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Density of a fluid - class-XI

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By cooling two liquids of equal volume from temperature $60 ^ { \circ } \mathrm { C }$  to $50 ^ { \circ } \mathrm { C }$ in same conditions time required are 324 and 810 sec respectively. If ratio of specific heat of both are 3:4. Then ratio of their. densities (water equivalent of calorimeter is negligible ):

  1. 3 /4

  2. 4 /9

  3. 8 /15

  4. 9 /20


Correct Option: A

The density of water at $4^oC$ in S.I. unit is $X\ kg m^{-3}$. Find $\dfrac{X}{250}$

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: D
Explanation:

Density of water at $4^{\circ}C$ in S.I. unit is $1000 kgm^{-3}$.

Hence, value of $X=1000$ 
$\therefore\dfrac{X}{250}=4$

The density of wood is 0.65 $ \displaystyle g\ cm^{3} $ in CGS system. Its density in SI system is

  1. 65 $ \displaystyle kgm^{3} $

  2. 6.5 $ \displaystyle kgm^{3} $

  3. 650 $ \displaystyle kgm^{3} $

  4. 0.65 $ \displaystyle kgm^{3} $


Correct Option: C
Explanation:

As 1 gm per cc  = 1000 kg per mc

so 0.65 gm per cc = 0.65 (1000) kg per mc = 650kg per mc       ( where cc = centimetre cube) and mc = metre cube)
hence option (C) is correct

A sphere is dropped under gravity through a fluid of viscosity $h$. If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is $(r=density\ of\ sphere,r=radius)$

  1. $\dfrac{4\rho r^{2}}{9 \eta}$

  2. $\dfrac{5\rho r}{9 \eta}$

  3. $\dfrac{4\rho r}{9 \eta}$

  4. $\dfrac{9\rho r}{9 \eta}$


Correct Option: A

The unit of density in MKS and CGS system respectively are:

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

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

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

  4. $\displaystyle { kg }/{ { cm }^{ 2 } }and\quad { g }/{ { m }^{ 2 } }$


Correct Option: A
Explanation:

Density is defined as the mass per unit volume of a substance. 

Density = $\cfrac{mass}{volume}$. 
In MKS system i.e. the SI system the unit of density is $\displaystyle { kg }/{ { m }^{ 3 } }$ and CGS system it is $\displaystyle { g }/{ { cm }^{ 3 } }$.

$1\ kg/m^3$ is equal to

  1. $10^3\ g/cm^3$

  2. $1\ g/cm^3$

  3. $10^{-3}\ g/cm^3$

  4. None of the above


Correct Option: C
Explanation:

$1\ kg/m^3 = 10^3g/10^6cm^3=10^{-3}g/cm^3$

A cylinder of radius R full of liquid of density $\rho$ is rotated about its axis at $\omega$ rad/s. The increase in pressure at the centre of the cylinder will be

  1. $\dfrac{\rho \omega^2 R^2}{2}$

  2. $\dfrac{\rho \omega^2 R}{2}$

  3. $\dfrac{\rho \omega R}{2}$

  4. $\dfrac{\rho^2 \omega^2 R^2}{2}$


Correct Option: B

Two solids A and B having same volume float in water. It is observed that A floats with half its volume immersed and B floats with $1/3^{rd}$ of its volume immersed. Compare the densities of A and B.

  1. $4: 3$

  2. $2: 3$

  3. $3: 4$

  4. $1: 3$


Correct Option: C

Equal masses of three liquid are kept in there identical cylindrical vessels $A, B$ and $C$. There densities are $\rho _{A}, \rho _{B}$ and $\rho _{C}$ with $\rho _{A} < \rho _{B} < \rho _{C}$. The force on the base will be

  1. Maximum in vessel $A$

  2. Maximum in vessel $B$

  3. Maximum in vessel $C$

  4. Same in all the three vessel


Correct Option: D
Explanation:

Since masses are same,(weight$=mg$) is same for all three.

The value of $g$ on the surface of earth is 9.8 $m / s ^ { 2 }$ and the radius of earth is $6400km$. The average density of earth in $k g / m ^ { 3 }$ will be

  1. $5.48 \times 10 ^ { 3 }$

  2. $2.64 \times 10 ^ { 3 }$

  3. $7.60 \times 10 ^ { 3 }$

  4. $1.46 \times 10 ^ { 3 }$


Correct Option: A

To convert density of $kg m^{-3} $ into $ g \ cm^{-3}$, we divide quantities with

  1. 100

  2. 10

  3. 200

  4. 1000


Correct Option: D
Explanation:

$1$ $\dfrac{kg}{m^3}  =1\times 1000 $ $g \times \dfrac{1}{10^6 (cm)^3}  = \dfrac{1}{1000}$  $\dfrac{g}{cm^3}$

The equation $\alpha =\dfrac { D-d }{ (n-1)d } $ is correctly matched for :
Where D= Theoretical vapour density
            d= Observed vapour density 

  1. $A\rightleftharpoons \dfrac { nB }{ 2 } +\dfrac { nC }{ 3 } $

  2. $A\rightleftharpoons \dfrac { nB }{ 3 } +\dfrac { 2n }{ 3 } C$

  3. $A\rightleftharpoons \left(\dfrac { n }{ 2 } \right)B+\left(\dfrac { n }{ 4 } \right)C$

  4. $A\rightleftharpoons \left(\dfrac { n }{ 2 } \right)B+C$


Correct Option: A

The density of an object is 62.3 in MKS unit. Express it in CGS unit.

  1. $\displaystyle 6.23{ kg }/{ { m }^{ 3 } }$

  2. $\displaystyle 0.623{ g }/{ { cm }^{ 3 } }$

  3. $\displaystyle 0.0623{ g }/{ { cm }^{ 3 } }$

  4. $\displaystyle 62300{ kg }/{ { m }^{ 3 } }$


Correct Option: C
Explanation:

Density is defined as the mass per unit volume of a substance. Density = mass (kg or g)/volume ($\displaystyle { { m }^{ 3 } }or\quad { { cm }^{ 3 } }$). In MKS system i.e. the SI system the unit of density is ${ kg }/{ { m }^{ 3 } }$ and in CGS system it is ${ g }/{ { cm }^{ 3 } }$. To convert MKS unit to CGS unit the number must be divided by 1000. Therefore, ${ 62.3 }/{ 1000 }=0.0623{ g }/{ { cm }^{ 3 } }$

A liquid column of height $80\ cm$ at $0^{\circ}$ balances the same liquid of height $80.4\ cm$ at $100^{\circ}C. \gamma _{R}$ is

  1. $4\times 10^{-5}/ ^{\circ}C$

  2. $3\times 10^{-5}/ ^{\circ}C$

  3. $5\times 10^{-5}/ ^{\circ}C$

  4. $6\times 10^{-5}/ ^{\circ}C$


Correct Option: C

If specific gravity of the plank is 0.5. then angle $\theta $ which plank make with horizontal its equilibrium is :

  1. $\dfrac { \pi }{ 4 } $

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

  3. $\dfrac { \pi }{ 6 } $

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


Correct Option: C

A block of solid insoluble in water weighs 24 gm in air and 21 gm when completely immersed in water. Its weight when completely immersed in liquid of specific gravity 1.1 is

  1. $20.7 gm$

  2. $27.3 gm$

  3. $24 gm$

  4. $3.3 gm$


Correct Option: A

Two infinite linear charges are placed parallel to each other at a distance 0.1 m from each other. if the linear charge density on each is 5 $ 5 \mu \ C /m $ , then the force acting on a unit length of each linear charge will be

  1. 2.5 N/m

  2. 3.25 N/m

  3. 4.5 N/m

  4. 7.5 N/m


Correct Option: C

A $10cm$ long needle can just rest on the surface of water without wetting, its weight is :

  1. $0.014N$

  2. $0.14N$

  3. $1.4N$

  4. $14N$


Correct Option: A
Explanation:

Surface Tension of Water $(T) = 0.00072 N/cm $


Here, Weight = Surface tension force = $2Tl$

$\Rightarrow Weight = 2 \times 0.00072 \times 10 = 0.014N$

Therefore, A is correct option.

A smooth rubber ball and a tennis ball some size are placed into two separated containers with the same amount of water in them. 
The ball will move more easily when both of them are spun at the same time?

  1. Smooth rubber ball

  2. Tennis ball

  3. Both will move more easily

  4. They do not move at all


Correct Option: A

Air streams horizontally past an air plane. The speed over the top surface is 60 m/s and that under the bottom surface is 45 m/s.The density of air is $1.293\;kg/{m^3}$, then the difference in pressure is.

  1. 1018 $N/{m^2}$

  2. 516 $N/{m^2}$

  3. 1140 $N/{m^2}$

  4. 2250 $N/{m^2}$


Correct Option: A
Explanation:

$\begin{array}{l} \dfrac { { { V^{ 2 } } } }{ 2 } +gh+\dfrac { p }{ d } ={ { constant } } \ d=density \ R=pressure \ \dfrac { { V _{ t }^{ 2 } } }{ 2 } +\dfrac { { { P _{ t } } } }{ d } =\dfrac { { V _{ B }^{ 2 } } }{ 2 } +\frac { { { P _{ B } } } }{ d }  \ { P _{ B } }-{ P _{ t } }=\dfrac { d }{ 2 } \left( { V _{ t }^{ 2 }-V _{ B }^{ 2 } } \right) =1018\, \, N/{ M^{ 2 } } \ where\, \, { V _{ t } }+{ V _{ B } }=speeds\, \, over\, \, top\, \, and\, \, bottom \ K.\in =\frac { 1 }{ 2 } \times 100\times { \left( { 11150 } \right) ^{ 2 } }m/s \ =6.2\times { 10^{ 9 } }\, \, joules \ \in scope\, \, velocity=11.15\, \, KM/s=11150\, \, m/s \end{array}$

As the pressure increases, density will

  1. remain same

  2. decrease

  3. increase

  4. may increase


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 :

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

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

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

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


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] $

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

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

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

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


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

  1. 6%

  2. 3%

  3. 10%

  4. 5%


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.

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


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

  1. $5\%$

  2. $10\%$

  3. $15\%$

  4. $25\%$


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 :

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


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

  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


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.

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