JEE 2012 :: General Questions :: Physics

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• General Questions - Mathematics

In the given circuit, a charge of +80 $\mu C$  is given to the upper plate of the $4 \mu F$  capacitor. Then in the steady-state, the charge  on the upper  plate of the $3 \mu F$  capacitor is 

Answer:

Explanation:

 Between $3 \mu F $ and $2 \mu F$ (in parallel) total charge of $80 \mu C$ will distribute in direct ratio of capacity

 $\therefore$   $\frac{q_{3}}{q_{2}}=\frac{3}{2}$

 or       $q_{2}=\left(\frac{3}{3+2}\right)(80)=48 \mu C$

A thin uniform cylindrical shell closed at both ends, is partially filled with water. It is floating vertically in water in a half-submerged state. If $\rho_{c}$ is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is

Answer:

Explanation:

Let $V_{1}$= total material volume of shell

$V_{2}$= total inside volume of shell and

x = fraction of $V_{2}$ volume filled with water.

ln floating condition,
Total weight = Upthrust

 $\therefore$     $V_{1} \rho_{c} g+(xV_{2})(1)g=\left(\frac{V_{1}+V_{2}}{2}\right)(1)g$

 or $  x=0.5+(0.5+\rho_{c}) \frac{V_{1}}{V_{2}}$

from here we can see that 

      $x>0.5$ if $\rho_{c} <0.5$

A lamina is made by removing a small disc of diameter 2.R from a bigger disc of uniform-rnass densiry and radius 2R, as shown in the figure.The moment of inertia of this lamina about axes passing through O and P is $I_{0}$ and $I_{p}$ respectively'.Both these axes are perpendicular to the plane  of the lamina. The ratio $\frac{I_{p}}{I_{0}}$ to the nearest  integer is 

Answer:

Explanation:

T= total portion

R= Remaining portion and

 C= cavity and

let $\sigma=$ mass per unit area

 then, $m_{T}=\pi (2R)^{2} \sigma=4 \pi R^{2} \sigma$

  $m_{C}=\pi (R)^{2} \sigma=\pi r^{2} \sigma$

For $I_{P}$

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

=$\frac{3}{2}m_{T}(2R)^{2}-\left[\frac{1}{2} m_{C}R^{2}+m_{C}r^{2}\right]$

 =$\frac{3}{2}(4\pi R^{2} \sigma)(4R^{2})-\left[\frac{1}{2}(\pi R^{2} \sigma)+(\pi R^{2} \sigma)(5R^{2})\right]$

=($18.5 \pi R^{4} \sigma)$

For $I_{0}$

=$\frac{1}{2}m_{T}(2R)^{2}-\frac{3}{2}m_{C}R^{2}$

  =$\frac{1}{2}(4 \pi R^{2}\sigma)(4R^{2})-\frac{3}{2}(\pi R^{2} \sigma)(R^{2})$

=$6.5  \pi R^{4} \sigma$

$\therefore$  $\frac{I_{P}}{I_{Q}}=\frac{18.5 \pi R^{4} \sigma}{6.5 \pi R^{4} \sigma}=2.846$

Therefore, the nearest integer is 3.

An infinitely long solid cylinder of radius. R has a uniform volume charge density $\rho$. it has a spherical cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure.The magnitude of the electric field at the point P, which is at a distance 2Rfrom the axis of the cylinder, is given by the expression  $\frac{23 \rho R}{16k \epsilon_{0}}$. The value of k is 

Answer:

Explanation:

Volume of cylinder per unit length (l=1)is

  $V= \pi R^{2}l=(\pi R^{2})$

 $\therefore$ Charge per unit length

 $\lambda=$  (volume per unit length ) x ( volume charge density)

   $=(\pi R^{2} \rho)$

 Now at P

 $E_{R}=E_{T}-E_{C}$

 R= remaining portion

T= total portion and

C= cavity

 $\therefore$   

$E_{R}=\frac{\lambda}{2\pi\epsilon_{0}(2R)}-\frac{1}{4 \pi\epsilon_{0}}\frac{Q}{(2R)^{2}}$

Q= charge on sphere

      $=\frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}\rho=\frac{\pi R^{3} \rho}{6}$

Substituting the values , we have

$E_{R}=\frac{(\pi R^{2} \rho)}{4 \pi\epsilon_{0}R}-\frac{1}{4\pi\epsilon_{0}}.\frac{(\pi R^{3} \rho/6)}{4R^{2}}$

=$\frac{23  \rho R}{96\epsilon_{0}}=\frac{23  \rho R}{(16)(6)\epsilon_{0}}$

$\therefore$   k=6

A circular wire loop of radius R is placed in the x-y plane centred at the origin O. A square loop of side a(a<< R) having two turns is placed with its centre at, $z=\sqrt{3} R$ along the axis of the circular wire loop, as shown in figure. The plane of the

the square loop makes an angle of $45^{0}$with respect to the z-axis. If the mutual inductance between the loops is given by 

 $\frac{\mu_{0}a^{2}}{2^{p/2}R}$, then the value of p is 

Answer:

Explanation:

If l current flows through the circular loop, then magnetic flux at the location of square loop is

$B=\frac{\mu_{0}IR^{2}}{2(R^{2}+Z^{2})^{3/2}}$

 Substituting the value of Z=$(\sqrt{3}R)$, we have

  $B= \frac{\mu_{0}I}{16 R}$

Now, total flux through the square loop is

$\phi_{T}=NBS \cos \theta$

=$=(2)\left(\frac{\mu_{0}T}{16R}\right)a^{2} \cos 45^{0}$

Mutual inductance

     $M=\frac{\phi_{T}}{l}=\frac{\mu_{0}a^{2}}{2^{7/2}R}$

 $\therefore$   p=7

• General Questions - Physics
• General Questions - Chemistry
• General Questions - Mathematics