JEE 2014 :: Advanced 2014 :: Physics 2014

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

 A transparent thin film of uniform thickness and refractive index n₁=1.4 is coated on the convex spherical surface of radius  R at one end of a long solid glass cylinder of refractive index n₂=1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f₁ from the film, while rays of light traversing from glass to air get focused at distance f₂ from the film, then

Answer:

Explanation:

$\frac{1}{f_{film}}=(n_{1}-1)\left(\frac{1}{R}-\frac{1}{R}\right)$

$\Rightarrow$  $f_{film}=\infty $           (infinite)

$\therefore$ There is no effect of presence  of  film.

From Air to Glass

  Using  the equation

                   $\frac{n_{2}}{v}-\frac{1}{u}=\frac{n_{2}-1}{R}$

  $\frac{1.5}{v}-\frac{1}{\infty}=\frac{1.5-1}{R}$

 $\Rightarrow$    v=3R

  $\therefore$ f₁=3R

From Glass to Air, Again using  same equation

$\frac{1}{v}-\frac{n_{2}}{u}=\frac{1-n_{2}}{-R}$

  $\Rightarrow$   $ \frac{1}{v}-\frac{1.5}{\infty}=\frac{1-1.5}{-R}$

$\Rightarrow$  v=2R

      $\therefore$ f₂= 2R

 In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of frictional between the wall and ladder is  $ \mu_{1}$  and that between the floor and the ladder is  $\mu_{2}$. The normal reaction of the wall on the ladder is N₁ and that of the floor is N₂. If the ladder is about to slip, then

Answer:

Explanation:

$ \mu_{2}$ can never be zero for maximum equilibrium.

when  $\mu_{1}$ = 0 we have

  $N_{1}=\mu_{2}N_{2}$    ..........(i)

     $N_{1}=m_{2}g$       ...........(ii)

$\tau_{B}=0\Rightarrow mg\frac{L}{2}\cos\theta=N_{1}L\sin\theta$

$\Rightarrow $   $N_{1}=\frac{mg \cot \theta}{2}$

$\Rightarrow$    $ N_{1}\tan \theta=\frac{mg }{2}$

 when  $\mu_{1}\neq 0$ we have

 $\mu_{1}N_{1}+N_{2}=mg$      ........(i)

  $\mu_{2}N_{2}$= N        .......(ii)

 $\Rightarrow$      $ N_{2}=\frac{mg}{1+\mu_{1}\mu_{2}}$

The heater of an electric kettle is made of a wire of length  L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d.The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40K?

Answer:

Explanation:

 Resistance of initially  given kettle3

   $R=\rho\frac{1}{A}=\rho$

$\frac{L}{\pi(d/2)^{2}}=\frac{4\rho L}{\pi d^{2}}$

Power , $P=\frac{V^{2}}{R}$

  or            $P\propto\frac{1}{R}$

   Resistance of two replaced kettles

             $R_{1}=\frac{\rho L}{\pi d^{2}}$

 and 

                 $R_{2}=\frac{\rho L}{\pi d^{2}}$

So    $R_{1}=R_{2}=\frac{R}{4}$

 if wires are in parallel then  equivalent resistanece

   $R_{p}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{R}{8}$

 Then power P_p=8p                   (as $P\propto\frac{1}{R}$]

 So, it will take 0.5 minute

   (as H=Pt  or      $t=\frac{H}{p}$      or  $t\propto\frac{1}{p}$ )

 if wires are in series then  equivalent resistanece

R_S=R₁+R₂=R/2

 tthen power P_s= 2P

 So, it will take 2 minutes.

A student is performing an experiment using a resonance column and a tuning fork of frequency  244 s^-1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum  height at which resonance occurs is   $(0.350\pm 0.005)$ m, the gas in the tube is

(Useful information)

$\sqrt{167RT}=640J^{1/2}mol^{-1/2}$

$\sqrt{140RT}=590J^{1/2}mol^{-1/2}$. The molar masses M in grams are given in the options. Take the value of 

$\sqrt{10/M}$  for each gas as given there.)

Answer:

Explanation:

Minimum  length  =  $\frac{\lambda}{4}\Rightarrow\lambda=4l$

 Now, v=f ,   $\lambda=(244)\times 4\times l$

  as        $l=0.350\pm0.005$

$\Rightarrow$    v lies between 336.7 m/s to 346.5 m/s

  Now,   $v=\sqrt{\frac{\gamma RT}{M\times 10^{-3}}}$  , here M is

 Molecular  mass in gram

                  $=\sqrt{100\gamma RT}\times\sqrt{\frac{10}{M}}$

 For  monoatomic gas 

       $\gamma=1.67$

 $\Rightarrow    $      $v=640\times \sqrt{\frac{10}{M}}$

For diatomic gas,

      $\gamma=1.4\Rightarrow v=590\times\sqrt{\frac{10}{M}}$

  $\therefore $     $ V_{Ne}=640\times \frac{7}{10}=448 m/s$

                        $ V_{Ar}=640\times \frac{17}{32}=340 m/s$

                           $ V_{O_{2}}=590\times \frac{9}{16}=331.8 m/s$

                                  $ V_{N_{2}}=590\times \frac{3}{5}=354 m/s$

$\therefore $   only possible answer is Argon

Let E₁(r), E₂(r) and E₃(r)  be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density λ, and an infinite plane with uniform surface charge density   $\sigma$.If   $E_{1}(r_{0})=E_{2}(r_{0})=E_{3}(r_{0})$    at a  given distance r₀ , then

Answer:

Explanation:

$\frac{Q}{4\pi \epsilon_{0}r_0^2}=\frac{\lambda}{2\pi \epsilon_{0}r_{0}}=\frac{\sigma}{2\epsilon_{0}}$

                          $Q=2\pi \sigma r_0^2$

 (a) is incorrect,   $r_{0}=\frac{\lambda}{\pi\sigma}$

   (b)    is incorrect,   $E_{1}(\frac{r_{0}}{2})=4E_{1}(r_{0})$

 As,            $E_{1}\propto\frac{1}{r^{2}}$

    $E_{2}\left(\frac{r_{0}}{2}\right)=2E_{2}(r_{0})$     as    $E_{2}\propto\frac{1}{r}$

$\Rightarrow  $  (c) is correct.

    $E_{3}\left(\frac{r_{0}}{2}\right)=E_{3}(r_{0})=E_{2}(r_{0})  $

 as          $E_{3}\propto r^{0}$

$\Rightarrow  $  (d) is in correct.   

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