There are two Vernier calipers both of which have 1 mathrm{~cm} divided into 10 equal divisions on m

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1.Units, Dimensions and Measurement

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There are two Vernier calipers both of which have $1 \mathrm{~cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left(C_1\right)$ has $10$ squal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $\left(C_2\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $\mathrm{cm}$ ) by calipers $C_1$ and $C_2$, respectively, are

$2.85$ and $2.82$

$2.87$ and $2.83$

$2.87$ and $2.86$

$2.87$ and $2.87$

(IIT-2016)

Solution

$C_1: 1$ Main scale division MSD $=0.1 cm$

1 vernier scale division VSD $=\frac{0.9}{10}=0.09 cm$

Least count $L C=1 M S D-1 V S D=0.1-0.09=0.01 cm$

Main scale reading MSR $=2.8 cm$

Number of vernier division matching main scale division $n=7$

Reading $R_1=M S R+n \times L C=2.8+7 \times 0.01=2.87 cm$

$C_2: 1$ Main scale division $MSD$ $=0.1 cm$

1 vernier scale division $V S D=\frac{1.1}{10}=0.11 cm$

Least count $L C=1 V S D-1 M S D=0.11-0.1=0.01 cm$

Main scale reading $MSR$ $=2.8 cm$

Here we should note that size of vernier scale division is bigger than main scale division, hence number of vernier division matching main scale division will be counted form back. In given figure we can see that 7th vernier scale division is matching so we shall consider $10-7=3$ rd.

Number of vernier division matching main scale division $(n)=3$

Reading $R_2=M S R+n \times L C=2.8+3 \times 0.01=2.83 cm$

Standard 11

Physics