JEE 2012 :: General Questions :: Physics

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A proton is fired from very far away towards a nucleus with charge Q=120 e, where e is the electronic charge. It makes the closest approach of 10 fm to the nucleus. The de-Broglie wavelength (in units of fm) of the proton at its start is 

[Take  the proton mass,

$m_{p}=(5/3)\times 10^{-27}$kg

$h/e=4.2 \times 10^{-15}J-s/C$

$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} m/F$; 1 fm= $10^{-15}$m]

Answer:

Explanation:

 r= closet distance =10fm

 from energy conservation , we have

 $K_{i}+U_{i}=K_{f}+U_{f}$

 or   $K+0=0+ \frac{1}{4\pi \epsilon_{0}}.\frac{q_{1}q_{2}}{r}$ 

   $K= \frac{1}{4\pi \epsilon_{0}}.\frac{(120e)(e)}{r}$.........(i)

 de-Broglie wavelength

 $\lambda=\frac{h}{\sqrt{2Km}}$...........(ii)

Substituting the given values in above two equations, we get

$\lambda=7 \times 10^{-15}m=7 fm$

A cylindrical cavity of diameter c exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by $\frac{N}{12} \mu_{0} aJ$ then the value of N is 

Answer:

Explanation:

$B_{R}=B_{T}-B_{C}$

 R= Remaining portion

T= Total portion and

C= cavity

 $B_{R}= \frac{\mu_{0}l_{T}}{2a \pi}- \frac{\mu_{0}l_{C}}{2(3a/2)\pi}$.......(i)

 $l_{T}=J(\pi a^{2})$

$l_{C}=J\left(\frac{\pi a^{2}}{4}\right)$

Substituting the values in Eq. (i), we
have

 $B_{R}=\frac{\mu_{0}}{a \pi}\left[\frac{l_{T}}{2}-\frac{l_{C}}{3}\right]$

$=\frac{\mu_{0}}{a \pi}\left[\frac{\pi a^{2}J}{2}-\frac{\pi a^{2} J}{12}\right]$

$=\frac{5\mu_{0}aJ}{12}$

 $\therefore$  N=5

A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle $\theta$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains  stationary if (take g=10m/s²)

Answer:

Explanation:

W=mg=0.1 x10=1N

 $F_{1}= $ component of weight

 =1$.\sin \theta= \sin \theta$

 $F_{2}=$ component of applied force

 =$1. \cos \theta= \cos \theta$

 now,  at $\theta= 45^{0}$ :$F_{1}=F_{2}$  and biock remains stationary without the help of friction.

 For  $\theta > 45^{0}$ ; $F_{2}>F_{1}$,  so friction will act

towards Q.

 For $\theta <45^{0}$ , $F_{2} >F_{1}$ ,

and friction will act towards P.

A cubical region of side a has its centre at the origin  .It encloses three  fixed point charges , -q at (0,-a/4,0),+3q at (0,0,0) and -q at (0,+a/4,0) . Choose the correct option(s)

Answer:

Explanation:

Option (a) is correct due to symmetry.

Option (b) is wrong again due to symmetry.
Option (c) is correct because as per Gauss's theorem, net electric flux passing through any closed surface = $\frac{q_{in}}{\varepsilon_{0}}$

here, $q_{in}=3q-q-q=q$

 $\therefore$  Net electric flux=$\frac{q}{\varepsilon_{0}}$

Option (d) is wrong because there is no symmetry in two given planes

Consider the motion of a  positive  point charge in a region where there are simultaneous uniform electric and magnetic fields $E= E_{0}\widehat{j}$ and $B=B_{0}\widehat{j}$. At time t=0 , this charge has velocity v in the x-y plane, making an angle $\theta$. $\theta$  with the x-axis, which of the following option(s)  is (are) correct for time t>0?

Answer:

Explanation:

(a) and (b)

The magnetic field will rotate the particle in a circular path {in x-z plane or perpendicular to B). The electric field will exert a constant force on the particle in Positive y-direction. Therefore, the resultant path is neither purely circular nor helical or the options (a) and (b) both are wrong

 (c) 

$v_{\perp}$ and B will rotate the particle in a circular path in the x- z plane (or perpendicular to B). Further $v_{||}$ and E will move the panicle (with increasing speed) along the positive y-axis (or along the axis of the above circular path). Therefore, the resultant path is helical with increasing pitch, along the y-axis (or along B and E). Therefore option (c) is correct.
(d)

Magnetic force is zero, as $\theta$  between B and v is zero. But electric force will act in the y-direction. Therefore. motion is 1-D  and uniformly accelerated (towards positive y-direction).

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