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

Correct Answer: 4.98 cm

Step I: Determine the Least Count (LC). It is given that $1~\text{MSD} = 0.1~cm$. Also, $10~\text{VSD} = 9~\text{MSD}$, so $1~\text{VSD} = 0.9~\text{MSD}$. $LC = 1~\text{MSD} - 1~\text{VSD} = 1~\text{MSD} - 0.9~\text{MSD} = 0.1~\text{MSD} = 0.1 \times 0.1~cm = 0.01~cm$.
Step II: Identify the Zero Error. When the jaws are closed, the zero of the Vernier scale is at $x = 0.1~cm$. This means the zero error is $+0.10~cm$.
Step III: Calculate the raw measured diameter. Main scale reading $M = 5~cm$ and coinciding vernier division $V = 8$. $\text{Measured value} = M + V \times LC = 5 + 8 \times 0.01 = 5.08~cm$.
Step IV: Apply the zero error correction. $\text{True diameter} = \text{Measured value} - \text{Zero Error} = 5.08~cm - 0.10~cm = 4.98~cm$.

सही उत्तर: 4.98 cm

Step I: अल्पतमांक (LC) निर्धारित करें। यह दिया गया है कि $1~\text{MSD} = 0.1~cm$। साथ ही, $10~\text{VSD} = 9~\text{MSD}$, इसलिए $1~\text{VSD} = 0.9~\text{MSD}$। $LC = 1~\text{MSD} - 1~\text{VSD} = 1~\text{MSD} - 0.9~\text{MSD} = 0.1~\text{MSD} = 0.1 \times 0.1~cm = 0.01~cm$।
Step II: शून्य त्रुटि (Zero Error) को पहचानें। जब जबड़े बंद होते हैं, तो वर्नियर स्केल का शून्य $x = 0.1~cm$ पर होता है। इसका मतलब है कि शून्य त्रुटि $+0.10~cm$ है।
Step III: कच्चे मापे गए व्यास की गणना करें। मुख्य स्केल रीडिंग $M = 5~cm$ और संपाती वर्नियर डिवीजन $V = 8$। $\text{मापा गया मान} = M + V \times LC = 5 + 8 \times 0.01 = 5.08~cm$।
Step IV: शून्य त्रुटि सुधार लागू करें। $\text{वास्तविक व्यास} = \text{मापा गया मान} - \text{शून्य त्रुटि} = 5.08~cm - 0.10~cm = 4.98~cm$।