JEE 2013 :: Advanced 2013 :: Physics 2013

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013

Match list I of the nuclear process with list II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists

Answer:

Explanation:

(p)  in $\alpha$ -decay  mass number  decreases by 4 and atomic number decreases by 2

(q)  in $\beta$⁺ decay mass number remains unchanged while atomic number decreases by 1

(r)  in fission, the parent nucleus breaks into all most two equal fragments.

(s)  in proton emission both mass number and atomic number decreases by 1

One mole of a monoatomic ideal gas is taken along two cyclic  processes E→ F→ G→ E  and

E→F→ H→ E as shown in the p-V diagram.

 The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

 Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

Answer:

Explanation:

in F→ G work done in isothermal process is 

$nRT ln\left(\frac{V_{f}}{V_{i}}\right)=32p_{0}V_{0}ln\left(\frac{32V_{0}}{V_{0}}\right)$

  $=32p_{0}V_{0}ln2^{5}=160p_{0}V_{0}ln 2$

$In G\rightarrow E, \triangle W=p_{0}\triangle V=p_{0}(31V_{0})=31p_{0}V_{0}$

 In G $\rightarrow$ H , work done is less than 31 p₀V₀ , i.e, 24 p₀V₀

 In F $\rightarrow$ H workdone is 36 p₀V₀

Match List I with List II  and select the correct answer using the codes given below the lists

Answer:

Explanation:

(p)$U=\frac{1}{2}kT$

 $\Rightarrow$      $[ML^{2}T^{-2}]=[k]K$

  $\Rightarrow$  $[K]= [ML^{2}T^{-2}K^{-1}]$

(q)    $F= \eta A\frac{dv}{dx}$

  $\Rightarrow [\eta]=\frac{[MLT^{-2}]}{[L^{2}LT^{-1}L^{-1}]}=[ML^{-1}T^{-1}]$

(r)   E=hv

  $\Rightarrow$    $ [ML^{2}T^{2}]=[h][T^{-1}]$

$\Rightarrow$    $[h]= [ML^{2}T^{-1}]$

(s)   $\frac{dQ}{dt}=\frac{kA\triangle \theta}{l}$

 $[k]= \frac{[ML^{2}T^{-3}L]}{[L^{2}K]}=[MLT^{-3}K^{-1}]$

A right-angled prism of refractive index  $\mu_{1}$ is placed in a rectangular block of refractive index $\mu_{2}$, which is surrounded by a medium of refractive index $\mu_{3}$ , as shown in the figure, A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between $\mu_{1}$, $\mu_{2}$  and $\mu_{3}$ , it takes one of the four possible paths 'ef', 'eg', 'eh', or 'ei'.

 Match the paths in the list I  with conditions of refractive indices in list II  and select the correct answer using the codes given below the lists.

Answer:

Explanation:

for e → i

 $45^{0}>\theta_{c}$

 $\sin 45^{0}>\sin\theta_{c}$

 $\frac{1}{\sqrt{2}}>\frac{\mu_{2}}{\mu_{1}}$

 $\mu_{1}> \sqrt{2\mu_{2}}$

  For e $\rightarrow$ f

 angle of refraction is lesser than angle of incidence , so   $\mu_{2}> \mu_{1}$ and $\mu_{2}> \mu_{3}$  

 for e $\rightarrow$  g $\mu_{1}= \mu_{2}$

 for e  $\rightarrow$  h  , $\mu_{2}< \mu_{1}$ <  $\sqrt{2}\mu_{2}$ and   $\mu_{2} >\mu_{3} $

The mass of a nucleus  $^{A}_{Z}X$ is less than the sum of the mass of (A-Z) number of neutrons and  Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy  of the nucleus.A heavy nucleus of mass M can break into two light nuclei of masses m₁ and m₂ only if (m₁+m₂) <M . Also two light nuclei of masses m₃  and m₄ can undergo complete fusion and form a heavy nucleus of mass M ' only if (m₃+m₄)>M'. The masses of some neutral  atoms are given in the table below:

  The kinetic energy (in KeV) of  the alpha particle  . when the nucleus   $^{210}_{84}Po$ at rest undergoes alpha decay, is 

Answer:

Explanation:

$_{84}Po^{210}\rightarrow_{2}He^{4}+_{84}Pb^{206}$

 Mass defect  $\triangle m=(m_{po}-M_{He}-m_{pb})$ =0.005818 u

$\therefore$    $Q=(\triangle m)(931.48)MeV$

=5.4193 MeV=5419 keV

$\alpha$ ←    →Pb

 from conservation of linear  momentum,

  p_pb= p_$\alpha$

$\therefore$   $\sqrt{2m_{pb}k_{pb}}=\sqrt{2m_{\alpha}k_{\alpha}}$

  or   

$\frac{k_{\alpha}}{k_{pb}}=\frac{m_{pb}}{m_{\alpha}}=\frac{206}{4}$

 $\therefore$  $k_{\alpha}=\left(\frac{206}{206+4}\right)(k_{total})$

$=\left(\frac{206}{210}\right)(5419)=5316 keV$

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013