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

• Physics (2018) - Solved questions 2018

A ring and a disc are initially at rest , side by side , at the top of an inclined plane which makes an angle 60° with the horizontal . They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is  $\frac{(2-\sqrt{3})}{\sqrt{10}}$  s, then the height of the top of the inclined plane, in meters is ...                   [Take, g=10 m s^-2 ]

Answer:

Explanation:

$a=\frac{g\sin\theta}{1+\frac{1}{MR^{2}}}$

$a_{ring}=\frac{g\sin\theta}{2}$             ($I= MR^{2}$)

$a_{disc}=\frac{2g\sin\theta}{3}$             ($I= \frac{MR^{2}}{2}$)

$s= \frac{h}{\sin\theta}=\frac{1}{2}at^{2}$

    $= \frac{1}{2}(\frac{g\sin\theta}{2})t_{1}^{2}$

    $\Rightarrow$                         $t_{1}=\sqrt{\frac{4h}{g\sin^{2}\theta}}=\sqrt{\frac{16h}{3g}}$

    $s= \frac{h}{\sin \theta}=\frac{1}{2}at^{2}=\frac{1}{2}(\frac{2g\sin\theta}{3})t_{2}^{2}$

$\Rightarrow$                   $t_{2}=\sqrt{\frac{3h}{g\sin^{2}\theta}}=\sqrt{\frac{4h}{g}}$

                       $t_{2}-t_{1}=\sqrt{\frac{16h}{3g}}-\sqrt{\frac{4h}{g}}$

                        $= \frac{2-\sqrt{3}}{\sqrt{10}}$

               $\sqrt{h}[\frac{4}{\sqrt{3}}-2] =2-\sqrt{3}$

Solving this equation we get, h=0.75m

Two men are walking along a horizontal straight line in the same direction.The man in front walks at speed 1.0 ms^-1 and the man behind  walks at a speed 2.0 ms^-1 . A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which  emit a sound of frequency 1430Hz. The speed of sound in air 330 ms^-1. At the instant , when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is....

Answer:

Explanation:

   $f_{A}= 1430 [\frac{330}{330-2\cos\theta}]$

   $= 1430 [\frac{1}{1-\frac{2\cos\theta}{330}}]\approx 1430[1+\frac{2\cos \theta}{330}]$

                                                          [ from binomial expansion]

                $f_{B}=1430[\frac{330}{330+1\cos\theta}]$

                       $\approx1430[1-\frac{\cos \theta}{330}]$

     Beat frequency = $f_{A-}-f_{B}$

                         $=1430[\frac{3\cos \theta}{330}] =13\cos\theta$

                          $=13[\frac{5}{13}] =5.00 Hz$

Two infinitely long straight wires lie in the xy-plane along the lines x=± R. The wire located at x=± R  carries a constant current I₁ and the wire located at x= -R carries a constant cureent I₂. A circular loop of radius R  is suspended with its centre at (0,0,√3R) and in a plane  parallel to the xy-plane . This loop carries a constant current I in the clockwise direction as seen from above the loop.The current in the wire taken to be postive , if it is in the +$\hat{j}$- direction . Which of the following statements regarding the magnetic field B is (are) true ?

Answer:

Explanation:

(a)At  origin, B=0 adue to two wires if  l₁=l₂, hence (B_net) at origin  is equal to B due to ring. which is non-zero.

(b) If l₁ >0 and l₂ <0, B at origin due to wires will be along $+\hat{k}$ . Direction of B  due to ring  $-\hat{k}$ direction and hence B can be zero at origin.

(c)  If l₁< 0 and l₂ > 0, B  at origin due to wires is along $-\hat{k}$ and also along $-\hat{k}$ due to ring, hence B cannot be zero.

(d) 

At centre of ring, B due to wires is along x-axis.

       Hence z-component is only because of ring which $B=\frac{\mu_{0}i}{2R}(-\hat{k})$

Two vectors A and B are defined as $A= a\hat{i}$  and   $B= a (\cos wt\hat{i}+\sin w\hat{j})$, where a is a constant  and $w=\frac{\pi}{6}rad $$s^{-1}$ . If  $\mid A+B\mid =\sqrt{3}\mid A-B\mid$  at time $t= \tau$ for the first time, the value of $\tau$ in seconds, is

Answer:

Explanation:

$A= a\hat{i}$  and  $B= a\cos\omega \hat{i}+a \sin \omega t\hat{j}$

  $A+B= (a+a\cos\omega t )\hat{i}+a \sin \omega t\hat{j}$

  $A-B= (a-a\cos\omega t )\hat{i}+a \sin \omega t\hat{j}$

    $\mid A+B\mid= \sqrt{3}\mid A-B\mid$

      $\sqrt{(a+a\cos\omega t)^{2}+(a\sin\omega t)^{2}}=\sqrt{3}\sqrt{(a-a\cos \omega t)^{2}+(a \sin \omega t)^{2}}$

  $\Rightarrow$      $2 \cos\frac{\omega t}{2}=\pm \sqrt{3}\times 2\sin \frac{\omega t}{2}$

  $\tan \frac{\omega t}{2}=\pm \frac{1}{\sqrt{3}}$

   $ \frac{\omega t}{2}=n\pi\pm \frac{\pi}{6}$

   $ \frac{\pi}{12}t=n\pi\pm \frac{\pi}{6}$

     $t=(12n\pm 2)s$

         = 2 s

One mole of a monoatomic ideal gas undergoes a cyclic process as shown in the figure (where ,V is the volume and T is the temperature ). Which of the statements below is (are) true?

Answer:

Explanation:

(b)   Process- I t is isothermal

  expansion   $\triangle U=0,W>0$

                    $\triangle Q=W>0$

(c)    Process - IV is isothermal compression

                  $\triangle U=0,W<0$

                 $\triangle Q=0,W<0$

(d)      Process - I and III are not isobaric beacause in isobaric process T $\alpha$ V  hence, T-V graph should be a straight line passing through origin.

• Physics (2018) - Solved questions 2018