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Question: The diameter of ball y is double that of x. The ratio of their terminal velocities inside water will...

The diameter of ball y is double that of x. The ratio of their terminal velocities inside water will be:
A) 1:4
B) 4:1
C) 1:2
D) 2:1

Explanation

Solution

Hint
Here, we will be using the formula of the terminal velocity VV of the body of radius rr, density ρ\rho falling through a medium of densityρo{\rho _o} ​ is given by V=2r2(ρρo)g9ηV = \dfrac{{2{r^2}\left( {\rho - {\rho _o}} \right)g}}{{9\eta }} where η\eta is the coefficient of viscosity of medium.

Complete step-by-step solution
Given, two balls xx and yy are flowing in water which have attained terminal velocity.
The diameter of ball yy is double that of ball xx. The radius is just half of the diameter so it implies that the radius of ball yy is also double that of xx.
Therefore, radius of ball yy is also double that of xx i.e. ry=2rx{r_y} = 2{r_x}
As we can clearly see from formula Vr2V\infty {r^2}
Therefore, the terminal velocity of ball x and y inside water will be VxVy=(rxry)2\dfrac{{{V_x}}}{{{V_y}}} = {\left( {\dfrac{{{r_x}}}{{{r_y}}}} \right)^2}
Now, as givenry=2rx{r_y} = 2{r_x}. Put this value in above equation, we get
VxVy=(rx2rx)2\Rightarrow \dfrac{{{V_x}}}{{{V_y}}} = {\left( {\dfrac{{{r_x}}}{{2{r_x}}}} \right)^2}
VxVy=(rx24rx2)\Rightarrow \dfrac{{{V_x}}}{{{V_y}}} = \left( {\dfrac{{{r_x}^2}}{{4{r_x}^2}}} \right)
VxVy=14\Rightarrow \dfrac{{{V_x}}}{{{V_y}}} = \dfrac{1}{4} or Vx:Vy{V_x}:{V_y}= 1:4
Hence the ratio of Vy:Vx{V_y}:{V_x} is 4: 1.
The ratio of terminal velocities of yy and xx inside water will be 4:1.
Thus, Option (B) is correct.

Note
In these types of questions, students may go wrong in identifying the correct ratio according to the question. One can do mistake by taking ratio of Vx:Vy{V_x}:{V_y} i.e. 1:4 and which is incorrect but according to question we need to write in form Vy:Vx{V_y}:{V_x} i.e. 4: 1 and which is correct answer.