Experimental Physics

JEE Main level, single correct, medium

30 of 70

Question 30 of 70

In a Vernier Caliper 1010 divisions of Vernier scale is equal to the 99 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th{4}^{th} Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1mm1mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 3030 and 3131 divisions of main scale reading and 6th{6}^{th} Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be:

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Answer: (C) 3.10cm3.10cm

Given that 1M.S.D=1mm1M.S.D=1mm
And
9M.S.D=10V.S.D9M.S.D=10V.S.D
⇒1V.S.D=0.9M.S.D=0.9mm\Rightarrow 1V.S.D=0.9M.S.D=0.9mm
L.C of vernier caliper
=1−0.9=0.1mm=0.01cm=1-0.9=0.1mm=0.01cm
Zero error
=(10−4)×0.1mm=0.6mm=\left(10-4\right)\times 0.1mm=0.6mm as at zero the, the zero of the vernier is shifted towards left zero error is negative. Therefore, zero error=−0.6mm=-0.6mm.
Reading
=M.S.R+V.S.R−=M.S.R+V.S.R- Zero error
Reading
=3+6×0.01−[−0.06]=3+6\times 0.01-\left[-0.06\right]
Reading
=3+0.06+0.06=3+0.06+0.06
Reading
=3.12cm=3.12cm
The closest option is
3.10cm3.10cm

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