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NEET Exam  >  NEET Tests  >  Test: Mechanical Properties of Fluids (October 9) - NEET MCQ

Test: Mechanical Properties of Fluids (October 9) - NEET MCQ


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10 Questions MCQ Test - Test: Mechanical Properties of Fluids (October 9)

Test: Mechanical Properties of Fluids (October 9) for NEET 2024 is part of NEET preparation. The Test: Mechanical Properties of Fluids (October 9) questions and answers have been prepared according to the NEET exam syllabus.The Test: Mechanical Properties of Fluids (October 9) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Mechanical Properties of Fluids (October 9) below.
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Test: Mechanical Properties of Fluids (October 9) - Question 1
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In rising from the bottom of a lake, to the top, the temperature of an air bubble remains unchanged, but its diameter gets doubled. If h is the barometric height (expressed in m of mercury of relative densityρ) at the surface of the lake, the depth of the lake is [1994]

Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 1

This gives H = 7hρ

Test: Mechanical Properties of Fluids (October 9) - Question 2
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The wetability of a surface by a liquid depends primarily on [NEET 2013]

Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 2

Wetability of a surface by a liquid primarily depends on angle of contact between the surface and liquid.
If angle of contact is acute liquids wet the solid and vice-versa.

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Test: Mechanical Properties of Fluids (October 9) - Question 3
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A wooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h are shown there. After some time the coin falls into the water. Then 

Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 3

Considering the fact that density of coin is very high in comparison of water and wood.
Density of coin is more so it was applying considerable force on wood before falling in water. 
But once it'll fall, wood can't apply same force alone so wood will come a little up as result L will decrease.
when coin falls in water wood goes up so displaced volume decreases and coin has high density so it's volume is low and it doesn't  displace liquid that much. As a result H will also decrease.

Test: Mechanical Properties of Fluids (October 9) - Question 4
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A wooden cylinder of diameter 4r, height h and density r/3 is kept on a hole of diameter 2r of a tank, filled with water of density r as shown in the figure. The height of the base of cylinder from the base of tank is H.

                         

Let the cylinder is prevented from moving up, by applying a force and water level is further decreased. Then, height of water level (h2 in figure) for which the cylinder remains in original position without application of force is 

 [JEE 2006]
Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 4


P0A1+ρ gHA1/3=P0A1+ρgh2A3
On solving, h2=4H/9

Test: Mechanical Properties of Fluids (October 9) - Question 5
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A glass tube of uniform internal radius (r) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End I has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve.

       

   [JEE 2008]
Test: Mechanical Properties of Fluids (October 9) - Question 6
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A small spherical monoatomic ideal gas double  is trapped inside a liquid of density r, (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T0, the height of the liquid is H and the atmospheric pressure is P0 (Neglect surface tension).

[JEE 2008]

Figure : 

 

                  

The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)
*Answer can only contain numeric values
Test: Mechanical Properties of Fluids (October 9) - Question 7
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 A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0 × 10 5 Nm-2 , density of water = 1000 kg m-3and g = 10 ms-2 . Neglect any effect of surface tension.] (Take temperature to be constant)

 [JEE 2009]

Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 7

Pressure due to falling water level at 200mm is 
 P+ρgh=P0​
or P=105−(1000)(10)(0.2)=98×103N/m2
 now, P0​V0​=PV
or 105[A(0.5−H)]=98×103[A(0.5−0.2)]    
where A= cross-sectional area of a vessel.
0.5−H=0.294 
⇒H=0.206m=206mm
 
The fall in height (in mm) of water level =206−200=6

 

*Multiple options can be correct
Test: Mechanical Properties of Fluids (October 9) - Question 8
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Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if 

[JEE 2011]

                   
Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 8

From the diagram we can clearly see that ball A tends to float and ball B tends to sink, but they are unable because of the tension in the string. Thus we get dA < dF and  dB > dF
Thus we get that  dA < dF < dB
Aslo making fbd of A and B gives 
 VdAg + T =  VdFg
and VdFg + T =  VdBg
Thus if we sun=btract above two equations we get
VdAg +VdBg  =  2VdFg

Test: Mechanical Properties of Fluids (October 9) - Question 9
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A rectangular block of mass m and area of crosssection A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes  oscillation with a time period T. Then [2006]
Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 9

Let the body be depressed by distance x from its equilibrium position. The extra upthrust created is xρAg which applies to whole body.
If a be acceleration created then,

Since, acceleration αx, so, it is equation of S.H.M.

Test: Mechanical Properties of Fluids (October 9) - Question 10
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The angle of contact between pure water and pure glass, is [1996]
Detailed Solution for Test: Mechanical Properties of Fluids (October 9) - Question 10

We know that angle of contact is the angle between the tangent to liquid surface at the point of contact and solid surface inside the liquid. In case of pure water and pure glass, the angle of contact is zero.

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