### JEE Mains Previous Years Questions with Solutions

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1

### 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

## Explanation

30 vernier scale divisions coincide with 29 main scale divisions.

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

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

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

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

= ${{1} \over {30}}$ $\times$ 0.5o

= ${{1} \over {30}}$ $\times$ ${1 \over 2}$o

= ${1 \over {60}}$o

= 1 min
2

### AIEEE 2008

An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by
A
a vernier scale provided on the microscope
B
a standard laboratory scale
C
a meter scale provided on the microscope
D
a screw gauge provided on the microscope

## Explanation

To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope
3

### AIEEE 2008

While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of $18$ $cm$ during winter. Repeating the same experiment during summer, she measures the column length to be $x$ $cm$ for the second resonance. Then
A
$18 > x$
B
$x > 54$
C
$54 > x > 36$
D
$36 > x > 18$

## Explanation

For first resonant length $v = {v \over {4{\ell _1}}} = {v \over {4 \times 18}}$ (in winter)

For second resonant length

$v' = {{3v'} \over {4{\ell _2}}} = {{3v'} \over {4x}}$ (in summer)

$\therefore$ ${v \over {4 \times 18}} = {{3v'} \over {4 \times x}}$

$\therefore$ $x = 3 \times 18 \times {{v'} \over v}$

$\therefore$ $x = 54 \times {{v'} \over v}cm$

$v' > v$ because velocity of light is greater in summer as compared to winter

$\left( {v \propto \sqrt T } \right)$

$\therefore$ $x > 54\,cm$
4

### AIEEE 2008

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
3.73 mm
C
3.67 mm
D
3.38 mm

## Explanation

Least count of screw gauge = ${{0.5} \over {50}}mm$ = 0.01mm

Main scale reading = 3 mm

$\therefore$ Reading = [Main scale reading + circular scale reading $\times$ L.C] - (zero error)

= [3 + 35 $\times$ 0.01] - (-0.03) = 3.38 mm