Physics MCQs for NEET — Practice Questions with Answers

U^{238
   92}
->Th²³⁸ + He⁴
        92        2
According to law of conservation of linear momentum, we have.

|P_Th|=|P_He|=P

=>As, kinetic energy of an element,

KE=P2/2m

where,m is mass of an element 

Thus,KE∝1/M

So, M_HE<M_Th=>K_He>K_Th

The binding energy for ${}_1{}^1\mathrm{H}$ is around zero and also not given in the question so we can ignore it.

Q=2(4x7.06)-7x(5.60)

=(8x7.06)-(7x5.60)

=(56.48-39.2)MeV

=17.28MeV≈17.3MeV

Ratio of X:Y is given=1:7 

m_x/m_y=1/7

=> 7m_x/m_y

=> Let the initial total mass is m.

=> m_x+m_y=m => m_y/7+m_y=m

=> 8m_y/7=m

=> m_y=7/8m

only 1/8 part remains 

=> 1->1/2->1/4->1/8
           T/2    T/2   T/2

So,time taken to become 1/8 unstable part

=3 x T_1/2 = 3 x 1.4 x 10⁹ =4.2 x 10⁹ y

(b) As N/N_o=(1/2)ⁿ 

N/N_o=(1/2)³=1/8

Number of half lices=3

=> T=20yr  

∴ T=t/n or t=T x n

=20 x 3yr=60yr

(c) Here,Δm=0.02866U

∴ Energy liberated per nucleon

=0.02866 x 931/4

=26.7/4 MeV

=6.675MeV

Nuclear radius $r\propto A^{1/3}$, where A is mass number

          r=$r_0{A}^{1/3}$

          r=$r_0{\left(27\right)}^{1/3}=3r_0$

          $r_0=\frac{3.6}{3}=1.2fm$

For ${}_{64}Cu$

       $r=r_0{A}^{1/3}$

        =1.2fm ${\left(64\right)}^{1/3}$

        =4.8 fm

For 40 g amount 

$40g\underset{half-life}{\overset{20 s}{\to }}20g\stackrel{20s}{\to }10g$

For 160g amount

$160g\stackrel{10s}{\to }80g\stackrel{10s}{\to }40g$

      $\stackrel{10s}{\to }20g\stackrel{10s}{\to }10g$

So, after 40s $A_{1}and A_2$ remains same.

$$\begin{gathered} At time t_1, \\ \frac{N}{3}=N{e}^{-\lambda t1} \left(1\right) \\ At time t_2, \\ \frac{2N}{3}=N{e}^{-\lambda t2} \left(2\right) \\ \left(2\right)/\left(1\right)- \\ 2=\frac{e^{-\lambda t2} }{e^{-\lambda t_2} } =e^{-\lambda t_1+\lambda t_2} \\ -\lambda t_1+-\lambda t_2=\log 2 \\ {t}_2-t_1=\frac{\log 2}{\lambda }=T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=50 days \end{gathered}$$

Let assume power P=1000 W

Energy per hour= 1000 x 3600 J

Energy per fission=200 MeV

                         =$200\times 1.6\times {10}^{-13} \mathrm{J}$

so, Number of fission per hour 

                     n=$\frac{1000\times 3600}{200\times 1.6\times {10}^{-13}}$

Number of mole per hour =$\frac{\mathrm{n}}{\mathrm{N}}$

so,   Mass per hour=$\frac{\mathrm{n}}{\mathrm{N}}\times 235$

                          =$$\begin{gathered} \frac{1000\times 3600\times 235}{200\times 1.6\times {10}^{-13}\times 6.02\times {10}^{23}} \\ =43.9\times {10}^{-6}\mathrm{g} \end{gathered}$$

This $43.9\times {10}^{-6}\mathrm{g}$ is nearest value of 40 micron 

so option (b) is correct.

We know that

            $$\begin{gathered} \frac{\mathrm{N}}{{\mathrm{N}}_0}={\left(\frac{1}{2}\right)}^{\mathrm{t}/{\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \frac{1}{16}={\left(\frac{1}{2}\right)}^{\mathrm{t}/50} \\ \mathrm{t}=4\times 50 \\ \mathrm{t}=200 \mathrm{yr} \end{gathered}$$