Terminal Velocity: Comprehensive NEET Physics Notes

1. Terminal Velocity

1.1 Definition of Terminal Velocity

When an object falls through a fluid (liquid or gas), it experiences two main forces: gravity pulling it downward and a viscous drag force exerted by the fluid acting in the opposite direction. As the object accelerates, the upward resistive force increases until it balances the downward gravitational force. At this equilibrium, the object falls at a constant speed known as Terminal Velocity.

1.2 Mathematical Derivation of Terminal Velocity

For a spherical object of radius $a$ falling through a fluid with viscosity $\eta$, density $\rho$, and with the object’s density being ${\rho }_s$, the forces acting on it are:

1. Gravitational Force (Weight): $F_g=\frac{4}{3}\pi a^3{\rho }_{s}g$
2. Buoyant Force: $F_b=\frac{4}{3}\pi a^3\rho g$
3. Viscous Drag Force: According to Stokes' Law, the resistive force is $F_v=6\pi \eta av$

At terminal velocity, the net force acting on the object becomes zero:

$F_g-F_b=F_v$

Substituting these values: $\frac{4}{3}\pi a^3({\rho }_s-\rho )g=6\pi \eta a{v}_t$

Solving for the terminal velocity $v_t$: $v_t=\frac{2a^2({\rho }_s-\rho )g}{9\eta }$

1.3 Factors Affecting Terminal Velocity

1. Radius of the object: Terminal velocity is proportional to the square of the radius ($a^2$).
2. Density Difference: It depends on the density difference between the object and the fluid (${\rho }_s-\rho$).
3. Viscosity of the fluid: Higher viscosity results in a lower terminal velocity.
4. Acceleration due to gravity: Terminal velocity increases with greater gravitational acceleration ($g$).

NEET Tip:

Always remember that terminal velocity occurs when the object reaches equilibrium between the downward gravitational force and the upward resistive force. This concept is commonly tested in NEET exams.

1.4 Real-life Application: Raindrops

Raindrops falling from clouds reach terminal velocity before hitting the ground. This is why raindrops, despite falling from a height, do not accelerate indefinitely and cause no injury upon impact.

Did You Know?

A human skydiver reaches terminal velocity of approximately 53 m/s (about 190 km/h) when in a belly-to-earth position. If the skydiver shifts to a head-first position, the terminal velocity increases due to reduced air resistance.

Common Misconception:

Misconception: Terminal velocity means the object stops falling. Clarification: An object at terminal velocity continues to fall but at a constant speed, with no further acceleration.

1.5 Visual Representation

Suggested Diagram:

• Include a diagram showing a falling object with forces acting on it: gravitational force downward, buoyant force, and viscous drag force upward. Clearly label the forces and indicate the point where terminal velocity is reached.

1.6 Quick Recap

• Terminal velocity is the steady speed an object reaches when gravitational force equals resistive forces in a fluid.
• Stokes' Law explains the viscous force experienced by spherical objects moving through a fluid.
• Terminal velocity depends on object radius, fluid viscosity, and the density difference between the object and fluid.

Practice Questions

1. Calculation-Based: A raindrop with a radius of 0.1 mm falls through the air (density = 1.2 kg/m³, viscosity = 1.8 × 10⁻⁵ Pa·s). Calculate its terminal velocity. (Given: density of water = 1000 kg/m³ and $g=9.8m/s^2$)
2. Application-Based: An iron ball (density = 7800 kg/m³) with a radius of 2 mm falls through oil with a viscosity of 0.2 Pa·s and density of 900 kg/m³. Find its terminal velocity.
3. Conceptual: Explain why larger objects reach a higher terminal velocity compared to smaller objects in the same fluid.
4. Relationship Analysis: Describe the relationship between fluid viscosity and terminal velocity.
5. Scenario-Based: A skydiver reaches terminal velocity of about 53 m/s in a belly-down position. How would this change if they shift to a head-first position?

Answers:

1. Substituting values into the formula: $v_t=\frac{2a^2({\rho }_s-\rho )g}{9\eta }$ Given $a=0.1mm=0.1\times 10^{-3}m$, ${\rho }_s=1000kg/m^3$, $\rho =1.2kg/m^3$, $\eta =1.8\times 10^{-5}Pa\cdot s$, and $g=9.8m/s^2$ Calculation yields: $v_t\approx 2.7m/s$
2. For the iron ball: $v_t=\frac{2(2\times 10^{-3}{)}^2(7800-900)9.8}{9(0.2)}\approx 1.38m/s$
3. Larger objects have greater gravitational force acting on them, resulting in a higher terminal velocity.
4. Terminal velocity decreases as viscosity increases.
5. In a head-first position, terminal velocity increases due to reduced air resistance.

Additional Features for Improved Learning

• Glossary:
• • Viscous Drag Force: The resistive force experienced by an object moving through a fluid.
  • Density: Mass per unit volume of a substance, denoted by $\rho$.
  • Terminal Velocity: The constant speed attained by an object when the net force acting on it is zero.

Supplementary Quick Reference Guide

• Key Formula: $v_t=\frac{2a^2({\rho }_s-\rho )g}{9\eta }$
• Important Points:
• • Terminal velocity occurs when gravitational and resistive forces balance.
  • Influenced by object radius, fluid viscosity, and density difference.

NEET Exam Strategy

• Time Management: Practice solving problems related to terminal velocity efficiently using the formula. Focus on the direct relationships between variables to quickly answer conceptual questions.
• Common NEET Question Patterns: Be prepared for questions that test your understanding of how changing one factor (e.g., fluid viscosity) affects terminal velocity.