The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale,

Question:

The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is $1 \mathrm{~mm}$. The main scale reading is $10 \mathrm{~mm}$ and $8^{\text {th }}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04 \mathrm{~cm}$, then

the radius of the bob is $\times 10^{-2} \mathrm{~cm}$

Solution:

$9 \mathrm{MSD}=10 \mathrm{VSD}$

$9 \times 1 \mathrm{~mm}=10 \mathrm{VSD}$

$\therefore 1 \mathrm{VSD}=0.9 \mathrm{~mm}$

$\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=0.1 \mathrm{~mm}$

Reading $=\mathrm{MSR}+\mathrm{VSR} \times \mathrm{LC}$

$10+8 \times 0.1=10.8 \mathrm{~mm}$

Actual reading $=10.8-0.4=10.4 \mathrm{~mm}$

radius $=\frac{\mathrm{d}}{2}=\frac{10.4}{2}=5.2 \mathrm{~mm}$

$=52 \times 10^{-2} \mathrm{~cm}$

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