JEE 2012 :: General Questions :: Physics

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A current-carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it.The correct statement(s) is(are)

Answer:

Explanation:

Due to the current in the straight wire, net magnetic flux from the circular loop is zero. Because in half of the circular magnetic field is inwards and in another hall magnetic field is outwards. Therefore, a change in current will not cause any change in magnetic flux from the loop. Therefore, induced emf under all conditions through the circular loop is zero

In the given circuit, the AC source has $\omega$ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

Answer:

Explanation:

Circuit I

  $X_{C}=\frac{1}{\omega C}=100 \Omega$

 $\therefore$ 

 $z_{1}=\sqrt{(100)^{2}+(100)^{2}}=100\sqrt{2} \Omega$

 $\phi_{1}=\cos^{-1}\left(\frac{R_{1}}{Z_{1}}\right)=45^{0}$

In this circuit current leads the voltage 

 $I_{1}=\frac{V}{Z_{1}}=\frac{20}{100 \sqrt{2}}=\frac{1}{5 \sqrt{2}}A$

$V_{100 \Omega}= (100) I_{1}=(100) \frac{1}{5 \sqrt{2}}V$

  $=10\sqrt{2}$V

 Circuit 2

  $X_{L}=\omega L=(100)(0.5)=50 \Omega$

 $Z_{2}=\sqrt{(50)^{2}+(50)^{2}}=50\sqrt{2}\Omega$

$\phi_{2}=\cos^{-1}\left(\frac{R_{2}}{Z_{2}}\right)=45^{0}$

in this circuit voltage leads the current

 $I_{2}=\frac{V}{Z_{2}}=\frac{20}{50\sqrt{2}}=\frac{\sqrt{2}}{5}$ A

$V_{50\Omega}=(50)I_{2}=50\left(\frac{\sqrt{2}}{5}\right)=10\sqrt{2}V$

Further $I_{1}$ and $I_{2}$ have a mutual phase  difference of  $90^{0}$

$\therefore$      $I=\sqrt{I_{1}^{2}+I_{2}^{2}}=\sqrt{\frac{1}{50}+\frac{4}{50}}=\frac{1}{\sqrt{10}}$

   =0.3 A 

The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed $\omega/2$. The ring and disc are separated by frictionless ball bearings. The system is in the x-z plane. The point P on the inner disc is at a
distance R from the origin, where OP  makes an angle of $30^{0}$ with the horizontal. Then with respect to the horizontal surface.

Answer:

Explanation:

 Velocity of point O is 

  $v_{0}=(3R\omega)\widehat{i}$

 $V_{PO}$ is $\frac{R.\omega}{2}$ in the direction shown in the figure. In vector form

 $v_{PO}=-\frac{R \omega}{2} \sin 30^{0}\widehat{i}+\frac{R\omega}{2}\cos 30^{0} \widehat{k}=-\frac{R\omega}{4}\widehat{i}+\frac{\sqrt{3}R\omega}{4}\widehat{i}$

 bu  $v_{PO}=v_{P}-v_{O}$

$\therefore$  $v_{p}=v_{po}+v_{o}$

 $=\left(-\frac{R\omega}{4}\widehat{i}+\frac{\sqrt{3}R\omega}{4}\widehat{k}\right)+3R \omega \widehat{i}$

 $=\frac{11}{4} R\omega \widehat{i}+\frac{\sqrt{3}}{4}R \omega \widehat{k}$

Two spherical planets P and Q have the same uniform density P, masses Mp and M_Q, and surface areas A and  4A, respectively. A spherical planet R also has uniform density ρ and its mass is (M_P + M_Q). The escape velocities from the planets P, Q and R, are V_P, V_Q^, and V_R, respectively. Then

Answer:

Explanation:

The surface area of Q is four times.
Therefore, radius of Q is two times. Volume is eight times. Therefore, mass of Q is also eight times.

 So, let $M_{p}=M$ and $R_{p}=r$

 Then, $M_{Q}=8M$ and $R_{Q}=2r$

 now, mass of  R is $(M_{p}+M_{Q})$ or 9M.

Therefore, radius of R is $(9)^{1/3}$ r. Now, escape velocity from the surface ofa planet  is given by

  $v=\sqrt{\frac{2GM}{r}}$

(r= radis of that planet)

$\therefore$   $v_{p}=\sqrt{\frac{2GM}{r}}$

$v_{Q}=\sqrt{\frac{2G(8M)}{(2r)}}$

$v_{R}=\sqrt{\frac{2G(9M)}{(9)^{1/3}r}}$

 from here we can see that,

  $\frac{v_{p}}{v_{Q}}=\frac{1}{2}$

 and  $V_{R}$ >$V_{Q}>$$V_{P}$

Two solid cylinders P and Q of the same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Answer:

Explanation:

$I_{P} >I_{Q}$

 In case of pure rolling,

   $a=\frac{g \sin \theta}{1+I/mR^{2}}$

$a_{Q} >a_{p}$  as its moment of inertia is less.

Therefore, Q reaches first with more linear speed and more translational kinetic energy.

 Further, $\omega=\frac{v}{R}$

 or   $\omega \propto V$

$\therefore$   $\omega_{Q} >\omega_{P}$                as   $v_{p} >v_{Q}$

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