Tag: upthrust in fluids, archimedes' principle and floatation

Quizzes (4) Questions (190)

Questions Related to upthrust in fluids, archimedes' principle and floatation

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

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. $A\rightleftharpoons \left(\dfrac { n }{ 2 } \right)B+C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The degree of dissociation alpha = (D-d) / ((n-1)d) corresponds to the dissociation reaction A -> nB + ... where the total number of moles increases by (n-1). Option A represents a reaction where the stoichiometry leads to this specific expression.

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

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. $\displaystyle 62300{ kg }/{ { m }^{ 3 } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, ${ 62.3 }/{ 1000 }=0.0623{ g }/{ { cm }^{ 3 } }$

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

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. $6\times 10^{-5}/ ^{\circ}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The coefficient of real expansion gamma_R = (V2 - V1) / (V1 * deltaT). Using the height change as a proxy for volume change: gamma_R = (80.4 - 80) / (80 * 100) = 0.4 / 8000 = 0.00005 = 5e-5 per degree Celsius.

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

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 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equilibrium of a floating plank depends on the balance of buoyancy and gravity. For a plank of specific gravity 0.5, the angle theta with the horizontal is determined by the geometry of the submerged portion. The correct angle is pi/6.

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

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. $3.3 gm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The loss in weight in water is 24 - 21 = 3 gm, which is the buoyant force. Since the volume is constant, the buoyant force in another liquid is proportional to its specific gravity. Thus, the loss in the new liquid is 3 * 1.1 = 3.3 gm, and the new weight is 24 - 3.3 = 20.7 gm.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The force per unit length between two infinite parallel wires is F/L = (mu_0 * lambda_1 * lambda_2) / (2 * pi * d). Using mu_0 = 4 * pi * 10^-7, lambda = 5 * 10^-6 C/m, and d = 0.1 m, the calculation yields 4.5 N/m.

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

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. $14N$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A smooth rubber ball has less surface friction than a tennis ball, which has a fuzzy texture. When spun in water, the tennis ball experiences more drag due to its surface texture, making the smooth ball move more easily.

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

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 $N/{m^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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