A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius 2r, falling through the same viscous medium will have terminal velocity:
- \(\frac{v}{2}\)
\(\frac{v}{4}\)
- $4v$
- $2v$
The Correct Option is A
Approach Solution - 1
To solve this problem, we need to understand the concept of terminal velocity in a viscous medium according to Stokes' Law. The terminal velocity \( v_t \) for a spherical object of radius \( r \) and density \( \rho \) falling through a medium with viscosity \( \eta \) is given by:
\(v_t = \frac{2}{9} \cdot \frac{r^2 (\rho - \rho_m)}{\eta} \cdot g\)
where:
- \( r \) is the radius of the sphere
- \( \rho \) is the density of the sphere
- \( \rho_m \) is the density of the medium (negligible here)
- \( \eta \) is the viscosity of the medium
- \( g \) is the acceleration due to gravity
For the first ball with radius \( r \), the terminal velocity is \( v \).
For the second ball with the same mass but a radius of \( 2r \), let's consider the terminal velocity \( v_2 \).
The mass of the sphere is proportional to the volume, which is proportional to \( r^3 \). Hence, keeping the mass constant implies:
\(\rho \cdot r^3 = \rho_2 \cdot (2r)^3\)
The density of the new sphere \( \rho_2\) will be:
\(\rho_2 = \frac{\rho}{8}\)
Now applying these changes to the terminal velocity formula for the second ball:
\(v_2 = \frac{2}{9} \cdot \frac{(2r)^2 (\rho_2 - \rho_m)}{\eta} \cdot g\)
Substitute \(\rho_2 = \frac{\rho}{8}\):
\(v_2 = \frac{2}{9} \cdot \frac{4r^2 \left(\frac{\rho}{8} - \rho_m\right)}{\eta} \cdot g\)
Thus:
\(v_2 = \frac{2}{9} \cdot \frac{r^2 (\rho - \rho_m)}{\eta} \cdot \frac{1}{2} \cdot g = \frac{v}{2}\)
Substituting the known relationship we find:
The terminal velocity for the second ball is \( \frac{v}{2} \).
Hence, the correct answer is \(\frac{v}{2}\).
Approach Solution -2
Since the density of the medium is negligible, the buoyancy force can be ignored. At terminal velocity, the gravitational force on the ball is balanced by the viscous drag force. The terminal velocity \( v \) is given by:
\[ v \propto \frac{1}{r}, \]for a sphere of constant mass.
Let the terminal velocity of the original ball (radius \( r \)) be \( v \) and the terminal velocity of the larger ball (radius \( 2r \)) be \( v' \).
Using the inverse proportionality:
\[ \frac{v}{v'} = \frac{r'}{r}. \]Since \( r' = 2r \):
\[ \frac{v}{v'} = 2 \implies v' = \frac{v}{2}. \]Thus, the terminal velocity of the larger ball is:
\[ \frac{v}{2}. \]Top JEE Main Physics Questions
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