 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2009

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $u$ and the image distance $v,$ from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of ${45^ \circ }$ with the $x$-axis meets the experimental curve at $P.$ The coordinates of $P$ will be :
A
$\left( {{f \over 2},{f \over 2}} \right)$
B
$\left( {f,f} \right)$
C
$\left( {4f,4f} \right)$
D
$\left( {2f,2f} \right)$

Explanation Here $u = - 2f,v = 2f$

As $|u|$ increases, $v$ decreases for $|u| > f.$ The graph between $|v|$ and $|u|$ is shown in the figure. A straight line passing through the origin and making an angle of ${45^ \circ }$ with the $x$-axis meets the experimental curve at $P\left( {2f,2f} \right).$
2

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
3

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
4

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

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

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