In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4^ 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 ~mm then how many main scale divisions are there in 1 ~cm ?

Units and Measurements PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q66 with Solution

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Correct answer: D

Since the zero of the vernier scale shifts towards left when the jaws are closed, the instrument has a negative zero error.

Also, the 4th vernier division coincides with a main scale division.

For a vernier caliper, if the $n^{\text{th}}$ vernier division coincides in case of negative zero error, then the magnitude of zero error is:

\text{Zero error} = n \times \text{Least count}

Given:

\text{Zero error} = 0.04\,\text{mm}, \quad n=4

So,

4 \times \text{LC} = 0.04

\text{LC} = \frac{0.04}{4} = 0.01\,\text{mm}

Given:

50\,\text{VSD} = 49\,\text{MSD}

So,

1\,\text{VSD} = \frac{49}{50}\,\text{MSD}

Least count is:

\text{LC} = 1\,\text{MSD} - 1\,\text{VSD}

\text{LC} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}

Thus,

\frac{1}{50}\,\text{MSD} = 0.01\,\text{mm}

1\,\text{MSD} = 50 \times 0.01 = 0.5\,\text{mm}

Since,

1\,\text{cm} = 10\,\text{mm}

Number of MSD in $1\,\text{cm}$ :

\frac{10}{0.5} = 20

Therefore, the correct option is:

\boxed{20}

So, Option D is correct.

View written solutionFree

Correct answer: D

Since the zero of the vernier scale shifts towards left when the jaws are closed, the instrument has a negative zero error.

Also, the 4th vernier division coincides with a main scale division.

For a vernier caliper, if the $n^{\text{th}}$ vernier division coincides in case of negative zero error, then the magnitude of zero error is:

\text{Zero error} = n \times \text{Least count}

Given:

\text{Zero error} = 0.04\,\text{mm}, \quad n=4

So,

4 \times \text{LC} = 0.04

\text{LC} = \frac{0.04}{4} = 0.01\,\text{mm}

Given:

50\,\text{VSD} = 49\,\text{MSD}

So,

1\,\text{VSD} = \frac{49}{50}\,\text{MSD}

Least count is:

\text{LC} = 1\,\text{MSD} - 1\,\text{VSD}

\text{LC} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}

Thus,

\frac{1}{50}\,\text{MSD} = 0.01\,\text{mm}

1\,\text{MSD} = 50 \times 0.01 = 0.5\,\text{mm}

Since,

1\,\text{cm} = 10\,\text{mm}

Number of MSD in $1\,\text{cm}$ :

\frac{10}{0.5} = 20

Therefore, the correct option is:

\boxed{20}

So, Option D is correct.