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 Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V^q, where V  is the Volume of the gas. The value of q is $\left(\gamma=\frac{C_{p}}{C_{V}}\right)$

Answer:

Explanation:

Central Idea  For an adiabtic  process $TV^{\gamma-1}$  = constant

 We know that average time of collision between molecules

  $\tau=\frac{1}{n\pi \sqrt{2}V_{rms}d^{2}}$

 where  , n= number of molecules per unit volume

     $V_{rms}$  = rms velocity of molecules

   As    $n\propto\frac{1}{V}$    and    $V_{rms}\propto \sqrt{T}$

     $\tau \propto \frac{V}{\sqrt{T}}$

  Thus, we can write  $n=K_{1}V^{-1}$      and   $V_{rms}=K_{2}T^{1/2}$

 where K₁    and K₂    are constants

   For adiabatic process,    $TV^{\gamma-1}$  = constant

 Thus , we can write

       $\tau \propto  VT^{-1/2}\propto V(V^{1-\gamma})^{-1/2}$

        or   $\tau \propto V^{\frac{\gamma+1}{2}}$

Consider a spherical shell of radius R at temperature T. The black body radiation  inside it can be considered as an ideal gas of photons with internal energy per unit volume  $u=\frac{U}{V}\propto T^{4}$ and pressure $p=\frac{1}{3}\left(\frac{U}{V}\right)$. If the shell now undergoes an adiabatic expansion, the relation between T and R is

Answer:

Explanation:

According to question

 $p=\frac{1}{3}\left(\frac{U}{V}\right)$

    $\Rightarrow$ $\frac{nRT}{V}=\frac{1}{3}\left(\frac{U}{V}\right)$                 [ $\because$  pV=nRT]

   or    $\frac{nRT}{V}\propto\frac{1}{3}T^{4}$

   or     $VT^{3}$  = constant

   or    $\frac{4}{3}\pi R^{3} T^{3}= constant$

  or      TR= constant

   $\Rightarrow$    $T\propto\frac{1}{R}$

A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its bob, the time period changes T_M. If the young's modulus of the material of the wire is Y, then   $\frac{1}{Y}$ is equal to (g= gravitational acceleration)

Answer:

Explanation:

We  know that  time period 

$T=2\pi \sqrt{\frac{L}{g}}$

 When additional  mass M is added to its bob

 $T_{M}=2\pi \sqrt{\frac{L+\triangle L}{g}}$

 where   $\triangle L$     is increase in length

 we know that

 $Y=\frac{Mg/A}{\triangle L/L}=\frac{MgL}{A \triangle L}$

 $\Rightarrow$       $\triangle L=\frac{MgL}{AY}$

 $\Rightarrow$   $\therefore$                       $T_{M}=2\pi\sqrt{\frac{L+\frac{MgL}{AY}}{g}}$

 $\Rightarrow$    $\left(\frac{T_{M}}{T}\right)^{2}=1+\frac{Mg}{AY}$

 or         $\frac{Mg}{AY}=\left(\frac{T_{M}}{T}\right)^{2}-1$

 or    $\frac{1}{Y}=\frac{A}{Mg}\left[\left(\frac{T_{M}}{T}\right)^{2}-1\right]$

From a solid sphere of mass M and radius R, a spherical  portion of radius  $\left(\frac{R}{2}\right)$   is  removed as shown in the figure. Taking gravitational potential V=0 at r= ∞ , the potential at the centre  of the cavity thus formed is (G= gravitational  constant)

Answer:

Explanation:

Central Idea      Consider cavity as negative mass and apply superposition of gravitatiional potential.

       Consider , the cavity formed in a solid sphere as shown in figure

                        V(∞)=0

According to the question, we can write potential at an internal point P due to the complete solid sphere.

     $V_{s}=-\frac{GM}{2R^{3}}\left[3R^{2}-\left(\frac{R}{2}\right)^{2}\right]$

      $=-\frac{GM}{2R^{3}}\left[3R^{2}-\frac{R^{2}}{4}^{}\right]$

    $=-\frac{GM}{2R^{3}}\left[\frac{11R^{2}}{4}^{}\right]=\frac{-11GM}{8R}$

 Mass of removed part

    $=\frac{M}{\frac{4}{3}\times\pi R^{3}}\times\frac{4}{3}\times\pi\left(\frac{R}{2}\right)^{3}=\frac{M}{8}$

  Potential at  point P due to removed part

  $V_{c}=\frac{-3}{2}\times\frac{GM/8}{\frac{R}{2}}=\frac{-3GM}{8R}$

Thus, potential due to remaining part at point P.

      $V_{p}=V_{s}-V_{c}=\frac{-11GM}{8R}-\left(-\frac{3GM}{8R}\right)$

   $=\frac{(-11+3)GM}{8R}=\frac{-GM}{R}$

A particle  of mass m moving in the x-direction which speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to 

Answer:

Explanation:

Central idea.    concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.

   Consider ,the stones thrown up simulaneously  as shown in the diagram below.

   Considering motion of the second particle with respect to the first we have realtive acceleration

       $|a_{21}|=|a_{2}-a_{1}|=g-g=0$

Thus, motion of first particle is straight line with respect to second particle till the first particle strikes ground at a time given by

                 $-240=10t-\frac{1}{2}\times 10\times t^{2}$

   or             $t^{2}-2t-48=0$

  or              $t^{2}-8t+6t-48=0$

   or    t=8,-6  (not possible)

Thus, distance  covered by second particle  with respect to first particle in 8 s is

    $s_{12}=(v_{21})t=(40-10)(8s)$

      $=30\times 8=240m$

  Similarly, time taken by second particle to strike the ground is given by

  $-240=40t-\frac{1}{2}\times10\times t^{2}$

 or           $-240=40t-5 t^{2}$

or         $5 t^{2}-40t-240=0$  

 or    $ t^{2}-8t-48=0$

    $ t^{2}-12t+4t-48=0$

  or  t (t-12)+4(t-12)=0

   or t=12,-4             (not possible)

Thus, after 8 s, magnitude of relative velocity will increase up to 12 s when second particle strikes the ground.

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