Tag: properties of substances

Questions Related to properties of substances

Calculate the mass of air enclosed in a room of length, breadth and height equal to $5m,3m$ and $4m$ respectively.
Density of air $=1.3kg/{m}^{3}$.
chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 60 kg

  2. 78 kg

  3. 18 kg

  4. 10 kg

Show answer
Correct Option: B
Explanation:

The density of a body is given by:

$density=\dfrac{mass}{volume}$

$1.3=\dfrac{mass}{(5 \times 3 \times 4)}$


$mass=1.3\times 60=78kg$

Density of solid sphere is varied by $\rho = \rho _0 \lgroup 1 + \frac{r}{R} \rgroup$ where $0 \leq r \leq R$, R is the radius of the sphere. Moment of inertia of sphere w.r.t. axis passing through its centre will be : ($\rho _0$ is constant)

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

  1. $\dfrac{44}{45} \pi \rho _0 R^5$

  2. $\dfrac{44}{45} \pi \rho _0 R^4$

  3. $\dfrac{44}{35} \pi \rho _0 R^5$

  4. $\dfrac{48}{45} \pi \rho _0 R^5$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} dM=s\left( { 4\pi { x^{ 2 } }dx } \right)  \ ={ \rho _{ 0 } }\left( { 1+\dfrac { x }{ R }  } \right) \left( { 4\pi { x^{ 2 } }dx } \right)  \ dI=\dfrac { 2 }{ 3 } dM{ x^{ 2 } } \ \dfrac { 2 }{ 3 } { \rho _{ o } }\int _{ 0 }^{ R }{ \left( { 4\pi { x^{ 4 } }dx+\dfrac { { 4\pi  } }{ R } { x^{ 5 } }dx } \right)  }  \ =\dfrac { { 8\pi { \rho _{ 0 } } } }{ 3 } \left[ { \dfrac { { { x^{ 5 } } } }{ 5 } +\dfrac { { { R^{ 5 } } } }{ 6 }  } \right]  \ =\dfrac { 8 }{ 3 } \pi { \rho _{ o } }\times \dfrac { { 11{ R^{ 5 } } } }{ { 30 } }  \ =\dfrac { { 44 } }{ { 45 } } \pi { \rho _{ 0 } }{ R^{ 5 } } \end{array}$

Hence,
option $(A)$ is correct answer.

The blades of a windmill sweep out a circle of area $A$. If the wind flows at a velocity $v$ perpendicular to the circle, then the mass of the air of density $\rho$ passing through it in time $t$ is:

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

  1. $Av\rho t$

  2. $2Av\rho t$

  3. $Av^{2}\rho t$

  4. $\dfrac {1}{2}Av\rho t$

Show answer
Correct Option: A
Explanation:

Volume of wind flowing per second $= Av$
Mass of wind flowing per second $= Av\rho$
Mass of air passing in time $t\ s = Av\rho t$.

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

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

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

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