 JEE Mains Previous Years Questions with Solutions

4.5
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1

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

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

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

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

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 $= 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$
3

AIEEE 2003

In the nuclear fusion reaction $${}_1^2H + {}_1^3H \to {}_2^4He + n$$
given that the repulsive potential energy between the two nuclei is $\sim 7.7 \times {10^{ - 14}}J$, the temperature at which the gases must be heated to initiate the reaction is nearly
[ Boltzmann's Constant $k = 1.38 \times {10^{ - 23}}\,J/K$ ]
A
${10^7}\,\,K$
B
${10^5}\,\,K$
C
${10^3}\,\,K$
D
${10^9}\,\,K$

Explanation

The average kinetic energy per molecule $= {3 \over 2}kT$

This kinetic energy should be able to provide the repulsive potential energy

$\therefore$ ${3 \over 2}kT = 7.7 \times {10^{ - 14}}$

$\Rightarrow T = {{2 \times 7.7 \times {{10}^{ - 14}}} \over {3 \times 1.38 \times {{10}^{ - 23}}}} = 3.7 \times {10^9}$
4

AIEEE 2003

Which of the following atoms has the lowest ionization potential ?
A
${}_7^{14}N$
B
${}_{55}^{133}\,Cs$
C
${}_{18}^{40}\,Ar$
D
${}_8^{16}\,O$

Explanation

The ionisation potential increases from left to right in a period and decreases from top to bottom in a group. Therefore ceasium will have the lowest ionisation potential.