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

• Physics (2018) - Solved questions 2018

A moving  coil galvnometer has 50 turns and each turn has an area 2 × 10^-4 m² .The magnetic field produced by the magnet inside the galvnometer is 0.02 T. The torsional constant of the suspension wire is 10^-4 N-m rad ^-1 . When a current flows through the galvanometer, a full scale deflection occurs, if the coil rotates by 0.2 rad. The resistance of the coil of the galvanometer is 50 Ω . This galvanometer is to be converted into an ammeter capable of measuring current in the range 0 - 10A , For this purpose , a shunt resistance is to be added in parallel to the galvanometer . The value of this shunt resistance in ohms, is................

Answer:

Explanation:

Given , N=50,

        A= 2 × 10^-4 m² , C=10 ^-4 , R= 50 Ω

              B= 0.02 T,  θ =0.2 rad

  N i_g AB= C θ

 $\Rightarrow$         $i_{g}=\frac{C\theta}{{N }{AB}}$

                       $i_{g}=\frac{10^{-4}\times 0.2}{50\times 2\times 10^{-4}\times0.02}=0.1A$

                       $V_{ab}=i_{g}\times G= (i - i_{g})S$

                       $V_{ab}=0.1\times 50= (1 - 0.1)\times S$

      5= 0.9 × S

              $S=\frac{50}{9}$Ω = 5.55 Ω

A particle of mass 10^-3 kg and charge 1.0 C is initially at rest. At time t=0, the particle comes under the influence of an electric field E (t)= E₀  sin ωt  $\hat{i}$, where E₀ = 1.0N C^-1  and   ω=10³ rad s^-1 . Consider the effect of only the electrical force on the particle . Then , the maximum speed in ms^-1 , attained by the particle at subsequent times is..........

Answer:

Explanation:

               $a=\frac{F}{m}=\frac{qE}{m}=10^{3}\sin(10^{3}t)$

                $\frac{\text{d}v}{\text{d}t}=10^{3}\sin(10^{3}t)$

   $\Rightarrow$      $\int_{0}^{v}dv =\int_{0}^{t} 10^{3}\sin(10^{3}t) dt$

$\therefore$     $v= \frac{10^{3}}{10^{3}}[1-\cos (10^{3}t)]$

    Velocity  will be maximum when $\cos(10^{3}t)=-1$

                        v_max = 2 m/s

A ball is projected from the ground at an angle of 45° with the horizontal surface . It reaches a maximum  height of 120m and  returns to the ground. Upon hitting the ground for the first time, it loses  half of its kinetic energy . Immediately after the bounce , the velocity of the ball makes an angle of 30° with the horizontal surface . The maximum height it reaches after the bounce, in metres, is.........

Answer:

Explanation:

$H=\frac{u^{2}\sin^{2}45^{0}}{2g}=120m$

$\Rightarrow$   $\frac{u^{2}}{4g}=120m$

  If speed  is v after the first collision, then speed should remain $\frac{1}{\sqrt{2}}$  times, as kinetic energy has reduced to half.

$\Rightarrow$    $v=\frac{u}{\sqrt{2}}$

$\therefore$     $h_{max}= \frac{v^{2}\sin^{2}30^{0}}{2g}$

                                       = $ \frac{(\frac{u}{\sqrt{2}})^{2}\sin^{2}30^{0}}{2g}$

                                      = $(\frac{u^{2}/4g}{4})=\frac{120}{4}=30m$

A solid horizontal surface is covered with  a thin layer of oil. A rectangular block of mass m= 0.4 kg  is at  rest on this surface . An impulse of 1.0 N  s is  applied to the block at time t=0, so  that it starts moving along the x- axis with a velocity $v_{t}=v_{0}e^{\frac{-t}{\tau}}$  where  v₀  is a constant  and   $\tau=4$ s. The displacement of the block, in metres at $t=\tau$ is...............  ( Take , e^-1  =0.37)

Answer:

Explanation:

Linear impluse, J= mv₀ 

   $v_{0}= \frac{J}{m}= 2.5m/s$

   $v_{t}=v_{0}e^{\frac{-t}{\tau}}$

  $\frac{\text{d}x}{\text{d}t}=v_{0}e^{-t/\tau}$

    $\int_{0}^{x}dx =v_{0}\int_{0}^{\tau}  e^{-t/\tau}dt$

  $x= v_{0}[\frac{e^{-t/\tau}}{\frac{-1}{\tau}}]_0^\tau$

       x= 2.5(-4)(e^-1 - e⁰ )

          = 2.5(-4)(0.37-1)

            x= 6.30 m

In an experiment to measure the speed of sound by a resonating air column, a turning fork of frequency 500Hz is used. The length of the air  column is varied by changing the level of water in the resonance tube . Two successive resonances are heard at air columns of length  50.7cm and 83.9 cm . Which of the following statements is (are) true?

Answer:

Explanation:

Let n th harmonic is  corresponding to 50.7 cm and (n+1)th harmonic is corresponding 83.9 cm.

               Their difference is $\frac{\lambda}{2}$

   $\frac{\lambda}{2}$   = (83.9 - 50.7) cm

                     $\lambda$    = 66.4 cm

  $\frac{\lambda}{4}$     = 16.6 cm

   Length corresponding to fundamental mode must be close to   $\frac{\lambda}{4}$ and 50.7 cm must be an odd multiple  of this length 16.6 × 3 =49.8 cm. Therefore, 50.7 is 3rd  harmonic.

   If end correction is e, then

           $e+50.7=\frac{3\lambda}{4}$

      e= 49.8 - 50.7 = -.09 cm

          speed of sound  ,  $v=f\lambda$

 v= 500 × 66.4 cm/s   =332 m/s

• Physics (2018) - Solved questions 2018