Tag: substances in the surroundings - their states and properties

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

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

The clouds float in the atmosphere because of their low.

  1. Pressure

  2. Velocity

  3. Temperature

  4. Density
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

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

  1. $\dfrac { 2 }{ 3 }$
  2. $\dfrac { 4 }{ 9 }$
  3. $\dfrac { 9 }{ 4 }$

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $J _0^d = \dfrac{2\pi V\in V _0}{d}$

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

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

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