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Units and Measurements question

2024 · 9 Apr · Shift 1 · Q73
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Units and Measurements question

2024 · 9 Apr · Shift 1 · Q73

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
One main scale division of a vernier caliper is equal to m\mathrm{m}m units. If nth \mathrm{n}^{\text {th }}nth  division of main scale coincides with (n+1)th (n+1)^{\text {th }}(n+1)th  division of vernier scale, the least count of the vernier caliper is :
  1. A
    1(n+1)\frac{1}{(\mathrm{n}+1)}(n+1)1​
  2. B
    m(n+1)\frac{m}{(n+1)}(n+1)m​
  3. C
    n(n+1)\frac{n}{(n+1)}(n+1)n​
  4. D
    mn(n+1)\frac{\mathrm{m}}{\mathrm{n}(\mathrm{n}+1)}n(n+1)m​
View written solutionFree

Correct answer: B

  1. Let one main scale division be 1 MSD=m units.1\,\text{MSD} = m\,\text{units}.1MSD=munits.

  2. Given: nth division of main scale=(n+1)th division of vernier scale.n\text{th division of main scale} = (n+1)\text{th division of vernier scale}.nth division of main scale=(n+1)th division of vernier scale. So, n (MSD)=(n+1) (VSD).n\,(\text{MSD}) = (n+1)\,(\text{VSD}).n(MSD)=(n+1)(VSD).

  3. Substitute 1 MSD=m1\,\text{MSD} = m1MSD=m units: nm=(n+1) (VSD).n m = (n+1)\,(\text{VSD}).nm=(n+1)(VSD). Hence, 1 VSD=nmn+1.1\,\text{VSD} = \frac{nm}{n+1}.1VSD=n+1nm​.

  4. For a vernier caliper, least count is L.C.=1 MSD−1 VSD.\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD}.L.C.=1MSD−1VSD.

  5. Therefore, L.C.=m−nmn+1.\text{L.C.} = m - \frac{nm}{n+1}.L.C.=m−n+1nm​.

  6. Simplify: L.C.=m(n+1)−nmn+1=mn+1.\text{L.C.} = \frac{m(n+1)-nm}{n+1} = \frac{m}{n+1}.L.C.=n+1m(n+1)−nm​=n+1m​.

  7. So the least count is mn+1.\boxed{\frac{m}{n+1}}.n+1m​​.

  8. Checking options:

    • A: 1n+1\frac{1}{n+1}n+11​ — incorrect, missing factor mmm
    • B: mn+1\frac{m}{n+1}n+1m​ — correct
    • C: nn+1\frac{n}{n+1}n+1n​ — incorrect
    • D: mn(n+1)\frac{m}{n(n+1)}n(n+1)m​ — incorrect

Hence, the correct option is B.

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