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  4. In an experiment to find out the diameter of wire using screw gauge, the following observation were noted: <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/7e77b99968dd48ab08831667926c7e9a-.png /> (a) Screw moves 0.5 mm on main scale in one complete rotation (b) Total divisions on circular scale =50 (c) Main scale reading is 2.5 mm (d) 45 text th division of circular scale is in the pitch line (e) Instrument has 0.03 mm negative error Then the diameter of wire is :

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Q. In an experiment to find out the diameter of wire using screw gauge, the following observation were noted :
image
(a) Screw moves $0.5 \,mm$ on main scale in one complete rotation
(b) Total divisions on circular scale $=50$
(c) Main scale reading is $2.5 \,mm$
(d) $45^{\text {th }}$ division of circular scale is in the pitch line
(e) Instrument has $0.03 \, mm$ negative error
Then the diameter of wire is :

JEE MainJEE Main 2022Physical World, Units and Measurements

Solution:

$ \text{MSR}=2.5 mm $
$ \text{CSR}=45 \times \frac{0.5}{50} mm $
$ =0.45 \,mm$
Diameter reading $=$ MSR $+$ CSR $-$ zero error
$ =2.5+0.45-(-0.03)$
$ =2.98 \,mm$