In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4^ th division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal

Correct answer: C. 20 aligned & 50 VSD=49 MSD & 1 VSD=(49)/(50) MSD & LC=1 MSD-1 VSD & LC=1 MSD · (49)/(50) MSD= MSD 50 . . . . . .(1) & Now, | zero error |=|^ 0 -4( L . C) | & 0.04=|-4( L . C)| L . C=0.01 mm & 0 . 0 4=|-4(L . C)| L . C . . .(2) & =0 . 0 1 mm aligned Using (1) and (2) aligned & (M S D)/(50)=0.01 & M S D=0.5 mm & M S D= 1 cm N & N= 1 cm 0.5 mm = 100 mm 5 mm & N=20 aligned

JEE Main 20246 AprilEvening Shift

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

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

In a vernier calliper, when both jaws touch each other, zero of the...

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its

4^{\text{th }}

division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is

0.04 \mathrm{ mm} then how many main scale divisions are there in

\mathrm{ cm}

Options

Correct answer

C. 20

Step-by-step solution

\begin{aligned} & 50 \mathrm{VSD}=49 \mathrm{MSD} \\ & 1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD} \\ & \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & \mathrm{LC}=1 \mathrm{MSD} \cdot \frac{49}{50} \mathrm{MSD}=\frac{\mathrm{MSD}}{50} \ldots . . . . . .(1) \\ & \text{Now, } | \text{ zero error }|=|^{0}-4(\mathrm{ L} . \mathrm{C}) | \\ & 0.04=|-4(\mathrm{ L} . \mathrm{C})| \Rightarrow \mathrm{L} . \mathrm{C}=0.01 \mathrm{ mm} \\ & \mathrm{0 . 0 4}=|-\mathrm{4}(\mathrm{L} . \mathrm{C})| \Rightarrow \mathrm{L} . \mathrm{C} \ldots . . .(2) \\ & =\mathrm{0 . 0 1} \mathrm{mm} \end{aligned}

Using (1) and (2)

\begin{aligned} & \frac{M S D}{50}=0.01 \\ & M S D=0.5 \mathrm{ mm} \\ & M S D=\frac{1 \mathrm{ cm}}{\mathrm{ N}} \\ & \mathrm{ N}=\frac{1 \mathrm{ cm}}{0.5 \mathrm{ mm}}=\frac{100 \mathrm{ mm}}{5 \mathrm{ mm}} \\ & \mathrm{ N}=20 \end{aligned}

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