### JEE Mains Previous Years Questions with Solutions

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1

### 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
2

### 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

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

$\therefore$ $x > 54\,cm$
3

### 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
4

### AIEEE 2005

In a potentiometer experiment the balancing with a cell is at length $240$ $cm.$ On shunting the cell with a resistance of $2\Omega ,$ the balancing length becomes $120$ $cm$. The internal resistance of the cell is
A
$0.5\Omega$
B
$1\Omega$
C
$2\Omega$
D
$4\Omega$

## Explanation

The internal resistance of the cell,

$r = \left( {{{{\ell _1} - {\ell _2}} \over {{\ell _2}}}} \right) \times R$

$= {{240 - 120} \over {120}} \times 2 = 2\Omega$