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 Multiple Choice QuestionsMultiple Choice Questions

1.

If two soap bubbles of different radii are connected by a tube,

  • air flows from the bigger bubble to the smaller bubble till the sizes are interchanged.

  • air flows from bigger bubble to the smaller bubble till the sizes are interchanged

  • air flows from the smaller bubble to the bigger.

  • there is no flow of air.


C.

air flows from the smaller bubble to the bigger.

The excess pressure inside the soap bubble is inversely proportional to the radius of soap bubble i.e. P ∝1/r, r being the radius of the bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.

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2.

Spherical balls of radius R are falling in a viscous fluid of viscosity η with a velocity v. The retarding viscous force acting on the spherical ball is

  • directly proportional to R but inversely proportional to v.

  • directly proportional to both radius R and velocity v.

  • inversely proportional to both radius R and velocity v.

  • inversely proportional to R but directly proportional to velocity v.


B.

directly proportional to both radius R and velocity v.

Retarding force acting on a ball falling into a viscous fluid
F = 6πηRv
where R = radius of the ball
v = velocity of ball and
η = coefficient of viscosity
∴ F ∝ R and F ∝ v
Or in words, retarding force is proportional to both R and v

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3.

A capillary tube (A) is dropped in water. Another identical tube (B) is dipped in a soap water solution.Which of the following shows the relative nature of the liquid columns in the two tubes? 


C.

Capillary rise h = 2T cosθ /ρgr. As soap solution has lower T, h will be low. 

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4.

A ball is made of a material of density ρ where ρoil < ρ < ρwater with ρoil and ρwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?


C.

 ρoil < ρ < ρwater 
Oil is the least dense of them, so it should settle at the top with water at the base. Now, the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So, it will stay at the oil -water interface.

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5.

A jar filled with two non-mixing liquids 1 and 2 having densities ρ1 and ρ2 respectively. A solid ball, made of a material of density ρ3, is dropped in the jar. It comes to equilibrium in the position shown in the figure.
Which of the following is true for ρ1, ρ2 and ρ3?

  • ρ3 < ρ1 < ρ2

  • ρ1 < ρ3 < ρ2

  • ρ1 < ρ2 < ρ3

  • ρ1 < ρ3 < ρ2


D.

ρ1 < ρ3 < ρ2



As liquid 1 floats above liquid 2,
ρ1 < ρ2
The ball is unable to sink into liquid 2, ρ3 < ρ2
The ball is unable to rise over liquid 1
ρ1 < ρ3 Thus, ρ1 < ρ3 < ρ2

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6.

Water is flowing continuously from a tap having an internal diameter 8 × 10-3 m. The water velocity as it leaves the tap is 0.4 ms-1. The diameter of the water stream at a distance 2 × 10-1 m below the tap is close
to:

  • 7.5 x 10-3 m

  • 9.6x 10-3 m

  • 3.6x 10-3 m

  • 5.0x 10-3 m


C.

3.6x 10-3 m

From Bernoulli's theorem
ρgh space equals space 1 half straight rho left parenthesis straight v subscript 2 superscript 2 minus straight v subscript 1 superscript 2 right parenthesis
gh space equals space 1 half straight v subscript 1 superscript 2 space open parentheses open parentheses straight v subscript 2 over straight v subscript 1 close parentheses minus 1 close parentheses space left parenthesis therefore straight A subscript 1 straight v subscript 1 space equals straight A subscript 2 straight v subscript 2 right parenthesis
rightwards double arrow space space open parentheses straight A subscript 1 over straight A subscript 2 close parentheses squared space equals space 1 space plus space fraction numerator 2 hg over denominator straight v subscript 1 superscript 2 end fraction
rightwards double arrow open parentheses straight D subscript 1 over straight D subscript 2 close parentheses to the power of 4 space equals space 1 space plus space fraction numerator 2 gh over denominator straight v subscript 1 superscript 2 end fraction
rightwards double arrow space straight D subscript 2 space equals space straight D subscript 1 over open parentheses 1 plus begin display style fraction numerator 2 gh over denominator straight v subscript 1 superscript 2 end fraction end style close parentheses to the power of 1 divided by 4 end exponent space equals space fraction numerator 8 space straight x space 10 to the power of negative 3 end exponent over denominator open parentheses begin display style fraction numerator 1 plus space 2 space straight x space 10 space straight x 0.2 over denominator left parenthesis 0.4 right parenthesis squared end fraction end style close parentheses to the power of 1 divided by 4 end exponent end fraction
space equals space 3.6 space straight x space 10 to the power of negative 3 end exponent space straight m

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7.

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90o angle at centre. Radius joining their interface makes an angle α with vertical. ratio d1/d2 is

  • fraction numerator 1 plus sin space straight alpha over denominator 1 minus sin space straight alpha end fraction
  • fraction numerator 1 plus space cosα over denominator 1 minus cosα end fraction
  • fraction numerator 1 plus space tan space straight alpha over denominator 1 minus tan space straight alpha end fraction
  • fraction numerator 1 plus cos space straight alpha over denominator 1 minus cos space straight alpha end fraction

C.

fraction numerator 1 plus space tan space straight alpha over denominator 1 minus tan space straight alpha end fraction
straight P subscript straight A space equals space straight P subscript straight B
straight P subscript 0 space plus space straight d subscript 1 straight g end subscript straight R space left parenthesis Cos space straight alpha space minus sin space straight alpha right parenthesis space equals straight P subscript 0 space plus straight d subscript 2 straight R space left parenthesis Cos space straight alpha plus sin space straight alpha right parenthesis
rightwards double arrow space straight d subscript 1 over straight d subscript 2 space equals space fraction numerator cos space straight alpha space plus sin space straight alpha over denominator cos space straight alpha minus sin space straight alpha end fraction space equals space fraction numerator 1 plus space tan space straight alpha over denominator 1 minus space tanα end fraction
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8.

If the terminal speed of a sphere of gold (density = 19.5 kg/m3 ) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3 ) of the same size in the same liquid.

  • 0.2 m/s

  • 0.4 m/s

  • 0.133 m/s

  • 0.1m/s


D.

0.1m/s

straight V subscript straight s over straight V subscript straight g space equals space fraction numerator left parenthesis straight p subscript straight s minus straight p subscript calligraphic l right parenthesis over denominator left parenthesis straight rho subscript straight g minus straight rho subscript calligraphic l right parenthesis end fraction
straight v subscript straight s space equals space 0.1 space straight m divided by straight s
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9.

A spherical solid ball of volume V is made of a material of density ρ1. It is falling through a liquid of density ρ221). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e., Fviscous = −kv2 (k>0). The terminal speed of the ball is

  • square root of fraction numerator Vg left parenthesis straight rho subscript 1 minus straight rho subscript 2 right parenthesis over denominator straight k end fraction end root
  • Vgρ subscript 1 over straight k
  • square root of Vgρ subscript 1 over straight k end root
  • fraction numerator Vg left parenthesis straight rho subscript 1 minus straight rho subscript 2 right parenthesis over denominator straight k end fraction

A.

square root of fraction numerator Vg left parenthesis straight rho subscript 1 minus straight rho subscript 2 right parenthesis over denominator straight k end fraction end root

ρ1Vg − ρ2Vg = kv2T

rightwards double arrow space straight V subscript straight T space equals space square root of fraction numerator Vg space left parenthesis straight rho subscript 1 minus straight rho subscript 2 right parenthesis over denominator straight k end fraction end root

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10.

A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be

  • 8 cm

  • 10 cm

  • 4 cm

  • 20 cm


D.

20 cm

Water will rise to the full length of capillary tube

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