$\begin{aligned}
& 50 \mathrm{VSD}=49 \mathrm{MSD} \
& 1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD} \
& \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \
& \mathrm{LC}=1 \mathrm{MSD} \cdot \frac{49}{50} \mathrm{MSD}=\frac{\mathrm{MSD}}{50} \ldots . . . . . .(1) \
& \text {Now, } | \text { zero error }|=| 0-4(\mathrm{L} . \mathrm{C}) | \
& 0.04=|-4(\mathrm{L} . \mathrm{C})| \Rightarrow \mathrm{L} . \mathrm{C}=0.01 \mathrm{mm} \
& \mathrm{0 . 0 4}=|-\mathrm{4}(\mathrm{L} . \mathrm{C})| \Rightarrow \mathrm{L} . \mathrm{C} \ldots . . .(2) \
& =\mathrm{0 . 0 1} \mathrm{mm}
\end{aligned}Using(1)and(2)\begin{aligned}
& \frac{M S D}{50}=0.01 \
& M S D=0.5 \mathrm{mm} \
& M S D=\frac{1 \mathrm{cm}}{\mathrm{N}} \
& \mathrm{N}=\frac{1 \mathrm{cm}}{0.5 \mathrm{mm}}=\frac{100 \mathrm{mm}}{5 \mathrm{mm}} \
& \mathrm{N}=20
\end{aligned}$
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…
JEE Main 2024 — Physics Mechanics
In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4th 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 mm then how many main scale divisions are there in 1 cm ?
Held on 6 Apr 2024 · Verified 6 Jul 2026.
Options
- A
10
- B
5
- C
20
- D
40
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