
The mass of a lead ball is M. It falls down in a viscous liquid with terminal velocity V. The terminal velocity of another lead ball of mass 8M in the same liquid will be:
\[
A.\;64V \\
B.4V \\
C.\;8V \\
D.V \\
\]
Answer
155.4k+ views
Hint: Terminal velocity is directly proportional to the square of the radius of the ball$V \propto {r^2}$. The mass of the second ball is given from which its radius can also be obtained by evaluating the respective cases. Now substitute the values in to find the terminal velocity of another lead ball.
Complete step-by-step answer

If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then becomes zero and it attains a constant velocity called terminal velocity. It is given by,
$V = \dfrac{{2{r^2}(\rho - \sigma )g}}{{9\eta }}$
From the above equation we know that
$V \propto {r^2}$
Let \[{r_1}\]and \[{r_2}\]be the radius of the first and second ball respectively.
Given that mass of the lead ball is M and for the second ball is 8M. The terminal velocity for the first case is V.
Case (1), since density is mass by volume,
$\rho \times \dfrac{4}{3}\pi {r_1}^3 = M$
${V_1} = \dfrac{{2{r_1}^2(\rho - \sigma )g}}{{9\eta }}$
Case (2), for the second ball,
$\rho \times \dfrac{4}{3}\pi {r_2}^3 = 8M$
Divide Case (1) by (2)
$\dfrac{{\rho \times \dfrac{4}{3}\pi {r_1}^3}}{{\rho \times \dfrac{4}{3}\pi {r_2}^3}} = \dfrac{M}{{8M}}$
\[{r_2}^3 = 8{r_1}^3\]
\[{r_2} = 2{r_1}\]
Since,
$V \propto {r^2}$
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{r_1}^2}}{{{r_2}^2}}$
Substitute the values of radius one first and second ball
${V_2} = \dfrac{{V \times 4r_1^2}}{{r_1^2}}$
${V_2} = 4V$
The terminal velocity for the second case is 4V and the correct option is B.
Note: Terminal velocity is directly proportional to the density of the ball. In the terminal velocity expression, if\[\rho > \sigma \] , \[V\] is positive and hence the body will attain constant velocity is the downward direction. If\[\rho < \sigma \] ,\[V\] is negative and the spherical body will attain constant velocity in upward direction.
Complete step-by-step answer

If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then becomes zero and it attains a constant velocity called terminal velocity. It is given by,
$V = \dfrac{{2{r^2}(\rho - \sigma )g}}{{9\eta }}$
From the above equation we know that
$V \propto {r^2}$
Let \[{r_1}\]and \[{r_2}\]be the radius of the first and second ball respectively.
Given that mass of the lead ball is M and for the second ball is 8M. The terminal velocity for the first case is V.
Case (1), since density is mass by volume,
$\rho \times \dfrac{4}{3}\pi {r_1}^3 = M$
${V_1} = \dfrac{{2{r_1}^2(\rho - \sigma )g}}{{9\eta }}$
Case (2), for the second ball,
$\rho \times \dfrac{4}{3}\pi {r_2}^3 = 8M$
Divide Case (1) by (2)
$\dfrac{{\rho \times \dfrac{4}{3}\pi {r_1}^3}}{{\rho \times \dfrac{4}{3}\pi {r_2}^3}} = \dfrac{M}{{8M}}$
\[{r_2}^3 = 8{r_1}^3\]
\[{r_2} = 2{r_1}\]
Since,
$V \propto {r^2}$
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{r_1}^2}}{{{r_2}^2}}$
Substitute the values of radius one first and second ball
${V_2} = \dfrac{{V \times 4r_1^2}}{{r_1^2}}$
${V_2} = 4V$
The terminal velocity for the second case is 4V and the correct option is B.
Note: Terminal velocity is directly proportional to the density of the ball. In the terminal velocity expression, if\[\rho > \sigma \] , \[V\] is positive and hence the body will attain constant velocity is the downward direction. If\[\rho < \sigma \] ,\[V\] is negative and the spherical body will attain constant velocity in upward direction.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Electrical Field of Charged Spherical Shell - JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
