Tag: option b: engineering physics

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Questions Related to option b: engineering physics

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

A gas is enclosed in a rectangular vessel. $20 \times 10^23$ molecules of the gas strike a well of the vessel normally per second, with a velocity of 250m/s and rebound with the same speed in the opposite direction. What is the force exerted by the gas on the wall if the mass of each molecule is $5 \times 10^-23$ g ? 

  1. 50 N

  2. 40 N

  3. 75 N

  4. 25 N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force = change in momentum per unit time. Delta p = m * v - (m * -v) = 2mv. Force = N * (2mv) / t. Given N/t = 20 * 10^23 molecules/s, m = 5 * 10^-26 kg (converting 5 * 10^-23 g), v = 250 m/s. F = 20 * 10^23 * 2 * 5 * 10^-26 * 250 = 20 * 10^23 * 10^-25 * 250 = 20 * 0.1 * 250 = 50 N.

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

A pump is required to lift 1000 kg of water per minute from a well 20 m deep and eject it at a rate of 20 $ms^-1$. What (horsepower engine is required for the purpose of lifting water)?

  1. $4.46 HP$
  2. $4.36 HP$
  3. $3.96 HP$
  4. $8.85 HP$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Power required to lift water = m*g*h / t. Given mass per minute m/t = 1000 kg/min = 1000/60 kg/s, height h = 20 m, g = 9.8 m/s^2. Potential energy rate = (1000/60) * 9.8 * 20 = 3266.67 W. Power required to eject water at velocity v = 20 m/s is (1/2) * (dm/dt) * v^2 = 0.5 * (1000/60) * 20^2 = 3333.33 W. Total power = 3266.67 + 3333.33 = 6600 W = 6.6 kW. Converting to horsepower: 6600 / 746 = 8.85 HP.

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

The branch of physics which deals with the study of fluids at rest is called :

  1. statistics

  2. hydrostatics

  3. hydrodynamics

  4. thermometry
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The branch of physics which deals with the study of fluids at rest is called hydrostatics.

Hydro stands for liquids and statics is for the fluid at rest.

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

KE per unit volume is E. The pressure exerted by the gas is given by:

  1. $\displaystyle \frac {E}{3}$
  2. $\displaystyle \frac {2E}{3}$
  3. $\displaystyle \frac {3E}{2}$
  4. $\displaystyle \frac {E}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure exerted by gas is given by $P=\dfrac{1}{3}\rho v^2$

where $v$ is the velocity of gas particles, $\rho $ is the density of gas.
Kinetic energy per unit volume$=E=\dfrac{\dfrac{1}{2}mv^2}{V}=\dfrac{1}{2}\rho v^2$
Thus $P=\dfrac{2}{3}(\dfrac{1}{2}\rho v^2)=\dfrac{2E}{3}$

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.