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 _______.
Pick an option to check your answer.
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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