Question

A screw gauge of pitch $0.5 \,mm$ is used to measure the diameter of uniform wire of length $6.8 \,cm$, the main scale reading is $1.5\, mm$ and circular scale reading is $7$ . The calculated curved surface area of wire to appropriate significant figures is : [Screw gauge has $50$ divisions on the circular scale]

• A. $6.8 \,cm ^2$
• B. $3.4 \,cm ^2$
• C. $3.9\, cm ^2$
• D. $2.4\, cm ^2$

Answer 1

Answer: (B)

L.C. $=\frac{ P }{ N }=\frac{0.5 mm }{50}=0.01\, mm$
Length of wire $=6.8 \, cm$
Diameter of wire $=1.5 \, mm +7 \times$ L.C
$=1.5\, mm +7 \times .01=1.57 \, mm$
Curved surface area $=\pi D \ell$
$=3.14 \times 6.8 \times 1.57 \times 10^{-1} cm ^2$
$=3.352 \, cm ^2=3.4\, cm ^2$