 JEE Mains Previous Years Questions with Solutions

4.5
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1

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
2

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
${3^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}}:1$
B
$1:{2^{1/3}}$
C
${2^{1/3}}:1$
D
$1:{3^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}}$

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 $\,\,{{{v_1}} \over {v{}_2}} = 2$

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

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

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

AIEEE 2004

The binding energy per nucleon of deuteron $\left( {{}_1^2\,H} \right)$ and helium nucleus $\left( {{}_2^4\,He} \right)$ is $1.1$ $MeV$ and $7$ $MeV$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
A
$23.6\,\,MeV$
B
$26.9\,\,MeV$
C
$13.9\,\,MeV$
D
$19.2\,\,MeV$

Explanation

The nuclear reaction of process is $2_1^2H \to {4 \over 2}$ He

Energy released $= 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$
4

AIEEE 2004

An $\alpha$-particle of energy $5$ $MeV$ is scattered through ${180^ \circ }$ by a fixed uranium nucleus. The distance of closest approach is of the order of
A
${10^{ - 12}}\,cm$
B
${10^{ - 10}}\,cm$
C
$1A$
D
${10^{ - 15a}}\,cm$

Explanation

KEY NOTE :

Distance of closest approach

${r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}$

Energy, $E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J$

$\therefore$ ${r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)} \over {5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$

$\Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm$