JEE 2018 :: Physics (2018) :: Solved questions 2018

• Physics (2018) - Solved questions 2018

Two conducting cyclinders of equal length but different radii are connected in series between two heat  baths kept at temperatures  T₁= 300K and T₂=100K, as shown in the figure. The radius of the bigger cyclinder is twice that of the smaller one and the thermal conductivites of the materials of the smaller and the larger cyclinders are K₁ and K₂ , respectively . If the temperature at the junction of the cyclinders in the steady state is 200K, then K₁/K₂=  .........

Answer:

Explanation:

Rate of heat flow will be same.

$\therefore$   $\frac{300-200}{R_{1}}=\frac{200-100}{R_{2}} $         

                                                                      ( as  $H= \frac{\text{d}Q}{\text{d}t}=\frac{T.D}{R}$)

 $\therefore$          $R_{1}=R_{2}\Rightarrow \frac{L_{1}}{K_{1}A_{1}}=\frac{L_{2}}{K_{2}A_{2}}$

$\Rightarrow$         $\frac{K_{1}}{K_{2}}=\frac{A_{2}}{A_{1}}=4$

Sunlight of intensity 1.3kWm^-2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss  of light due to the lens and assume  that the lens aperture size is much smaller than its focal length. The average intensity of light , in kW m^-2, at a distance 22 cm from the lens on the other side is .....

Answer:

Explanation:

$\frac{A_{0}^{'}}{A_{0}}=(\frac{2}{20})^{2}=\frac{1}{100}$

$\Rightarrow$       $A_{0}'=\frac{A_{0}}{100}$

                       $P= I_{0}A_{0}=I_{0}'A_{0}'$

$\Rightarrow$     $I_{0}'= \frac{I_{0}A_{0}}{\frac{A_{0}}{100}}=100I_{0}=130kW/m^{2}$

In the XY-plane , the region y >0 has a uniform magnetic field $B_{1}\hat{K}$  and the region y<0 has another uniform magnetic field $B_{2}\hat{K}$ . A positively charged particle is projected from origin along the positive Y-axis  with speed $v_{0}=\pi$ms^-1 at t=0, as shown in figure. Neglect gravity in this problem. Let t=T be the time  when the particle crosses the X-axis from below for the first time . If B₂ =4B_1, the average speed of the particle , in ms^-1 , along the  X-axis in the time interval T is........

Answer:

Explanation:

If average speed is considered along X-axis

          $R_{1}=\frac{mv_{0}}{qB_{1}}$ , $R_{2}=\frac{mv_{0}}{qB_{2}}=\frac{mv_{0}}{4qB_{1}}$

                                       $R_{1}>R_{2}$

Distance travelled along X-axis 

    $\triangle x=2(R_{1}+R_{2})=\frac{5mv_{0}}{2qB_{1}}$

    Total time = $\frac{T_{1}}{2}+\frac{T_{2}}{2}= \frac{\pi m}{qB_{1}}+\frac{\pi m}{qB_{2}}$

                    $= \frac{\pi m}{qB_{1}}+\frac{\pi m}{4qB_{1}}=\frac{5\pi m}{4qB_{1}}$

   Magnitude of average speed  $= \frac{\frac{5mv_{0}}{2qB_{1}}}{\frac{5\pi m}{4qB_{1}}}$

                                                 = 2m/s

Three identical capacitors  C₁ ,C₂ and C₃ have a capacitance of 1.0 μF each and they are uncharged initially. They are connected in a circuit as shown  in the figure and  C₁ is then filled completely with a dielectric material of relative permittivity ε_r. The cell electromotive force (emf) V₀=8V. First the switch S₁ is closed while the switch S₂ is kept open. When the capacitor C₃ is fully charged, S₁ is opened and S₂ is closed simultaneously. When all the capacitors reach equilibrium, the acitors reach equilibrium, the charge on C₃ is found to be 5μC . The value of ε_r= .......

Answer:

Explanation:

           C= ε_r C₁ =(ε_r )μF

     Applying loop rule,

   $\frac{5}{1}-\frac{3}{\epsilon_{r}}-\frac{3}{1}=0\Rightarrow\frac{3}{\epsilon_{r}}=2$

          ε_r = 1.50

A spring  block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 Nm^-1 and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially ,the spring is in an unstretched condition. Another block of mass 1.0kg moving with a speed of 2.0 ms^-1 collides elastically with the first block. The collision is such that the 2.0kg block does not hit the wall. The distance , in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is...

Answer:

Explanation:

Just Before Collision

Just After Collision

Let velocities of 1 kg and 2 kg blocks just after collision be v₁ and v₂ respectively.

     From momentum conservation principle,

     $1\times 2= 1v_{1}+2v_{2}$       .............(i)

   Collision is elastic. Hence e=1 or relative velocity of separation = relation velocity of approach

               $v_{2}-v_{1}=2$      ..........(ii)

From Eqs.(i) and (ii),

           $v_{2}=\frac{4}{3}m/s$  , $v_{1}=\frac{-2}{3}m/s$

2 kg block will perform SHM after collision,

         $t=\frac{T}{2}=\pi\sqrt{\frac{m}{k}}=3.14s$

Distance = $\mid v_{1}\mid t=\frac{2}{3}\times 3.14$

         =2.093= 2.093 m

• Physics (2018) - Solved questions 2018