Question

The main scale of vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and $4^{th}$ division of the vernier scale coincides with a main scale division. The radius of the cylinder is $1.76\times {10}^{-x}m$. Find the value of x.

Answer 1

Hint: We measure the diameter of a cylinder using a vernier caliper. We are given the radius of the cylinder. We have to find the missing term ‘x’ in the given radius. To find ‘x’ we will first find the radius of the cylinder. By finding the radius, we can find the value of ‘x’.

Formula Used:
Least count,
$\text{LC=}{\displaystyle \frac{\text{Pitch}}{\text{Vernier}\text{ scale}\text{ divisions}}}$

Complete step by step answer:
We have a vernier scale which is calibrated in mm.
It is said that 19 divisions of the main scale are equal in length to 20 divisions of the vernier scale.
Now, let us find the least count of the vernier scale.
 Least count is given by,
$\text{LC=}{\displaystyle \frac{\text{value of 1 m}\text{.s}\text{.d}}{\text{No}\text{ of}\text{ divisions}\text{ on}\text{ vernier}\text{ scale}}}$
Total reading $\text{=M}\text{.S}\text{.R+(L}\text{.C }\times \text{ V}\text{.S}\text{.D)}$
Value of 1 M. S. D, i.e. the value of 1 main scale division = 1 mm = 0.1 cm.
We know that, number of divisions on the vernier scale = 20.
Therefore,
$\begin{array}{rl}& \text{LC=}{\displaystyle \frac{0.1cm}{20cm}}\\ & \text{LC=0}\text{.005cm}\end{array}$
We are measuring the diameter of a cylinder.
It is said that the main scale reads 35 divisions.
Therefore main scale reading (M. S. D) = 35 mm = 3.5 cm
It is also said that the $4^{th}$ division of the vernier scale coincides with the main scale.
Therefore the vernier scale reading (V. S. D) = 4
Hence, the total reading on the vernier calipers, i.e. the diameter of the cylinder,
Total reading = main scale reading + ((vernier scale reading) (Least count))
Total reading $=3.5+(4\times 0.005)$
Total reading $=3.5+0.02$
Therefore, diameter of cylinder $=3.52cm$
Radius of cylinder $={\displaystyle \frac{Diameter}{2}}$
$\begin{array}{rl}& Radius={\displaystyle \frac{3.52}{2}}\\ & Radius=1.76cm\end{array}$
Therefore,
Radius of cylinder $=1.76\times {10}^{-2}m$
Hence the value of ${}^{\prime }{x}^{\prime }=-2$

Note:
Vernier calipers are used to measure length, breadth, depth, diameter etc of small objects accurately. It consists of a main scale and a vernier scale. The least distance that a vernier caliper can measure accurately is its least count.
To convert centimeter into meter, we divide the value in centimeter by 100, or
$1m=100cm$