Question

The pitch of a screw gauge is 1 mm and its circular
scale has 100 divisions. In measurement of the
diameter of a wire, the main scale reads
2 mm and 45th mark on circular scale coincides with
the base line. Find :
(i) the least count, and
(ii) the diameter of the wire.
Ans. (i) 0·001cm (ii) 0·245 cm.​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with

the base line.

Find :

(i) the least count, and

(ii) the diameter of the wire.

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Pitch of screw gauge = 1mm
• Circular scale division = 100
• Main scale reading = 2mm or 0.2cm
• 45 marks on circular scale coincides with base line.

$\large\underline{\underline{\sf To\:Find:}}$

• Least Count
• Diameter of wire

❏ Least Count ➝

✪${\boxed{\bf \blue{Least\:Count(LC)=\dfrac{Pitch}{No.\:of\: division}} }}$

$\implies{\sf LC = \dfrac{1}{100}}$

$\implies{\sf LC = 0.01mm}$

$\implies{\bf \red{Least\: Count=0.001cm}}$

❏ Diameter of wire ➝

45 marks on circular scale coincides with base line. ( Given )

Circular Scale Reading

$\implies{\sf 45×LC }$

$\implies{\sf 45×0.001 }$

$\implies{\sf 0.045cm}$

Total Reading = Main Scale Reading + Circular Scale Reading

$\implies{\sf 0.2+0.045 }$

$\implies{\bf \red{Diameter\:of\:wire= 0.245cm}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Least Count ${\bf \red{0.001cm}}$

Diameter of wire ${\bf \red{0.245cm}}$

Answer 2

Answer:

least count is 0.001

Diameter of wire is 0.245