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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.

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

A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting the viscosity, find the velocity of the water flow, if the thickness of the water layer is equal to $\mathrm { h } _ { 1 } = 30 \mathrm { cm }$ and that of the kerosene layer to $h _ { 2 } = 20 \mathrm { cm }$.Density of kerosene $= 600 \operatorname { kg } / m ^ { 3 }$

  1. $5.74 \mathrm { ms } ^ { - 1 }$
  2. $1.91 \mathrm { ms } ^ { - 1 }$
  3. $2.87 \mathrm { ms } ^ { - 1 }$
  4. $3.82 \mathrm { ms } ^ { - 1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Bernoulli's principle at the hole: P_atm + rho_k*g*h2 + rho_w*g*h1 = P_atm + 0.5*rho_w*v^2. v = sqrt(2*(rho_k*g*h2 + rho_w*g*h1)/rho_w) = sqrt(2*(600*10*0.2 + 1000*10*0.3)/1000) = sqrt(2*(1200+3000)/1000) = sqrt(8.4) = 2.89 m/s.

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

A pump motor is used to deliver water at a certain rate from the given pipe. To obtain 'n' times water from the same pipe in the same time, the amount of power of the motor should be increased to:

  1. np
  2. ${n^3}$
  3. ${n^2}$

  4. 2np
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = (1/2)mv^2/t = (1/2)(rho*A*v*t)v^2/t = (1/2)rho*A*v^3. If flow rate Q = Av is increased by n, then v must increase by n (assuming area constant). Thus P is proportional to v^3, so power increases by n^3.

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

Equal amount of same gas in two similar cylinders $A \text { and } B$,compressed to same final volume from same initial volume one adiabatically and another isothermally, respectively then  

  1. final pressure in $A$ is more than in $B$
  2. final pressure in $B$ is greater than in $A$

  3. final pressure in both able equal
  4. for the gas, value of $\gamma = \frac { C _ { p } } { C _ { V } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Adiabatic compression (PV^gamma = constant) results in a higher final pressure than isothermal compression (PV = constant) for the same volume change, because gamma > 1.

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

Water flows into a large tank with flat bottom at the rate of $  10-4 \mathrm{m}^{3} \mathrm{s}-1  $ . Water is also leaking out
of a hole of area 1 $  \mathrm{cm}^{2}  $ at its bottom. If the height of the water in the tank remains steady, then this height is:

  1. 4 cm

  2. 2.9 cm

  3. 1.7 cm

  4. 5.1 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel is equal to:

  1. half of the radius of the vessel.

  2. radius of the vessel.

  3. one-fifth of the radius of the vessel.

  4. three-fourth of the radius of the vessel.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $h$ is the height of liquid in cylinder, $r$ be the radius of the cylinder and $\rho$ be the density of the liquid. then we have
weight of the liquid $=\pi { r }^{ 2 }h\rho g          ....... (I)$
Mean pressure on the wall $=\frac{1}{2} \rho g h $
force on the wall $=\frac{1}{2} \rho g h \times 2 \pi r h=\pi r \rho g h^2          ....... (II)$
On equating $(I)$ and $(II)$ we have
$\pi { r }^{ 2 }h\rho g=\pi r \rho g h^2$
$\Rightarrow r=h$
i.e. the liquid should be filled up-to a height equal to the radius of the cylinder.

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

A thermally insulated cylinder is divided into two equal halves by a thermally insulated wall. One half contains He at 600$\mathrm { K }$ and the other contains $\mathrm { H } _ { 2 }$ at 800$\mathrm { K }$ , pressure being same in two parts equal to $P _ { 0 }$ . Now, the wall is removed and two gases mix. The resulting pressure is

  1. $P _ { 0 }$
  2. 2$P _ { 0 }$
  3. $\frac { 28 P _ { 0 } } { 54 }$
  4. $\frac { 28 P _ { 0 } } { 27 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Dalton's law and energy conservation: P_final = (n_He*R*T_He + n_H2*R*T_H2) / (V_total). Since P0 = nRT/V, n = P0*V/RT. P_final = (P0*V/R*600 * R*600 + P0*V/R*800 * R*800) / (2V) is incorrect. Correct: P_final = (n1+n2)RT_mix / 2V. Using internal energy: (n1Cv1T1 + n2Cv2T2) = (n1Cv1 + n2Cv2)T_mix. For monoatomic (He) Cv=3/2R, diatomic (H2) Cv=5/2R. Result is 28P0/27.

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

Two identical cylinders contain Helium at 2.5atmosphere and Argon at 1 atmosphere respectively. If both gases are transferred in one ofthe cylinders, what is the new pressure? 

  1. $3.5$ atmosphere
  2. $1.5$ atmosphere
  3. $1.75$ atmosphere
  4. $1$ atmosphere.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the cylinders are identical and we are transferring gases into one, we use Dalton's Law of partial pressures. P_total = P1 + P2 = 2.5 + 1 = 3.5 atm.

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

By sucking through a straw, a student can reduce the pressure in his lungs to $750mm$ of $Hg$ (Density $=13.6g/{cm}^{3}$). Using the straw, he can drink water from a glass up to a maximum depth of

  1. $10cm$
  2. $75cm$
  3. $13.5cm$
  4. $1.36cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,
$\rho _H=13.6g/cm^3$
$\rho _w=1g/cm^3$
$P _0=760mm\,  of\,  Hg$
$P _l=750mm\,   of\,  Hg$
The pressure difference between the lungs of student and the atmosphere is given by
$\Delta P=P _0-P _l$
$\Delta P=760-750=10mm \,  of\,   Hg$
$\Delta P=1cm\,  of\,   Hg$
This pressure can be used for the drinking water.
$1cm\,   of\,   Hg=$ Pressure difference due to water column
$\rho _H gh _H=\rho _w gH$
$1\times 13.6\times g=1\times g\times H$
$H=13.6cm $
The correct option is C.

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

The pressure and temperature of two different gases is $P$ and $T$ having the volume $V$ for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be

  1. $P/2$
  2. $P$
  3. $2P$
  4. $4P$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If two gases at same P, V, T are mixed into the same volume V, the total pressure is the sum of partial pressures. Since n = PV/RT, total moles = n1 + n2 = 2PV/RT. New pressure = (2PV/RT) * RT/V = 2P.