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Vernier Calliper And Screw Gauge With Solutions - JEE MCQ


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15 Questions MCQ Test - Vernier Calliper And Screw Gauge With Solutions

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Vernier Calliper And Screw Gauge With Solutions - Question 1
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A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line ?

Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 1

Least count
= pitch / no. of division on circular scale
= 0.5 mm / 50
= 0.01 mm

Negative zero error = – 5 × LC = -5 × 0.01 = -0.05 mm

Measured value
= main scale reading + screw gauge reading – zero error
= 0.5mm + 25 × 0.01mm – (-0.05mm)
= 0.75mm + 0.05mm
= 0.80mm

*Multiple options can be correct
Vernier Calliper And Screw Gauge With Solutions - Question 2
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Consider a Vernier callipers 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:

Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 2

Vernier calliper:

M.S.D =1/8 cm = 0.125 cm = 1.25 mm

5 V.S.D. = 4 M.S.D

⇒ V.S.D. = 4/5 M.S.D. = 4/5 × 1.25mm = 1 mm

Least count = M.S.D. – V.S.D = 1.25 – 1 = 0.25 mm

Screw gauge:

No. of circular scale divisions = 100

1 rotation (pitch) = 2 linear scale divisions

» If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –

Pitch = 2 × Least Count of vernier callipers = 2 × 0.25mm = 0.5 mm

Least count of the screw gauge

= Pitch/no. of circular scale divisions = 0.5/100 =0.005mm

Option (B) is correct.

» If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –

Linear scale division of screw gauge

= 2 × Least count of vernier callipers

= 2 × 0.25 mm

= 0.5 mm

Now,

Pitch = 2 linear scale divisions = 2 × 0.5 = 1mm

Least count of the screw gauge

= Pitch/ No. of circular divisions

= 1/100 mm

= 0.01 mm

Option (C) is correct.

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Vernier Calliper And Screw Gauge With Solutions - Question 3
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Diameter of a steel ball is measured using a vernier callipers which has divisions of 0.1cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as :If the zero error is – 0.03 cm, then mean corrected diameter is:Q11 JEE 2015

Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 3

MSD = 0.1 cm

VSD = 9/10 MSD = 0.9 × 0.1= 0.09 cm

Least count = MSD – VSD = 0.1 – 0.09 = 0.01 cm

Total reading = Main Scale Reading + Vernier coincidence × Least Count

1st reading = 0.5 + 8 × 0.01 = 0.58 cm

2nd reading = 0.5 + 4 × 0.01 = 0.54 cm

3rd reading = 0.5 + 6 × 0.01 = 0.56 cm

Average reading = (0.58 + 0.54 + 0.56)/3 = 0.56 cm

Corrected reading = reading – zero error  = 0.56 – (-0.03) = 0.56 + 0.03 = 0.59 cm

Vernier Calliper And Screw Gauge With Solutions - Question 4
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The diameter of a cylinder is measured using a Vernier callipers 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 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is:
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 4

Smallest main scale division = 0.05 cm (5.15 – 5.10)
Main scale reading = 5.10 cm (comes just before the zero of the Vernier scale)
Vernier coincidence = 24
Least count =0.05 – 2.45/50 = 0.001 cm

Diameter
= Main Scale Reading + Vernier coincidence × Least Count
= 5.10 + 24 × 0.001 = 5.124 cm

Vernier Calliper And Screw Gauge With Solutions - Question 5
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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 (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) 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 C1 and C2, respectively, are:Q JEE 2016 Vernier
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 5

MSD = 1/10 cm = 0.1 cm

For first vernier caliper,
10 VSD = 9 MSD
⇒ VSD = 9/10 MSD = 0.9 × 0.1 = 0.09 cm

Reading
= main scale reading upto coinciding main scale division − n × VSD
= 3.5 – 7 × 0.09 = 3.5 – 0.63 = 2.87 cm

For second vernier caliper,
10 VSD = 11 MSD
⇒ VSD = 11/10 MSD = 1.1 × 0.1 = 0.11 cm
Reading = 3.6 – 7 × 0.11 = 3.6 – 0.77 = 2.83 cm

Why we not used the formula here that we were using in previous problems,
Here is the formula we were using,
Least count = MSD – VSD
Reading = Main Scale Reading + Vernier coincidence × Least Count
We derived these formula for Vernier callipers having n divisions of the Vernier scale equal to n-1 divisions of the main scale.
If we apply these formula on first vernier caliper then we will get the same result, but if we apply these formula on the second vernier caliper then we will get negative least count (0.1 − 0.11 = −0.01cm), which is not possible, that is why we cannot apply these formula in this case. 

*Answer can only contain numeric values
Vernier Calliper And Screw Gauge With Solutions - Question 6
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The main scale of a vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of vernier scale coinsides with a main scale division. Find:
(i). Least count(in cm)

Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 6

(i).
MSD = 0.1 cm
20 VSD = 19 MSD
⇒ VSD = 19/20 MSD = 19/20 × 0.1 cm = 0.095cm
Least Count = MSD – VSD = 0.1 – 0.095 cm = 0.005 cm

*Answer can only contain numeric values
Vernier Calliper And Screw Gauge With Solutions - Question 7
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The main scale of a vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of vernier scale coinsides with a main scale division. Find:
(ii). Radius of cylinder(in cm)

Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 7

Main scale reading = 35 mm = 3.5 cm
Diameter
= Main scale reading + vernier coincidence × Least count
= 3.5 + 4 × 0.005
= 3.5 + 0.02
= 3.52 cm
Radius = Diameter/2 = 3.52/2 = 1.76 cm

Vernier Calliper And Screw Gauge With Solutions - Question 8
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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°) then the least count of the instrument is:
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 8

MSD = 0.5°
30 VSD = 29 MSD
Least Count
= Length of 1 main scale division / No. of divisions of Vernier scale
= 0.5 ° / 30
= 0.5 × 60 min / 30
= 1 min 

NOTE:to convert degree into minute multiply it with 60 and for seconds multiply it with 3600.

Vernier Calliper And Screw Gauge With Solutions - Question 9
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A vernier callipers 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 callipers, the least count is:
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 9

MSD = 1 mm
20 VSD = 16 MSD
⇒ VSD = 16/20 MSD = 0.8 MSD = 0.8 mm
Least Count = MSD – VSD = 1 – 0.8 = 0.2 mm

Vernier Calliper And Screw Gauge With Solutions - Question 10
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A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 10

As the measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm
For vernier callipper with MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD – 1 VSD = 0.1 mm

Vernier Calliper And Screw Gauge With Solutions - Question 11
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A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degreeVernier scale reading : 09 divisionsGiven 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
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 11

MSD = 0.5°
LC = MSD/n = 0.5 / 30 = 1/60
Total Reading
= Main scale reading + vernier coincidence × LC
= 58.5 + 9 × 1/60
= 58.5 + 0.15
= 58.65°

Vernier Calliper And Screw Gauge With Solutions - Question 12
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The jaws of a vernier callipers touch the inner wall of calorimeter without any undue pressure. The position of zero of vernier scale on the main scale reads 3.48. The 6th of vernier scale division is coinciding with any main scale division. Vernier constant of callipers is 0.01 cm. Find actual internal diameter of calorimeter, when it is observed that the vernier scale has a zero error of – 0.03 cm.
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 12

Main scale reading = 3.48 cm
LC = 0.01 cm
Vernier coincidence = 6

Reading
= Main scale reading + Vernier coincidence × LC
= 3.48 + 6 × 0.01
= 3.48 + 0.06
= 3.54 cm
Corrected reading
= reading – zero error
= 3.54 – (-0.03)
= 3.54 + 0.03
= 3.57 cm

Vernier Calliper And Screw Gauge With Solutions - Question 13
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The thin metallic strip of vernier callipers moves downward from top to bottom in such a way that it just touches the surface of beaker. Main scale reading of calliper is 6.4 cm whereas its vernier constant is 0.1 mm. The 4th of vernier scale division is coinciding with main scale division. The actual depth of beaker is (when zero of vernier coincides with zero of main scale)
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 13

Reading = 6.4 + 4 × 0.01 cm = 6.44 cm

Vernier Calliper And Screw Gauge With Solutions - Question 14
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In an instrument, there are 25 divisions on the vernier scale which coincides with 24th division of the main scale. 1 cm on main scale is divided into 20 equal parts. The least count of the instrument is:
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 14

MSD = 1/20 = 0.05 cm
25 VSD = 24 MSD
⇒ VSD = 24/25 MSD = 24/25 × 0.05 = 0.048 cm
Least Count = MSD – VSD = 0.05 – 0.048 = 0.002 cm

Vernier Calliper And Screw Gauge With Solutions - Question 15
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In a vernier calliper, there are 10 divisions on the vernier scale and 1 cm on main scale is divided in 10 parts. While measuring a length, the zero of the vernier scale lies just ahead of 1.8 cm mark and 4th division of vernier scale coincides with a main scale division. The value of length is:
Detailed Solution for Vernier Calliper And Screw Gauge With Solutions - Question 15

Length of 1 main scale division = 1/10 = 0.1 cm
No. of vernier scale divisions = 10
Main scale reading = 1.8 cm
Coinciding vernier scale division = 4th

Least Count
= Length of smallest main scale division / No. of vernier scale divisions
= 0.1/10
= 0.01 cm
(vernier calliper has a least count 0.01 cm. So, measurement is accurate only upto
three significant figures.)

length
= Main scale reading + Coinciding vernier scale division × LC
= 1.8 + 4 × 0.01
= 1.8 + 0.04
= 1.84 cm

 

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