JEE 2014 :: Advanced 2014 :: Physics 2014

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

A point source S is placed at the bottom of the transparent block of height 10 mm and refractive index 2.72 . It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block . The refractive index of the liquid is 

Answer:

Explanation:

At point Q angle of incidence is critical angle  $\theta_{C}$   , where

$\sin\theta_{C}=\frac{\mu_{1}}{\mu_{block}}$

 In   $\triangle PQS,$      $\sin \theta_{C}=\frac{r}{\sqrt{r^{2}+h^{2}}}$

 $\therefore$     $ \frac{\mu_{l}}{\mu_{block}}=\frac{r}{\sqrt{r^{2}+h^{2}}}$

$\Rightarrow$        $ \mu_{l}=\frac{r}{\sqrt{r^{2}+h^{2}}}\times2.72$

   $=\frac{5.77}{11.54}\times2.72=1.36$

Charges Q,2Q and 4Q  are uniformly distributed in three dielectric solid spheres  1,2 and 3 of radii  R/2, R and 2R respectively, as shown in the figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1,2 and 3 are E₁,E₂  and E₃

respectively, then 

Answer:

Explanation:

$E_{1}=\frac{kQ}{R^{2}}$   ,where    $ k=\frac{1}{4\pi\epsilon_{0}}$

  $E_{2}=\frac{k(2Q)}{R^{2}}\Rightarrow E_{2}=\frac{2kQ}{R^{2}}$

$E_{3}=\frac{k(4Q)R}{2R^{3}}\Rightarrow E_{3}=\frac{kQ}{2R^{2}}$

 $E_{3}$ < $E_{1}$< $E_{2}$

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω . as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is 

Answer:

Explanation:

 For balanced meter bridge

    $\frac{X}{R}=\frac{l}{(100-l)}$   ( where , R=90 Ω)

  $\therefore $     $\frac{X}{90}=\frac{40}{(100-40)}$

  $\therefore $     $X=60$

  $X=R\frac{l}{(100-l)}$

 $\frac{\triangle X}{X}=\frac{\triangle l}{l}+\frac{\triangle l}{100-l}$

                        $=\frac{0.1}{40}+\frac{0.1}{60}$

  $\triangle X=0.25$

   So,   X= $60\pm 0.25$ Ω

A wire, which passes through the hole in a small bead, is bent in the form pf quarter of a circle . The wire is fixed vertically on ground as shown in figure. The bead is released from near the top of the wire and it slides along the wire without friction. Aa the bend moves from A to B, the force it applies on the wire is

Answer:

Explanation:

h= R -Rcos Ω

 Using conservation energy

$mgR(1-\cos\theta)=\frac{1}{2}mv^{2}$

  Radial force equation  is 

  $mg\cos\theta-N=\frac{1}{R}mv^{2}$

 = mg (  3$\cos\theta$-2)

Here, $N=mg \cos\theta-\frac{mv^{2}}{R}$

 N= 0 at   $\cos\theta=\frac{2}{3}$

 $\Rightarrow$    Normal force act radially outward on bead if cos θ  >2/3   and normal  force act radially inward on bead if 

 cos θ < 2/3

$\therefore$  Force ring is opposite to normal force on bead.

 A thermodynamic system is taken from an initial state i with internal energy  U_i=100J  to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the path af, ib and bf are  W_af=200J, W_ib=50J and  W_bf= 100 J respectively. The heat supplied to the system along the path iaf,ib and bf are Q_iaf, Q_ib and Q_bf respectively. If the internal energy  of the system in the state b is U_b=200J and Q_iaf =500J, the ratio Q_bf/Q_ib    is 

Answer:

Explanation:

  $W_{ibf}=W_{ib}+W_{bf}$

   =50J +100J=150J

$W_{iaf}=W_{ia}+W_{af}$

  =0+200 J= 200J

  Q_iaf= 500J

  So,   $\triangle U_{iaf}=Q_{iaf}-W_{iaf}$

                    =500 J-200J=300J

           =U_f-U_i

 So,     $U_{f}=U_{iaf}+U_{i}$

            =300 J+100J

              =400J

 $\triangle U_{ib}=U_{b}-U_{i}$

            =200J-100J

              =100 J

 $Q_{ib}=\triangle U_{ib}+W_{ib}$

             =100J +50J=150J

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

   $=\triangle U_{iaf}+W_{ibf}$

        =300J+150J

      =450 J

So, the required ratio,

 $\frac{Q_{bf}}{Q_{ib}}=\frac{Q_{ibf}-Q_{ib}}{Q_{ib}}$

    =  $\frac{450-150}{150}=2$

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