JEE 2015 :: Advanced 2015 :: Physics 2015

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

 A conductor (shown in the figure) carrying constant current I in kept in the x-y plane in a uniform magnetic field B. If F is the magnitude of the total magnetic force acting on the conductor, then the correct statements is /are

Answer:

Explanation:

Force on the complete wire = force on straight wire PQ carrying a current I.

    $F=I(PQ \times B)$

              $=I\left[\left\{2(L+R)\hat{i}\right\} \times B\right]$

This force is zero if B  is along  $\hat{i}$  direction or x-direction . If magnetic  field is along $\hat{j}$  direction or $\hat{k}$  direction,

                  $|F|=F=(I)(2)(L+R)B\sin 90^{0}$

         or              F= 2I(L+R)B

 or            $F \infty(L+R)$

 $\therefore$     Options (a),(b) and (c) are correct.

 Two identical glass rod S₁ and S₂ (refractive index=1.5)  have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in figure, with their axes (shown by the dashed line)  aligned. When a point source of light P is placed inside rod S₁ on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S₂. The distance d is 

Answer:

Explanation:

R=10cm

       Applying   $\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$  two

 times

                             $\frac{1}{v}-\frac{1.5}{-50}=\frac{1-1.5}{-10}$

                              $\frac{1}{v}+\frac{1.5}{50}=\frac{0.5}{10}$

                                    $\frac{1}{v}=\frac{0.5}{10}-\frac{1.5}{50}$

                                                   $=\frac{2.5-1.5}{50}$

  $\Rightarrow$                                               v=50

                                               MN=d ,  MI₁  =50 cm,

                          NI₁     =(d-50) cm

Again,       $=\frac{1.5}{\infty}-\frac{1}{-(d-50)}=\frac{1.5-1}{10}$

               $\frac{1}{d-50}=\frac{1}{20}$

      d=70

The figure below depict, two situations  in which two infinitely  long static  line charges of constant positive line charge density λ are kept  parallel to each other

 In their resulting electric field, point charges q and -q are kept in equilibrium between them. The point charges are confined to move in the x-direction only. If they are given a small displacement about their equilibrium  positions, then the correct statements is/are

Answer:

Explanation:

 At the shown position net force on both charges is zero. Hence they are in equilibrium. But the equilibrium of +q is a stable equilibrium. So, it will start oscillations when displaced from this position, These small oscillations are simple harmonic in nature. While the equilibrium of -q is unstable. So, it continues to move in the direction of its displacement.

 A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing through its centre O with two-point masses each of mass  $\frac{M}{8}$  at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure.

At some instant , the angular speed of the system is  $\frac{8}{9}\omega$  and one of the maases is at a distance of $\frac{3}{5}R$   from O.  At this instant , the distance  of the other mass from O is

Answer:

Explanation:

 Let the other mass at this instant is at a distance of x from the centre O.

  Applying law of conservation of angular momentum, we have

                     $I_{1}\omega_{1}=I_{2}\omega_{2}$

$\therefore$       $(MR^{2})(\omega)$

      $= \left[MR^{2} +\frac{M}{8}\left(\frac{3}{5}R\right)^{2}+\frac{M}{8}x^{2}\right]\left(\frac{8}{9}\omega\right)$

 Solving this equation , we get ,   $x=\frac{4}{5}R$

Note  If we take  identical situations with both point masses, then answer will be (c) .But in that case angular momentum is not conserved.

 A nuclear power plant supplying electrical power to a village uses a radioactive material of half-life  T years as fuel.The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time.  If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is

Answer:

Explanation:

 Let initial power available from the plant is P₀ . After time t=nT or n half livers , this will become

  $\left(\frac{1}{2}\right)^{n}P_{0}$  , Now , it is given that,

$\left(\frac{1}{2}\right)^{n}P_{0}=12.5$  % of  $P_{0}=(0.125)P_{0}$

    Solving this equation we get ,

                 n=3

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