Tag: buoyancy

Questions Related to buoyancy

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.

physics option b: engineering physics buoyancy floatation fluid pressure

  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

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Correct Option: C
Explanation:

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

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

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

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

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

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

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 :-

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $Zero$

  2. $hdg$

  3. $\frac{h}{2}dg$

  4. $2 hdg$

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Correct Option: D

The pressure at the bottom of a water tank is 4P, where P is atmospheric pressure. If water is drawn out till the water level decrease by $\frac{3}{5}$ the, then pressure at the bottom of the tank is 

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $\frac{3P}{8}$

  2. $\frac{7P}{6}$

  3. $\frac{11P}{5}$

  4. $\frac{9P}{4}$

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Correct Option: C

The side of glass aquarium is $1m$ high and $2m$ long. When the aquarium is filled to this is the total force against the side-

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $980 \times {10^3}N$

  2. $9.8 \times {10^3}N$

  3. $0.98 \times {10^3}N$

  4. $0.098 \times {10^3}N$

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Correct Option: C

Find the force exerted by water on the bottom

physics option b: engineering physics buoyancy floatation fluid pressure

  1. 303 N

  2. 211 N

  3. 102 N

  4. 10 N

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Correct Option: C

A gas cylinder containing cooking gas can withstand a pressure of  $14.9 atm. $ The pressure gauge of cylinder indicates  $12 atm $ at  $27 ^ { \circ } \mathrm { C } . $  Due to sudden fire in building the temperature starts rising. The temperature at which the cylinder explodes is

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $42.5 ^ { \circ } C$

  2. $67.8 ^ { \circ } C$

  3. $99.5 ^ { \circ } C$

  4. $25.7 ^ { \circ } C$

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Correct Option: C

At the mouth of the tap, the area of cross-section is $2.0cm^{2}$ and the speed of water is $3m/s$. The area of cross-section of the water column $80cm$ below the tap is $(use g=10m/s^{2})$

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $0.6cm^{2}$

  2. $1.2cm^{2}$

  3. $1.5cm^{2}$

  4. $2.0cm^{2}$

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Correct Option: A

oil bath (density of oil$=0.85\times { 10 }^{ 3 }kg/m^{ 3 })$ has a spherical cavity of diameter $26\times { 10 }^{ -6 }$ m at a depth of 0.2 face tension of oil is $26\times { 10 }^{ -3 }$ N/m and the pressure of air over the surface of oil is 76 cm of mercury, the 

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $1.03\times 105N/m^{ 2 }$

  2. $1.17\times { 10 }^{ 5 }N/m^{ 2 }$

  3. $3.07\times { 10 }^{ 5 }N/m^{ 2 }$

  4. $1.07\times { 10 }^{ 5 }N/m^{ 2 }$

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Correct Option: C

A jet of water with cross section of $6{ cm }^{ 2 }$ strikes a wall at an angle of ${ 60 }^{ \circ  }$ to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is $12 m/s$, the force acting on the wall is

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $0.864N$

  2. $86.4N$

  3. $72N$

  4. $7.2N$

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Correct Option: B
Explanation:
F = $\dfrac{dP}{dt}$
   = $\dfrac{2 dm V\cos 60^o}{dt}$
   = $\dfrac{2 (\rho A dx) V\cos 60^o}{dt}$
   = $2 \rho A {V}^2\cos 60^o$
   = ${10}^3\times 6\times {10}^{-4} \times{12}^2$
   =$86.4N$

The Kinetic energy per cubic metre of a perfect gas at N.T.P. is ( Take atmospheric pressure $ = 1 \times {10^5}N/{m^2})$)

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $1.5 \times {10^5}J/{m^3}$

  2. $2 \times {10^5}J/{m^3}$

  3. $0.75 \times {10^5}J/{m^3}$

  4. $2.5 \times {10^5}J/{m^3}$

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Correct Option: B