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

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2

3

4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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]

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

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2

3

4

(B)

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

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

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

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

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(B)

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

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

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

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

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

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]

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]

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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]

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]

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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]