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The internal jaws of a Vernier calliper is used to measure the diameter of a sphere.
(A) True
(B) False

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Last updated date: 26th Jul 2024
Total views: 405k
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Answer
VerifiedVerified
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Hint The internal jaws are designed in such a way that it can rest on the internal walls of a hollow tube or cylinder usually slipped in through an open end. A completely spherical solid even though hollow will not be opened enough for the internal jaws of the calliper to rest on the opposite walls simultaneously.

Complete step by step answer
A Vernier calliper is a measuring instrument which is designed and calibrated to measure linear lengths (as opposed to a tape rule or other instrument which has the capability to measure lengths along a curve) mostly in diameter. The Vernier calliper has the capability to measure both internal and external diameter of a shape. The Vernier calliper has different parts designed for different purposes. Two of these parts are
The external jaws: the external jaws of a calliper are built to hold and rest on the external surfaces of shapes to their external diameter. It is built so it can actually clamp the object to hold it still and thus measure accurate diameter.
The internal jaws: the internal jaws of a Vernier calliper is built to rest on the internal surfaces of hollow shapes opened at least at one end to allow the jaw to slip into the internal part of the shape. The diameter of the open end has to be equal (or almost) to the internal diameter of the shape so the walls don't interfere with the internal jaws.
From these definitions, we can conclude that the internal jaws of a Vernier calliper cannot measure the diameter of a sphere. Thus, statement is false
Hence, the correct option is B.

Note
In application, although Vernier callipers are popularly used, internal micrometre, digital calliper and other better instruments are used for a higher accuracy and precision. A vernier calliper, for example, has an accuracy of about $ 0.05mm $ . Some instruments like internal micrometre have accuracy of about $ 0.00005mm $ .