Vernier Calipers and Screw Gauge

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LEVELJEE Main

The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2.

The diameter of a cylinder is measured using a vernier calipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 cm$ and $5.15 cm$ of the main scale. The vernier scale has $50$ division equivalent to $2.45 cm$ . The $2 4^{th}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

(A)

5.112 cm

(B)

5.124 cm

(C)

5.136 cm

(D)

5.148 cm

JEE Main 2010

LEVELJEE Main

A vernier calipers has $1 mm$ marks on the main scale. It has $20$ equal divisions on the vernier scale which match with $16$ main scale divisions. For this vernier calipers, the least count is

(A)

0.02 mm

(B)

0.05 mm

(C)

0.1 mm

(D)

0.2 mm

LEVELJEE Main

$N$ divisions on the main scale of a vernier calipers coincide with $(N + 1)$ divisions on the vernier scale. If each division on the main scale is of $a$ units, determine the least count of instrument.

JEE Advanced 2015

LEVELJEE Advanced

Consider a vernier caliper in which each $1 cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the vernier callipers, $5$ divisions of the vernier scale coincide with $4$ divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then

* Multiple Correct Options

(A)

if the pitch of the screw gauge is twice the least count of the vernier caliper, the least count of the screw gauge is

0.01 mm

(B)

if the pitch of the screw gauge is twice the least count of the Vernier caliper, the least count of the screw gauge is

0.005 mm

(C)

if the least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers, the least count of the screw gauge is

0.01 mm

(D)

if the least count of the linear scale of the screw gauge is twice the least count of the vernier caliper, the least count of the screw gauge is

0.005 mm

JEE Advanced 2016

LEVELJEE Advanced

There are two vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The vernier scale of one of the calipers ( $C_{1}$ ) has 10 equal divisions that correspond to 9 main scale divisions. The vernier scale of the other caliper ( $C_{2}$ ) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers $C_{1}$ and $C_{2}$ respectively, are

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $D$ of a tube. The measured value of $D$ is:

Two thin wires, Wire-1 of diameter $0.650 mm$ and Wire-2 of unknown diameter $d$ are given. To obtain the value of $d$ , the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of $0.5 mm$ and there are $100$ divisions on the circular scale (CS). The smallest division on the linear scale (LS) is $0.5 mm$ . The table shows the readings of LS and CS for the measurements. The value of $d$ (in $μ m$ ) is: $\cline 2 - 3 Wire-1 Wire-2 \multicolumn 2 c ∣ Readings LS (mm) 0.5 1.5 CS 4295$

JEE Advanced 2022

LEVELJEE Advanced

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$ . The circular scale has $100$ divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below. \begin{array}{|l|c|c|} \hline \textbf{Measurement condition} & \textbf{Main scale reading} & \textbf{Circular scale reading} \\ \hline \text{Two arms of gauge touching} & & \\ \text{each other without wire} & 0\text{ division} & 4\text{ division} \\ \hline \text{Attempt-1: With wire} & 4\text{ divisions} & 20\text{ divisions} \\ \hline \text{Attempt-2: With wire} & 4\text{ divisions} & 16\text{ divisions} \\ \hline \end{array} What are the diameter and cross-sectional area of the wire measured using the screw gauge?

(A)

2.22 \pm 0.02 mm, π (1.23 \pm 0.02) mm^{2}

(B)

2.22 \pm 0.01 mm, π (1.23 \pm 0.01) mm^{2}

(C)

2.14 \pm 0.02 mm, π (1.14 \pm 0.02) mm^{2}

(D)

2.14 \pm 0.01 mm, π (1.14 \pm 0.01) mm^{2}

JEE Main 2021

LEVELJEE Advanced

The smallest division on the main scale of a Vernier calipers is $0.1 cm$ . Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

One main scale division of a vernier callipers is $a$ cm and $n$ th division of the vernier scale coincide with $(n - 1)$ th division of the main scale. The least count of the callipers (in mm) is

(A)

\frac{10 na}{( n - 1 )}

(B)

\frac{10 a}{( n - 1 )}

(C)

(\frac{n - 1}{10 n}) a

(D)

\frac{10 a}{n}

LEVELJEE Main

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : $58. 5^{\circ}$ Vernier scale reading : $09$ divisions Given that $1$ division on main scale corresponds to $0. 5^{\circ}$ . Total divisions on the vernier scale is $30$ and match with $29$ divisions of the main scale. The angle of the prism from the above data is

(A)

58.5 9^{\circ}

(B)

58.7 7^{\circ}

(C)

58.6 5^{\circ}

(D)

5 9^{\circ}

LEVELJEE Main

In an experiment, the angles are required to be measured using an instrument. $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree ( $= 0. 5^{\circ}$ ), then the least count of the instrument is

Assertion (A) If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5 mm$ and there are $50$ total divisions on circular scale, then least count is $0.001 cm$ . Reason (R) Least count = $\frac{Pitch}{Total divisions on circular scale}$ In the light of the above statements, choose the most appropriate answer from the options given below.

(A)

Both A and R are correct and R is the correct explanation of A.

(B)

Both A and R are correct and R is not the correct explanation of A.

(C)

A is correct but R is not correct.

(D)

A is not correct but R is correct.

JEE Main 2020

LEVELJEE Main

The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10th division and coincide with 9th division of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale lies between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division)

The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of a wire is

(A)

50

(B)

200

(C)

500

(D)

100

JEE Main 2021

LEVELJEE Advanced

Student A and student B used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322 cm$ . The absolute value of the difference between the final circular scale readings observed by the students A and B is ......... . [Figure shows position of reference O when jaws of screw gauge are closed] Given, pitch $= 0.1 cm$ .

LEVELJEE Main

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading $0 mm$ Circular scale reading $52$ divisions Given that $1 mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is

(A)

0.052 cm

(B)

0.026 cm

(C)

0.005 cm

(D)

0.52 cm

JEE Main 2021

LEVELJEE Main

The pitch of the screw gauge is $1 mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while $7 2^{nd}$ division on circular scale coincides with the reference line. The radius of the wire is

(A)

1.64 mm

(B)

0.82 mm

(C)

1.80 mm

(D)

0.90 mm

JEE Main 2021

LEVELJEE Main

In a screw gauge, 5th division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.

A student measured the length of a rod and wrote it as $3.50 cm$ . Which instrument did he use to measure it?

(A)

A meter scale

(B)

A vernier calliper where the

10

divisions in vernier scale matches with

9

divisions in main scale and main scale has

10

divisions in

1 cm

(C)

A screw gauge having

100

divisions in the circular scale and pitch as

1 mm

(D)

A screw gauge having

50

divisions in the circular scale and pitch as

1 mm

JEE Main 2021

LEVELJEE Main

The vernier scale used for measurement has a positive zero error of $0.2 mm$ . If while taking a measurement, it was noted that '0' on the vernier scale lies between $8.5 cm$ and $8.6 cm$ , vernier coincidence is $6$ , then the correct value of measurement is ……… $cm$ .

(A)

8.36

(B)

8.54

(C)

8.58

(D)

8.56

LEVELJEE Main

Two full turns of the circular scale of a screw gauge cover a distance of $1 mm$ on its main scale. The total number of divisions on the circular scale is $50$ . Further, it is found that the screw gauge has a zero error of $- 0.03 mm$ . While measuring the diameter of a thin wire, a student notes the main scale reading of $3 mm$ and the number of circular scale divisions in line with the main scale as $35$ . The diameter of the wire is

(A)

3.32 mm

(B)

3.72 mm

(C)

3.67 mm

(D)

3.38 mm

JEE Main 2021

LEVELJEE Main

The diameter of a spherical bob is measured using a Vernier callipers. 9 divisions of the main scale, in the vernier calipers, are equal to 10 divisions of vernier scale. One main scale division is $1 mm$ . The main scale reading is $10 mm$ and 8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04 cm$ , then the radius of the bob is .......... $\times 1 0^{- 2} cm$ .

JEE Main 2020

LEVELJEE Main

Using screw gauge of pitch $0.1 cm$ and $50$ divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

(A)

2.121 cm

(B)

2.124 cm

(C)

2.125 cm

(D)

2.123 cm

The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5μm diameter of a wire is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of a wire is

A student measured the length of a rod and wrote it as $3.50 cm$ . Which instrument did he use to measure it?

Using screw gauge of pitch $0.1 cm$ and $50$ divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as