Tag: density of a fluid

Questions Related to density of a fluid

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

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

  1. 100

  2. 10

  3. 200

  4. 1000

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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 

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative 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. $A\rightleftharpoons \left(\dfrac { n }{ 2 } \right)B+C$

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Correct Option: A

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

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

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

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

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

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

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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

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

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

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

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

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

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Correct Option: C

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

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

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

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

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

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

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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

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

  1. $20.7 gm$

  2. $27.3 gm$

  3. $24 gm$

  4. $3.3 gm$

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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

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

  1. 2.5 N/m

  2. 3.25 N/m

  3. 4.5 N/m

  4. 7.5 N/m

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Correct Option: C

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

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

  1. $0.014N$

  2. $0.14N$

  3. $1.4N$

  4. $14N$

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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?

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

  1. Smooth rubber ball

  2. Tennis ball

  3. Both will move more easily

  4. They do not move at all

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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.

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

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

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

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

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

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Correct Option: A
Explanation:

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