Question

Least count of vernier calipers is $0.01cm$. When the two jaws of the instrument touch each other, the 5th division of the vernier scale lies to the left of the zero of the main scale. Furthermore, while measuring the diameter of a sphere, the zero mark of the vernier scale lies between $2.4cm$ and $2.5cm$ and the 6th vernier division coincides with a main scale division. Calculate the diameter of the sphere.

Answer 1

Here, we are given the least count of the instrument. Also, we are given that when the jaws touch each other, the 5th division of the vernier scale coincides with the main scale, that is we are given information about the zero error. Use the above provided data to find the net reading of the instrument or the diameter of the sphere.

Complete step by step answer:
When the jaws are touching each other, the 5th division of the vernier scale lies to the left of zero of the main scale. Therefore, the zero error will be $\left( {LC} \right)\left( {VSR} \right) = 0.01 \times 5 = 0.05cm$. The zero error of the instrument is $0.05cm$. As it is given that the 5th division of the vernier scale lies to the left of zero, the zero error must be added in order to compensate for the loss of the length.

Now, when the sphere is kept in between the jaws, the main scale reading is $2.4cm$. As the 6th vernier division coincides with a main scale division, the overall reading of the instrument will be,
$MSR + \left( {LC} \right)\left( {VSR} \right) = 2.4 + \left( {0.01} \right)\left( 6 \right) \\\ \Rightarrow MSR + \left( {LC} \right)\left( {VSR} \right) = 2.46cm$.
In order to get the diameter of the sphere, we add the zero error in the above obtained reading, we get, $2.46 + 0.05 = 2.51cm$.

Therefore, the diameter of the sphere is $2.51cm$.

Note: In these types of questions, always read the whole question and pick up each and every data available in the question, it will help you to solve the question quickly and easily. Keep in mind the zero error that we have considered while solving the question. You must know when the zero error is supposed to be added or subtracted. Also keep in mind the overall reading is given by sum of MSR and product of least count and VSR.