JEE 2015 :: Advanced 2015 :: Physics 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

An electron in an excited state of Li²⁺ ion has angular momentum  $\frac{3h}{2\pi}$. The de Broglie wavelength of the electron in this state is $p\pi a_{0}$ (where $a_{0}$ is the Bohr radius). The value of p is 

Answer:

Explanation:

Angular momentum

                     $=n\left(\frac{h}{2\pi}\right)=3\left(\frac{h}{2\pi}\right)$

  $\therefore$   n=3

  Now,     $r_{n}\propto \frac{n^{2}}{z}$

 $\therefore$           $r_{3}= \frac{(3)^{2}}{3}(a_{0})=3a_{0}$

 Now,           $mv_{3}r_{3}=3\left( \frac{h}{2\pi}\right)$

$\therefore$   $mv_{3}(3a_{0})=3\left(\frac{h}{2\pi}\right)$

or            $\frac{h}{mv_{3}}=2\pi a_{0}$

or            $\frac{h}{P_{3}}=2\pi a_{0}$           (P=mv)

or      $\lambda_{3}=2\pi a_{0}$            $\lambda=\frac{h}{p}$

  $\therefore$     Answer is 2.

In the following circuit, the current through the resistor R(=2Ω) is I  amperes. The value of I is 

Answer:

Explanation:

A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n  and emerges from the opposite face making an angle θ  (n) with the normal (see figure). For n=$\sqrt{3}$ the value of θ is 60° and $\frac{d\theta}{dn}=m,$   The value of m is 

Answer:

Explanation:

Appluing Snell's law at M and N,

  $ \sin60^{0}=n\sin r$          ............(i)

    $\sin\theta=\pi\sin(60-r)$     ............(ii)

 Differentiating we get

         $\cos\theta\frac{d\theta}{dn}=-n\cos(60-r)\frac{dr}{dn}+\sin(60-r)$

Differentiating Eq.(i),

  $n\cos r\frac{dr}{dn}+\sin r=0$

or         $\frac{dr}{dn}=-\frac{\sin r}{n\cos r}=\frac{-\tan r}{n}$

   $\Rightarrow$               $\cos \theta \frac{d\theta}{dn}=-n\cos(60^{0}-r)\left[\frac{-\tan r}{n}\right]+\sin(60^{0}-r)$

                $\frac{d\theta}{dn}=\frac{1}{\cos\theta}\left[ \cos(60^{0}-r\right)\tan r+\sin(60^{0}-r)]$

  From Eq.(i)  , r =30°  for n=$\sqrt{3}$

                      $\frac{d\theta}{dn}=\frac{1}{\cos 60}(\cos 30\times \tan 30+\sin 30)$

                         $=2(\frac{1}{2}+\frac{1}{2})=2$

For a radioactive material , its activity A and rate of change of its activity R are defined as $A=-\frac{dN}{dt}$  and    $A=-\frac{dA}{dt}$ , where N(t)  is the number of nuclei at time  t. Two radioactive source P(mean life τ )  and Q (mean life 2τ)  have the same activity at t=0. Their rate of change of activities at t=2τ is R_p and R_Q, respectively. If  $\frac{R_{P}}{R_{Q}}=\frac{n}{e}$, then the value of n is 

Answer:

Explanation:

For  initial numbers are N₁   and N₂.

      $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\tau_{2}}{\tau_{1}}=\frac{2\tau}{\tau}$

                              $=2=\frac{T_{2}}{T_{1}}$               (T= Half life)

                                     $A= \frac{-dN}{dt}=\lambda N$

 initial activity is same

$\therefore$               $\lambda_{1}N_{1}=\lambda_{2}N_{2}$          ........(i)

Activity at time t,

                                               $A=\lambda N=\lambda N_{0}e^{-\lambda t}$

                                     $A_{1}=\lambda_{1}N_{1}e^{-\lambda_{1}t}$

$\Rightarrow$            $R_{1}=-\frac{dA_{1}}{dt}=\lambda_{1}^{2}N_{1}e^{-\lambda_{1}t}$

similarly,                     $R_{2}=\lambda^{2}_{2}N_{2}e^{-\lambda_{2}t}$

After t= 2τ                        

                       $\lambda_{1}t=\frac{1}{\tau_{1}}(t)=\frac{1}{\tau}(2\tau)=2$

                          $\lambda_{2}t=\frac{1}{\tau_{2}}(t)=1$

                                      $\frac{1}{2\tau}(2\tau)=1$

                      $\frac{R_{P}}{R_{Q}}=\frac{\lambda^{2}_{1}N_{1}e^{-\lambda_{1}t}}{\lambda^{2}_{2}N_{2}e^{-\lambda_{2}t}}$

   $\frac{R_{P}}{R_{Q}}=\frac{\lambda_{1}}{\lambda_{2}}\left(\frac{e^{-2}}{e^{-1}}\right)=\frac{2}{e}$

Four harmonic waves of equal frequencies and equal intensities I₀   have phase angles    $0,\frac{\pi}{3},\frac{2\pi}{3}$ and $\pi$. When they are superposed, the intensity of the resulting wave is  nI₀. The value of n is

Answer:

Explanation:

Let individual amplitudes are A₀ each. Amplitude can be added by vector method.

 A₁ = A₂ = A₃ = A₄ = A₀ 

Resultant  of  A₁  and  A₄ is zero, Resultant   of  A₂  and  A₃  is   

    $A=\sqrt{A_{0}^{2}+A_{0}^{2}+2A_{0}A_{0}\cos 60^{0}}=\sqrt{3}A_{0}$

 This is also the net resultant.

Now,               $I\propto A^{2}$

    $\therefore$                    Net intensity will become 3I₀

         $\therefore$          Answer is 3.

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015