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What is the least count of commonly available vernier? (A) \(0.01 \mathrm{~cm}\) (B) \(0.001 \mathrm{~cm}\) (C) \(0.0001 \mathrm{~cm}\) (D) \(0.1 \mathrm{~cm}\)

Short Answer

Expert verified
The least count of a commonly available vernier is (A) \(0.01 \,cm\).

Step by step solution

01

Recognize the information given

In this problem, we are given four choices for the least count of a commonly available vernier. We need to find the correct one by knowing the actual least count of the vernier scale.
02

Calculate the least count of the vernier scale

To find the least count, we need to know the difference between the main scale and the vernier scale. The least count of the vernier scale can be calculated using the formula: \(Least \, Count = \frac{1 \, MSD}{10}\) where MSD is the least count of the main scale, which is usually 1 mm or 0.1 cm.
03

Compare the calculated least count with the given choices

Using the formula from Step 2, we can calculate the least count: \(Least \, Count = \frac{1 \, MSD}{10} = \frac{0.1 \, cm}{10} = 0.01 \, cm\)
04

Choose the correct answer

By comparing the calculated least count with the given choices, we can see that the correct answer is: (A) \(0.01 \,cm\).

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Most popular questions from this chapter

In an unusual vernier, 9 vernier scale divisions coincide with 8 main scale division, then what is the least count of the vernier? (A) \((8 / 9) \mathrm{mm}\) (B) \((1 / 9) \mathrm{mm}\) (C) \((1 / 17) \mathrm{mm}\) (D) \((1 / 8) \mathrm{mm}\)

When the jaws of a standard vernier are together, the \(6^{\text {th }}\) main scale division coincides with the \(7^{\text {th }}\) vernier scale division, then what is the zero error? (A) \(-0.7 \mathrm{~mm}\) (B) \(+0.3 \mathrm{~mm}\) (C) \(-0.3 \mathrm{~mm}\) (D) \(+0.7 \mathrm{~mm}\)

When the zero mark on the vernier scale lies towards the right side of the zero mark of the main scale, when the jaws are in contact, then what will be the zero error? (A) zero correction in positive (B) zero correction is negative (C) zero error in positive (D) zero error does not exist

The edge of a cube is measured using a vernier caliper \((9\) divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is \(1 \mathrm{~mm}\) ). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is \(2.736 \mathrm{~g}\). What will be the density in $\left\\{\mathrm{g} /\left(\mathrm{cm}^{3}\right)\right\\}$ upto correct significant figures? (A) $2.66 \times 10^{-3}\left\\{\mathrm{~g} /\left(\mathrm{cm}^{3}\right)\right\\}$ (B) $2.66 \times 10^{3}\left\\{\mathrm{~g} /\left(\mathrm{cm}^{3}\right)\right\\}$ (C) \(2.66\left\\{\mathrm{~g} /\left(\mathrm{cm}^{3}\right)\right\\}\) (D) $2.66 \times 10^{-6}\left\\{\mathrm{~g} /\left(\mathrm{cm}^{3}\right)\right\\}$

In an usual vernier, 10 vernier scale divisions, coin side with 8 main scale divisions, then what is the least count of the vernier? (A) \(0.1 \mathrm{~mm}\) (B) \(0.2 \mathrm{~mm}\) (C) \(0.8 \mathrm{~mm}\) (D) \((1 / 8) \mathrm{mm}\)

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