Question

A gas bubble of $ 2cm $ diameter rises through a liquid of density $ 1.75gc{m^{ - 3}} $ with a fixed speed of $ 0.35cm{s^{ - 1}} $ . Neglect the density of the gas. The coefficient of viscosity of the liquid is
(A) 870 poise
(B) 1120 poise
(C) 982 poise
(D) 1089 poise

Answer 1

Hint: As the bubble rises with a constant speed, so the buoyant force will be equal to the viscous drag force. Now equating the values of the buoyant force and the drag force, we will get the value of the coefficient of viscosity of the liquid.

Formula Used: In the solution, we will be using the following formula,
 $\Rightarrow {F_b} = V\rho g $
where $ {F_b} $ is the buoyant force, $ V $ is the volume of the bubble, $ \rho $ is the density of the liquid and $ g $ is the acceleration due to gravity.
 $\Rightarrow {F_v} = 6\pi \eta rv $
where $ {F_v} $ is the viscous drag force, $ \eta $ is the coefficient of viscosity, $ r $ is the radius and $ v $ is the speed of the bubble.

Complete step by step answer:
 In the question we are provided that the bubble rises through the liquid with a constant speed. So since there is no acceleration, the net force that is acting on the bubble is zero. The forces that are acting on the bubble are the buoyant force and the viscous drag force. So these two forces cancel each other out. Therefore we can write,
 $\Rightarrow {F_b} = {F_v} $
Now the value of the buoyant force is given by the formula,
 $\Rightarrow {F_b} = V\rho g $ and the value of the viscous drag force is given by the formula,
 $\Rightarrow {F_v} = 6\pi \eta rv $
Now equating these two we get,
 $\Rightarrow V\rho g = 6\pi \eta rv $
Now, we can keep the $ \eta $ in the LHS and take the rest of the terms to the RHS. So we get,
 $\Rightarrow \eta = \dfrac{{V\rho g}}{{6\pi rv}} $
In the question we are provided that the density of the liquid is $ \rho = 1.75gc{m^{ - 3}} $ . The diameter of the bubble is $ d = 2cm $ . So the radius will be half of the diameter. Hence, we get $ r = 1cm $ .
The speed of the bubble in the liquid is $ v = 0.35cm{s^{ - 1}} $ . We can take the value of the acceleration due to gravity as, $ g = 980cm{s^{ - 2}} $ . The bubble is spherical in shape, so the volume of the bubble will be,
 $ V = \dfrac{4}{3}\pi {r^3} $ . Substituting the volume we get,
 $\Rightarrow \eta = \dfrac{4}{3}\pi {r^3}\dfrac{{\rho g}}{{6\pi rv}} $
So we get,
 $\Rightarrow \eta = \dfrac{{4\pi {r^2}\rho g}}{{18\pi v}} $
So substituting the rest of the values we get,
 $\Rightarrow \eta = \dfrac{{4\pi {{\left( 1 \right)}^2} \times 1.75 \times 980}}{{18 \times \pi \times 0.35}} $
On calculating the values in the numerator and the denominator we get,
 $\Rightarrow \eta = \dfrac{{6860\pi }}{{6.3\pi }} $
We can cancel the $ \pi $ from the numerator and denominator and get,
 $\Rightarrow \eta = \dfrac{{6860}}{{6.3}} $
On calculating we have,
 $\Rightarrow \eta = 1088.88poise $
This is equivalent to, $ \eta = 1089poise $
Hence the correct answer is the option D.

Note:
The viscous drag force is a type of friction or the fluid friction. It is the force that is acting opposite to the relative motion of any object with respect to a surrounding fluid. It is directly proportional to the velocity of the object and decreases the velocity of the object.