Tag: floatation

Quizzes (12) Questions (74)

Questions Related to floatation

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

A uniform solid cylinder of density $ 0.8 g/cm^3 $ floats in equilibrium in a combination of teo non-mixing liquid A and B with its axis vertical. the densities of liquid A ad B with its axis vertical. the densities of liquid A and B are $ 0.7 g /cm^3 $ and $ 1.2 \times gm/cm^3 $. the height of liquid A is $ h _A = 1.2 cm $ and the length of the part of cylinder immersed in liquid B is $ h _B = 0.8 cm $ then the length of the cylinder in air is

  1. 0.21 m

  2. 0.25 cm

  3. 0.35 m

  4. 0.4 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

When the volume of gas is reduced at constant temperature, the pressure exerted by the gas on the walls of the container increases because

  1. each molecules hits the walls with greater speed

  2. each molecule loses more energy when it strikes the wall

  3. each molecule loses momentum when it strikes the wall

  4. the number of molecules striking the wall per unit time increase.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to the kinetic theory of gases, pressure is caused by molecular collisions with the container walls. When the volume is reduced at constant temperature, the molecular density increases, causing the number of molecular collisions per unit area per unit time to increase, thereby raising the pressure.

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

A container with insulating walls is divided into equal parts by a partition fitted with a value.One part is filled with an ideal gas at a pressure P and temperature T, whereas the other part is completely evacuted.If the value is suddenly opened,the pressure and temperature of the gas will be

  1. $ \dfrac {p}{2}, T $
  2. $ \dfrac {p}{2} , \frac {T}{2} $

  3. p,T
  4. $ p, \dfrac {T}{2}, $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a free expansion of an ideal gas into a vacuum. Since the walls are insulating (adiabatic) and no work is done (expansion into vacuum), the internal energy remains constant, meaning the temperature T remains constant. The volume doubles, so the pressure halves.

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

A cylindrical vessel of $100\ cm$ height is kept filled upto the brim. It has four holes $1, 2, 3, 4$ which are respectively at heights of $27\ cm, 30\ cm, 50\ cm$ and $80\ cm$ from the horizontal floor. The water falling at the maximum horizontal distance from the vessel comes from

  1. Hole number $4$
  2. Hole number $3$
  3. Hole number $2$
  4. Hole number $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The horizontal range of liquid issuing from a hole is given by R = 2 * sqrt(y(H - y)), where y is the depth of the hole from the top surface and H is the total height of the liquid (100 cm). The range is maximized when the hole is located at the midpoint, y = H / 2 = 50 cm from the top. Looking at the options, hole number 3 is at a height of 50 cm from the floor, meaning its depth from the top is 100 - 50 = 50 cm, giving the maximum range.

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

Water is falling in a cylindrical tank at the rate of $ \pi m^3 / s. $ If the radius of the tank is 2 m, the rate of increases in the level of water in the tank is

  1. 1 m/s

  2. 0.25 m/s

  3. 0.5 m/s

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

The rate of change of volume in the tank is given by dV/dt = A * (dh/dt) = pi * r^2 * (dh/dt). Given dV/dt = pi m^3/s and radius r = 2 m, we have pi = pi * (2^2) * (dh/dt), which yields dh/dt = 1 / 4 = 0.25 m/s.

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

The pressure and temperature of an ideal gas in a closed vessel are $720$ kpa and $40^oC$ respectively. If - th of the gas is released from the vessel and the temperature of the remaining gas is raised to $353^oC$, the final pressure of the gas is 

  1. $ 1440$ kPa
  2. $1080$ kPa
  3. $720$ kPa
  4. $ 540$ kPa
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial: P1 = 720 kPa, T1 = 313 K. After releasing 1/3 of gas, remaining gas is 2/3 of original: n2 = (2/3)n1. Temperature rises to T2 = 353 + 273 = 626 K. Using P2V = n2RT2: P2 = (n2/n1)(T2/T1)P1 = (2/3)(626/313)(720) = (2/3)(2)(720) = 960 kPa. Wait, this doesn't match. Let me recalculate: if 1/3 is released, remaining is 2/3. Temperature ratio: (353+273)/(40+273) = 626/313 = 2. So P2 = (2/3) × 2 × 720 = 960 kPa. This doesn't match option B (1080 kPa). Perhaps 1/3 remaining means released 2/3? Then P2 = (1/3) × 2 × 720 = 480 kPa. Still no match. Checking if -th means 1/4 released (3/4 remaining): P2 = (3/4) × 2 × 720 = 1080 kPa. Yes! This matches.

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

Two metal plates $'A'$ and $'B'$ having the same breadth but different lengths $\ell 1$ and $\ell _2 $ respectively are placed at same depth inside water such that their breadth is held exactly in vertical positions. Then, the ratio of the pressure acting on $'A'$ and $'B'$ by water is ____.

  1. $1:1$
  2. $\ell _1:\ell _2$
  3. $\ell _2:\ell _1$
  4. $\ell _1 b:\frac{\ell _2}{b}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hydrostatic pressure depends solely on the vertical depth below the free surface of the liquid. Since both plates A and B are placed at the same depth and have their breadths in vertical positions, the pressure acting on them is identical, giving a ratio of 1:1.

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

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 

  1. $ 4 ms^{-1} $
  2. $5 ms^{-1} $
  3. $ 1 ms^{-1} $
  4. $ 0.5 ms^{-1} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The volume flow rate is Q = 3 pi litres / minute = 3 pi * 10^-3 m^3 / 60 s = (pi / 20) * 10^-3 m^3/s. Also, Q = A * v = pi * r^2 * v, where r = 1 cm = 10^-2 m. Equating the two expressions: pi * (10^-2)^2 * v = (pi / 20) * 10^-3, which simplifies to v = 0.5 m/s.

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

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.

  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, $
Reveal answer Fill a bubble to check yourself
C Correct answer
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