A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0\ mm$
Circular scale reading : $52\ divisions$
Given that $1\ mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is:
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0\ mm$
Circular scale reading : $52\ divisions$
Given that $1\ mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is:
- [AIEEE 2011]
- A
$0.052$ $cm$
- B
$0.026$ $ cm$
- C
$0.005$ $ cm$
- D
$0.52$ $ cm$
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Answer the following :
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
$(b)$ A screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only ?
Answer the following :
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
$(b)$ A screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only ?