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Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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-

  1. $h=d$
  2. $h=2d$
  3. $h=3d$
  4. $h=d/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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:

  1. $d$
  2. $2d$
  3. $4d$
  4. $\cfrac{d}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Force on bottom = P * Area = (rho * g * h) * (pi * (d/2)^2). Force on vertical surface = Integral of (rho * g * y) * (pi * d) dy from 0 to h = (rho * g * h^2 / 2) * (pi * d). Setting these equal: h/4 = h^2/2d => h = d/2.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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,

  1. $d$
  2. $2d$
  3. $4d$
  4. $\dfrac{d}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is identical to the previous question. The force on the bottom is rho * g * h * Area, and the force on the side is the integral of pressure over the depth, resulting in h = d/2.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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?

  1. Equal to the radius.

  2. Less than radius

  3. More than radius

  4. Four times of radius
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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 :

  1. $115 ms^{-1} $
  2. $11.5 ms^{-1} $
  3. $0.115 ms^{-1} $
  4. $1.15 ms^{-1} $
Reveal answer Fill a bubble to check yourself
D Correct answer
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}}} $

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

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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).

  1. $\dfrac {2T}{\rho g R}$
  2. $\dfrac {T}{\rho g R}$
  3. $\dfrac {T}{4\rho g R}$
  4. $\dfrac {4T}{\rho g R}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Water enters the vessel when the pressure due to depth (rho * g * h) exceeds the capillary pressure (2T / R) at the hole. Thus, h = 2T / (rho * g * R).

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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$^{-2}$)

  1. 10 N

  2. 20 N

  3. 40 N

  4. 80 N
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A vessel, whose bottom has round holes with diameter 0.1 mm, is filled with water. The maximum height up to which water can be filled without leakage is:

  1. 100 cm

  2. 75 cm

  3. 50 cm

  4. 30 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Sln :
For equilibrium,
Total upward force by surface tension 
= Weight of the water in tube
$\Rightarrow \, \pi \, \times \, D \, \times$ surface tension (circumference)
$= \, \pi(D/2)^2 \, \times \, h \, \times \, density \, \times \, g(cross \, section)$
where D (diameter) = 0.1 mm = 0.01 cm
Density of water = 1 $\times \, 10^{-3} \, gcm^3$
$\Rightarrow \, \pi \, \times \, (0.01) \, \times \, 75 \, \times \, 10^{-3}$
$= \, \pi \times \, \left(\dfrac{0.01}{2} \right)^2 \, \times \, h \, \times \, 1 \, \times \, 10^{-3} \, \times \, 1000$
$\therefore \, h = \, \dfrac{0.75 \, \times \, 0.01 \, \times \, 4}{0.01 \, \times \, 0.01}$ = 0.3 m = 30 cm

Multiple choice physics light, shadows and reflections characteristics of an image formed by a plane mirror formation of image in a plane mirror characteristics of the image formed by a plane mirror

A plane mirror is made up of:

  1. a few metre thick glass plate

  2. a few mm thick glass plate

  3. a few cm thick glass plate

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A plane mirror is made from a few mm thick glass plate. The front part is highly polished and the back part is silvered.