Every question from past board papers on Vernier calipers and screw gauge, newest paper first. Open any question for its full worked answer.
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at \(x = 0.1\) cm when the jaws of Vernier callipers are closed.
If the main scale reading for the diameter is M = 5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
In a vernier calipers, \((N+1)\) divisions of vernier scale coincide with \(N\) divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is :
A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is :
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitch of the screw gauge is :
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of \(-0.004\) cm, the correct diameter of the ball is
