VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
To measure the diameter of small spherical cylindrical body using Vernier Callipers is considered one the most difficult concept.
39 Questions around this concept.
A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading: 58.5 degree.
Vernier scale reading: 09 divisions
Given that 1 division on main scale corresponds to 0.5 degree. 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:
The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of $(-0.05)_{\mathrm{cm}}$. While measuring the diameter of a sphere, the main scale reading is 1.7 cm, and the coinciding vernier division is 5. The corrected diameter will be________$\times 10^{-2} \mathrm{~cm}$.
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
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The diameter of a cylinder is measured using 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 divisions equivalent to 2.45 cm. The division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is :
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 of 0.1mm. The main scale divisions are of
In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $4^{\text {th }}$ 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 ?
In a vernier calliper, N divisions of the vernier scale coincide with (N-1) divisions of the main scale (in which 1 division represents 1mm). The least count of the instrument in cm should be
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
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One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is
Given below are two statements :
Statement I : In a vernier calliper, 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 divisions.
In the light of the above statements, choose the correct answer from the options given below.
A Vernier caliper is an instrument that measures internal or external dimensions and distances. It allows you to take more precise measurements than you could with regular rulers.
Important terminologies -
LEAST COUNT AND ZERO ERROR
The magnitude of the smallest measurement that can be measured by an instrument accurately is called its least count (L.C.).
The difference between one main scale division (M.S.D.) and one vernier scale division is called least count.
i.e. L.C. = One M.S.D. – One V.S.D.
ZERO ERROR
If there is no object between the jaws (i.e., jaws are in contact), the vernier should give zero reading. But due to some extra material on jaws, even if there is no object between the jaws, it gives some excess reading. This excess reading is called zero error.
Zero correction : Zero correction is invert of zero error.
Zero correction = – (Zero error)
Actual reading = observed reading – excess reading (zero error)
= observed reading + zero correction
Aim of the experiment
To measure diameter of a small sphericallcylindrical body using Vernier Callipers.
Apparatus
Vernier callipers, a spherical body (pendulum bob) or a cylinder and a magnifying lens.
Theory
If with the body between the jaws, the zero of vernier scale lies ahead of Nth division of main scale, then main scale reading (M.S.R.) = N
If with division of vernier scale coincides with any division of main scale, then vernier scale reading (V.S.R.)
= n × (L.C.) ( Here, L.C. is least count of vernier callipers)
= n × (V.C.) ( Here, V.C. is vernier constant of vernier callipers)
And total reading, T.R. = M.S.R. + V.S.R. = N + n × (V.C.)
Procedure
1. Determine the vernier constant (V.C.) i.e. least count (L.C.) of the vernier callipers and record it stepwise.
2. Bring the movable jaw BD in close contact with the fixed jaw AC and find the zero error. Do it three times and record them. If there is no zero error, record zero error nil.
3. Open the jaws, place the sphere or cylinder between the two jaws A and B and adjust the jaw DB, such that it gently grips the body without any undue pressure on it. Tight the screw S attached to the vernier scale V.
4. Note the position of the zero mark of the vernier scale on the main scale. Record the main scale reading just before the zero mark of the vernier scale. This reading ( 1 ST) is called main scale reading (M.S.R.).
(Mote the number ( n ) of the vernier scale division which coincides with some division of the main scale.
6. Repeat steps 4 and 5 after rotating the body by for measuring the diameter in a perpendicular direction.
7. Repeat steps 3,4,5 and 6 for three different positions. Record the observations in each set in a tabular form.
8. Find total reading and apply zero correction.
9. Take mean of different values of diameter and show that in the result with proper unit.
Observations:
1. Vernier constant (least count) of the Vernier Callipers:
$
1 \mathrm{M} . \mathrm{S} . \mathrm{D} .=1 \mathrm{~mm}
$
10 vernier scale divisions $=9$ main scale divisions
i.e. $\quad 10$ V.S.D. $=9$ M.S.D.
$
\therefore \quad \text { 1 V.S.D. }=\frac{9}{10} \text { M.S.D. }
$
Vernier Constant $(\mathrm{V} . \mathrm{C})=1$ M.S.D. - 1 V.S.D. $=1$ M.S.D. $-\frac{9}{10}$ M.S.D.
$
\begin{aligned}
& =\left(1-\frac{9}{10}\right) M . S . D .=\frac{1}{10} \times 1 M . S . D . \\
& =\frac{1}{10} \times 1 \mathrm{~mm}=0.1 \mathrm{~mm}=0.01 \mathrm{~cm}
\end{aligned}
$
2. Zero error: (i) cm (ii) $\qquad$ cm (iii) $\qquad$ .cm
Mean Zero Error (e) = $\qquad$ cm
Mean Zero Correction $(\mathrm{c})=-($ Mean Zero Error $)$
$=$ cm
PRECAUTIONS
Make the motion of vernier scale on main scale very smooth (by oiling).
Find the vernier constant and zero error carefully and record them properly.
Grip the body between the jaws formerly but gently.
Take atleast observation at three different places at right angles at each place.
No parallax should be there.
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