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

Correct answer: C. 71 aligned & MSR=1 mm, CSR=42, pitch =1 mm & LC= pitch No. of CSD =((1)/(100))=0.01 mm & Diameter =MSR+LC × CSD & Diameter =1+(0.01) × 42 mm & Diameter =1.42 mm= x 50 & x=71 aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Physics
Chapter: Experimental Physics

JEE Main 2024 · 6 April · Morning Shift · Physics · Experimental Physics

While measuring diameter of wire using screw gauge the following...

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 :

Options

A21
B142
C71
D42

Correct answer

C. 71

Step-by-step solution

\begin{aligned} & \mathrm{MSR}=1 \mathrm{ mm}, \mathrm{CSR}=42, \text{ pitch }=1 \mathrm{ mm} \\ & \mathrm{LC}=\frac{\text{ pitch }}{\text{ No.\text{ 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}

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