Tag: substances in the surroundings - their states and properties

Questions Related to substances in the surroundings - their states and properties

The clouds float in the atmosphere because of their low.

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. Pressure

  2. Velocity

  3. Temperature

  4. Density

Show answer
Correct Option: D
Explanation:
As the rule of thumb, you can assume that the things with less density float on things with higher density. And water do have less density than air that's the reason clouds float in the atmosphere.

Sea water at frequency $\nu \  =\  4\  x\  { 10 }^{ 8 }$ Hz has permittivity $\varepsilon  \  \approx \  80\  { \varepsilon   } _{ 0 }$, permeability $\mu \  \approx \  { \mu  } _{ 0 }$ and resistivity $\rho \  =\  0.25\  \Omega m$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) = ${ V } _{ 0 }\  \sin { \  (2\pi \nu t) }$. The of amplitude of the displacement current density to the conduction current density is

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $\dfrac { 2 }{ 3 }$

  2. $\dfrac { 4 }{ 9 }$

  3. $\dfrac { 9 }{ 4 }$

  4. 2

Show answer
Correct Option: B
Explanation:

Suppose distance between the parallel plates is $D$ and applied voltage $V _{(t)} = V _02\pi vt$.

thus electric field
$E = \dfrac{V _0}{d} \sin (2\pi vt)$
Now using Ohm's law 
$J _c = \dfrac{1}{\phi} \dfrac{V _0}{d}\sin (2\pi vt)$

$\dfrac{V _0}{\phi d}\sin  (2 \pi vt) = J _0^c \sin  2 \pi vt$

Here $J _0^c = \dfrac{V _0}{pd}$
Now the displacement current density is given as
$Jd = \in \dfrac{\delta E}{dt} =\dfrac{\in \delta}{dt}$    $\left[\dfrac{V _0}{dt} \sin (2\pi vt)\right]$

$= \dfrac{\in 2\pi v V _0}{d} \cos (2\pi vt)$

$\Rightarrow = J^d _0 \cos (2\pi vt)$

Where $J _0^d = \dfrac{2\pi V\in V _0}{d}$

$\Rightarrow \dfrac{J^d _0}{J^c _0} = \dfrac{2\pi v \in V _0}{d}. \dfrac{pd}{V _0} = 2\pi v \in \rho$

$= 2\pi \times 80\in _0v\times 0.25 = 4\pi \in _0v \times 10$ 

$= \dfrac{10v}{9\times 10^9} = \dfrac{4}{9}$