Problem 80 - Master Resource Book in JEE Main Physics

Question

A rain drop of radius \(1.5 \mathrm{~mm}\), experiences a drag force \(F=\left(2 \times 10^{-5} \mathrm{v}\right) \mathrm{N}\), while falling through air from a height \(2 \mathrm{~km}\), with a velocity \(\bar{v} .\) The terminal velocity of the rain drop will be nearly (use \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(200 \mathrm{~ms}^{-1}\) (b) \(80 \mathrm{~ms}^{-1}\) (c) \(7 \mathrm{~ms}^{-1}\) (d) \(3 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The terminal velocity is approximately 7 m/s, option (c).

Step 1: Understand Terminal Velocity

Terminal velocity is defined as the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. At this velocity, the force of gravity is balanced by the drag force.

Step 2: Express Forces Acting on Raindrop

The gravitational force on the raindrop can be calculated using the formula: \[ F_{gravity} = mg \]where \( m \) is the mass of the raindrop, and \( g = 10 \mathrm{~m/s}^2 \) is the acceleration due to gravity.

Step 3: Find the Mass of the Raindrop

To find the mass \( m \) of the raindrop, we need its density and volume. Assume the density of water \( \rho = 1000 \mathrm{~kg/m}^3 \). The volume \( V \) of the raindrop (a sphere) is:\[ V = \frac{4}{3}\pi r^3 \]where \( r = 1.5 \mathrm{~mm} = 0.0015 \mathrm{~m} \). Plug in these values to find the volume and mass of the raindrop.

Step 4: Calculate Volume and Mass

Calculate the volume \( V \):\[ V = \frac{4}{3} \pi (0.0015)^3 \approx 1.4137 \times 10^{-8} \mathrm{~m}^3 \]Now calculate the mass \( m \):\[ m = \rho \times V = 1000 \times 1.4137 \times 10^{-8} \approx 1.4137 \times 10^{-5} \mathrm{~kg} \]

Step 5: Set Up the Drag Force Equation

The drag force \( F_{drag} \) is given by:\[ F_{drag} = 2 \times 10^{-5}v \]At terminal velocity, the drag force equals the gravitational force:\[ 2 \times 10^{-5}v = mg \]

Step 6: Solve for Terminal Velocity

Substitute the mass \( m \) and \( g \) into the equation:\[ 2 \times 10^{-5} v = 1.4137 \times 10^{-5} \times 10 \]Solve for \( v \):\[ v = \frac{1.4137 \times 10^{-4}}{2 \times 10^{-5}} \approx 7.0685 \mathrm{~m/s} \].

Step 7: Compare With Options

The calculated terminal velocity is approximately \( 7 \mathrm{~m/s} \). So, the closest option given is (c) \( 7 \mathrm{~ms}^{-1} \).

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept in physics, describing the force by which planet Earth attracts objects towards its center. In simpler terms, it's what keeps us grounded. In any falling object, including raindrops, this force acts downwards.

For a raindrop with mass \( m \), the gravitational force can be calculated using:

• \( F_{gravity} = mg \)

Here, \( g \) is the acceleration due to gravity. It is standard to use \( 10 \, \mathrm{m/s^2} \) as its value, especially for simplifying calculations in problems. This force helps determine how fast the raindrop can fall and interact with the air.

When the gravitational force equates to the drag force acting upward, the terminal velocity is achieved, at which point the raindrop's speed becomes constant.

Drag Force

Drag force is essentially air resistance that opposes the motion of an object moving through a fluid, such as air or water. It's crucial in determining how fast an object like a raindrop can fall because it counteracts the gravitational force pulling the object downward.

For the raindrop in our problem, the drag force is given by:

• \( F_{drag} = 2 \times 10^{-5}v \)

Here, \( v \) is the velocity of the raindrop. As the velocity increases, so does the drag force. This relationship allows the calculation of terminal velocity when the drag force equals the gravitational force, resulting in no net force and hence no further acceleration. The object then falls at a constant speed.

Sphere Volume Calculation

Calculating the volume of a sphere is an essential step in this exercise as it helps determine the mass of the raindrop, which is pivotal to finding both the gravitational force and eventually the terminal velocity.

The formula to find the volume \( V \) of a sphere with radius \( r \) is:

• \( V = \frac{4}{3} \pi r^3 \)

In our specific scenario, the raindrop's radius is \( 1.5 \, \mathrm{mm} \). Converted to meters, this is \( 0.0015 \, \mathrm{m} \). Plugging this into the formula gives the volume, which can then be multiplied by the density to find the mass.

By determining the volume accurately, we ensure that the subsequent calculations for mass and gravitational force are precise and reliable.

Density of Water

Density is a property that refers to the mass per unit volume of a substance. For water, this density is a standard value of \( 1000 \, \mathrm{kg/m^3} \), which is used to determine the mass of the raindrop based on its volume.

The process involves multiplying the volume of the sphere, calculated in the previous section, by the density:

• \( m = \rho \times V \)

Given the volume from the sphere volume calculation and \( \rho = 1000 \, \mathrm{kg/m^3} \), the raindrop's mass can be calculated. This mass is essential for determining the gravitational force acting on the raindrop.

Understanding these core concepts allows for a comprehensive grasp of how water and air interact with falling objects like raindrops, influencing their speed and behavior.