Question

The pitch of the screw gauge is 1mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in $c{{m}^{2}}$) of the wire in an appropriate number of significant figures.
(A). 2.6
(B). 3.2
(C). 4.8
(D). 1.3

Answer 1

Hint: Screw gauge is used to find the exact diameter or width of an object. Least count is defined as pitch per number of divisions on the circular scale. Diameter is obtained by using the total reading formula i.e. $\text{main scale reading + circular scale reading }\times \text{ least count}$. Then substitute the value in the curved surface area formula to get the desired result.

Formula Used:
$L.C.=\dfrac{\text{pitch}}{\text{number of division on the circular scale}}=\dfrac{1}{100}=0.01mm$
For a screw gauge,
$\text{total reading = main scale reading + circular scale reading }\!\!\times\!\!\text{ least count}$

Complete step-by-step answer:
Screw gauge is an instrument that can be used to measure the exact diameter of a thin rod or exact width of a thin metal sheet.
Least count of a screw gauge can be defined as the smallest value that can be measured by the screw gauge. Least can be defined as the ratio of the pitch of the screw gauge to the number of divisions in the circular scale.
Pitch of the screw gauge is 1mm and the number of divisions in the circular scale is 100.
Least count of the screw gauge is given by,
$L.C.=\dfrac{\text{pitch}}{\text{number of division on the circular scale}}=\dfrac{1}{100}=0.01mm$
The main scale reading is 1mm and the circular scale reading is 47
Diameter(d) of the wire is given by,
$\begin{align}
  & \text{total reading = main scale reading + circular scale reading }\!\!\times\!\!\text{ least count} \\
 & \text{d = 1mm + 0}\text{.01}\times \text{47 mm} \\
 & \text{d = 1}\text{.47 mm = 0}\text{.147 cm} \\
\end{align}$
Length of the wire is given as \[l\text{ }=\text{ }5.6\text{ }cm.\]
Now, the curved surface area of the wire is given by,
$\begin{align}
  & \text{area = }\pi \times d\times l \\
 & \text{area = }\pi \times 0.147cm\times 5.6cm \\
 & \text{area = 2}\text{.586 c}{{\text{m}}^{2}} \\
 & \text{area }\approx \text{ 2}\text{.6 c}{{\text{m}}^{2}} \\
\end{align}$
The area of the curved surface of the wire is, $\text{area }\approx \text{ 2}\text{.6 c}{{\text{m}}^{2}}$
The correct option is (A)

Note: The pitch and number of circular divisions in the screw gauge may be different, but still the least count may be the same. For example, if the pitch of the screw gauge is 2 and the number of divisions in the circular scale is 50, then also the least count will be 0.01mm.