Mechanical Properties of Fluids

More Topic from Physics

Question 1

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

Solution

D.

20 cm

Water will rise to the full length of capillary tube

Question 2

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?

Solution

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.

Question 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? 

Solution

C.

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

Question 4

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

Solution

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

Question 5

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

Solution

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