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 Multiple Choice QuestionsMultiple Choice Questions

1.

A liquid is allowed into a tube of truncated cone shape. Identify the correct statement from the following.

  • The  speed is high at the wider end and low at the narrow end

  • The speed is low at the wider end and high at the narrow end.

  • The speed is same at both ends in a streamline flow.

  • The liquid flows with the uniform velocity in the tube.


B.

The speed is low at the wider end and high at the narrow end.

For an incomressible liquid equation of continuity.

AV = = constant or A ∝(1/V)

Therefore, at the wider end speed below and at the narrow end speed will be high.


2.

Water rises to a height 'h' in a capillary tube. If the length of capillary tube above the surface of water is made less than 'h' then

  • water rises upto the tip of the capillary tube and then starts overflowing like a fountain

  • water rises  upto the top of capillary tube and stays there without overflowing 

  • water rises upto a point little below the top and stays there

  • Water does not rise at all


A.

water rises upto the tip of the capillary tube and then starts overflowing like a fountain

It is given that water rises to a height 'h' in a capillary tube. so, the length of the capillary tube above the surface water is made less than 'h' then the height of water > length of capillary tube ⇒ So, the liquid will be staying there.

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3.

A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at one litre per second. On increasing the pressure. this increases to two litres per second. The firman has now to

  • push forward twice as hard

  • push forward four times as hard

  • push backward four times as hard

  • push backward twice as hard


B.

push forward four times as hard

The rate of change of mass, dm/dt = avρ

Therefore, F =vdm/dt

= (avρ)v = av2ρ

A volume of liquid flowing per second = av

This is proportional to velocity. By doubling this volume of liquid flow per second the force becomes four times. when liquid flows forward, the hosepipe tens to come backwards. So to keep it intact, it should be pushed forward. Thus, the hosepipe should be pushed forward four times.


4.

Mercury boils at 367° C. However, mercury thermometers are made such that they can measure temperature upto 500°C. This is done by

  • maintaining vacuum above mercury column in the stem of the thermometer.

  • filling nitrogen gas at high pressure above the mercury column.

  • filling oxygen gas at a high pressure above mercury column.

  • filling nitrogen gas at a low pressure above the mercury column.


B.

filling nitrogen gas at high pressure above the mercury column.

If we fill nitrogen gas at a high pressure above mercury level, the boiling point of mercury is increased which can extend to the range upto 500°C


5.

If dimensions of critical velocity vc of a liquid flowing through  a tube are expressed as left square bracket straight eta to the power of straight x space straight rho to the power of straight y space straight r to the power of straight z right square bracket comma space where space straight eta comma space straight rho space and space straight r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by

  • 1,-1,-1

  • -1,-1,1

  • -1,-1,-1

  • 1,1,1


A.

1,-1,-1

According to the principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct if the dimension of all the terms occurring on both sides of the equations is same.

Given critical velocity of liquid flowing through a tube are expressed as
straight v subscript straight c space equals space straight eta to the power of straight n straight rho to the power of straight n straight r to the power of straight z
Coefficient space of space viscosity space of space liquid comma space straight eta space equals space left square bracket ML to the power of negative 1 end exponent straight T to the power of negative 1 end exponent right square bracket
Density space of space liquid comma space straight rho space equals space left square bracket ML to the power of negative 3 end exponent right square bracket
Radius space of space straight a space tube comma space straight r space equals space left square bracket straight L right square bracket
critical space velcity space of space liquid space straight v subscript straight c space equals space left square bracket straight M to the power of straight o LT to the power of negative 1 end exponent right square bracket

equals space left square bracket straight M to the power of straight o straight L to the power of 1 straight T to the power of negative 1 end exponent right square bracket space equals space left square bracket straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 1 end exponent right square bracket to the power of straight x. left square bracket ML to the power of negative 3 end exponent right square bracket to the power of straight y. left square bracket straight L right square bracket to the power of straight z

left square bracket straight M to the power of 0 straight L to the power of 1 straight T to the power of negative 1 end exponent right square bracket space equals space left square bracket straight M to the power of straight x plus straight y end exponent straight L to the power of negative straight x minus 3 straight y plus straight z end exponent straight T to the power of negative straight x end exponent right square bracket

Comparing space exponents space of space straight M comma space straight L space space and space straight L comma space we space space get comma

x+ y = 0, - x-3y+z = 1, -x = -1

z = -11 x = 1, y = -1


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6.

Two non-mixing liquids of densities straight rho and nstraight rho (n >1) are put in a container. The height of each liquid is h. a solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to,

  • open curly brackets 2 plus left parenthesis straight n plus 1 right parenthesis straight p close curly brackets rho
  • open curly brackets 2 plus left parenthesis straight n minus 1 right parenthesis straight p close curly brackets rho
  • {1+(n-1)p}straight rho

  • {1+(n+1)p}straight rho


C.

{1+(n-1)p}straight rho

The situation can be depicted as follows:

According to Archimedes principle,

Weight of the cylinder = (upthrust)1 + (upthrust)2

i.e.,

space space space space ALdg space equals space left parenthesis 1 minus straight P right parenthesis LAρg space plus space left parenthesis PLA right parenthesis nρg

rightwards double arrow space straight d space equals space left parenthesis 1 minus straight P right parenthesis straight rho space plus space Pnρ
space space space space space space space space space equals space straight rho space minus space Pρ space plus space nPρ
space space space space space space space space space equals space straight rho space plus space left parenthesis straight n minus 1 right parenthesis Pρ
space space space space space space space space space equals straight rho space left square bracket space 1 plus left parenthesis straight n minus 1 right parenthesis straight rho right square bracket

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7.

The wettability of a surface by a liquid depends primarily on

  • viscosity

  • surface tension

  • density

  • angle of contact between the surface and the liquid


D.

angle of contact between the surface and the liquid

The value of angle of contact determines whether a liquid will spread on the surface or not.

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8.

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

  • Energy = 4VT open parentheses 1 over straight r minus 1 over straight R close parentheses is released.

  • Energy = 3VTopen parentheses 1 over straight r plus 1 over R close parentheses is absorbed

  • Energy = 3 VT open parentheses 1 over straight r minus 1 over straight R close parentheses is released

  • Energy is neither absorbed nor released.


C.

Energy = 3 VT open parentheses 1 over straight r minus 1 over straight R close parentheses is released

If the surface area changes, it will change the surface energy also.

When the surface area decreases, it means energy is released ad vice-versa.

Change in surface energy is increment straight A space straight x space straight T   ... (i)
Let, there be n number of drops initially.

So, increment straight A space equals space 4 πR squared space minus space straight n space left parenthesis 4 πr squared right parenthesis       ... (ii)
Volume is constant.

So, straight n space 4 over 3 πr cubed space equals space 4 over 3 πr cubed space equals space straight V         ... (iii)
From equations (ii) and (iii), we have

increment straight A space equals space 3 over straight R fraction numerator 4 straight pi over denominator 3 end fraction xR cubed space minus space 3 over straight r open parentheses straight n fraction numerator 4 straight pi over denominator 3 straight r cubed end fraction close parentheses
space space space space space space space space equals space space 3 over straight R xV space minus space 3 over straight r straight V
increment straight A space space equals space 3 straight V space open parentheses 1 over straight R minus 1 over straight r close parentheses
space space space space space space space space equals space left parenthesis negative right parenthesis ve space value

Since, R > r, incrementA is negative.

That is, the surface area is decreased.

Hence, energy must be released.

Energy released = increment straight A space straight x space straight T space equals space minus 3 VT open parentheses 1 over straight r minus 1 over straight R close parentheses
In the above expression, (-)ve sign shows that amount of energy is released.

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9.

A steam of a liquid of density ρ flowing horizontally with speed v rushes out of a tube of radius r and hits a vertical wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of the liquid is given by

  • πrρv

  • πrρv2

  • πr2ρv

  • πr2ρv2


D.

πr2ρv2

Cross-sectional area A = πr2

Volume of liquid flowing per second = AV = πr2v

Mass of the liquid flowing out per second = πr2

Initial momentum of liquid per second = mass of liquid flowing x speed of liquid

= πr2vρ x v = πr2v2ρ

Since the liquid does not rebound after impact, the momentum after impact is zero.

Therefore, the rate of change of momentum = πr2v2ρ

According to Newton's second law, the force exerted on wall = rate of change of momentum

=  πr2ρv2


10.

A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is

  • 650 kg m–3

  • 425 kg m–3

  • 800 kg m–3

  • 928 kg m–3


D.

928 kg m–3

hoil ρoil g = hwater ρwater g
140 × ρoil= 130 × ρwater
straight rho subscript oil space equals space 13 over 14 space straight x space 1000 space kg divided by straight m cubed
straight rho subscript oil space equals space 928 space kg space straight m to the power of negative 3 end exponent

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