Question 35: Units , Dimensions and Error Analysis

Question

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $=0.05 \mathrm{~mm}$, then the least count of the vernier callipers is $\_\_\_\_$ mm.

Accepted Answer

Value of one Main Scale Division, $1 \mathrm{MSD}=0.05 \mathrm{~mm}$
Number of Vernier Scale Divisions $\mathrm{m}=50$
Number of Main Scale Divisions is $\mathrm{n}=48$
$$ 50 \mathrm{VSD}=48 \mathrm{MSD} $$
$$\Rightarrow $$ $$ 1 \mathrm{VSD}=\frac{48}{50} \mathrm{MSD} $$
The Least Count of a Vernier instrument is the smallest measurement that can be taken accurately.
Least Count (LC) of Vernier scale $=1$ MSD -1 VSD
Using the least count formula,
$$ \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} $$
$$\Rightarrow $$ $\mathrm{LC}=1 \mathrm{MSD}-\left(\frac{48}{50}ight) \mathrm{MSD}$
$$\Rightarrow $$ $\mathrm{LC}=\left(1-\frac{48}{50}ight) \mathrm{MSD}$
$$\Rightarrow $$ $\mathrm{LC}=\left(\frac{50-48}{50}ight) \mathrm{MSD}$
$$\Rightarrow $$ $$ \mathrm{LC}=\frac{2}{50} \mathrm{MSD}=\frac{2}{50} 	imes 0.05 \mathrm{~mm}=0.002 \mathrm{~mm} $$
Therefore, the smallest measurement that this Vernier calliper can record is 0.002 mm .
Hence, the correct option is (A).

Answer

0.002

Answer

0.02

Answer

0.05

Answer

0.005

Units , Dimensions and Error Analysis

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division =0.05 \mathrm{~mm}, then the least count of the vernier callipers is \_\_\_\_ mm.

Choose the correct answer: