Question 1 |

A pressure measurement device fitted on the surface of a submarine, located at a depth H below the surface of an ocean, reads an absolute pressure of 4.2 MPa. The density of sea water is 1050 kg/m^3, the atmospheric pressure is 101 kPa, and the acceleration due to gravity is 9.8 m/s^2. The depth H is _______ m (round off to the nearest integer).

128 | |

398 | |

478 | |

256 |

Question 1 Explanation:

\begin{aligned} &P_{A}=4.2 \mathrm{MPa}\;\;\;\;\text { [Absolute pressure] }\\ P_{\text {atm }} &=101 \mathrm{kPa} \\ \rho &=1050 \mathrm{~kg} / \mathrm{m}^{3} \\ g &=9.8 \mathrm{~m} / \mathrm{s}^{2} \\ P_{A} &=P_{\text {atm }}+\rho \mathrm{gH} \\ 4.2 \times 10^{6} &=\left(101 \times 10^{3}\right)+[1050 \times 9.81 \times \mathrm{H}] \\ H &=397.94 \text { or } 398 \mathrm{~m} \end{aligned}

Question 2 |

In the space above the mercury column in a barometer tube, the gauge pressure of the
vapour is

positive, but more than one atmosphere | |

negative | |

zero | |

positive, but less than one atmosphere |

Question 2 Explanation:

In space above mercury column in barometer, ideally perfect vaccum is expected. But due to evaporation of mercury , the pressure in that space is equal to vapour pressure of mercury [nearly 0.2 Pa (abs)] . As absolute pressure in that space is less than atmospheric pressure, gauge pressure is negative.

Question 3 |

Which of the following conditions is used lo determine the stable equilibrium of all partially
submerged floating bodies?

Centre of buoyancy must be above the centre of gravity | |

Centre of buoyancy must be below the centre of gravity | |

Metacentre must be at a higher level than the centre of gravity | |

Metacentre must be at a lower level than the centre of gravity |

Question 3 Explanation:

Metacentre must be higher level than the centre of gravity

Question 4 |

For the stability of a floating body the

centre of buoyancy must coincide with the centre of gravity | |

centre of buoyancy must be above the centre of gravity | |

centre of gravity must be above the centre of buoyancy | |

metacenter must be above the centre of gravity |

Question 4 Explanation:

GM=BM-BG

(i) GM \gt 0, stable

(ii) GM = 0, Neutral

(iii) GM \gt 0, unstable

NOTE:: Metacentre (M) must always be above the centre of gravity.

Question 5 |

Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2R x L at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semi-cylinder of radius R. The tank is filled with a liquid of density \rho up to the tip of the plug. The gravitational acceleration is g. Neglect the effect of the atmospheric pressure.

The force F required to hold the plug in its position

The force F required to hold the plug in its position

2\rho R^{2}gL\left( 1-\frac{\pi }{4}\right ) | |

2\rho R^{2}gL\left( 1+\frac{\pi }{4}\right ) | |

\pi R^{2}\rho gL | |

\frac{\pi }{2}\rho R^{2}gL |

Question 5 Explanation:

\begin{aligned} \text{Volume, } V_{total}&=2R \times L \times R\\ &=2R^2L\;\;...(i)\\ \text{Volume: }V_1&=\frac{1}{2}\times \pi R^2 \times L\;\;...(ii)\\ F&= \text{Weight of fluid}\\ &=[V_{total}-V_{1}]\rho g\\ &=\left [ 2R^2L-\frac{\pi R^2L}{2} \right ]\rho g\\ &=\left [ 1-\frac{\pi}{4} \right ]2R^2Lg\rho \\ &=2\rho R^2gL\left [ 1-\frac{\pi}{4} \right ] \end{aligned}

Question 6 |

The large vessel shown in the figure contains oil and water. A body is submerged at the interface of oil and water such that 45 percent of its volume is in oil while the rest is in water. The density of the body is _________ kg/m^{3}.

The specific gravity of oil is 0.7 and density of water is 1000 kg/m^{3} .

Acceleration due to gravity g = 10 m/s^{2}

The specific gravity of oil is 0.7 and density of water is 1000 kg/m^{3} .

Acceleration due to gravity g = 10 m/s^{2}

540 | |

258 | |

741 | |

865 |

Question 6 Explanation:

Let V= Volume of metalic body

45% of volume is in oil

i.e. V_{oil}= 45% of V=0.45V

and V_{ater}= 0.55V

For equilibrium condition,

Net buoyant force =(F_B)_{water}+(F_B)_{oil}

\begin{aligned} M_bg&=\rho _w V_{water}\;g+\rho _{oil}V_{oil}\;g \\ M_b&= \rho _w V_{water}+\rho _{oil}V_{oil}\\ \rho _b V&= \rho _w \times 0.55V+\rho _{oil} \times 0.45V\\ \rho _b&= \rho _w \times 0.55+\rho _{oil} \times 0.45\\ &=1000 \times 0.55+700 \times 0.45 \\ &= 865\;kg/m^3 \end{aligned}

45% of volume is in oil

i.e. V_{oil}= 45% of V=0.45V

and V_{ater}= 0.55V

For equilibrium condition,

Net buoyant force =(F_B)_{water}+(F_B)_{oil}

\begin{aligned} M_bg&=\rho _w V_{water}\;g+\rho _{oil}V_{oil}\;g \\ M_b&= \rho _w V_{water}+\rho _{oil}V_{oil}\\ \rho _b V&= \rho _w \times 0.55V+\rho _{oil} \times 0.45V\\ \rho _b&= \rho _w \times 0.55+\rho _{oil} \times 0.45\\ &=1000 \times 0.55+700 \times 0.45 \\ &= 865\;kg/m^3 \end{aligned}

Question 7 |

An inverted U-tube manometer is used to measure the pressure difference between two pipes A and B, as shown in the figure. Pipe A is carrying oil (specific gravity = 0.8) and pipe B is carrying water. The densities of air and water are 1.16kg/m^{3} and 1000kg/m^{3}, respectively. The pressure difference between pipes A and B is __________kPa.

Acceleration due to gravity g = 10 m/s^{2} .

Acceleration due to gravity g = 10 m/s^{2} .

-2.199 | |

-1.895 | |

-5.264 | |

-2.978 |

Question 7 Explanation:

Given data:

Specific gravity of oil,

s^{oil} = 0.8

\therefore \quad \rho_{\mathrm{oil}}=0.8 \times 1000=800 \mathrm{kg} / \mathrm{m}^{3}

Density of air: \rho_{\text {air }}=1.16 \mathrm{kg} / \mathrm{m}^{3}

Density of water,

\rho_{\text {water }}=1000 \mathrm{kg} / \mathrm{m}^{3}

Acceleration due to gravity,

\begin{aligned} g &=10 \mathrm{m} / \mathrm{s}^{2} \\ h_{1} &=80 \mathrm{mm}=0.08 \mathrm{m} \\ h_{2} &=200 \mathrm{mm}=0.2 \mathrm{m} \\ h_{3} &=100 \mathrm{mm}=0.1 \mathrm{m} \end{aligned}

Pressure at section 1 on left limb = Pressure at

section 1 on right limb

\begin{aligned} P_{A}-\rho_{\text {oil }} g h_{2}-\rho_{\text {air }} g h_{1}&=p_{B}-\rho_{\text {water }} g\left(h_{1}+h_{2}+h_{3})\right. \\ p_{A}-800 \times 10 \times 0.2&-1.16 \times 10 \times 0.08 =p_{B}-\\ &1000 \times 10(0.08+0.2+0.1) \\ p_{A}-1600-0.928&=p_{B}-3800\\ p_{A}-p_{B} &=-2199.07 \mathrm{Pa} \\ &=-2.199 \mathrm{kPa} \end{aligned}

Question 8 |

For a floating body, buoyant force acts at the

centroid of the floating body | |

center of gravity of the body | |

centroid of the fluid vertically below the body | |

centroid of the displaced fluid |

Question 8 Explanation:

For floating body Buoyancy force acts through the centre of buoyancy which is C.G. for displaced volume.

Question 9 |

A spherical balloon with a diameter of 10 m, shown in the figure below is used for advertisements. The balloon is filled with helium (R_{He}=2.08 kJ/kg.K ) at ambient conditions of 15^{\circ}C and 100 kPa. Assuming no disturbances due to wind, the maximum allowable weight (in newton) of balloon material and rope required to avoid the fall of the balloon (R_{air}=0.289 kJ/kg.K ) is _______

5303.688N | |

1548.586N | |

4859.256N | |

4859.654N |

Question 9 Explanation:

Step-I: Mass of balloon

\begin{aligned} \text { Volume: } \quad V &=\frac{4}{3} \pi r^{3}=\frac{4 \times 3.14 \times 5^{3}}{3} \\ &=523.33 \mathrm{m}^{3} \\ \rho_{He} &=\frac{P}{R T}=\frac{100 \times 10^{3}}{2080 \times 288} \\ &=0.16693 \mathrm{kg} / \mathrm{m}^{3} \\ \text { Total mass } &=0.16693 \times 523.33 \\ &=87.359 \mathrm{kg} \end{aligned}

Step-II : Mass of air

\begin{aligned} \rho &=\frac{P}{R T}=\frac{100 \times 10^{3}}{289 \times 288} \\ &=1.2 \mathrm{kg} / \mathrm{m}^{3} \end{aligned}

Mass of displaced volume of air

=1.2 \times 523.33=628 \mathrm{kg}

Step-III : Net buoyant mass

=628-87.359=540.64 \mathrm{kg}

\therefore Maximum allowable weight

\begin{aligned} =540.64 \times 9.81 \\ =5303.688 \mathrm{N} \end{aligned}

\begin{aligned} \text { Volume: } \quad V &=\frac{4}{3} \pi r^{3}=\frac{4 \times 3.14 \times 5^{3}}{3} \\ &=523.33 \mathrm{m}^{3} \\ \rho_{He} &=\frac{P}{R T}=\frac{100 \times 10^{3}}{2080 \times 288} \\ &=0.16693 \mathrm{kg} / \mathrm{m}^{3} \\ \text { Total mass } &=0.16693 \times 523.33 \\ &=87.359 \mathrm{kg} \end{aligned}

Step-II : Mass of air

\begin{aligned} \rho &=\frac{P}{R T}=\frac{100 \times 10^{3}}{289 \times 288} \\ &=1.2 \mathrm{kg} / \mathrm{m}^{3} \end{aligned}

Mass of displaced volume of air

=1.2 \times 523.33=628 \mathrm{kg}

Step-III : Net buoyant mass

=628-87.359=540.64 \mathrm{kg}

\therefore Maximum allowable weight

\begin{aligned} =540.64 \times 9.81 \\ =5303.688 \mathrm{N} \end{aligned}

Question 10 |

The difference in pressure (in N/m^{2} ) across an air bubble of diameter 0.001 m immersed in water (surface tension = 0.072 N/m) is _______

445 | |

248 | |

985 | |

288 |

Question 10 Explanation:

Forbuble, \quad \Delta P=\frac{2 \sigma}{r}

Where

\Delta P =pressure difference

Surface tension : [latex]\sigma=0.072 \mathrm{N} / \mathrm{m}

Radiu Ills of bubble: r=\frac{0.001}{2} \mathrm{m}=0.0005 \mathrm{m}

\therefore \quad \Delta P=\frac{2 \times 0.072}{0.0005}=288 \mathrm{N} / \mathrm{m}^{2}

Where

\Delta P =pressure difference

Surface tension : [latex]\sigma=0.072 \mathrm{N} / \mathrm{m}

Radiu Ills of bubble: r=\frac{0.001}{2} \mathrm{m}=0.0005 \mathrm{m}

\therefore \quad \Delta P=\frac{2 \times 0.072}{0.0005}=288 \mathrm{N} / \mathrm{m}^{2}

There are 10 questions to complete.

ANSWER FOR QN:20 IS B

WHY BECAUSE IN MOHR’S CIRCLE DIAGRAM NORMAL STRESS IS POINTED ON NEGATIVE X-AXIS.