JEE 2014 :: Advanced 2014 :: Physics 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

The figure shows a circular loop of radius   $\alpha$  with two long parallel wires (numbered 1 and 2)  all in the plane of the paper. The distance of each wire from the centre of the loop is d, The loop and the wires are carrying the same current I. The  current in the loop is in the counter-clockwise direction if seen from above

 When d≈a  but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case

Answer:

Explanation:

B_R =B due to ring

B₁ = B due to wire-1

B₂= B due to wire -2

 In magnitudes

     B₁  =B₂ =  $\frac{\mu_{0}l}{2\pi r}$

 Resultant of B₁ and B₂

$=2B_{1}\cos \theta =2\left(\frac{\mu_{0}l}{2\pi r}\right)\left(\frac{h}{r}\right)$

   = $\frac{\mu_{0}lh}{\pi r^{2}}$

   $B_{R}=\frac{\mu_{0}lR^{2}}{2(R^{2}+X^{2})^{3/2}}$

$=\frac{2\mu_{0}l\pi a^{2}}{4\pi r^{3}}$

   As , R=a ,x=h  and a²+h²=r²

 For zero magnetic field at P.

  $\frac{\mu_{0}lh}{\pi r^{2}}=\frac{2\mu_{0}l\pi a^{2}}{4\pi r^{3}}$

    $\Rightarrow$       $   \pi a^{2}=2rh\Rightarrow$ h≈1.2 a

A spray gun is shown in the figure where a piston pushes air out of the nozzle. A thin tube of uniform cross-section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out.

 For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1mm respectively. The upper end of the container is open to the atmosphere.

If the density of air is   $\rho_{a}$    and that of  the liquid $\rho_{l}$, then for a given piston  speed the rate (volume per unit time)  at which the liquid  is sprayed will be proportional to 

Answer:

Explanation:

$p_{1}-p_{2}=\frac{1}{2}\rho_{a}v_{a}^{2}$

$p_{3}-p_{2}=\frac{1}{2}\rho_{l}v_{l}^{2}$

p₃= p₁

   $\therefore$     $\frac{1}{2}\rho_{l}v_{l}^{2}$= $ \frac{1}{2}\rho_{a}v_{a}^{2}$

          $\Rightarrow$           $  v_{l}=\sqrt{\frac{\rho_{a}}{\rho_{l}}}v_{a}$

     $\therefore $     Volume flow rate  $\propto  \sqrt{\frac{\rho_{a}}{\rho_{l}}}$

A spray gun is shown in the figure where a piston pushes air out of the nozzle. A thin tube of the uniform cross-section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out.

 For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1mm respectively. The upper end of the container is open to the atmosphere.

  If the piston is pushed  at a speed of 5 mms^-1, the air comes out of the nozzle with a speed of 

Answer:

Explanation:

From continuity equation

 $A_{1}v_{1}=A_{2}v_{2}$

 Here, A₁  =400 A₂ because

r₁=20r₂  and   $A=\pi r^{2}$

$\therefore $   $  v_{2}=\frac{A_{1}}{A_{2}}(v_{1})=400v_{1}$

  = 400(5)  mm/s

    =2000 mm/s= 2 m/s

In the figure, a container is shown to have a movable (without friction) position on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400K.  The heat capacities per mole of an ideal monoatomic gas are   $C_{v}=\frac{3}{2}R,C_{p}=\frac{5}{2}R,$  , and those for an ideal diatomic  gas are  $C_{v}=\frac{5}{2}R,C_{p}=\frac{7}{2}R,$

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments in the same. Then total work done by the gases till the time they achieve  equilibrium will be

Answer:

Explanation:

  $\triangle W_{1}+\triangle U_{1}=\triangle Q_{1}$       ...........(i)

                   $\triangle W_{2}+\triangle U_{2}=\triangle Q_{2}$

               $\triangle Q_{1}+\triangle Q_{2}=0$

   $\therefore$     $ (nC_{p}\triangle T)_{1}+(nC_{p}\triangle T)_{2}=0$

      But  n₁ =n₂=n

$\therefore $    $\frac{5}{2} R(T-700)+\frac{7}{2} R(T-400)=0$

 Solving , we get T=525 k

Now, from equatins (i) and (ii) , we get

               $\triangle W_{1}+\triangle W_{2}=-\triangle U_{1}- \triangle U_{2}$

        as   $\triangle Q_{1}+\triangle Q_{2}=0$

  $\therefore$     $ \triangle W_{1}+\triangle W_{2}=-[(nC_{v}\triangle T)_{1}+(nC_{v}\triangle T)_{2}]$

     =  $-\left[2 \times\frac{3}{2}R\times (525-700)+2\times\frac{5}{2}R\times  (525-400)\right]$

           =-100 R

In the figure a container is shown to have a movable (without friction) position on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400K.  The heat capacities per mole of an ideal monoatomic gas are   $C_{v}=\frac{3}{2}R,C_{p}=\frac{5}{2}R,$  , and those for an ideal diatomic  gas are  $C_{v}=\frac{5}{2}R,C_{p}=\frac{7}{2}R,$

Consider the partition to be rigidly fixed so that it does not move. When equilibrium  is achieved , the final temperature of the gases will be

Answer:

Explanation:

 Let final equilibrium temperature of gases is T 

 Heat rejected by gas by lower cmpartment

          $=nC_{v}\triangle T$

  $=2\times \frac{3}{2}R(700-T)$

  Heat  received by the gas in above compartment

  =  $=nC_{p}\triangle T$

      $=2\times \frac{7}{2}R(T-400)$

 Equating the two , we get

   2100-3T= 7T-2800

 $\Rightarrow$            $ T=490 K$

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014