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 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 :

Question

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 : (A) 2.92 mm (B) 2.54 mm (C) 2.98 mm (D) 3.45 mm

Tardigrade Admin · Accepted Answer

textMSR=2.5 mm textCSR=45 × (0.5/50) mm =0.45 mm Diameter reading = MSR + CSR - zero error =2.5+0.45-(-0.03) =2.98 mm

$2.92 mm$

$2.54 mm$

$2.98 mm$

$3.45 mm$

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

2.92mm

2.54mm

2.98mm

3.45mm

MSR=2.5mm CSR=45×500.5​mm =0.45mm Diameter reading = MSR + CSR − zero error =2.5+0.45−(−0.03) =2.98mm

$2.92 mm$

$2.54 mm$

$2.98 mm$

$3.45 mm$

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