JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

In an experiment for determination of refractive index of glass of a prism by $i - \delta ,$ plot it was found thata ray incident at angle ${35^ \circ }$, suffers a deviation of ${40^ \circ }$ and that it emerges at angle ${79^ \circ }.$ In that case which of the following is closest to the maximum possible value of the refractive index?
A
$1.7$
B
$1.8$
C
$1.5$
D
$1.6$

Explanation

We know that $i + e - A = \delta$

${35^ \circ } + {79^ \circ } - A = {40^ \circ }$

$\therefore$ $A = {74^ \circ }$

But $\mu = {{\sin \left( {{{A + {\delta _m}} \over 2}} \right)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} \right)} \over {\sin {{74} \over 2}}}$

$= {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} \right)$

${\mu _{\max }}\,$ can be ${5 \over 3}.$ That is ${\mu _{\max }}$ is less than ${5 \over 3} = 1.67$

But ${\delta _m}$ will be less than ${40^ \circ }$ so

$\mu < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,$ $\sin {60^ \circ } \Rightarrow \mu = 1.45$
2

AIEEE 2012

A spectrometer gives the following reading when used to measure the angle of a prism.
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

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

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

Given that,

Vernier scale reading = 09 division

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

= 58.65o
3

AIEEE 2011

A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.
The diameter of wire from the above date is:
A
0.052 cm
B
0.026 cm
C
0.005 cm
D
0.52 cm

Explanation

Least count of screw gauge

= ${{Pitch} \over {Number\,of\,division\,on\,circular\,scale}}$

= ${1 \over {100}}mm$

= 0.01 mm

Diameter of the wire = M.S.R + C.S.R $\times$ L.C

= 0 + 52 $\times$ 0.01

= 0.52 mm

= 0.052 cm
4

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 $P\left( {2f,2f} \right).$