The L.C of a vernier callipers is 0.01 cm and its zero error is - 0.03 cm while measuring the length of a rod. The main scale reading is 4.6 cm and the sixth division on the vernier is in line with a marking on the main scale, then what will be the length of the rod ?

4.95

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4.69

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5.03

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4.99

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Solution

The correct option is B $4.69$ cm
$\begin{matrix} L . C = 0.01 c m zero error = - 0.03 c m MSR = 4.6 c m coinciding division = 6 \begin{matrix} VSR & = L \cdot C \times 6 = 0.06 c m \end{matrix} \end{matrix}$

$\begin{matrix} \begin{matrix} Final reading = & M S R + V S R - zero error = & 4.6 + 0.06 + 0.03 = 4.69 \end{matrix} \end{matrix}$

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Q. Consider the following data for a vernier calliper:
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6^{t h}

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The actual reading for length measurement is 4.3 cm with 2nd vernier division coinciding with main scale graduation.

Estimate the length.

In a vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided in 10 parts. While measuring a length, the zero of the vernier lies just ahead of 1.8 cm mark and 4th division of vernier coincides with a main scale division.

(a) Find the length.

(b) If zero error of vernier callipers is - 0.02 cm, what is the correct length ?

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The L.C of a vernier callipers is 0.01 cm and its zero error is - 0.03 cm while measuring the length of a rod. The main scale reading is 4.6 cm and the sixth division on the vernier is in line with a marking on the main scale, then what will be the length of the rod ?

The correct option is B 4.69 cmL.C=0.01 cm zero error =−0.03 cm MSR =4.6 cm coinciding division =6 VSR =L⋅C×6=0.06 cm Final reading =MSR+VSR− zero error =4.6+0.06+0.03=4.69