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Chapter 21: Problem 2768

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) \((1 / 8) \mathrm{mm}\)

Short Answer

Expert verified
The least count of the unusual vernier scale is \(\frac{1}{9}\) mm, which corresponds to option (B).

Step by step solution

01

Determine the values of the main scale and vernier scale divisions

We know that 9 vernier scale divisions coincide with 8 main scale divisions. Let's assume the length of one main scale division is M, and one vernier scale division is V. Since 8 main scale divisions are equivalent to 9 vernier scale divisions, we can write the equation: \(8M = 9V\)
02

Find the least count (LC) of the vernier scale

The least count of a vernier scale is defined as the difference between the values of one main scale division (M) and one vernier scale division (V). Least Count (LC) = M - V We will now use the equation from step 1 to express V in terms of M: \(V = \frac{8}{9}M\)
03

Substitute the value of V in the least count formula

Now that we have found the value of V in terms of M, we can substitute it back into the least count formula: LC = M - \(\frac{8}{9}M\)
04

Calculate the least count

To calculate the least count, we need to solve for LC: LC = \(\frac{9}{9}M - \frac{8}{9}M\) LC = \(\frac{1}{9}M\) Since the main scale is given in millimeters (mm), the least count will also be in millimeters. LC = \(\frac{1}{9}\) mm The least count of the vernier scale is \(\frac{1}{9}\) mm, which corresponds to option (B).

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Most popular questions from this chapter

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

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) \((1 / 8) \mathrm{mm}\)

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) \(+0.7 \mathrm{~mm}\)

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) $2.66 \times 10^{-6}\left\\{\mathrm{~g} /\left(\mathrm{cm}^{3}\right)\right\\}$

N divisions on the main scale of a vernier calipers coin sides with \((\mathrm{N}+1)\) divisions on the vernier scale. If each division on the main scale is of a units, the least of count of instrument is................ (A) \(\\{\mathrm{a} /(\mathrm{N}+1)\\}\) (B) \(\\{\mathrm{a} /(\mathrm{N}-1)\\}\) (C) \(\\{(N+1) / a\\}\) (D) \(\\{(N-1) / a\\}\)
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