Tag: buoyancy

Quizzes (0) Questions (109)

Questions Related to buoyancy

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

How much work is done by an agent fcn forcing 3$\mathrm { m } ^ { 3 }$ of water through a pipe of radius 2$\mathrm { cm }$ , It the difference in pressure at the two ends of the pipe is $10 ^ { 4 } \mathrm { N } \mathrm { m } ^ { 2 } \mathrm { ? }$

  1. $3 \times 10 ^ { 6 } J$
  2. $2 \times 10 ^ { 6 } J$
  3. $4 \times 10 ^ { 5 } J$
  4. $1 \times 10 ^ { 5 } 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A box is divided into two equal compartments by a thin partition and they are filled with gases $  P  $and $  Q  $ respectively. The two compartments have a pressure of 250 torr each. The pressure after removing the partition will be equal to

  1. $125 torr$
  2. $2.5 torr$
  3. $250 torr$
  4. $500 torr$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the partition is removed, the total volume doubles (V + V = 2V). Since the pressure in each was 250 torr, the total moles are proportional to 250*V + 250*V = 500*V. New pressure = 500V / 2V = 250 torr.

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

Two containers $A$ and $B$ are partly filled with water and closed. The volume of $A$ is twice that of $B$ and it contains half the amount of water in $B$. If both are at the same temperature the water vapour in the containers will have pressure in the ratio of

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

The vapour pressure of a liquid depends only on its temperature and is independent of the volume of the container or the amount of liquid present, as long as some liquid remains to maintain vapor-liquid equilibrium. Since both containers are at the same temperature, their water vapour pressures will be equal, resulting in a 1:1 ratio.

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

A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water. The quater of water flowing out per second from both holes are the same. Then radius R, is equal to :

  1. $\dfrac { L }{ \sqrt { 2\pi } } $
  2. $2\pi L$

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

The volume of water flowing out per second is given by the product of the hole area and the speed of efflux, Q = A * sqrt(2gh). For the square hole of side L, area is L^2 and depth is y. For the circular hole of radius R, area is pi * R^2 and depth is 4y. Equating the two flow rates: L^2 * sqrt(2gy) = pi * R^2 * sqrt(2g(4y)). Simplifying this gives R = L / sqrt(2pi).

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

The hydrostatic pressure on a diver $100m$ below the surface of an ocean is

  1. $1$ atm
  2. $2$ atm
  3. $11$ atm
  4. $20$ atm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Hydrostatic pressure P = P_atm + rho*g*h. P_atm = 1 atm. rho*g*h = 1000 * 10 * 100 = 10^6 Pa. Since 1 atm approx 10^5 Pa, rho*g*h = 10 atm. Total pressure = 1 + 10 = 11 atm.

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

Two vessels of volume 100 c.c and 150 c.c contains gases at pressure of 1 atm and 2 atm. When they are joined the common pressure is 

  1. 2 atm

  2. 1.5 atm

  3. 1.6 atm

  4. 1 atm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Boyle's law and conservation of mass for ideal gases at constant temperature, the final common pressure P is given by (P1V1 + P2V2) / (V1 + V2). Substituting the values: (1 * 100 + 2 * 150) / (100 + 150) = (100 + 300) / 250 = 400 / 250 = 1.6 atm.

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

A tank of height H is fully filled with water.If the water rushing from a made in the tank below the free surface,strikes the floor at maximum horizontal distance then depth of the hole from the free surface must be.

  1. $ \left( \frac { 3 }{ 4 } \right) H $
  2. $ \left( \frac { 2 }{ 3 } \right) H $
  3. $ \left( \frac { 1 }{ 4 } \right) H $
  4. $ \left( \frac { 1 }{ 2 } \right) H $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The horizontal range of a fluid stream from a hole at depth y in a tank of height H is given by R = 2 * sqrt(y(H - y)). To maximize this range, the derivative with respect to y must be zero, which occurs when y = H / 2.

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

A conical portion is cut out of a solid hemisphere of radius R and the remaining portion is held in a liquid of density $ \rho $ through a string as shown in the figure.What is the net force exerted by the liquid on the body.

  1. $ \frac {1}{6} \pi R^3 \rho g $
  2. $ \frac {1}{3} \pi R^3 \rho g $
  3. $ \frac {1}{4} \pi R^3 \rho g $
  4. $ \frac {1}{2} \pi R^3 \rho g $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

Mark out the correct statement(s)

  1. Net force acting on the base of the vessel > weight of the liquid inside the vessel
  2. Net force acting on the base of the vessel $=$ weight of the liquid inside the vessel
  3. Net pressure force acting on the liquid $=$ weight of the vessel

  4. Both (a) and (c) are correct
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Weight of the liquid inside the vessel,

$W=\rho(A _1\times \dfrac{5}{100}+A _2\times \dfrac{1}{100})g=1N$
So, $F>W$
Net force on the liquid is zero.

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

Pressure on a swimmer at depth H below free surface of water is 3 atm.Then H is

  1. 10 m

  2. 30 m

  3. 20 m

  4. 50 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total pressure at depth H is given by P = P_0 + rho * g * H, where P_0 is atmospheric pressure (1 atm). Given that the total pressure is 3 atm, the gauge pressure due to the water column is 2 atm. Since 10 meters of water column corresponds to 1 atm of pressure, a pressure of 2 atm requires a depth of 2 * 10 = 20 meters.