Tag: fluid pressure

Quizzes (8) Questions (185)

Questions Related to fluid pressure

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

Water is floating smoothly through a closed-pipe system. At one point $A$, the speed of the water is $3.0\ m$ while at another point $B$, $1.0\ m$ higher, the speed is $4.0\ m/s$. The pressure at $A$ is $20\ kPa$ when the flowing $18\ kPa$ when the water flow stop. Then

  1. the pressure at $B$ when water is flowing is $6.5\ kPa$
  2. the pressure at $B$ when water is flowing is $8.0\ kPa$
  3. the pressure at $B$ when water is flowing is $10\ kPa$

  4. None of these
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A cylindrical container is filled with water upto the brim. If the pressure exerted by the water at the bottom of the container is 1000 Pa the height of the container is ..........cm.(take $g = 10 m s^{-2}$)

  1. 10

  2. 100

  3. 1

  4. 20
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

pressure inside a fluid = $\rho gh$ , in this case $\rho$ = 1000 kg $  m^{3}$

so , 
1000 =  (1000)(10) (h)
hence h = 0.1 m = 10 cm
hence option (A) is correct

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

Two vessels have different base area. They are filled with water to the same height. If the amount of water in one be $4$ times that in the other, then the ratio of pressure on their bottom will be :

  1. $16:1$
  2. $8:1$
  3. $4:1$
  4. $1:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hydrostatic pressure at the bottom depends only on the density of the liquid, acceleration due to gravity, and the height of the liquid column. Since both vessels are filled with water to the same height, the pressure at their bottoms is equal, giving a ratio of 1:1.

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

A cylinder at a certain ternperature has a gas at a pressure ot $50cm$ of Hg.Then it is divided into three equal parts so that the gas in the central part is completely transferred to either equally. Find the pressure of the gas in each portion.

  1. $10cm$
  2. $20cm$
  3. $75cm$
  4. $100cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The length of vacuum above mercury column in a barometer is $10cm/cc$ of air from outside where the pressure is $76cm$ of Hg is passed into the barometer tube. The area of cross section of the tube is $1{ cm }^{ 2 }$.The height of the mercury column then will be

  1. $71cm$
  2. $72cm$
  3. $73cm$
  4. $74cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

a circular tank has a hole of 1 cm^2 in its bottom. if the watre is allowed to flow into tank from a tube above it at the rate of $70 cm^3/sec$ then the max height upto which water can rise in the tank

  1. 2.5 cm

  2. 5 cm

  3. 10 cm

  4. 0.25 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At the maximum height, the rate of water flowing into the tank equals the rate of water leaking out through the hole. Using Torricelli's law, the outflow velocity is sqrt(2gh), so the volume flow rate is A * sqrt(2gh). Equating this to 70 cm^3/s allows solving for h, which yields 2.5 cm.

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

A beaker is filled with a liquid of density $\rho$ upto a height h. If the beaker is at rest , the mean pressure on the walls is 

  1. $0$
  2. $h \rho g$
  3. $\dfrac{h \rho g}{2}$
  4. $2 h \rho g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Pressure varies linearly with depth from 0 at the surface to rho*g*h at the bottom. The mean pressure is the average of these two values: (0 + rho*g*h) / 2 = (rho*g*h) / 2.

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

Water in a storage tank stands $2.5m$ above the level of a value in the side of the tank. With what speed the water will rush out of the value (neglecting friction)

  1. $5m/s$
  2. $7m/s$
  3. $9m/s$
  4. $11m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Torricelli's law, the speed of efflux of liquid from a hole at a depth h below the free surface is given by v = sqrt(2gh). Substituting g = 9.8 m/s^2 (or approximately 9.8) and h = 2.5 m gives v = sqrt(2 * 9.8 * 2.5) = sqrt(49) = 7 m/s.

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

Water enters a house through a pipe with an inside diameter of $2 \,cm$ at an absolute pressure of $4 \times 10^5 \,Pa$. A pipe of diameter $1 \,cm$ leads to the second floor room $5 \,m$ above the entry point. When the flow speed at the inlet is $1.5 \,m/s$. Which of the following statements are correct.

  1. Flow speed on the second floor room is $6 \,m/s$
  2. Volume flow rate in the second floor room is nearly $0.47 \,L/s$
  3. Water pressure in the second floor room is approximately $3.33$ atmosphere
  4. Water pressure in the second floor room is $3.50$ atmosphere
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$4 \times 10^5 + \dfrac{1}{2} \times 1000 \times (1.5)^2 = P _2 + \dfrac{1}{2} \times 1000 \times (6)^2 + 1000 \times 10 \times 5$

$10^3 \left(400 + \dfrac{g}{\theta} \right) = P _2 + 10^3 (10 + 50)$

$P _2 = 10^3 \left(\dfrac{320 \,g}{\theta} - 6\theta \right)$

$= 10^3 \left(\dfrac{320 \,g - 544}{\theta}\right)$

= $10^3 \dfrac{2665}{\theta}$

$= 10^3 \times 333$
$= 3.33 \,atm$

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

A cylinder is filled with a liquid of density d upto a height h.if the beaker is at rest , then the mean pressure on the wall is :-

  1. $Zero$
  2. $hdg$
  3. $\frac{h}{2}dg$
  4. $2 hdg$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The pressure due to a liquid column increases linearly with depth from zero at the free surface to hdg at the bottom. The mean pressure on the wall is therefore the average of the pressure at the top and the pressure at the bottom, which is (0 + hdg) / 2 = (h/2)dg.