FLUID MECHANICS QUIZ-12
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.
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Q1. The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed v with which the liquid is crossing points at a distance X from O along a radius XO would look like
Solution
When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero
When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero
Q2.A sphere of mass M and radius R is dropped in a liquid, then terminal velocity of sphere is proportional to
Solution
If a sphere of mass M and radius R is dropped in a liquid, its weight Mg(=4/3 π R3 ρg) acts vertically downwards. While upthrust 4/3 π R^3 σg and viscous force 6πɳRv acts vertically upwards. Initially the body will be accelerated down. At a certain instant when viscous force F will balance the net downward force, acceleration will become zero and the body will fall with constant velocity.
If a sphere of mass M and radius R is dropped in a liquid, its weight Mg(=4/3 π R3 ρg) acts vertically downwards. While upthrust 4/3 π R^3 σg and viscous force 6πɳRv acts vertically upwards. Initially the body will be accelerated down. At a certain instant when viscous force F will balance the net downward force, acceleration will become zero and the body will fall with constant velocity.
∴ 6πɳ RvT=4/3 π R3 (ρ-σ)g
ie,vr=2/9 R2 ((ρ-σ)/ɳ)g
or vT∝R2
Q3. Streamline flow is more likely for liquid with
Solution
Streamline flow is more likely for non-viscous and incompressible liquid. So low density and low viscosity is the correct answer.
Streamline flow is more likely for non-viscous and incompressible liquid. So low density and low viscosity is the correct answer.
Q4. Water stands at level A in the arrangement shown in the figure. What will happen if a jet of air is gently blown into the horizontal tube in the direction shown in the figure?
Solution
When air is blown in the horizontal tube, thepressure of air decreases in the tube. Due to which the water will rise above the tube A
When air is blown in the horizontal tube, thepressure of air decreases in the tube. Due to which the water will rise above the tube A
Q5. Work done forming a liquid drop of radius R is W1 and that of radius 3R is W2. The ratio of work done is
Solution
W=T×4πR2 ⟹ W1/W1 =(T×4πR2)/(T×4π(3R)2 ) =(T×4πR2)/(T×36πR2 )=1/9 ∴ W1:W2=1:9
W=T×4πR2 ⟹ W1/W1 =(T×4πR2)/(T×4π(3R)2 ) =(T×4πR2)/(T×36πR2 )=1/9 ∴ W1:W2=1:9
Q6. Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density 9 g/cm^3. If the mass of the other is 48 g, its density in g/cm3 is
Solution
Apparent weight =V(ρ-σ)g=m/ρ (ρ-σ)g Where m= mass of the body ρ= density of the body σ= density of water If two bodies are in equilibrium then their apparent weight must be equal ∴m1/ρ1 (ρ1-σ)=m2/ρ2 (ρ2-σ)⇒36/9 (9-1)=48/ρ2 (ρ2-1) By solving we get ρ2=3
Apparent weight =V(ρ-σ)g=m/ρ (ρ-σ)g Where m= mass of the body ρ= density of the body σ= density of water If two bodies are in equilibrium then their apparent weight must be equal ∴m1/ρ1 (ρ1-σ)=m2/ρ2 (ρ2-σ)⇒36/9 (9-1)=48/ρ2 (ρ2-1) By solving we get ρ2=3
Q7. The working of an atomizer depends upon
Solution
Bernoulli’s theorem
Bernoulli’s theorem
Q8. An L-shaped tube with a small orifice is held in a water stream as shown in fig. The upper end of the tube is 10.6 cm above the surface of water. What will be the height of the jet of water coming from the orifice? Velocity of water stream is 2.45 m/s
Solution
According to Bernoulli’s theorem, h=v2/2g ⇒h=(2.45)2/(2×10)=0.30m=30.0cm ∴ Height of jet coming from orifice =30.0-10.6=19.4cm≅20cm
According to Bernoulli’s theorem, h=v2/2g ⇒h=(2.45)2/(2×10)=0.30m=30.0cm ∴ Height of jet coming from orifice =30.0-10.6=19.4cm≅20cm
Q9. The surface energy of a liquid drop is u. It is sprayed into 1000 equal droplets. Then its surface energy becomes
Solution
Given : u=S×4 p R2; when droplet is splitted into 1000 droplets each of radius r, then 4/3 πR3=1000×4/3 πr3or r=R/10 ∴ Surface energy of all droplets =S×1000×4πr2=S×1000×4π(R/10)2 =10 (S 4π/R2)=10u
Given : u=S×4 p R2; when droplet is splitted into 1000 droplets each of radius r, then 4/3 πR3=1000×4/3 πr3or r=R/10 ∴ Surface energy of all droplets =S×1000×4πr2=S×1000×4π(R/10)2 =10 (S 4π/R2)=10u
Q10. In making an alloy, a substance of specific gravity s1 and mass m1 is mixed with another substance of specific gravity s2 and mass m2 : then the specific gravity of the alloy is
Solution
Specific gravity of alloy =(Density of alloy)/(Density of water) =(Mass of alloy)/(Volume of alloy×density of water) =(m1+m2)/((m1/ρ1 +m2/ρ2)×ρw )=(m1+m2)/(m1/(ρ1/ρw)+m2/(ρ2/ρw ))=(m1+m2)/(m1/s1 +m2/s2 ) [As specific gravity of substance=(density of substance)/(density of water)]
Specific gravity of alloy =(Density of alloy)/(Density of water) =(Mass of alloy)/(Volume of alloy×density of water) =(m1+m2)/((m1/ρ1 +m2/ρ2)×ρw )=(m1+m2)/(m1/(ρ1/ρw)+m2/(ρ2/ρw ))=(m1+m2)/(m1/s1 +m2/s2 ) [As specific gravity of substance=(density of substance)/(density of water)]
