JEE 2015 :: Advanced 2015 :: Physics 2015

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

 Two spheres P and Q   for equal radii have densities  ρ₁ and ρ₂, respectively, The spheres are connected by a massless string and placed in liquids L₁ and L₂ of densities  σ₁   and σ ₂  and viscosities η₁ and η₂, respectively. They float in equilibrium with the sphere P  in L₁  and sphere Q  in L₂   and the string being taut (see figure). If  sphere  P alone in L₂  has  terminal  velocity  v_p  and Q alone in L₁ has terminal velocity v_Q, then

Answer:

Explanation:

 For floating , net weight of system=  net upthrust.

  $\Rightarrow  (\rho_{1}+\rho_{2})V_{g}=(\sigma_{1}+\sigma_{2})V_{g}$

           Since string is taut,   $\rho_{1}<\sigma_{1}$     and  $\rho_{2}>\sigma_{2}$

                      $v_{p}=\frac{2r^{2}g}{2\eta_{2}}(\sigma_{2}-\rho_{1})$

                                                    (upward terminal velocity)

                          $v_{Q}=\frac{2r^{2}g}{9\eta_{1}}(\rho_{2}-\sigma_{1})$

                                                 ( downward  terminal velocity)

                      $|\frac{v_{P}}{v_{Q}}|= \frac{\eta_{1}}{\eta_{2}}$

   Further ,  $\overline{v_{P}}.\overline{v_{Q}}$   will negative  as they are opposite  to each other.

A fission  reaction is given by   $_{92}^{236}U\rightarrow_{54}^{140}Xe +_{38}^{94} Sr+x+y$  , where  x and y are two paricles.  Considering   $_{92}^{236}U$ to be at rest, the kinetic energies of the products are denoted bt K_xe , K_sr , K_x  (2MeV) and  K_y   (2 MeV), respectively, Let the binding energies per nucleon of  $_{92}^{236}U$ . $_{54}^{140}xe$  and  $_{38}^{94}sr$ be 7.5 MeV , 8.5 MeV and 8.5 MeV, respectively. Considering different vconservation laws , the correct options is/are

Answer:

Explanation:

From conservation laws of mass number and atomic number, we can say that x= n, y=n

         $(x=_0^1n, y=_0^1n)$

  $\therefore$   Only (a) and (d)  options may be correct

   From the conservation of momentum.

            $|p_{xe}|= |p_{sr}|$

    From $K=\frac{P^{2}}{2m}\Rightarrow K\propto\frac{1}{m}$

              $\frac{K_{sr}}{K_{xe}}=\frac{m_{xe}}{m_{sr}}$

$\therefore$           K_sr  =129MeV

                     K_xe      =86MeV

Note:  There is no need of finding the total energy released in the process.

An ideal monoatomic gas is confined ina horizontal cylinder by a spring-loaded piston  (as shown in the figure). Initially, the gas is at temperature T₁,  pressure P₁  and volume V₁  and the spring is in its relaxed state.  The gas is then heated very slowly to temperature T₂, pressure P₂  and volume V₂  . during this process the piston moves out  by a distance of x, Ignoring the friction  between the piston and the cylinder, the correct statements is/are

Answer:

Explanation:

 Note:   This question can be solved if the right-hand side chamber is assumed to open so that its pressure remains constant even if the piston shifts towards night.

(a)    $pV=n RT \Rightarrow p \propto\frac{T}{V}$

 Temperature  is  made three times and volume is doubled

$\Rightarrow$       $p_{2}=\frac{3}{2}p_{1}$

     Further        $x=\frac{\triangle V}{A}=\frac{V_{2}-V_{1}}{A}$

                                  $=\frac{2V_{1}-V_{1}}{A}=\frac{V_{1}}{A}$

                     $p_{2}=\frac{3p_{1}}{2}=p_{1}+\frac{kx}{A}$

$\Rightarrow$     $KX=\frac{P_{1}A}{2}$

    Energy of spring

              $\frac{1}{2}kx^{2}=\frac{p_{1}A}{4}x=\frac{p_{1}V_{1}}{4}$

(b)       $\triangle U=nc_{v}\triangle T=n\left( \frac{3}{2}R\right) \triangle T$

                           = $\frac{3}{2}(p_{2}V_{2}-p_{1}V_{1})$

                    $\frac{3}{2}\left[ \left(\frac{3}{2}P_{1}\right) (2V_{1})-p_{1}V_{1}\right]=3p_{1}V_{1}$

  (c)                        $p_{2}=\frac{4p_{1}}{3}\Rightarrow p_{2}=\frac{4}{3}p_{1}=p_{1}+\frac{kx}{A}$

            $\Rightarrow$           $kx=\frac{p_{1}A}{3}\Rightarrow x=\frac{\triangle V}{A}=\frac{2V_{1}}{A}$

                                     $W_{gas}=(p_{0}\triangle V +W_{spring})$

                  $=(p_{1}Ax+\frac{1}{2}kx.x)$

$=+\left( p_{1}A.\frac{2V_{1}}{A}+\frac{1}{2}.\frac{p_{1}A}{3}.\frac{2V_{1}}{A}\right)$

                                  $=2p_{1}V_{1}+\frac{p_{1}V_{1}}{3}=\frac{7p_{1}V_{1}}{3}$

(d)

                 $\triangle Q=W+\triangle U$

  =      $\frac{7p_{1}V_{1}}{3}+\frac{3}{2}\left( p_{2}V_{2}-p_{1}V_{1}\right)$

           =   $\frac{7p_{1}V_{1}}{3}+\frac{3}{2}\left(\frac{4}{3}p_{1}.3V_{1}-p_{1}V_{1} \right)$

               =  $\frac{7p_{1}V_{1}}{3}+\frac{9}{2}p_{1}V_{1} =\frac{41p_{1}V_{1}}{6}$

 Note:  $\triangle U=\frac{3}{2}(p_{2}V_{2}-p_{1}V_{1})$   , has been obtained in part (b).

 A parallel plate capacitor having plates of area S and plate separation d, has capacitance  C₁   in air. When two dielectrics of different relative permittivities (ε₁= 2 and ε₂  =4) are introduced between the two plates as shown in the figure, the capacitance becomes C₂. The ratio   $\frac{C_{2}}{C_{1}}$  is

Answer:

Explanation:

$C_{1}=\frac{\epsilon_{0}s}{d},C=\frac{2\epsilon_{0}\frac{s}{2}}{\frac{d}{2}}=\frac{2\epsilon_{0}s}{d}$

$C'=\frac{4\epsilon_{0}\frac{s}{2}}{\frac{d}{2}}=\frac{4\epsilon_{0}s}{d}$

and    $C"=\frac{2\epsilon_{0}\frac{s}{2}}{d}=\frac{\epsilon_{0}s}{d}$

  $C_{2}=\frac{CC'}{C+C'}+C"=\frac{4}{3}\frac{\epsilon_{0}s}{d}+\frac{\epsilon_{0}s}{d}$

=   $\frac{7}{3}\frac{\epsilon_{0}s}{d}\frac{C_{2}}{C_{1}}=\frac{7}{3}$

 A spherical  body of radius R consists  of a fluid  of constant density and is in equilibrium  under  its own gravity . If  P(r)  is the pressure  at r(r<R) , then  the correct options  is /are

Answer:

Explanation:

 Gravitational  field at a distance  r due to mass 'm'

        $E=\frac{G\rho \frac{4}{3}\pi r^{3}}{r^{2}}=\frac{4G\rho \pi r}{3}$

 Consider a small element  of width  dr and are  $\triangle A$    at a distance r.

  Pressure force on this element  outwards = gravitational force on 'dm'  from 'm'  inwards

        $\Rightarrow$          $(dp)\triangle A=E(dm)$

  $\Rightarrow$    $-dp.\triangle A=\left\{\frac{4}{3}G \pi \rho r\right\}(\triangle A.dr.\rho)$

$-\int_{0}^{P}dp=\int_{R}^{r}\left(\frac{4G\rho^{2}\pi}{3}  \right)rdr$

   $-p=\frac{4G\rho^{2}\pi}{3\times 2}[r^{2}-R^{2}]$

$\Rightarrow$     $P=c(R^{2}-r^{2})$

$r=\frac{3R}{4},p_{1}=c\left(R^{2}-\frac{9R^{2}}{16}\right)=c\left(\frac{7R^{2}}{16}\right)$

  $r=\frac{2R}{4},p_{2}=c\left(R^{2}-\frac{4R^{2}}{9}\right)=c\left(\frac{5R^{2}}{9}\right)$

    $\frac{p_{1}}{p_{2}}=\frac{63}{80}$

    $r=\frac{3R}{5},p_{3}=c\left(R^{2}-\frac{9}{25}R^{2}\right)=c\left(\frac{16R^{2}}{25}\right)$

  $r=\frac{2R}{5},p_{4}=c\left(R^{2}-\frac{4R^{2}}{25}\right)$

      $=c\left(\frac{21R^{2}}{25}\right)\Rightarrow\frac{p_{3}}{p_{4}}=\frac{16}{21}$

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