JEE 2015 :: Advanced 2015 :: Physics 2015

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

 Two independent harmonic oscillators of equal  masses are oscillating about the origin with angular frequencies   $\omega_{1}$ and  $\omega_{2}$ and have  total energies  $E_{1}$  and $E_{1}$, respectively, The variations  of their momenta p with  positions  x are shown  in the figures, If  $\frac{a}{b}=n^{2}$  and $\frac{a}{R}=n$, then the correct equations is/are

Answer:

Explanation:

 Ist particle

               P=0 at x=a   $\Rightarrow$ 'a'  in the amplitude of oscillation 'A₁'

At x=0, P=b         (at mean position)

$\Rightarrow$   $mv_{max}=b\Rightarrow v_{max}=\frac{b}{m}$

  $E_{1}=\frac{1}{2}mv^{2}_{max}=\frac{m}{2}\left[\frac{b}{m}\right]^{2}=\frac{b^{2}}{2m}$

$A_{1}\omega_{1}=v_{max}=\frac{b}{m}$

$\Rightarrow$      $\omega_{1}=\frac{b}{ma}=\frac{1}{mn^{2}}\left(A_{1}=a, \frac{a}{b}=n^{2}\right)$

  II nd particle

  P=0 at x= R  $\Rightarrow$ A₂ = R

At   x=0, P=R $\Rightarrow$  $v_{max}=\frac{R}{m}$

  $E_{2}=\frac{1}{2}mv_{max}^{2}=\frac{m}{2}\left[\frac{R}{m}\right]^{2}=\frac{R^{2}}{2m}$

   $A_{2}\omega_{2}=\frac{R}{m}\Rightarrow \omega_{2}=\frac{R}{mR}=\frac{1}{m}$

  (b)  

                $\frac{\omega_{2}}{\omega_{1}}=\frac{1/m}{1/mn^{2}}=n^{2}$

(c)           $\omega_{1}{\omega_{2}}=\frac{1}{mn^{2}}\times\frac{1}{m}=\frac{1}{m^{2}n^{2}}$

  (d)  $\frac{E_{1}}{\omega_{1}}=\frac{b^{2/}2m}{1/mn^{2}}=\frac{b^{2}n^{2}}{2}=\frac{a^{2}}{2n^{2}}=\frac{R^{2}}{2}$

             $\frac{E_{2}}{\omega_{2}}=\frac{R^{2}/2m}{1/m}=\frac{R^{2}}{2}$

   $\Rightarrow$   $\frac{E_{1}}{\omega_{1}}=\frac{E_{2}}{\omega_{2}}$

 Note:    It is not given that the second figure is a circle . But from the figure and as per the requirement of question , we consider it is a circle

 Plank's constant h, speed of light c and gravitational constant G are used to form a unit of length  L and a unit of mass M . Then, the correct option is /are

Answer:

Explanation:

$M\propto h^{a}c^{b}G^{c}$

 $M^{-1}\propto(ML^{2}T^{-1})^{a}(LT^{-1})^{b}(M^{-1}L^{3}T^{-2})^{c}$

                              $\propto M^{a-c}L^{2a+b+3c}T^{-a-b-2c}$

              a-c=1            ......(i)

   2a+b+3c=0 ........(ii)

  a+b+2c=0    ......(iii)

 Onsolving (i), (ii) (iii) ,  $a=\frac{1}{2},b=+\frac{1}{2},c=-\frac{1}{2}$

$\therefore$   $M\propto \sqrt{c}$  only →  (a) is correct.

  In the same way we can find that,

 $L\propto  h^{1/2}h^{-3/2}G^{1/2}$

  $L\propto \sqrt{h}$ ,  $L\propto \sqrt{G}$ →  (c), (d) are also correct.

Consider Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale  coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular  scale moves it by two divisions on the linear scale, then

Answer:

Explanation:

 For Vernier Callipers

     $1MSD=\frac{1}{8}cm$

       5  VSD=4 MSD

  $\therefore$                  $1VSD=\frac{4}{5}MSD$

                      = $\frac{4}{5}\times\frac{1}{8}=\frac{1}{10}cm$

 Least count of Vernier callipers

  = 1MSD-1VSD

      =  $\frac{1}{8}cm-\frac{1}{10}cm=0.025 cm$

 (a) and (b)

   Pitch of screw gauge

            =   $2\times0.025 =0.05 cm$

 Least count of screw gauge

                     $=\frac{0.05}{100}cm=0.005mm$

  (c) and (d)

 Least count of linear  scale of screw gauge=0.05

  Pitch= $0.05\times 2=0.1cm$

      Least count of screw gauge

                                         $=\frac{0.1}{100}cm=0.01 mm$

 In an aluminium (Al) bar of square cross-section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are  $2.7\times 10^{-8}$ Ω m and  $10\times 10^{-7}$ Ω m, respectively. The electrical  resistance between  the two faces P and Q of the composite bar is

Answer:

Explanation:

  $\frac{1}{R}=\frac{1}{R_{Al}}+\frac{1}{R_{Fe}}=\left(\frac{A_{Al}}{\rho_{Al}}+\frac{A_{Fe}}{\rho_{Fe}}\right)\frac{1}{l}$

   $= \left[\frac{(7^{2}-2^{2})}{2.7}+\frac{2^{2}}{10}\right]\frac{10^{-6}}{10^{-8}}\times\frac{1}{50\times 10^{-3}}$

  Solving we get,

            $R=\frac{1875}{64}\times 10^{-6}$Ω

              $=\frac{1875}{64}\mu$ Ω

 A container of fixed volume has a mixture  of one mole of hydrogen  and one mole  of helium  in equilibrium at temperature  T. Assuming the gases are ideal, the correct statement is/are

Answer:

Explanation:

(a)  Total  Internal energy

          $U=\frac{f_{1}}{2}nRT+\frac{f_{2}}{2}nRT$

                    $(U_{ave})_{per mole}=\frac{U}{2n}$

  =     $\frac{1}{4}\left[ 5RT+3RT\right]=2RT$

  (b)                    $\gamma_{mix}=\frac{^{n_{1}}C_{p_{1}}+^{n_{2}}C_{p_{2}}}{^{n_{2}}C_{v_{1}}+^{n_{2}}C_{v_{2}}}$

      $=\frac{(1)\frac{7R}{2}+(1)\frac{5R}{2}}{(1)\frac{5R}{2}+(1)\frac{3R}{2}}=\frac{3}{2}$

              $M_{mix}=\frac{n_{1}M_{1}+n_{2}M_{2}}{n_{1}+n_{2}}$

                   $\frac{M_{1}+M_{2}}{2}=\frac{2+4}{2}=3$

 speed of sound          $V=\sqrt{\frac{\gamma RT}{M}}$

        $\Rightarrow$                $V\propto\sqrt{\frac{Y}{M}}$

                 $\frac{V_{mix}}{V_{He}}=\sqrt{\frac{\gamma_{mix}}{\gamma_{He}}\times\frac{M_{He}}{M_{mix}}}$

                   =    $\frac{V_{mix}}{V_{He}}=\sqrt{\frac{3/2}{5/3}\times\frac{4}{3}}=\sqrt{\frac{6}{5}}$

 (d)        $V_{rms}=\sqrt{\frac{3RT}{M}}\Rightarrow V_{rms} \propto \frac{1}{\sqrt{M}}$

                    $\frac{V_{He}}{V_{H}}=\sqrt{\frac{M_{H}}{M_{He}}}=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}$

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