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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2024
  • Hard
  • MCQ

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 _______.

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Answer & Solution

Correct Answer

(B) \(71\)

Step-by-step Solution

Detailed explanation

\(\mathrm{MSR}=1 \mathrm{~mm}, \mathrm{CSR}=42, \text { pitch }=1 \mathrm{~mm}\)

\(\mathrm{LC}=\frac{\text { pitch }}{\text { No. of } \mathrm{CSD}}=\left(\frac{1}{100}\right)=0.01 \mathrm{~mm}\)

\(\text { Diameter }=\mathrm{MSR}+\mathrm{LC} \times \mathrm{CSD}\)

\(\text { Diameter }=1+(0.01) \times 42 \mathrm{~mm}\)

\(\text { Diameter }=1.42 \mathrm{~mm}=\frac{\mathrm{x}}{50}\)

\(\therefore \mathrm{x}=71\)

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