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****FLUID MECHANICS QUIZ-18**

**FLUID MECHANICS QUIZ-18**

**Dear Readers,**

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

**Q1.**The onset of turbulence in a liquid is determined by

Solution

(i) FBD of the cube with respect to the container B=(m/σ)ρg_eff=m(ρ/σ) √(a

(i) FBD of the cube with respect to the container B=(m/σ)ρg_eff=m(ρ/σ) √(a

^{2}+g^{2}) tanθ=a/g sinθ=a/√(a^{2}+g^{2});cosθ=g/√(a^{2}+g^{2}) Let acceleration of the cube w.r.t. container in horizontal direction in a_{x}' mg+B sinθ-mg/2=ma_{x}a_{x}=g/2+ρ/σ a=a_{x}⇒g/2+ρ/σ∙g/2=a_{x}Cart: (T+B) sinθ'mg (i) (T+B) cosθ'=mg (ii) tanθ'=a/g (iii) Hence, tanθ'/tanθ =1/1=θ'/θ=1:1**Q2.**An object weighs m_1 in a liquid of density d

_{1}and that in liquid of density d

_{2}is m

_{2}. The density d of the object is

Solution

V(d-d

V(d-d

_{1})g=m_{1}g V(d-d_{2})g=m_{2}g (d-d_{1})/(d-d_{2})=m_{1}/m_{2}∴ d=(m_{1}d_{2}-m_{2}d_{1})/(m_{1}-m_{2})**Q3.**Water flows steadily through a horizontal pipe of variable cross-section. If the pressure of water is p at a point where flow speed is v, the pressure at another point where the flow of speed is 2v, is (take density of water as ρ)

Solution

From Bernoulli’s equation, the sum of all forms of energy in a fluid flowing along an enclosed path (a streamline) is the same at any two points in the path. Therefore, p+1/2 ρv

From Bernoulli’s equation, the sum of all forms of energy in a fluid flowing along an enclosed path (a streamline) is the same at any two points in the path. Therefore, p+1/2 ρv

_{1}^{2}=p'+1/2 ρv_{2}^{2}Given,v_{2}=2v, v_{1}=v ∴ p+1/2 ρv^{2}=p'+1/2 ρ(2v)^{2}⟹ p^'= p-3/2 ρv^{2}

**Q4.**A cylindrical vessel is filled with equal amounts of weight of mercury on water. The overall height of the two layers is 29.2 cm, specific gravity of mercury is 13.6. Then the pressure of the liquid at the bottom of the vessel is

Solution

Let A be the area of cross-section of the cylindrical vessel and x cm be the height of mercury in vessel. The height of water in the vessel =(29.2×x) cm As per question Ax×13.6=(29.2-x)×1 or x=2 cm ∴ Height of water column =(29.2-2)=27.2 cm ∴ Pressure of the liquids at the bottom =27.2 cm of water column +2 cm of Hg column =27.2/13.6 of Hg column +2 cm of Hg column =4 cm of Hg column

Let A be the area of cross-section of the cylindrical vessel and x cm be the height of mercury in vessel. The height of water in the vessel =(29.2×x) cm As per question Ax×13.6=(29.2-x)×1 or x=2 cm ∴ Height of water column =(29.2-2)=27.2 cm ∴ Pressure of the liquids at the bottom =27.2 cm of water column +2 cm of Hg column =27.2/13.6 of Hg column +2 cm of Hg column =4 cm of Hg column

**Q5.**When a glass capillary tube of radius 0.015 cm is dipped in water, the water rises to height of 15 cm within it. Assuming contact angle between water and glass to be 0°, the surface tension of water is [ρ_(water )=1000 kg m

^{(-3)},g=9.81 ms

^{(-2)}]

Solution

2πr×T cosθ=πr

2πr×T cosθ=πr

^{2}hρg ⟹ T=rhρg/2=0.11Nm^{(-1)}**Q6.**The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (S= surface tension of solution)

Solution

Change in surface area 2×4π[(D/2)

Change in surface area 2×4π[(D/2)

^{2}-(d/2)^{2}]=2π(D^{2}-d^{2}) Work done = surface tension × change in area =S×2π(D^{2}-d^{2})**Q7.**A monometer connected to a closed tap reads 4.5×〖10〗

^{5}pascal. When the tap is opened the reading of the monometer falls to 4×〖10〗^5 pascal. Then the velocity of flow of water is

Solution

(P

(P

_{1}-P_{2})/ρg=v^{2}/2g⇒(4.5×〖10〗^{5}-4×〖10〗^{5})/(〖10〗^{3}×g)=v^{2}/2g ∴v=10m/s**Q8.**Soap bubbles can be formed floating in air by blowing soap solution in air, with the help of a glass tube but not water bubbles. It because

Solution

The excess pressure inside water bubble being more due to large surface tension

The excess pressure inside water bubble being more due to large surface tension

**Q9.**A large open tank has two holes in its wall. One is a square hole of side a at a depth of x from the top and the other is a circular hole of radius r at a depth 4x from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then r is equal to

Solution

We have v

We have v

^{2}=ρgh a^{2}√ρgh=πr^{2}√ρgh×2 r=a/2π**Q10.**If the atmospheric pressure is P

_{a}, then the pressure P at depth h below the surface of a liquid of density ρ open to the atmosphere is

Solution

P=P

P=P

_{a}+ρgh