Tag: option b: engineering physics

Questions Related to option b: engineering physics

The water flowing from a garden hose fills a container $ 3 \pi $ litre in one minute.Then speed of the water coming from that pipe with opening of radius 1 cm is 

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $ 4 ms^{-1} $

  2. $5 ms^{-1} $

  3. $ 1 ms^{-1} $

  4. $ 0.5 ms^{-1} $

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Correct Option: A

A container holds $ 10^{26} molecules / m^3 $ each of mass $ 3 \times 10^{-27} $ Kg. Assume that 1/6 of the ,molecule move with  velocity 2000 m/s directly towards one wall of the container while the remaining 5/6 of the molecules move either away from the wall or in perpendicular direction, and all collision of the molecules with the wall or in perpendicular direction, and all collision of the molecules with the wall are elastic.

physics option b: engineering physics buoyancy floatation fluid pressure

  1. Number of molecules hitting $ 1 m^2 $ of the wall every second is $ 3 .33 \times 10^{28} $

  2. Number of molecules hitting $ 1 m^2 $ of the wall every second is $ 2 \times 10^{29} $

  3. Pressure exerted on the wall by molecules is $ 24 \times 10^5 Pa. $

  4. Pressure exerted on the wall by moleculaes is $ 4 \times 10^5 Pa, $

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Correct Option: C

To what height h should a cylindrical vessel of diameter d be filled with a liquid so that the total force on the vertical surface of the vessel be equal to the force on the bottom-

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $h=d$

  2. $h=2d$

  3. $h=3d$

  4. $h=d/2$

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Correct Option: D
Explanation:

If we fill the cylinder upto a height h, the force exerted on at the bottom of the cylinder would be equal to F = PA


$ F = \rho gh \times \pi d^2/4 $

Similarly, the average force exerted along the sides of the cylinder will be because of half the height filled for the cylinder.

Therefore, $ F = \rho g h/2 \times \pi d h $

Equating the 2 forces, and solving for h, gives h = d/2

The height of liquid in a cylindrical vessel of diameter $d$ so that the total force on the vertical surface of the vessel be equal to the force on the bottom, will be:

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $d$

  2. $2d$

  3. $4d$

  4. $\cfrac{d}{2}$

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Correct Option: D

What should be the height of liquid in a cylindrical vessel of diameter d so that the total force on the vertical surface of the vessel be equal to the force on the bottom,

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $d$

  2. $2d$

  3. $4d$

  4. $\dfrac{d}{2}$

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Correct Option: D

To what height should a cylindrical vessel be filled with a homogeneous liquid to make the force with which the liquid pressure on the sides of the vessel equal to the force exerted by the liquid on the bottom of the vessel?

physics option b: engineering physics buoyancy floatation fluid pressure

  1. Equal to the radius.

  2. Less than radius

  3. More than radius

  4. Four times of radius

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Correct Option: A
Explanation:
If  $h$ is the height of liquid in cylinder, $r $ be the radius of the cylinder and $ ρ$ be the density of the liquid.

Then we have
Weight of the liquid $=\pi r^2h \rho g$........................................(I)
Mean pressure on the wall $=\dfrac12 \rho  gh$

The total force on the wall =  $ 2\pi rh \times \dfrac12 \rho  gh= \pi rh^2\rho g $....................................(2)

On equating (I) and (2) we have
$\pi r^2h \rho g=\pi rh^2\rho g $
$r=h$
$\therefore$ The liquid should be filled up-to a height equal to the radius of the cylinder.

The efflux velocity of a liquid of density $1500 kg m^{-3} $ from a tank in which the pressure of liquid is $1000pa$ above the atmosphere is :

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $115 ms^{-1} $

  2. $11.5 ms^{-1} $

  3. $0.115 ms^{-1} $

  4. $1.15 ms^{-1} $

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Correct Option: D
Explanation:

The velocity of a liquid is given as,

$v = \sqrt {2gh} $

$v = \sqrt {2g \times \frac{{\Delta P}}{{\rho g}}} $

$v = \sqrt {2 \times \frac{{1000}}{{1500}}} $

$v = 1.15\;{\rm{m/s}}$

A small hollow vessel open to atmosphere having a small circular hole radius $R\ mm$  in its base is immersed in a tank of water. To what depth should the base of vessel be immersed in water so that water will start coming into the vessel through the hole. ($TT$ is surface tension of water) ($\rho=$density of water).

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $\dfrac {2T}{\rho g R}$

  2. $\dfrac {T}{\rho g R}$

  3. $\dfrac {T}{4\rho g R}$

  4. $\dfrac {4T}{\rho g R}$

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Correct Option: A

A cylindrical vessel filled with water up to the height H becomes empty in time $ t _0 $ due to a small  hole at the bottom of the vessel. if water is filled to a height 4 H it will flow out in time 

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $ t _0 $

  2. $ 4t _0 $

  3. $ 8t _0 $

  4. $ 2t _0 $

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Correct Option: B

A tank with a square base of area 2 m$^2$ is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area 20 cm$^2$ at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.53 x 10 kg m$^{-3}$ in the other, both to a height of 4 m. The force necessary to keep the door closed is (Take g = 10 m s$^{-2}$)

physics option b: engineering physics buoyancy floatation fluid pressure

  1. 10 N

  2. 20 N

  3. 40 N

  4. 80 N

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Correct Option: C
Explanation:

The situation is as shown in the figure.
For compartment contain water,
$h = 4 m, \rho _w = 10^3\, kg \,m^{-3}$
Pressure exerted by the water at the door at the bottom is
$P _w=\rho _w hg  $

$=10^3\,kg \,m^{-3} \times 4 \,m \times 10 \,m s^{-2}$
$= 4 \times 10^4\,N\,m^{-2}$
For compartment containing acid.
$\rho _a =1.5 \times 10^3\, kg\, m^{-3}, h = 4 \,m$
Pressure exerted by the acid at the door at the bottom is
$P _a=\rho _ahg $
$= 1.5 \times 10^3\,kg \,m^{-3} \times 4\,m \times 10\,m\,s^{-2} $
$= 6 \times 10^4\,N\,m^{-2}$
$\therefore$ Net pressure on the door =$P _a - P _w = (6  \times 10^4 - 4 \times 10^4) N \,m^{-2}$

$= 2 \times 10^4 \,N \,m^{-2}$
Area of the door $= 20 cm^2 = 20 \times 10^{-4} m^2$
$\therefore$ Force on the door$= 2 \times 10^4 N m^{-2} \times 20 \times 10^{-4} m^2 = 40 N$
Thus, to keep the door dosed the force of $40 N$ must be applied horizontally from the water side.