 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2006

The $'rad'$ is the correct unit used to report the measurement of
A
the ability of a beam of gamma ray photons to produce ions in a target
B
the energy delivered by radiation to a target
C
D
the rate of decay of radioactive source

Explanation

The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue. Absorbed dose is expressed in rad. A rad is equal to $100$ $ergs$ of energy absorbed by $1$ gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist $L.$ $H.$ Gray); one gray equals $100$ rad.
2

AIEEE 2006

If the binding energy per nucleon in ${}_3^7Li$ and ${}_2^4He$ nuclei are $5.60$ $MeV$ and $7.06$ $MeV$ respectively, then in the reaction $$p + {}_3^7Li \to 2\,{}_2^4He$$
energy of proton must be
A
$28.24$ $MeV$
B
$17.28$ $MeV$
C
$1.46$ $MeV$
D
$39.2$ $MeV$

Explanation

Let $E$ be the energy of proton, then

$E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]$

$\Rightarrow E = 56.48 - 39.2 = 17.28MeV$
3

AIEEE 2006

When ${}_3L{i^7}$ nuclei are bombarded by protons, and the resultant nuclei are ${}_4B{e^8}$, the emitted particles will be
A
alpha particles
B
beta particles
C
gamma photons
D
neutrons

Explanation

${}_3^7Li + {}_1^1p \to {}_4^8Be + {}_0^0\gamma$
4

AIEEE 2006

An alpha nucleus of energy ${1 \over 2}m{v^2}$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
A
${v^2}$
B
${1 \over m}$
C
${1 \over {{v^2}}}$
D
${1 \over {Ze}}$

Explanation

Work done to stop the $\alpha$ particle is equal to $K.E.$

$\therefore$ $qV = {1 \over 2}m{v^2} \Rightarrow q \times {{K\left( {Ze} \right)} \over r} = {1 \over 2}m{v^2}$

$\Rightarrow r = {{2\left( {2e} \right)K\left( {Ze} \right)} \over {m{V^2}}} = {{4KZ{e^2}} \over {m{v^2}}}$

$\Rightarrow r \propto {1 \over {{v^2}}}$ and $r \propto {1 \over m}.$