JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2005

The intensity of gamma radiation from a given source is $L$. On passing through $36$ $mm$ of lead, it is reduced to ${{\rm I} \over 8}.$ The thickness of lead which will reduce the intensity to ${{\rm I} \over 2}$ will be
A
$9mm$
B
$6mm$
C
$12mm$
D
$18mm$

Explanation

KEY CONCEPT : Intensity $I = {I_0}.{e^{ - \mu d}},$

Applying logarithm on both sides,

$- \mu d = \log \left( {{I \over {{I_0}}}} \right)$

$- \mu \times 36 = \log \left( {{{I/8} \over I}} \right).........\left( i \right)$

$- \mu \times d = \log \left( {{{I/2} \over I}} \right).........\left( {ii} \right)$

Dividing $(i)$ by $(ii),$

${{36} \over d} = {{\log \left( {{1 \over 8}} \right)} \over {\log \left( {{1 \over 2}} \right)}}$

$= {{3\log \left( {{1 \over 2}} \right)} \over {\log \left( {{1 \over 2}} \right)}} = 3$

or $\,\,\,d = {{36} \over 3} = 12\,mm$
2

AIEEE 2005

Starting with a sample of pure ${}^{66}Cu,{7 \over 8}$ of it decays into $Zn$ in $15$ minutes. The corresponding half life is
A
$15$ minutes
B
$10$ minutes
C
$7{1 \over 2}$ minutes
D
$5$ minutes

Explanation

${7 \over 8}$ of $Cu$ decays in $15$ minutes.

$\therefore$ $Cu$ undecayed $= N = 1 - {7 \over 8} = {1 \over 8} = {\left( {{1 \over 2}} \right)^3}$

$\therefore$ No. of half lifes $=3$

$n = {t \over T}$ or $3 = {{15} \over T}$

$\Rightarrow T =$ half life period $= {{15} \over 3} = 5\,\,$ minutes
3

AIEEE 2005

If radius of the $\matrix{ {27} \cr {13} \cr }$ $Al$ nucleus is estimated to be $3.6$ fermi then the radius of $\matrix{ {125} \cr {52} \cr } \,Te$ nucleus is estimated to be nearly
A
$8$ fermi
B
$6$ fermi
C
$5$ fermi
D
$4$ fermi

Explanation

KEY CONCEPT : $R = {R_0}{\left( A \right)^{1/3}}$

Here A = Mass number

$\therefore$ ${{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{1/3}}$

$= {\left( {{{27} \over {125}}} \right)^{1/3}} = {3 \over 5}$

${R_2} = {5 \over 3} \times 3.6 = 6$ fermi
4

AIEEE 2004

A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of $2:1.$ The ratio of their nuclear sizes will be
A
${3^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}:1$
B
$1:{2^{1/3}}$
C
${2^{1/3}}:1$
D
$1:{3^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$

Explanation

From conservation of momentum ${m_1}{v_1} = {m_2}{v_2}$

$\Rightarrow \left( {{{{m_1}} \over {{m_2}}}} \right) = \left( {{{{v_2}} \over {{v_1}}}} \right)\,\,$ given $\,\,{{{v_1}} \over {v{}_2}} = 2$

$\Rightarrow {{{m_1}} \over {{m_2}}} = {1 \over 2}$

$\Rightarrow {{r_1^3} \over {r_2^3}} = {1 \over 2}$

$\Rightarrow \left( {{{{r_1}} \over {{r_2}}}} \right) = {\left( {{1 \over 2}} \right)^{1/3}}$