Question

Least count of a vernier callipers is $0.01cm$. When the two jaws of the instrument touch each other the $5^{th}$ division of the vernier scale coincide with a main scale division and the zero 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 $6^{th}$ vernier division coincides with a main scale division. Calculate the diameter of the sphere.

• A. $2.51cm$
• B. $5.12cm$
• C. $2.51mm$
• D. $5.12mm$

Answer 1

The correct option is A $2.51cm$

Least count, $LC=0.01cm$

Here the negative zero error of callipers $=5\times 0.01=0.05cm$ and it should be added to final reading.

Main scale reading, $MSR=2.4cm$ and $VSD=6$

So total reading means the diameter of sphere $R=MSR+(VSD\times LC)+$ zero error $=2.4+(6\times 0.01)+0.05=2.51cm$