Question

A Vernier calliper has a positive error of 0.2 mm. While finding the diameter of a brass rod the MSR is 6 mm and the VSD is 5. Similarly, the MSR and VSD while finding the length of the rod are 43 and 6. If the least count of the Vernier caliper is 0.01 cm, the volume of the cylinder is ____ $\text{m}{{\text{m}}^{3}}$.

Answer 1

Hint: We must substitute the values of the Main scale reading and Vernier scale reading in the Vernier equation of measurement in order to find the length and diameter of the cylinder. Zero error should be taken into account while calculating.

Complete step by step answer:
The equation to take readings from the Vernier calipers is given by,
$\text{Reading}=\text{M}\text{.S}\text{.R}+\text{V}\text{.S}\text{.D }\!\!\times\!\!\text{ L}\text{.C}$
Where,
$\text{M}\text{.S}\text{.R}$ is the main scale reading.
$\text{V}\text{.S}\text{.D}$ is the Vernier scale division.
$\text{L}\text{.C}$ is the least count of the Vernier caliper.
The actual reading is the reading we got from the equation corrected by the zero error.
$\text{Actual Reading}=\text{Reading}-\text{Zero Error}$
So the zero error associated with the Vernier calliper in the problem is a positive zero error of 0.2 mm and the least count of the Vernier calliper is 0.01 cm which is 0.1 mm.
In measuring the diameter of the cylinder, we got the MSR reading as 6 mm and VSD as 5. Substituting the values of MSR and VSD in the equation we get,
$\text{Reading}=\text{M}\text{.S}\text{.R}+\text{V}\text{.S}\text{.D }\!\!\times\!\!\text{ L}\text{.C}$
$\text{Reading}=6\text{mm}+5\text{ }\!\!\times\!\!\text{ 0}\text{.1mm}$
$\text{Reading}=6.5\text{ mm}$
The actual reading is,
$\text{Actual Reading}=\text{6}\text{.5 mm}-0.2\text{ mm}$
$\therefore \text{Actual Reading}=6.3\text{ mm}$
In measuring the length of the cylinder, we got the MSR reading as 43 mm and VSD as 6. Substituting the values of MSR and VSD in the equation we get,
$\text{Reading}=\text{M}\text{.S}\text{.R}+\text{V}\text{.S}\text{.D }\!\!\times\!\!\text{ L}\text{.C}$
$\text{Reading}=43\text{mm}+6\text{ }\!\!\times\!\!\text{ 0}\text{.1mm}$
$\text{Reading}=43.6\text{ mm}$
The actual reading is,
$\text{Actual Reading}=43.6\text{ mm}-0.2\text{ mm}$
$\therefore \text{Actual Reading}=43.4\text{ mm}$
So the volume of the cylinder is given by, $V=\pi {{\left( \dfrac{d}{2} \right)}^{2}}l$
From our measurement, we have found the diameter of the cylinder (d) as 6.3 mm and length of the cylinder (l) as 43.4 mm. Substituting these values in the above equation we get,
$V=\pi {{\left( \dfrac{6.3mm}{2} \right)}^{2}}\times 43.4mm$
$\therefore \text{V}=1353.4\text{ m}{{\text{m}}^{3}}$
So the volume of the cylinder is $\text{V}=1353.4\text{ m}{{\text{m}}^{3}}$.

Note: The Vernier constant is the least count of the Vernier callipers, it is defined as the difference between one main scale division and one Vernier scale division. So Vernier constant can be written as
$\text{Vernier Constant}=\text{One Main Scale Division}-\text{One Vernier Scale Division}$
There can be errors associated with the Vernier callipers when the main scale zero is not coinciding with Vernier scale zero when the jaws of the Vernier are in contact.
The errors can be positive or negative depending on the main scale zero coincidences with the point on the Vernier scale.