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

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) \(0.1 \mathrm{~cm}\)

What is the least count of the vernier calipers? (A) Smallest division on the vernier scale. (B) difference of the smallest division on the main scale and the smallest division on the vernier scale. (C) sum of the smallest division on the main scale and the smallest division on the vernier scale. (D) smallest division on the main scale.

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) \(\\{(N-1) / a\\}\)

If observed reading is OR, corrected reading is CR, zero error in \(\mathrm{ZE}\) and zero correction in \(\mathrm{ZC}\), then what will be the possibility? (A) \(\mathrm{CR}=\mathrm{OR}+\mathrm{ZC}\) and \(\mathrm{ZE}=\mathrm{CR}-\mathrm{OR}\) (B) \(\mathrm{CR}=\mathrm{OR}+\mathrm{ZE}\) and \(\mathrm{ZC}=\mathrm{CR}-\mathrm{OR}\) (C) \(\mathrm{CR}=\mathrm{OR}-\mathrm{ZC}\) and \(\mathrm{ZE}=\mathrm{OR}-\mathrm{CR}\) (D) \(\mathrm{CR}=\mathrm{OR}-\mathrm{ZE}\) and \(\mathrm{ZC}=\mathrm{CR}-\mathrm{OR}\)
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