JEE Main Units and Measurements Previous year Question

21.(JEE Main 2021 (Online) 16th March Morning Shift )

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) $\frac{10 a}{n}$

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

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

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

Correct answer is (A)

n VSD = (n $-$ 1) MSD

1 VSD = $(\frac{n - 1}{n})$ MSD

L.C. = 1 MSD $-$ 1 VSD

= 1 MSD $-$ $(\frac{n - 1}{n})$ MSD

= 1 MSD $-$ 1 MSD + $\frac{M S D}{n}$

= $\frac{M S D}{n}$

= $\frac{a}{n}$ cm

= $\frac{10 a}{n}$ mm

22. (JEE Main 2021 (Online) 31st August Evening Shift )

The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1 mm. The main scale reading is 10 mm and 8^th 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 ___________ $\times$ 10^{$-$ 2} cm.

Correct answer is (52)

9 MSD = 10 VSD

9 $\times$ 1 mm = 10 VSD

$∴$ 1 VSD = 0.9 mm

LC = 1 MSD $-$ 1 VSD = 0.1 mm

Reading = MSR + VSR $\times$ LC

10 + 8 $\times$ 0.1 = 10.8 mm

Actual reading = 10.8 $-$ 0.4 = 10.4 mm

radius = $\frac{d}{2} = \frac{10.4}{2}$ = 5.2 mm

= 52 $\times$ 10^{$-$ 2} cm

23.(JEE Main 2020 (Online) 2nd September Morning Slot )

The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7^th 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 between 3.1 cm and 3.2 cm and 4^th VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division)

(A) 3.21 cm

(B) 2.99 cm

(C) 3.07 cm

(D) 3.2 cm

Correct answer is (C)

Least count = 1 mm or 0.01 cm

Zero error = 0 + 0.01 × 7 = 0.07 cm

Reading = 3.1 + (0.01 × 4) – 0.07

= 3.1 + 0.04 – 0.07

= 3.1 – 0.03

= 3.07 cm

24.(AIEEE 2012 )

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

(A) 58.59 degree

(B) 58.77 degree

(C) 58.65 degree

(D) 59 degree

Correct answer is (C)

30 vernier scale divisions coincide with 29 main scale divisions.

Therefore 1 V.S.D = $\frac{29}{30}$ M.S.D

Least count = 1 M.S.D - 1 V.S.D

= 1 M.S.D - $\frac{29}{30}$ M.S.D

= $\frac{1}{30}$ M.S.D

= $\frac{1}{30}$ $\times$ 0.5^o

Reading of Vernier = Main Scale Reading + Vernier scale reading $\times$ Least count

Given that,

Main Scale Reading = 58.5

Vernier scale reading = 09 division

$∴$ Reading of Vernier = 58.5^o + 9 $\times$ $\frac{{0.5}^{\circ}}{30}$

= 58.65^o

25.(AIEEE 2009 )

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:

(A) one minute

(B) half minute

(C) one degree

(D) half degree

Correct answer is (A)

30 vernier scale divisions coincide with 29 main scale divisions.

Therefore 1 V.S.D = $\frac{29}{30}$ M.S.D

Least count = 1 M.S.D - 1 V.S.D

= 1 M.S.D - $\frac{29}{30}$ M.S.D

= $\frac{1}{30}$ M.S.D

= $\frac{1}{30}$ $\times$ 0.5^o

= $\frac{1}{30}$ $\times$ $\frac{1}{2}$ ^o

= $\frac{1}{60}$ ^o

= 1 min