 70

The Vernier constant of Vernier callipers is 0.1mm0.1mm and it has zero error of (−0.05cm)\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.

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Answer: 180

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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