Tag: floatation

Quizzes (12) Questions (74)

Questions Related to floatation

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.

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

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 

  1. $\frac{3P}{8}$
  2. $\frac{7P}{6}$
  3. $\frac{11P}{5}$
  4. $\frac{9P}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial pressure at bottom is P_atm + rho*g*H = 4P. Thus, rho*g*H = 3P. If level decreases by 3/5, the new depth is 2/5*H. New pressure = P_atm + rho*g*(2/5*H) = P + 2/5*(3P) = P + 6P/5 = 11P/5.

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

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-

  1. $980 \times {10^3}N$
  2. $9.8 \times {10^3}N$
  3. $0.98 \times {10^3}N$
  4. $0.098 \times {10^3}N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total force on a vertical wall of a container is calculated by multiplying the pressure at the centroid (half the height) by the total area of the wall. Force F = (1/2) * rho * g * h * (h * w), which equals (1/2) * 1000 * 9.8 * 1 * (1 * 2) = 9.8 * 10^3 N.

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

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

  1. $42.5 ^ { \circ } C$
  2. $67.8 ^ { \circ } C$
  3. $99.5 ^ { \circ } C$
  4. $25.7 ^ { \circ } C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Use Gay-Lussac's law: P₁/T₁ = P₂/T₂. Convert to Kelvin: 27°C = 300K. (12+1)atm /300K = 14.9atm/T₂, so T₂ = (14.9×300)/13 ≈ 343.8K = 70.8°C. Wait: pressure gauge reads 12, so absolute pressure is 13 atm. T₂ = (14.9×300)/13 ≈ 343.8K = 70.8°C. This doesn't match 99.5°C. Let me recalculate: For 99.5°C = 372.5K to be correct, we'd need P₁ to be different. Actually, if gauge reads relative to atmospheric, then absolute P₁ = 13 atm. At explosion P₂ = 14.9 atm. T₂ = (14.9/13)×300K = 343.8K = 70.8°C. Answer should be B, not C. However, the claimed answer is C (99.5°C). There might be different interpretation. If initial absolute P = 12 atm (not 13), then T₂ = (14.9/12)×300K = 372.5K = 99.5°C. This suggests gauge already shows absolute pressure, which is unusual. Given the answer key claims C, the question likely treats 12 atm as absolute pressure.

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

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 

  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. $1.07\times { 10 }^{ 5 }N/m^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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

  1. $0.864N$
  2. $86.4N$
  3. $72N$
  4. $7.2N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
F = $\dfrac{dP}{dt}$
   = $\dfrac{2 dm V\cos 60^o}{dt}$
   = $\dfrac{2 (\rho A dx) V\cos 60^o}{dt}$
   = $2 \rho A {V}^2\cos 60^o$
   = ${10}^3\times 6\times {10}^{-4} \times{12}^2$
   =$86.4N$
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

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})$)

  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. $2.5 \times {10^5}J/{m^3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The kinetic energy per unit volume (energy density) of an ideal gas is given by E = (3/2)P, where P is the pressure of the gas. At N.T.P., P = 1 * 10^5 N/m^2, so E = (3/2) * (1 * 10^5) = 1.5 * 10^5 J/m^3.

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

Power of a water pump is 2 kW. If $g=m/{ sec }^{ 2 }$, The amount of water it can raise in one minute to a height of 10 m/s 

  1. 100 Litre

  2. 1200 Litre

  3. 1000 Litre

  4. 2000 Litre
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = 2 kW = 2000 W. Work needed to lift mass m to height h in time t: W = mgh. P = mgh/t, so m = Pt/(gh) = (2000×60)/(10×10) = 120000/100 = 1200 kg. This equals 1200 liters of water.