JEE 2014 :: Main 2014 :: Physics 2014

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014

A green light is an incident from the water to the air-water interface at the critical angle (θ). Select the correct statement

Answer:

Explanation:

The total internal reflection of light takes place. following conditions must be obeyed 

(i) The ray must travel from denser to rarer medium.

(ii) Angle of incidence (θ) must be  greater than or equal to critical angle (C) i.e,

   $C= \sin^{-1}\left(\frac{\mu_{rarer}}{\mu_{denser}}\right)$

Here,  $\sin C=\frac{1}{n_{water}}$

 and      $n_{water}=a+\frac{b}{\lambda^{2}}$

  If frequency is less = λ is greater and hence,  RI $n_{water}$ is less and therefore, critical angle increases. So, they do not suffer reflection and come out at an angle less than 90°

A thin convex lens made from crown glass (μ =3/2) has focal length f. When it is measured in two different liquids having refractive index 4/3 and 5/3. It has the focal length f₁  and f₂ respectively. The correct relation between  the focal length is 

Answer:

Explanation:

It is based on lens makers formula and its magnification

 i.e,   $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

According to lens Maker's formula,when the lens in the air.

$\frac{1}{f}=\left( \frac{3}{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\frac{1}{f}=\frac{1}{2x}$

$\Rightarrow$            $ f=2x$

    $\left(\frac{1}{x}=\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

In case of liquid , where  refractive index is  $\frac{4}{3}$ and   $\frac{5}{3}$, we get

focal length in first liquid 

$\frac{1}{f_{1}}=\left(\frac{\mu_{s}}{\mu_{l1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\Rightarrow$          $\frac{1}{f_{1}}=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)\frac{1}{x}$

$\Rightarrow$            f₁  is positive

$\frac{1}{f_{1}}=\frac{1}{8x}=\frac{1}{4(2x)}=\frac{1}{4f}$

$\Rightarrow$             f₁=4f

  focal length in second liquid

            $\frac{1}{f_{2}}=\left(\frac{\mu_{s}}{\mu_{l1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\Rightarrow$          $\frac{1}{f_{2}}=\left(\frac{\frac{3}{2}}{\frac{5}{3}}-1\right)\frac{1}{x}$

$\Rightarrow$            f₂ is negative

During the propagation elecromagnetic waves in a medium,

Answer:

Explanation:

Both the energy densities are equal, i.e, energy is equally divided between electric and magnetic field

In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterwards, suddenly point C is disconnected from point A and connected to point B at time t=0, Ration of the voltage  across resistance and the inductor at t=L/R will be equal to 

Answer:

Explanation:

 After connecting  C to B hanging the switch, the circuit will act like an L-R discharging circuit.

Applying Kirchhoff's loop  equation,

  $V_{R}+V_{L}=0\Rightarrow V_{R}=-V_{L}$

$\frac{V_{R}}{V_{L}}=-1$

The coercivity of a small magnet where the ferromagnet gets demagnetised is  $3\times 10^{3}Am^{-1}$. The current required to be passed ina solenoid  of length 10cm and number of turns 100, so that the magnet gets demagnetised  when  inside the solenoid is 

Answer:

Explanation:

 For solenoid, the magnetic field needed to be magnetised the magnet.

           $B=\mu_{0}nI$

[where,

   n=100,I=10cm= $\frac{10}{100}m=0.1m$]

 $\Rightarrow 3\times 10^{3}=\frac{100}{0.1}\times I$

    I=3A

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014