Home
Class 11
PHYSICS
In a vernier callipers, N divisions of v...

In a vernier callipers, N divisions of vernier scale coincide with (N-1) divisions of main scale (in which 1 division represents 1 mm). The least count of the instrument in cm should be

A

`N`

B

`N-1`

C

`(1)/(10N)`

D

`(1)/(N-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the least count of the vernier calipers in centimeters, we can follow these steps: ### Step 1: Understand the relationship between the divisions of the scales In the given problem, we know that N divisions of the vernier scale coincide with (N - 1) divisions of the main scale. This means that: - N divisions of the vernier scale = (N - 1) divisions of the main scale ### Step 2: Calculate the value of one vernier scale division The length of one main scale division (MSD) is given as 1 mm. Therefore, the total length of (N - 1) main scale divisions is: \[ \text{Total length of (N - 1) MSD} = (N - 1) \times 1 \text{ mm} = (N - 1) \text{ mm} \] Since N divisions of the vernier scale coincide with this length, the length of one vernier scale division (VSD) can be calculated as: \[ \text{Length of one VSD} = \frac{(N - 1) \text{ mm}}{N} = \frac{(N - 1)}{N} \text{ mm} \] ### Step 3: Use the formula for least count The least count (LC) of the instrument is given by the formula: \[ \text{Least Count (LC)} = \text{1 MSD} - \text{1 VSD} \] Substituting the values we have: \[ \text{LC} = 1 \text{ mm} - \frac{(N - 1)}{N} \text{ mm} \] ### Step 4: Simplify the expression Now, we simplify the expression for LC: \[ \text{LC} = 1 \text{ mm} - \frac{(N - 1)}{N} \text{ mm} = \frac{N}{N} \text{ mm} - \frac{(N - 1)}{N} \text{ mm} \] \[ \text{LC} = \frac{N - (N - 1)}{N} \text{ mm} = \frac{1}{N} \text{ mm} \] ### Step 5: Convert the least count to centimeters Since we need the least count in centimeters, we convert mm to cm: \[ \text{LC in cm} = \frac{1}{N} \text{ mm} \times \frac{1 \text{ cm}}{10 \text{ mm}} = \frac{1}{10N} \text{ cm} \] ### Final Answer Thus, the least count of the vernier calipers in centimeters is: \[ \text{Least Count} = \frac{1}{10N} \text{ cm} \] ---

To find the least count of the vernier calipers in centimeters, we can follow these steps: ### Step 1: Understand the relationship between the divisions of the scales In the given problem, we know that N divisions of the vernier scale coincide with (N - 1) divisions of the main scale. This means that: - N divisions of the vernier scale = (N - 1) divisions of the main scale ### Step 2: Calculate the value of one vernier scale division ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN|Exercise EXERCISE-II AIPMT 2006|1 Videos
  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN|Exercise EXERCISE-II AIIMS 2006|1 Videos
  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN|Exercise EXERCISE-I ERRORS|18 Videos
  • MISCELLANEOUS

    ALLEN|Exercise SUBJECTIVE QUESTION|9 Videos
  • SEMICONDUCTORS

    ALLEN|Exercise Part-3(Exercise-4)|51 Videos

Similar Questions

Explore conceptually related problems

In a Vernier calipers, N divisions of Vernier scale coincides with (N-1 divisions of main scale (in which one division represent 2mm ).The least count of the instrument (in cm )

The side of a cube is measured by vernier callipers ( 10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1mm ). The main scale reads 10mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g . find the density of the cube in appropriate significant figures.

Knowledge Check

  • In a vernier calliper, N divisions of vernier scale coincide with (N-1) divisions of main scale (in which division represent 1mm). The least count of the instrument in cm. should be

    A
    N
    B
    `N-1`
    C
    `(1)/(10 N)`
    D
    `(1)/(N-1)`
  • A vernier callipers has 20 divisions on the vernier scale which coincide with 19 divisions on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are of

    A
    0.5 mm
    B
    1 mm
    C
    2 mm
    D
    `1//4` mm
  • The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n-1) divisions of main scale. The least count of the vernier callipers is,

    A
    `(1)/(n(n+1))cm`
    B
    `(1)/((n+1)(n-1))cm`
    C
    `(1)/(n)cm`
    D
    `(1)/(n^2)cm`
  • Similar Questions

    Explore conceptually related problems

    In a vernier callipers, N divisions of the main scale coincide with N + m divisions of the vernier scale. what is the value of m for which the instrument has minimum least count.

    A vernier calliper has 20 divisions on the vernier scale. One centimetre on the main scale is divided into 20 equal parts. Find the least count of this instrument.

    N divisions on the main scale of a vernier callipers coincide with (N+1) divisions of the vernier scale if each division of the main scale is 'a' units, then the least count of the instrument is

    A vernier callipers has 20 divisions on the vernier scale which coincides with 19mm on the main scale. Its least count is

    In a vernier callipers, one main scale division is x cm and n division of the vernier scale coincide with (n-1) divisions of the main scale. The least count (in cm) of the callipers is: