JEE 2015 :: Advanced 2015 :: Physics 2015

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

 In a thin rectangular metallic strip, a constant current I flows along the positive x-direction, as shown in the figure. The length,  width and thickness of the strip are l, w and d respectively. A uniform magnetic field  B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in the accumulation of charge carries on the surface PQRS and the appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues untill the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

 Consider two different metallic strips (1  and 2)  of same dimensions (length l, width w and thickness d)  with carrier densities n₁  and n₂, respectively. Strip  1 is placed in magnetic field  B₁  and strip 2 is placed in magnetic field B₂, both along positive y- directions. Then V₁  and V₂ are the potential differences developed between K and M in strips  1 and 2, respectively. Assuming that the current I is the  same for both the strips, the correct options is/are

Answer:

Explanation:

   $V=\frac{BI}{ned}$

   $ \Rightarrow$                  $\frac{V_{1}}{V_{2}}=\frac{B_{1}}{B_{2}}\times\frac{n_{2}}{n_{1}}$

  If  If $B_{1}=B_{2}$  and $n_{1}=2n_{2}$ , then $V_{2}=2V_{1}$

   If $B_{1}=2B_{2}$  and $n_{1}=n_{2}$ , then $V_{2}=0.5V_{1}$

$\therefore$   Correct answers are (a) and (c).

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length,  width and thickness of the strip are l, w and d respectively. A uniform magnetic field  B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in the accumulation of charge carries on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues untill the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section  of the strip and carried by electrons.

Consider two different metallic strips (1 and 2 )  of the same material. Their lengths are the same , width  are w₁   and w₂  and thickness  are d₁ and d₂ , respectively. Two points K and M are symmetrically  located on the opposite  faces parallel to the x-y plane  (see figure) .V₁ and V₂  are the potential  differences  between K and M in strips 1 and 2 respectively. Then , for  a given  current I flowing through  them in a given  magnetic field  strength  B, the correct statements is/are

Answer:

Explanation:

$ F_{B}=B_{ev}=B_{e}\frac{1}{nAe}=\frac{BI}{nA}$

     Fe= eE

   $\Rightarrow$           F_e    =F_B

                               $eE=\frac{BI}{nA}$

  $\Rightarrow$                $E=\frac{B}{nAe}$

                                     $V=Ed=\frac{BI}{nAe}.w=\frac{BIw}{n(wd)e}=\frac{BI}{ned}$

                                        $\frac{V_{1}}{V_{2}}=\frac{d_{2}}{d_{1}}$

  $\Rightarrow$                        $E=\frac{B}{nAe}$

                              $V=Ed=\frac{BI}{nAe}.w=\frac{BIw}{n(wd)e}=\frac{BI}{ned}$

                          $\frac{V_{1}}{V_{2}}=\frac{d_{2}}{d_{1}}$

    $\Rightarrow$  if w₁    =2w₂   and d₁  =d₂

                                                  V₁  =V₂

$\therefore$   Correct answes are (a) and (d)

 Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n₁  surrounded by a medium of lower refractive index n₂. The light guidance in  the structure takes place due to successive total internal reflections at the interface of the media n₁  and n₂ as shown in the figure. All rays with the angle of incidence i less than a particular value i_m are confined in the medium of refractive index n₁ . The numerical aperture  (NA) of the structure is defined as sin i_m.

If two structures of same cross-sectional area, but different numerical  apertures NA₁  and NA₂  (NA₂ <NA₁) are joined  longitudinally, the numerical aperture  of the combined structure is 

Answer:

Explanation:

       $\sin i_{m}=n_{1}\sin (90-\theta_{c})$ 

$\Rightarrow$   $\sin i_{m}=n_{1}\cos\theta_{c}$

$\Rightarrow$                               $NA=n_{1}\sqrt{1-\sin^{2}\theta_{c}}$

                                            $NA=n_{1}\sqrt{1-\frac{n_{2}^{2}}{n_{1}^{2}}}=\sqrt{n^{2}_{1}-n^{2}_{2}}$

 Substituting the values we get,

       $NA_{1}=\frac{3}{4}$   and   $NA_{2}=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}}$

  NA₂  <NA₁  

   Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical  aperture. Which is NA₂

Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n₁  surrounded by a medium of lower refractive index n₂. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n₁  and n₂ as shown in the figure. All rays with the angle of incidence i less than a particular value i_m are confined in the medium of refractive index n₁. The numerical aperture  (NA) of the structure is defined as sin I'm.

For two structures namely S₁   with $n_{1}=\frac{\sqrt{45}}{4}$  and   $n_{2}=\frac{3}{2}$   and S₂    with $n_{1}=\frac{8}{5}$  and $n_{2}=\frac{7}{5}$  and taking the  refractive index of water to be $\frac{4}{3}$  and that to air to be 1. the correct options is/are

Answer:

Explanation:

  $\frac{4}{3}\sin i=\frac{\sqrt{45}}{4}\sin(90-\theta_{c})=\frac{\sqrt{45}}{4}\cos \theta_{c}$

                                       $\sin \theta_{c}=\frac{n_{2}}{n_{1}}$

  $\therefore$                    $\cos \theta_{c}=\sqrt{1-\left(\frac{n_{2}}{n_{1}}\right)^{2}}$

            $\Rightarrow$                   $\frac{4}{3} \sin i=\frac{\sqrt{45}}{4}\frac{3}{\sqrt{45}}$

                                               $\sin i=\frac{9}{16}$

 In second case,

                          $\sin \theta_{c}=\frac{n_{2}}{n_{1}}=\frac{7}{8}$

 $\Rightarrow$                  $\cos \theta_{c}=\frac{\sqrt{15}}{8}$   

                      $\frac{16}{3\sqrt{15}}\sin i=\frac{5}{8}\sin (90-\theta)$

 Simplifying we get,   $\sin i=\frac{9}{16}$

  (a) is correct.

 Same approach can be adopted for other options. Correct answers are (a)  and (c).

 In terms of potential difference V, electric current i, Permittivity  $\epsilon_{0}$ , permeability  $\mu_{0}$ and speed of light c, the dimensionally correct equations is /are

Answer:

Explanation:

 (A)  Energy of inductor

                    $\frac{1}{2}LI^{2}=\frac{1}{2}\frac{\mu_{0}N^{2}A}{l}I^{2}e$

 Energy of capacitor

                  $\frac{1}{2}CV^{2}=\frac{1}{2}\epsilon_{0}\frac{A}{d}V^{2}\mu_{0}\frac{A}{l}I^{2}$            and

$\epsilon_{0}\frac{A}{d}V^{2}$     have same dimensions.

  So, $\mu_{0}I^{2}$   and    $\epsilon_{0}V^{2}$    have same dimensions.

  (C)     $Q=CV\Rightarrow\frac{Q}{t}=\frac{CV}{t}$

                                  $I=\epsilon_{0}\frac{A}{l}\frac{V}{t}$

   $\frac{A}{lt}$   have unit of speed

   so,                      $I=\epsilon_{0}cV$          

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