In an experiment to find out the diameter of wire using screw gauge, the following observation were noted: (a) Screw moves 0 . 5 mm on main scale in one complete rotation (b) Total divisions on circul

Question

In an experiment to find out the diameter of wire using screw gauge, the following observation were noted:

(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) $45^{th}$ division of circular scale is in the pitch line
(e) Instrument has $0.03 mm$ negative error Then the diameter of wire is :

TestHub · Accepted Answer

From the data given in the question,
Pitch of the screw gauge $= 0.5 mm$ .
Total divisions on the circular scale $n = 50$ .
Therefore, the least count of the screw gauge is $LC = \frac{pitch}{n} = \frac{0.5}{50} = 0.01 mm$
Now, $MSR = 2.5 mm$ , $CSR = 45$
Diameter reading $=$ $MSR + LC \times CSR -$ zero error
$\Rightarrow d = 2.5 + 0.45 - (- 0.03)$
$\Rightarrow d = 2.98 mm$

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