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PhysicsPractical physicsExperimental & Practical PhysicsMedium2 minPYQ_2022
PhysicsMediumsingle choice

In a Vernier Caliper10divisions of Vernier scale is equal to the9divisions 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 and4thVernier scale division exactly coincides with the main scale reading. One main scale division is equal to1 mm. 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 between30and31divisions of main scale reading and6th Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be:

Options:

Answer:
C
Solution:

Given that 1 M.S.D=1 mm

And 9 M.S.D =10 V.S.D

1 V.S.D=0.9M.S.D=0.9 mm

L.C of vernier caliper =1-0.9=0.1 mm=0.01 cm

Zero error =10-4×0.1 mm=0.6 mm as at zero the, the zero of the vernier is shifted towards left zero error is negative. Therefore, zero error=-0.6 mm.

Reading =M.S.R+V.S.R-  Zero error

Reading=3+6×0.01--0.06

Reading=3+0.06+0.06

Reading=3.12 cm

The closest option is 3.10 cm

Stream:JEESubject:PhysicsTopic:Practical physicsSubtopic:Experimental & Practical Physics
2mℹ️ Source: PYQ_2022

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