Tag: density of a fluid

Questions Related to density of a fluid

As the pressure increases, density will

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. remain same

  2. decrease

  3. increase

  4. may increase

Show answer
Correct Option: C
Explanation:

If pressure increases, density generally increases in response. The pressure reduces the volume taken up while the mass remains the same. That follows from the definition of density as mass per unit volume. This effect varies greatly between materials however, gases compress readily, such that density increases nearly in proportion to pressure, liquids and solids much less so.

The density of aluminium is 2.7 $ \displaystyle g/cm^{3} $. Its density in $ \displaystyle kg/m^{3} $ will be :

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. 27 $ \displaystyle kg/m^{3} $

  2. 2700 $ \displaystyle kg/m^{3} $

  3. 270 $ \displaystyle kg/m^{3} $

  4. 27000 $ \displaystyle kg/m^{3} $

Show answer
Correct Option: B
Explanation:

density of aluminium = $2.7g/{ cm }^{ 3 }$

                                    = $\dfrac { 2.7\times { 10 }^{ -3 }kg }{ { 10 }^{ -6 }{ m }^{ 3 } } $
density of aluminium = $2700kg/{ m }^{ 3 }$

A vessel contains a mixture of $7g$ of nitrogen and $8g$ of oxygen at temperature $T=300K$. If the pressure of the mixture is $1atm$, its density is 
$\left[ R=\cfrac { 25 }{ 3 } J/mol\quad K \right] $

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $0.6kg/{m}^{3}$

  2. $1.2kg/{m}^{3}$

  3. $1.5kg/{m}^{3}$

  4. $2kg/{m}^{3}$

Show answer
Correct Option: A
Explanation:

Molar mass of mixture $=\cfrac{total mass}{total mole}$

$M=\cfrac{15}{\cfrac{4}{14}+\cfrac{8}{16}}=15\M=15\ PM=\rho RT\ \rho=\cfrac{1(15)\times10^{-3}}{(\cfrac{25}{3}300)}\ \rho=\cfrac{15}{25}=\cfrac{3}{5}=0.6kg/m^3$

The mass and volume of a body are found to be 5.00 $\pm$ 0.05 kg and 1.00 $\pm$ 0.05 $m^3$  respectively. Then the maximum possible percentage error in its density is

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. 6%

  2. 3%

  3. 10%

  4. 5%

Show answer
Correct Option: A

The volume of a cube is $\displaystyle 2.5{ cm }^{ 3 }$ and its mass is 20g. Calculate the density of the cube in MKS and CGS systems.

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $\displaystyle 0.008{ g }/{ { cm }^{ 3 } }and\quad 8{ kg }/{ { m }^{ 3 } }$

  2. $\displaystyle 8000{ g }/{ { cm }^{ 3 } }and\quad 8{ kg }/{ { m }^{ 3 } }$

  3. $\displaystyle 8{ g }/{ { cm }^{ 3 } }and\quad 8000{ kg }/{ { m }^{ 3 } }$

  4. $\displaystyle 0.8{ g }/{ { cm }^{ 3 } }and\quad 800{ kg }/{ { m }^{ 3 } }$

Show answer
Correct Option: C
Explanation:

Explanation: In MKS system i.e. the SI system the unit of density is $\displaystyle { kg }/{ { m }^{ 3 } }$ and in CGS system it is $\displaystyle { g }/{ { cm }^{ 3 } }$. Density = mass/volume = $\displaystyle 20/2.5=8{ g }/{ { cm }^{ 3 } }$ in CGS system. In MKS system density = $\displaystyle 8\times 1000=8000{ kg }/{ { m }^{ 3 } }$

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weights $0.25\ kg$ in air and is observe to displace $0.015$ litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, $19.3$ and $8.9$. The approximate percentage of copper in the ornament is

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $5\%$

  2. $10\%$

  3. $15\%$

  4. $25\%$

Show answer
Correct Option: D
Explanation:

Let volume of gold in ornament $V _1$ and that of copper $=V _2$ and density of gold $=\rho g$ and that of copper$=\rho _c$

$\Rightarrow \rho g V _1+\rho _cV-2=0.25\rightarrow (1)$
Volume of ornament $=$Volume of water displaced
$\Rightarrow V _1+V-2=0.15\times10^{-3}ms\Rightarrow V _2=(0.015\times10^{-3}-V _1)$
According to question-
$\cfrac{\rho _g}{\rho _w}=19.3$ and $\cfrac{\rho _c}{\rho _w}=8.9\ \rho _g=19.3\times10^3 kg/m^3$
$\rho _c=8.9\times10^3 kg/m^3$
Putting these value in equation $(1)$
$(19.3\times 10^{ 3 })V _{ 1 }+(1.9\times 10^{ 3 })(0.0015\times 10^{ -3 }-V _{ 1 })0.25\ [(19.3\times 10^{ 3 })-(1.9\times 10^{ 3 })]V _{ 1 }+(8.9\times 0.015)=0.25\ 10.4\times 10^{ 3 }V _{ 1 }=0.12\ V _{ 1 }=\cfrac { 0.12 }{ 10.4\times 10^{ 3 } } =0.0115\times 10^{ -3 }=1.15\times 10^{ -5 }m^{ 3 }$
$V _2=(0.015\times10^3-0.0115\times10^{-3})=0.0035\times10^{-3}\ \% \quad of\quad copper=(\cfrac{V _2}{V _1+V _2})\times100=[\cfrac{0.0035\times10^{-3}}{(0.0115+0.0035)\times10^{-3}}]\times100$
$\approx 23.34\%\ \approx 25\%$

A liquid mixture of volume $V$ has two liquids as its ingredients with densities $\alpha  \; and\; \beta $. If the density of the mixture is $\sigma $, then the mass of the first liquid in the mixture is :

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $\dfrac{\alpha V[\sigma \beta +1]}{\beta [\alpha +\sigma ]}$

  2. $\dfrac{\alpha V[\sigma -\beta ]}{ [\sigma +\beta]}$

  3. $\dfrac{\alpha V[\beta-\sigma ]}{ \beta-\alpha }$

  4. $\dfrac{\alpha V[1-\sigma\alpha ]}{ \beta[\alpha-\sigma ] }$

Show answer
Correct Option: C
Explanation:
Let mass of liquid with density $\alpha =M _1$
mass of liquid with density $\beta =M _2$
Total volume$=V$
Net density of mixture$=\sigma$
Total mass$=M _1+M _2$
$\Rightarrow V\sigma =M _1+M _2$
$\Rightarrow M _2=V\sigma -M _1$ ......$(1)$

$\left[\because \dfrac{Total \, Mass}{v}=\sigma\right]$

$T=\dfrac{Total \,mass}{Total \, volume}=\dfrac{M _1+M _2}{\dfrac{M _1}{\alpha}+\dfrac{M _2}{\beta}}$ .......$(2)$
sub $(1)$ in $(2)$

$\Rightarrow \sigma =\dfrac{M _1+(v\sigma -M _1)}{\dfrac{M _1}{\alpha}+\left(\dfrac{v\sigma -M _1}{\beta}\right)}$

$\Rightarrow M _1=\dfrac{\alpha V(\beta -\sigma)}{\beta -\alpha}$.

A long straight cable of length l is placed symmetrically along z-axis and has radius $a(<<l)$. the cable consists of a thin wire and co-axial conducting tube. An alternating current I(t) = ${ I } _{ 0 }\  \sin { \  (2\pi \nu t) }$flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is E(s, t) = ${ \mu  } _{ 0 }{ I } _{ 0 }\cos { (2\pi \nu t)\  ln \left( \dfrac { s }{ a }  \right) \hat{ k } }$. The displacement current density inside the cable is

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $\dfrac { 2\pi }{ { \lambda }^{ 2 } } { I } _{ 0 }ln \left( \dfrac { a }{ s } \right) \sin { (2\pi \nu t) } \hat{ k }$

  2. $\dfrac { 1 }{ { \lambda }^{ 2 } } { I } _{ 0 }ln \left( \dfrac { a }{ s } \right) \sin { (2\pi \nu t) } \hat{ k }$

  3. $\dfrac { \pi }{ { \lambda }^{ 2 } } { I } _{ 0 }ln \left( \dfrac { a }{ s } \right) \sin { (2\pi \nu t) } \hat{ l }$

  4. Zero

Show answer
Correct Option: A
Explanation:

Given :  The induced electric field at a distance from wire is $E(s,t)=\mu _0I _0v\,\cos(2\pi v t)ln(\dfrac{s}{a})\hat{k}$

Displacement current density is given by :
$\vec{J} _d=\epsilon _0\dfrac{dE}{dt}$

$\Rightarrow \epsilon _0\mu _0I _0v \begin{matrix}\partial  \\partial(t)  \end{matrix}(\cos 2\pi vt)(\begin{matrix}  \dfrac{s}{a} \end{matrix})\hat{k}$

Substitute for $\dfrac{1}{\sqrt{\mu _0\epsilon _0}}$

$\Rightarrow \dfrac{1}{c^2}I _0 2\pi v^2(-\sin(2\pi vt))ln(\begin{matrix}  \dfrac{s}{a} \end{matrix})\hat{k}$

$\Rightarrow (\begin{matrix} \dfrac{v} {c}\end{matrix})^2  2\pi I _0 \sin(2\pi vt))ln(\begin{matrix}  \dfrac{a}{s}\end{matrix})\hat{k}$

$\Rightarrow \dfrac{2\pi}{\lambda^2}I _0 ln (\begin{matrix} \dfrac{a}{ s} \end{matrix})\sin (2\pi vt)\hat{k}$

Option (A) is correct.