 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

An object $2.4$ $m$ in front of a lens forms a sharp image on a film $12$ $cm$ behind the lens. A glass plate $1$ $cm$ thick, of refractive index $1.50$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?
A
$7.2$ $m$
B
$24$ $m$
C
$3.2$ $m$
D
$5.6$ $m$

Explanation

The focal length of the lens

${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$

$= {{20 + 1} \over {240}} = {{21} \over {240}}$

$f = {{240} \over {21}}cm$

Shift$= t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$

$= 1 \times {1 \over 3}$

Now $v' = 12 - {1 \over 3} = {{35} \over 3}cm$

Now the object distancce $u.$

${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$

${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$

$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$
2

AIEEE 2012

An electromagnetic wave in vacuum has the electric and magnetic field $\mathop E\limits^ \to$and $\mathop B\limits^ \to$, which are always perpendicular to each other. The direction of polarization is given by $\mathop X\limits^ \to$ and that of wave propagation by $\mathop k\limits^ \to$. Then
A
$\mathop X\limits^ \to ||\mathop B\limits^ \to$ and $\mathop X\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to$
B
$\mathop X\limits^ \to ||\mathop E\limits^ \to$ and $\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to$
C
$\mathop X\limits^ \to ||\mathop B\limits^ \to$ and $\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to$
D
$\mathop X\limits^ \to ||\mathop E\limits^ \to$ and $\mathop k\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to$

Explanation

as The $E.M.$ wave are transverse in nature i.e.,

$= {{\overrightarrow k \times \overrightarrow E } \over {\mu \omega }} = \overrightarrow H \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

where $\overrightarrow H = {{\overrightarrow B } \over \mu }$

and ${{\overrightarrow k \times \overrightarrow H } \over {\omega \varepsilon }} = - \overrightarrow E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

$\overrightarrow k$ is $\bot \,\,\overrightarrow H$ and $\overrightarrow k$ is also $\bot$ to $\overrightarrow E$

or In other words $\overrightarrow X ||\overrightarrow E$ and $\overrightarrow k ||\overrightarrow E \times \overrightarrow B$
3

AIEEE 2012

In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If ${{\rm I}_m}$ be the maximum intensity, the resultant intensity ${\rm I}$ when they interfere at phase difference $\phi$ is given by :
A
${{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)$
B
${{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)$
C
${{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)$
D
${{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)$

Explanation

Let ${a_1} = a,\,{I_1} = a_1^2 = {a^2}$

${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$

Therefore ${{\rm I}_2} = 4{{\rm I}_1}$

${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi }$

${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi$

$\Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Now, ${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$

${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$

Substituting in equation $\left( 1 \right)$

${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi$

${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$

${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$

${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$
4

AIEEE 2011

A car is fitted with a convex side-view mirror of focal length $20$ $cm$. A second car $2.8m$ behind the first car is overtaking the first car at a relative speed of $15$ $m/s$. The speed of the image of the second car as seen in the mirror of the first one is :
A
${1 \over {15}}\,m/s$
B
$10\,m/s$
C
$15\,m/s$
D
${1 \over {10}}\,m/s$

Explanation

From mirror formula

${1 \over v} + {1 \over u} = {1 \over f}\,\,\,$

so, $\,\,\,{{dv} \over {dt}} = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$

$\Rightarrow {{dv} \over {dt}} = - {\left( {{f \over {u - f}}} \right)^2}{{du} \over {dt}}$

$\Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$