Tag: option b: engineering physics

Quizzes (17) Questions (184)

Questions Related to option b: engineering physics

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.

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