Q62 / 90 · JEE Main 2024 (06 Apr Shift 1) · Physics

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is $1 \mathrm{~mm}$ and circular scale reading is equal to 42 divisions. Pitch of screw gauge is $1 \mathrm{~mm}$ and it has 100 divisions on circular scale. The diameter of the wire is $\frac{x}{50} \mathrm{~mm}$. The value of $x$ is :

  • A
    21
  • B
    142
  • Correct answer
    C
    71
  • D
    42

Solution

$\begin{aligned} & \mathrm{MSR}=1 \mathrm{~mm}, \mathrm{CSR}=42, \text { pitch }=1 \mathrm{~mm} \\ & \mathrm{LC}=\frac{\text { pitch }}{\text { No. of CSD }}=\left(\frac{1}{100}\right)=0.01 \mathrm{~mm} \\ & \text { Diameter }=\mathrm{MSR}+\mathrm{LC} \times \mathrm{CSD} \\ & \text { Diameter }=1+(0.01) \times 42 \mathrm{~mm} \\ & \text { Diameter }=1.42 \mathrm{~mm}=\frac{\mathrm{x}}{50} \\ & \therefore \mathrm{x}=71\end{aligned}$

Hence, the correct answer is option: 'C'.

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