VITEEE 2016 :: Physics :: General questions

• Physics - General questions

The current in an L-R  circuit builds up to (3/4)^th of its steady-state value in 4 seconds . The time constant of this circuit is 

Answer:

Explanation:

$I= I_{0}(1-e^{-t / \tau}$)

  Where $\tau$ → time constant 

 $\therefore$     $\frac{3}{4}I_{0}=I_{0}(1-e^{-t/\tau})$

 $\Rightarrow\frac{3}{4}=1-e^{-t/\tau}\Rightarrow e^{-t/\tau}=\frac{1}{4}$

 $\Rightarrow\frac{t}{\tau}In e=In \frac{1}{4}\Rightarrow\frac{-4}{\tau}=-2 In 2$

 $\Rightarrow $     $\tau= \frac{2}{In 2}$

The equation of AC voltage  is  $E= 220 sin (\omega t+\frac{\pi}{6})$ and AC current $I=10\sin (\omega t+\frac{\pi}{6})$ . The average  power dissipated is 

Answer:

Explanation:

 We know that ,   $Z=\frac{E_{0}}{I_{0}}$

 E₀= 220 V and I₀=10 A so Z= 220/10=  22 ohm

 $\phi =\left[ \frac{\pi}{6}-(-\frac{\pi}{6})\right]=\frac{\pi}{3}$

  $P_{a}=\frac{E_{0}}{\sqrt{2}}\times\frac{I_{0}}{\sqrt{2}}\times\cos\phi$

  $=\frac{220}{\sqrt{2}}\times\frac{10}{\sqrt{2}}\times\cos\frac{\pi}{3}=550 W$

 if i₁=3 sin ω t and i₂= 4 cos ω t, the i₃ is 

Answer:

Explanation:

From Kirchhoff's current law,

 $i_{3}=i_{1}+i_{2}=3\sin\omega t+4\sin (\omega t+90^{0})$

$=\sqrt{3^{2}+4^{2}+2(3)(4)\cos 90^{0}\sin (\omega t+\phi)}$

 where  $\tan\phi= \frac{4\sin 90^{0}}{3+4\cos 90^{0}}=\frac{4}{3}$

 $\therefore$    i₃ = $5 (\sin\omega t+53^{0})$

 A long straight wire of radius R carries a current I. The magnetic field inside the wire at distance r from its centre is expressed as

Answer:

Explanation:

Using Ampere's law, we have

 $\oint \overrightarrow{B}.\overrightarrow{dl}=\mu_{0}i_{in}$   

or   $B \times 2\pi r =\mu_{0}\frac{i}{\pi R^{2}}\pi r^{2}$

  $\therefore$      $B= \frac{\mu_{0}}{2\pi}.\frac{ir}{R^{2}}$

 A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Answer:

Explanation:

 The angular momentum L of the particle is given by

 $L=mr^{2}\omega $   where  $ \omega =2\pi n$

 $\therefore$  Freequency n= $\frac{\omega}{2\pi}$

 Further  $i= q\times n= \frac{\omega q}{2\pi}$

 Magnetic momemt ,  $M=i A= \frac{\omega q}{2\pi}\times \pi r^{2}$

 $\therefore$    $M= \frac{\omega q r^{2}}{2}$

So,  $\frac{M}{L}= \frac{\omega q r^{2}}{2mr^{2}\omega}=\frac{q}{2m}$

• Physics - General questions