JEE 2014 :: Advanced 2014 :: Physics 2014

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

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure.The total capacitance of the capacitor is C while that of the portion with dielectric in between is C₁. When the capacitor is charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q₂. The electric field in the dielectric is E₁ and that in the other portion is E₂. Choose the correct option/options. Ignoring edge effects.

Answer:

Explanation:

C= C₁+C₂

$C_{1}=\frac{K\epsilon_{0}A/3}{d},C_{2}=\frac{\epsilon_{0}2A/3}{d}$

$\Rightarrow$      $ C_{1}=\frac{(K+2)\epsilon_{0}A}{3d}$

$\Rightarrow$       $ \frac{C}{C_{1}}=\frac{(K+2)}{K}$

   Also, E₁  =E₂  =V/d , where V is potential difference between the plates.

One end of a taut string of length 3m along the x-axis is fixed at x=0. The speed of the waves in the string is 100ms^-1. The order end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible waveform (s)  of these stationary wave is (are)

Answer:

Explanation:

 There should be a node at x=0 and antinode at x=3m

Also,    $v=\frac{\omega}{k}=100 m/s$

 $\therefore$    y=0 at x=0

 and   $y=\pm A$     at  x=3m

 Only  (a).(c)  and (d) satisfy condition.

A light source, which emits two wavelengths $\lambda_{1}$= 400nm and $\lambda_{2}$=600 nm  . is used in Young's double-slit experiment. If recorded fringe widths for  $\lambda_{1}$  and $\lambda_{2}$ are  $\beta_{1}$  and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central  maximum are   $m_{1}$ and $m_{2}$ respectively, then

Answer:

Explanation:

Fringe width $\beta= \frac{\lambda D}{d}$   or $   \beta\propto\lambda $

 $\therefore$     $ \lambda_{2}$   >  $ \lambda_{1}$ 

 So     $\beta_{2}$  >  $\beta_{1}$

Number of fringes ina given width

   $m=\frac{y}{\beta}$    or   $m\propto\frac{1}{\beta}$

  $\Rightarrow$     $  m_{2}$   <  $m_{1}$   as    $\beta_{2}$ >   $\beta_{1}$

Distance of 3 rd maximum of  $\lambda_{2}$  from central maximum

       $=\frac{3\lambda_{2}D}{d}=\frac{1800D}{d}$

 Distance of 5 th minimum of  $\lambda_{1}$ from central maximum

                     $=\frac{9\lambda_{1}D}{2d}=\frac{1800D}{d}$

So,  3rd maximum of  $\lambda_{2}$ will overlap with 5th minimum of $\lambda_{1}$

 Angular seperation (or angular fringe width)

   $=\frac{\lambda}{d}\propto \lambda$

    $\Rightarrow $   aangular separation for $\lambda_{1}$  will be lesser.

At time t=0 terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t)= I₀ $\cos (\omega t) $ with I₀ =I A and $\omega $= 500 rad s^-1 starts flowing in it with the initial direction shown in the figure. At t= $\frac{7\pi}{6\omega}$, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C= 20μ F, R=10Ω and the battery is ideal with emf of 50 V, identify the correct statement (s)

Answer:

Explanation:

$\frac{dQ}{dt}=I\Rightarrow Q=\int_{}^{} I dt=\int_{}^{} (I_{0}\cos \omega t)dt$

 $\therefore $        $ Q_{max}=\frac{I_{0}}{\omega}=\frac{1}{500}=2 \times 10^{-3}C$

  Justrt after switching

 In steady state,

At     $t=\frac{7\pi}{6\omega}$    or    $\omega t=\frac{7\pi}{6} $

   Current comes out to be negative from the given expression. So , current is anti-clockwise.

  Charge supplied by source from t=0 to t =  $\frac{7\pi}{6\omega}$

  $Q=\int_{0}^{\frac{7\pi}{6\omega}}  \cos (500t) dt$

    $=\left[\frac{\sin 500t}{500}\right]^{\frac{7\pi}{6\omega}}_{0}=\frac{\sin \frac{7\pi}{6}}{500}=-1mC$

  Apply Kirchhoff's loop law  just after changing the switch to position D

    $ 50+\frac{Q_{1}}{C}-IR=0$

substituting  the values of Q1 , C and R we get,

   I= 10A

   In steady state Q₂= CV=1mC

 $\therefore$    Net charge flown from battery= 2mC

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