Experimental Physics

JEE Main level, single correct, medium

15 of 70

Question 15 of 70

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4th 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 mm0.04 \mathrm{~mm} then how many main scale divisions are there in 1 cm1 \mathrm{~cm} ?

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Answer: (C) 20

\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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