A screw gauge of pitch 0 . 5mm is used to measure the diameter of...

Least Count (L.C)$= \dfrac{P}{N} = \dfrac{0 . 5mm}{50} = 0 . 01mm$

Length of wire $= 6 . 8cm$

Diameter of wire $= 1 . 5mm + 7 \times$ L.C

$= 1 . 5mm + 7 \times . 01 = 1 . 57mm$

Curved surface area $= \piDl$

$= 3 . 14 \times 6 . 8 \times 1 . 57 \times {10}^{- 1}{cm}^{2}$

$= 3 . 352{cm}^{2} = 3 . 4{cm}^{2}$