JEE 2015 :: Advanced 2015 :: Physics 2015

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

Two spherical stars A and B emit blackbody radiation. The radius of A  is 400 times that of B and A emits 10⁴  times the power emitted from B. The ratio $\left(\frac{\lambda_{A}}{\lambda_{B}}\right)$   of their  wavelengths  $\lambda_{A}$  and $\lambda_{B}$  at which the peaks  occur in their respective radiation curves is 

Answer:

Explanation:

 Power,    $P=(\sigma T^{4}A)=\sigma T^{4}(4\pi R^{2})$

   or                  $P \propto T^{4}R^{2}$  .........(i)

  According to Wien's law,

                       $\lambda\propto \frac{1}{T}$

    ($\lambda$ is the wavelength at which peak occurs)

       $\therefore$      Eq (i) will become,

              $P\propto \frac{R^{2}}{\lambda^{4}}$

    or                          $\lambda\propto \left[\frac{R^{2}}{P}\right]^{1/4}$

  $\Rightarrow$         $\frac{\lambda_{A}}{\lambda_{B}}=\left[\frac{R_{A}}{R_{B}}\right]^{1/2} \left[\frac{P_{A}}{P_{B}}\right]^{1/4}$

$=[400]^{1/2}\left[\frac{1}{10^{4}}\right]^{1/4}=2$

 Two identical uniform discs roll without slipping on two different surfaces AB and CD  (see figure) starting at A and C with linear speeds  v₁ and v₂, respectively, and always remain in contact with the surfaces. If they  reach B and D with the same linear speed  and v₁ =3 m/s, then v₂   in m/s is (g=10 m/s² )

Answer:

Explanation:

 In case of pure rolling, mechanical energy remains constant (as work-done by friction is zero) . Further in case of a disc.

  translational kinetic energy/  rotational kinetic energy =   $\frac{K_{T}}{K_{R}}$

              $=\frac{\frac{1}{2}mv^{2}}{\frac{1}{2}I\omega^{2}}$

         =$\frac{mv^{2}}{\left(\frac{1}{2}mR^{2}\right)\left(\frac{V}{R}\right)^{2}}=\frac{2}{1}$

  or      $K_{r}=\frac{2}{3}$                              ( Total kinetic energy)

 or Total Kinetic energy

                           $K=\frac{3}{2}K_{r}=\frac{3}{2}\left(\frac{1}{2}mv^{2}\right)=\frac{3}{4}mv^{2}$

Decrease in potential energy = increase in kinetic energy

      or        $mgh=\frac{3}{4}m\left(v_{f}^{2}-v_i^2\right)$

  or      $v_{f}=\sqrt{\frac{4}{3}gh+v_{i}^{2}}$

  As final velocity in both cases is same

 So, value of   $\sqrt{\frac{4}{3}gh+v_{i}^{2}}$ should be same in both cases.

$\therefore$   $\sqrt{\frac{4}{3}\times10\times 30+(3)^{2}}=\sqrt{\frac{4}{3}\times10\times 27+(v_{2})^{2}}$

                        $v_{2}=7 m/s$

 A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is  $\frac{1}{4}$th of its value at the surface of the planet. If the escape  velocity from the planet is  $v_{sec}=v\sqrt{N}$, then the value of N is  (ignore energy loss due to atmosphere)

Answer:

Explanation:

 At height h

                    $g'=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}$      ......(i)

 Given, $g'=\frac{g}{4}$

 Substituting  in Eq .(i) we get,

                                    h=R

   Now, From A to B,

  Decrease in kinetic energy= increase in potential energy

$\Rightarrow$    $\frac{1}{2}mv^{2}=\frac{mgh}{1+\frac{h}{R}}$

$\Rightarrow$    $\frac{v^{2}}{2}=\frac{gh}{1+\frac{h}{R}}=\frac{1}{2}gR$              (h=R)

$\Rightarrow$    $v^{2}= gR$    or   $v= \sqrt{gR}$

  Now,      $v_{esc}=\sqrt{2gR}=v\sqrt{2}$

$\Rightarrow$                N=2

 Consider a hydrogen atom with its electron in the n^th orbital. An electromagnetic radiation of wavelength 90mm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is  (hc= 1242 eV nm)

Answer:

Explanation:

 The kinetic energy of ejected  electron

                   =    Energy of incident photon - energy required to ionize  the electron from nth orbit (all in eV)

      $\therefore$     $10.4=\frac{1242}{90}-|E_{n}|=\frac{1242}{90}-\frac{13.6}{n^{2}}$

                                                (as   $E_{n}\propto \frac{1}{{n^2}}$  and E₁   =-13.6 eV)

             Solving this equation , we get

                                 n=2

An infinitely long uniform line charge distribution of charge per unit length λ  lies parallel to the y-axis in the y-z plane at   

$z=\frac{\sqrt{3}}{2}a$    ( see figure). If the magnitude of the flux of the electric field through  the rectangular  surface ABCD lying in the x-y plane with its centre at the origin is   $\frac{\lambda L}{n \epsilon_{0}}$

                   ($ \epsilon_{0}=$  = permittivity of free space) then the value of n is 

Answer:

Explanation:

 ANBP is cross-section of a cylinder of length L. The line charges pass through  the centre O and perpendicular to paper

$AM=\frac{a}{2},MO=\frac{\sqrt{3}a}{2}$

$\therefore$    $\angle AOM=\tan ^{-1}\left(\frac{AM}{OM}\right)$

                              $=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{0}$

 Electric Flux passing from the whole cylinder

                                                 $\phi_{1}=\frac{q_{in}}{\epsilon_{0}}=\frac{\lambda L}{\epsilon_{0}}$

  $\therefore$ Electric flux passing through ABCD plane surface  (shown only AB)= Electric flux  passing through cylindrical  surface ANB

                             =  $\left( \frac{60^{0}}{360^{0}}\right)(\phi_{1})=\frac{\lambda L}{6\epsilon_{0}}$

   $\therefore$              n=6

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