Start free

Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 6 April, Shift 1 — Question 37

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm and it has 100 divisions on circular scale. The diameter of the wire is x50 mm\frac{x}{50} \mathrm{~mm}. The value of xx is :

  1. Option A:

    142

  2. Option B:

    71

    Correct
  3. Option C:

    42

  4. Option D:

    21

Answer: B

Step-by-step solution

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

LC= pitch  No. of CSD=(1100)=0.01 mm\mathrm{LC}=\frac{\text { pitch }}{\text { No. of } \mathrm{CSD}}=\left(\frac{1}{100}\right)=0.01 \mathrm{~mm}

Diameter =MSR+LC×CSD=\mathrm{MSR}+\mathrm{LC} \times \mathrm{CSD}

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

Diameter =1.42 mm=x50=1.42 \mathrm{~mm}=\frac{\mathrm{x}}{50}

∴x=71\therefore \mathrm{x}=71

Answer key and solution verified before publishing.

Practise Units, Dimensions & Error Analysis

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
6 April, Shift 1
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge

More Units, Dimensions & Error Analysis questions from this paper

While measuring diameter of wire using screw gauge the following… | JEE Main 2024 PYQ with Solution · DhiX AI