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

(A) Screw moves $0.5 \mathrm{~mm}$ on main scale in one complete rotation
(B) Total divisions on circular scale $=50$
(C) Main scale reading is $2.5 \mathrm{~mm}$
(D) $45^{\text {th }}$ division of circular scale is in the pitch line
(E) Instrument has $0.03 \mathrm{~mm}$ negative error
Then the diameter of wire is :
Option: 1
$2.92\: \mathrm{mm}$

Option: 2
$2.54\: \mathrm{mm}$

Option: 3
$2.98\: \mathrm{mm}$

Option: 4
$3.45\: \mathrm{mm}$

Answers (1)

best_answer

$\mathrm{pitch =0.5 \mathrm{~mm}}$

$\mathrm{No. of \: C S D=50}$

$\mathrm{M S R=2.5 \mathrm{~mm}}$

$\mathrm{C S D=45}$

Zero error $\mathrm{=-0.03 \mathrm{~mm}}$
$\mathrm{C S R=C S D \times L C=45 \times \frac{0.5}{50} \mathrm{~mm}=0.45 \mathrm{~mm}}$

$\mathrm{Diameter\: of \: wire =M S R+C S R+Zero\: correction}$

$\mathrm{=2.5\: mm+0.45\: mm(0.03\: mm)}$

$\mathrm{Diameter \: of\: wire =2.98 \mathrm{~mm}}$

Hence 3 is correct option

Posted by

HARSH KANKARIA

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