Experimental Skills | Physics JEE previous year questions with Solutions

Q 1 :

A Vernier callipers has 20 divisions on the Vernier scale, which coincides with 19th division on the main scale. The least count of the instrument is 0.1 mm. One main scale division is equal to _______ mm. [2024]

1
5
2
0.5

1

2

3

4

(3)

$20$ $V S D = 19$ $V S D \Rightarrow 1$ $V S D = \frac{19}{20}$ $M S D$

$L . C . = 1$ $M S D - 1$ $V S D = 1$ $M S D - \frac{19}{20}$ $M S D$

$0.1$ $mm = \frac{1}{20}$ $M S D$

$Hence, 1$ $M S D = 2 mm$

Q 2 :

In a Vernier calliper, when both jaws touch each other, zero or the Vernier scale shifts towards left and its 4th division coincides exactly with a certain division on main scale. If 50 Vernier scale divisions are equal to 49 main scale divisions and zero error in the instrument is 0.4 mm, then how many main scale divisions are there in 1 cm? [2024]

5
20
10
40

1

2

3

4

(2)

$50$ $V S D = 49$ $M S D \Rightarrow 1$ $V S D = \frac{49}{50}$ $M S D$

$L C = 1$ $M S D - 1$ $V S D = 1$ $M S D - \frac{49}{50}$ $M S D = \frac{M S D}{50}$

$N o w, | z e r o e r r o r | = | 0 - 4 (L . C) |$

$0.04 = | - 4 (L . C) | \Rightarrow L . C . = 0.01 m m$

$Using (i) and (i i), \frac{M S D}{50} = 0.01$

$M S D = 0.5 mm \Rightarrow M S D = \frac{1 cm}{N}$

$N = \frac{1 cm}{0.5 mm} = \frac{100 mm}{5 mm} = 20$

Q 3 :

The diameter of a sphere is measured using a Vernier calliper whose 9 divisions of main scale are equal to 10 divisions of Vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of Vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is [2024]

$2.5$ $g / c m^{3}$
$1.7$ $g / c m^{3}$
$2.0$ $g / c m^{3}$
$2.2$ $g / c m^{3}$

1

2

3

4

(3)

$Given : 9$ $M S D = 10$ $V S D, m = 8.635$ $g$

$L C = 1$ $M S D - V S D$

$L C = 1$ $M S D - \frac{9}{10} M S D = \frac{1}{10} M S D \Rightarrow LC = 0.01$ $cm$

$Reading of diameter = M S R + L C \times V S R$

$= 2 cm + (0.01) \times (2) = 2.02$ $c m$

$Volume of sphere = \frac{4}{3} π {(\frac{d}{2})}^{3} = \frac{4}{3} π {(\frac{2.02}{2})}^{3} = 4.32 {cm}^{3}$

$Density = \frac{mass}{volume} = \frac{8.635}{4.32} = 1.998 ~ 2.00$ ${g / cm}^{3}$

Q 4 :

Least count of a vernier calliper is $\frac{1}{20 N}$ cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is [2024]

$(\frac{2 N - 1}{2 N})$
$(2 N - 1)$
$(\frac{2 N - 1}{2})$
$(\frac{2 N - 1}{20 N})$

1

2

3

4

(3)

$Least count of vernier callipers = \frac{1}{20 N} cm$

$∵$ $Least count = 1$ $M S D - 1$ $V S D$

$Let x number of divisions of the main scale coincides with N divisions of the vernier scale, then$

$1$ $V S D = \frac{x \times 1 mm}{N}$

$∴ \frac{1}{20 N} c m = 1 m m - \frac{x \times 1 m m}{N}$

$\frac{1}{2 N}$ $m m = 1$ $m m - \frac{x}{N} m m$

$x = (1 - \frac{1}{2 N}) N = \frac{2 N - 1}{2}$

Q 5 :

One main scale division of a vernier calliper is equal to $m$ units. If $n^{t h}$ division of main scale coincides with ${(n + 1)}^{t h}$ division of vernier scale, the least count of the vernier calliper is: [2024]

$\frac{n}{(n + 1)}$
$\frac{m}{(n + 1)}$
$\frac{1}{(n + 1)}$
$\frac{m}{n (n + 1)}$

1

2

3

4

(2)

Q 6 :

If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of travelling microscope is [2024]

0.1 mm
0.1 cm
0.01 cm
0.01 mm

1

2

3

4

(4)

$50$ $V S D = 49$ $M S D$

$1$ $V S D = \frac{49}{50}$ $M S D$

Vernier constant = L.C.

$L . C . = 1$ $M S D - 1$ $V S D$

$= 1$ $M S D - \frac{49}{50}$ $M S D$

$L . C . = \frac{M S D}{50}$

$1$ $M S D = 0.5$ $mm (g iven)$

So, $L . C . = \frac{0.5 mm}{50} = 0.01 mm$

Q 7 :

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.

Reason (R): The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling.

In the light of the above statements, choose the correct answer from the given below: [2024]

Both (A) and (R) are correct and (R) is the correct explanation of (A)
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(A) is true but (R) is false
(A) is false but (R) is true

1

2

3

4

(2)

Q 8 :

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is [2024]

$\frac{1}{2}$
$\frac{10}{11}$
$\frac{50}{11}$
$\frac{5}{11}$

1

2

3

4

(4)

$10$ $M S D = 11$ $V S D$

$1$ $V S D = \frac{10}{11}$ $M S D$

$L C = 1$ $M S D - 1$ $V S D = 1$ $M S D - \frac{10}{11}$ $M S D$

$= \frac{1 M S D}{11} = \frac{5}{11} units$

Q 9 :

In finding out refractive index of glass slab, the following observations were made through travelling microscope 50 vernier scale division = 49 MSD; 20 divisions on main scale in each cm

For mark on paper
$M S R = 8.45 c m, V C = 26$

For mark on paper seen through slab
$M S R = 7.12 c m, V C = 41$

For powder particle on the top surface of the glass slab
$M S R = 4.05 c m, V C = 1$

(MSR = Main Scale Reading, VC = Vernier Coincidence)

Refractive index of the glass slab is : [2024]

1.24
1.35
1.42
1.52

1

2

3

4

(3)

$MSD = \frac{1}{20} cm = 0.05 cm$

$50 VSD = 49 MSD \Rightarrow 1 VSD = \frac{49}{50} MSD$

$So, L . C = 1 MSD - 1 VSD = 1 MSD - \frac{49}{50} MSD$

$L . C = \frac{MSD}{50} = \frac{0.05 cm}{50} = 0.001 cm$

$t = 8.45 + 26 \times (0.001) = 8.476 cm$

$t' = 7.12 + 41 \times (0.001) = 7.161 cm$

$Width of powder particle = 4.05 + 1 \times (0.001) = 4.051 cm$

$t_{real} = 8.476 - 4.051 = 4.425 cm$

$t_{app} = 7.161 - 4.051 = 3.11 cm$

$μ_{glass} = \frac{t_{real}}{t_{app}} = \frac{4.425}{3.11} = 1.42$

Q 10 :

Given below are two statements:

Statement I: In a Vernier callipers, one Vernier scale division is always smaller than one main scale division.

Statement II: The Vernier constant is given by one main scale division multiplied by the number of Vernier scale division.

In the light of the above statements, choose the correct answer from the options given below: [2025]

Both Statement I and Statement II are false.
Statement I is true but Statement II is false.
Both Statement I and Statement II are true.
Statement I is false but Statement II is true.

1

2

3

4

(2)

In general, one Vernier scale division is smaller than one main scale division.

Vernier constant = 1 main scale division – 1 vernier scale division

Statement-II is not true

Q 11 :

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm): [2025]

0.001
0.002
0.0005
0.0025

1

2

3

4

(2)

$300 msd = 15 cm$

$1 msd = \frac{15}{300} = 0.05 cm$

$25 vsd = 24 msd$

$1 vsd = \frac{24}{25} msd$

$LC = 1 msd - 1 vsd$

$LC = 1 msd - \frac{24}{25} msd = \frac{1}{25} msd$

$LC = \frac{1}{25} \times 0.05 = 0.002 cm$

Q 12 :

In an experiment with Vernier callipers of least count 0.1 mm, when two jaws are joined together the zero of Vernier scale lies right to the zero of the main scale and 6th division of Vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between 3.2 cm and 3.3 cm marks, and 4th division of vernier scale coincides with the main scale division. The diameter of bob is measured as [2023]

3.18 cm
3.25 cm
3.26 cm
3.22 cm

1

2

3

4

(1)

$L C = 0 . 1 mm$

$Zero Error = 6 \times L C = 0.6 mm$

$Reading = MSR + VSR \times L C - Zero Error$

$= [32 mm + (0.1) 4 mm] - 0.6 mm$

$= 31.8 mm = 3.18 cm$

Q 13 :

In an experiment, a set of reading are obtained as follows – 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is ______ mm. [2026]

0.05
0.01
0.001
0.1

1

2

3

4

(2)

Least count will be 0.01 mm.

Q 14 :

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm. [2026]

0.05
0.02
0.002
0.005

1

2

3

4

(3)

$LC = 1 MSD - 1 MSD = 1 MSD - \frac{48}{50} MSD$

$= \frac{2}{50} MSD = \frac{2}{50} \times 0.05 mm = 0.002 mm$

Q 15 :

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, 4th mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and 5th division of vernier scale coincides with a main scale division. Measured length of cylinder is ______ mm.
(Least count of Vernier calliper = 0.1 mm) [2026]

15.4
15.1
15.9
15.5

1

2

3

4

(2)

$Reading = MSR + (VSR \times LC) - (zero error)$

$= 15 mm + (5 \times 0.1 mm) - (4 \times 0.1 mm)$

$Reading = 15.1 mm$

$∴$ $ℓ = 15.1 mm$

Topic Question Set

Q 1 :

A Vernier callipers has 20 divisions on the Vernier scale, which coincides with 19th division on the main scale. The least count of the instrument is 0.1 mm. One main scale division is equal to _______ mm. [2024]

Q 4 :

Least count of a vernier calliper is $\frac{1}{20 N}$ cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is [2024]

Q 5 :

One main scale division of a vernier calliper is equal to $m$ units. If $n^{t h}$ division of main scale coincides with ${(n + 1)}^{t h}$ division of vernier scale, the least count of the vernier calliper is: [2024]

(2)

Q 6 :

If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of travelling microscope is [2024]

(2)

Q 8 :

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is [2024]

Q 13 :

In an experiment, a set of reading are obtained as follows – 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is ______ mm. [2026]

Q 14 :

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm. [2026]

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Topic Question Set

Q 1 : A Vernier callipers has 20 divisions on the Vernier scale, which coincides with 19th division on the main scale. The least count of the instrument is 0.1 mm. One main scale division is equal to _______ mm. [2024]

Q 4 : Least count of a vernier calliper is 120N cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is [2024]

Q 5 : One main scale division of a vernier calliper is equal to m units. If nth division of main scale coincides with (n+1)th division of vernier scale, the least count of the vernier calliper is: [2024]

(2)

Q 6 : If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of travelling microscope is [2024]

(2)

Q 8 : 10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is [2024]

Q 13 : In an experiment, a set of reading are obtained as follows – 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is ______ mm. [2026]

Q 14 : In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm. [2026]

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Q 1 :

A Vernier callipers has 20 divisions on the Vernier scale, which coincides with 19th division on the main scale. The least count of the instrument is 0.1 mm. One main scale division is equal to _______ mm. [2024]

Q 4 :

Least count of a vernier calliper is $\frac{1}{20 N}$ cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is [2024]

Q 5 :

One main scale division of a vernier calliper is equal to $m$ units. If $n^{t h}$ division of main scale coincides with ${(n + 1)}^{t h}$ division of vernier scale, the least count of the vernier calliper is: [2024]

Q 6 :

If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of travelling microscope is [2024]

Q 8 :

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is [2024]

Q 13 :

In an experiment, a set of reading are obtained as follows – 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is ______ mm. [2026]

Q 14 :

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm. [2026]