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QJEEM22Ph290a: JEE Main 2022 Physics - Class 11 JEEM22Ph290a Solution: In a Vernier Calipers, 10 divisions of Vernier scale is equal…

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th 4^{\text {th }} Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm1 \mathrm{~mm} . While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6th 6^{\text {th }} Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :

Multiple Choice Options

  • A 3.02 cm
  • B 3.06 cm
  • C 3.10 cm
  • D 3.20 cm

Step-by-Step Verified Solution

Given, In Vernier calipers, 10 VSD =9=9 MSD

⇒1\Rightarrow \quad 1 VSD =910=\frac{9}{10} MSD

∴\therefore Least count of vernier scale,

 LC amp;=1MSD−1VSDamp;=1MSD−910MSD=MSD⁡(1−910)amp;= MSD 10=1 mm10[∵1MSD=1 mm]amp;=0.1 mm=0.01 cm\begin{aligned}\text { LC } &=1 \mathrm{MSD}-1 \mathrm{VSD} \\\\&=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}=\operatorname{MSD}\left(1-\frac{9}{10}\right) \\\\&=\frac{\text { MSD }}{10}=\frac{1 \mathrm{~mm}}{10} \quad[\because 1 \mathrm{MSD}=1 \mathrm{~mm}] \\\\&=0.1 \mathrm{~mm}=0.01 \mathrm{~cm}\end{aligned}

According to given situation,

Negative error == Main scale reading - Least count ×\times Number of coinciding main scale division

=0.1−0.01×4=0.1−0.04=0.06 cm=0.1-0.01 \times 4=0.1-0.04=0.06 \mathrm{~cm}

∴\therefore Diameter of spherical body

amp;=30×0.1+6×0.01+0.06amp;=3.0+0.06+0.06=3.12 cm\begin{aligned}&=30 \times 0.1+6 \times 0.01+0.06 \\\\&=3.0+0.06+0.06=3.12 \mathrm{~cm}\end{aligned}

Which is closest to 3.10 cm3.10 \mathrm{~cm} .

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