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A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charge on the 4 µF and 9 µF capacitors ) at a point distant 30m from it, would equal to

Answer:

Explanation:

3 μ and 9 μ =12 μF

4μF and 12 μ F = $\frac{4\times 12}{4 \times 12}=3\mu F$

Q =CV =3× 8 =24 μC(on 4 μF and 3 μF )

Now, this 24 μC distributes in direct ratio of capacity between 3μ F and 9 μ F

  Therefore ,Q_9μF =18μC

   $Q_{4\mu F}+Q_{9\mu F}=24+18$

= 42μ C =Q            (say)

$E=\frac{kQ}{R^{2}}=\frac{9\times 10^{9}\times42\times 10^{-6}}{30^{2}}=420 N/C$

The region between two concentric spheres of radii a and b, respectively (see the figure), has  volume charge density 

 $\rho =\frac{A}{r}$  where A is a constant and r is distance from the center. At the centre of the spheres is a point charge Q. The value of A, such that the electric field in the region between the sphere will be constant is

Answer:

Explanation:

As  E is constant

 Hence  , E_a ≡ E_b

A s per Guass theorem, only Q contributes in electric field

$\therefore    \frac{kQ}{a^{2}}=\frac{k[Q+\int_{a}^{b}4\pi r^{2}dr.\frac{A}{r} }{b^{2}}$

Here,   $k= \frac{1}{4\pi\epsilon_{0}}$

$\Rightarrow Q\frac{b^{2}}{a^{2}}=Q+4\pi A \left[\frac{r^{2}}{2}| _{a}^{b} \right]$

$=Q+4\pi A [\frac{b^{2}-a^{2}}{2}]$

$\Rightarrow  Q (\frac{b^{2}}{a^{2}})=Q+2\pi A (b^{2}-a^{2})$

$\Rightarrow  Q (\frac{b^{2}-a^{2}}{a^{2}})=2\pi A (b^{2}-a^{2})$

$\Rightarrow  A= \frac{Q}{2\pi a^{2}}$

A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (Take , g=10 ms^-2)

Answer:

Explanation:

At distance x from the bottom

      $v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{(\frac{mgx}{L})}{(\frac{m}{L})}}=\sqrt{gx}$

    $\therefore    \frac{\text{d}x}{\text{d}t}=\sqrt{x}\sqrt{g}$

$\Rightarrow  \int_{0}^{L}x^{-\frac{1}{2}} dx=\sqrt{g}\int_{0}^{t} dt$

 $\Rightarrow [\frac{x^{-\frac{1}{2}}}{(\frac{1}{2})}\mid_0^L]=\sqrt{g}.t$

$\Rightarrow  t=\frac{2\sqrt{L}}{\sqrt{g}}$

$\Rightarrow  t=2\sqrt{\frac{20}{10}}=2\sqrt{2} s$

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that  it is at  a distance 

$\frac{2}{3}$  A   from the equilibrium position. The new amplitude  of the  motion is 

Answer:

Explanation:

$v= \omega \sqrt{A^{2}-x^{2}}$

At  $x=\frac{2A}{3}$

$v=\omega\sqrt{A^{2}-(\frac{2A^{}}{3})^{2}}=\frac{\sqrt{5}}{3}\omega A$

As, velocity is trebled , hence v'=$\sqrt{5}A\omega$

This leads to new amplitude A'

$\omega\sqrt{A'^{2}-(\frac{2A}{3})^{2}}=\sqrt{5}A\omega$

$\Rightarrow \omega^{2}[A'^{2}-\frac{4A^{2}}{9}]=5A^{2}\omega^{2}$

$\Rightarrow A'^{2}= 5A^{2}+\frac{4}{9}A^{2}=\frac{49}{9}A^{2}$

$A'=\frac{7}{3}A$

n moles of an ideal gas undergoes a process A and B shown in the figure. The maximum temperature of the gas during the process will be

Answer:

Explanation:

p-V equation for path AB

$p= -(\frac{p_{0}}{V_{0}})V+3p_{0}$

$\Rightarrow pV=3p_{0}V -\frac{p_{0}}{V_{0}}V^{2}$

OR   $T=\frac{pV}{nR}=\frac{1}{nR}(3p_{0}V-\frac{p_{0}}{V_{0}}V^{2})$

For maximum temperature

     $\frac{dT}{dV}=0$

  $\Rightarrow  3p_{0}-\frac{2p_{0}V}{V_{0}}=0\Rightarrow V=\frac{3}{2}V_{0}$

and         $p= 3p_{0}-\frac{p_{0}}{V_{0}}=\frac{3p_{0}}{2}$

   Therefore, at these values,

$T_{max}=\frac{(\frac{3p_{0}}{2})(\frac{3V_{0}}{2})}{nR}=\frac{9p_{0}V_{0}}{4nR}$

• Main 2016 - Physics 2016
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