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

$28\, gm$ of $N _2$ gas is contained in a flack at a pressure of $10\, atm$ and at a temperature of $57^0$. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to $27^0 C$. The quantity of $N _2$ gas that leaked out is :-

  1. $\dfrac{11}{20}\, gm$
  2. $\dfrac{20}{11}\, gm$
  3. $\dfrac{5}{63}\, gm$
  4. $\dfrac{63}{5}\, gm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the ideal gas law PV = nRT, we calculate the initial moles of N2 (n1 = P1V1 / RT1) and final moles (n2 = P2V2 / RT2). Since the volume is constant, the ratio of moles is (P2/P1) * (T1/T2). Calculating the difference in moles and converting to mass gives 63/5 grams.

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
C 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
C Correct answer
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
C Correct answer
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
C Correct answer
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