Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm$$^3$$ of oleic acid per cm$$^3$$ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm$$^2$$ by considering 100 spherical drops of radius $$\left(\frac{3}{40\pi}\right)^{1/3} \times 10^{-3}$$ cm. Then the thickness of oleic acid layer will be $$x \times 10^{-14}$$ m. Where $$x$$ is ________.

We need to determine the magnitude of the downward negative lift $$F_L$$ acting on a racing car moving along a flat circular track.

1. Identify the Forces Acting on the Car

From the problem statement page, the car is moving in a horizontal circle of radius $$R$$ with a speed $$v$$. Let's analyze the forces acting in both vertical and horizontal directions:

• Vertical Direction:
  The forces keeping the car vertically balanced are:
  • Downward gravitational force: $$mg$$
  • Downward negative lift force (aerodynamic downforce): $$F_L$$
  • Upward normal reaction force from the track: $$N$$
  Since there is no vertical acceleration, the net vertical force is zero:

  $$N = mg + F_L \quad \text{--- (Equation 1)}$$

• Horizontal Direction:
  The centripetal force required to keep the car moving in a circle of radius $$R$$ at speed $$v$$ is provided entirely by the static friction ($$f_s$$) between the tires and the track:

  $$f_s = \frac{m v^2}{R}$$

2. Relate Friction to the Normal Force

To safely navigate the turn at speed $$v$$, the required centripetal friction must be provided by the maximum limiting static friction available ($$f_s = \mu_s N$$):

$$\mu_s N = \frac{m v^2}{R}$$

Solving for the normal force $$N$$ gives:

$$N = \frac{m v^2}{\mu_s R} \quad \text{--- (Equation 2)}$$

3. Solve for the Negative Lift ($$F_L$$)

Now, equate Equation 1 and Equation 2 to eliminate $$N$$:

$$mg + F_L = \frac{m v^2}{\mu_s R}$$

Isolate the negative lift force $$F_L$$:

$$F_L = \frac{m v^2}{\mu_s R} - mg$$

Factoring out the mass $$m$$ from both terms yields the final expression:

$$F_L = m \left( \frac{v^2}{\mu_s R} - g \right)$$

Conclusion

The magnitude of the negative lift acting downwards on the car is $$m \left( \frac{v^2}{\mu_s R} - g \right)$$, which corresponds to Option B.