Mechanical Properties Of Fluid

Calculation of Tension (T)

We are given the equation for work:

The change in area is given by:

Substituting the values:

Now, using the formula to calculate the tension:

Simplifying the equation:

Conclusion:

The tension, T, is 0.125 N/m.

To solve this problem, we need to understand the forces acting on the ball as it moves through the glycerine.

The small ball, when falling through a viscous fluid like glycerine, experiences three forces:

1. Gravitational Force : This is the force due to gravity acting downward, given by , where is the mass of the ball, and is the acceleration due to gravity.
2. Buoyant Force : This is the upward force exerted by the fluid on the ball, given by the Archimedes' principle as , where is the volume of the ball, is the density of the fluid (glycerine, in this case), and is the acceleration due to gravity.
3. Viscous Force : This is the resistive force due to the viscosity of the fluid, which acts upward against the motion of the ball.

After some time, the velocity of the ball becomes constant, meaning it reaches terminal velocity. At terminal velocity, the net force acting on the ball is zero, as the forces are in equilibrium. Thus,

Substituting the expressions for the forces, we get:

The volume of the ball can be expressed in terms of its mass and density as:

Plugging this back into the buoyant force equation, we have:

Now substitute this into the equilibrium equation:

Solve this equation for the viscous force :

Therefore, the viscous force acting on the ball is .

Thus, the correct answer is , matching Option B.

The problem involves understanding the speed-time relationship for a spherical ball dropped in a highly viscous liquid. This is a classic physics problem involving terminal velocity. To determine the correct curve, we need to consider the forces acting on the ball. Initially, gravity accelerates the ball downwards, increasing its speed. As the ball speeds up, viscous drag force, which is proportional to speed in most fluids, opposes this motion.
Formally, the net force on the ball can be expressed as:

Net Force:

where is the mass of the ball, the acceleration due to gravity, the drag coefficient (a constant for a particular fluid-ball system), and the velocity.

As time progresses, the viscous force increases until it balances the gravitational force , resulting in a net force of zero. At this point, the ball achieves terminal velocity , where the speed remains constant. Hence:

Terminal Velocity:

The graph of velocity versus time will show an initial increase in velocity steeply, eventually leveling off as the terminal velocity is achieved. This results in an exponential growth curve that asymptotically approaches .

By comparing graph options, the curve representing this behavior is option B.

The phenomenon relating to the pressure inside a soap bubble can be understood through the principles of fluid mechanics. A soap bubble is characterized by two layers of soap film with air trapped between them. The pressure inside the bubble is determined by both the surface tension of the soap film and the radius of the bubble according to the Young-Laplace equation:

P = P₀ +

where:

• P is the internal pressure of the bubble.
• P₀ is the atmospheric pressure outside the bubble.
• γ is the surface tension of the soap solution.
• R is the radius of the bubble.

When a soap bubble expands, the radius (R) of the bubble increases. According to the equation, increasing the value of R while keeping surface tension constant results in the term decreasing. Consequently, the internal pressure P also decreases if P₀ is constant (atmospheric pressure). Thus, the correct answer is that the pressure inside the bubble decreases as it expands.

To determine the energy required to form a soap bubble of radius 2 cm, we use the formula for surface energy of a bubble: , where is the radius, and is the surface tension. The factor of 2 accounts for the inner and outer surfaces of the bubble.

Given:

• Radius,
• Surface tension,

Substitute these values into the equation:

(approximating as 3.1416)

J

Thus, the energy required is approximately J.

Therefore, the correct answer is:

3.01 x 10^-4 J