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]

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

Tardigrade Admin · Accepted Answer

L.C. =( P / N )=(0.5 mm /50)=0.01 mm Length of wire =6.8 cm Diameter of wire =1.5 mm +7 × L.C =1.5 mm +7 × .01=1.57 mm Curved surface area =π D ℓ =3.14 × 6.8 × 1.57 × 10-1 cm 2 =3.352 cm 2=3.4 cm 2

$6.8 c m^{2}$

$3.4 c m^{2}$

$3.9 c m^{2}$

$2.4 c m^{2}$

L.C. $= \frac{P}{N} = \frac{0.5 mm}{50} = 0.01 mm$
Length of wire $= 6.8 c m$
Diameter of wire $= 1.5 mm + 7 \times$ L.C
$= 1.5 mm + 7 \times .01 = 1.57 mm$
Curved surface area $= π D ℓ$
$= 3.14 \times 6.8 \times 1.57 \times 1 0^{- 1} c m^{2}$
$= 3.352 c m^{2} = 3.4 c m^{2}$

Q. A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm 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]

6.8cm2

3.4cm2

3.9cm2

2.4cm2

L.C. =NP​=500.5mm​=0.01mm Length of wire =6.8cm Diameter of wire =1.5mm+7× L.C =1.5mm+7×.01=1.57mm Curved surface area =πDℓ =3.14×6.8×1.57×10−1cm2 =3.352cm2=3.4cm2

$6.8 c m^{2}$

$3.4 c m^{2}$

$3.9 c m^{2}$

$2.4 c m^{2}$

L.C. $= \frac{P}{N} = \frac{0.5 mm}{50} = 0.01 mm$
Length of wire $= 6.8 c m$
Diameter of wire $= 1.5 mm + 7 \times$ L.C
$= 1.5 mm + 7 \times .01 = 1.57 mm$
Curved surface area $= π D ℓ$
$= 3.14 \times 6.8 \times 1.57 \times 1 0^{- 1} c m^{2}$
$= 3.352 c m^{2} = 3.4 c m^{2}$