Question

A 31.4 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 2 cm. The pressure exerted by the heel on the horizontal floor is (Acceleration due to gravity = 10 m/s\(^2\)):

• A. \( 10^6 \, \text{Pa} \)
• B. \( 2.5 \times 10^5 \, \text{Pa} \)
• C. \( 10^5 \, \text{Pa} \)
• D. \( 2.5 \times 10^6 \, \text{Pa} \)

Hint:

For calculating pressure, use the formula \( P = \frac{F}{A} \), where \( A \) is the area of contact. For circular areas, \( A = \pi r^2 \).

Answer 1

The Correct Option is A

Step 1: The pressure is given by the formula: \[ P = \frac{F}{A} \] where \( F = \text{force} = \text{weight of the girl} = mg \), and \( A \) is the area of contact of the heel with the floor. Step 2: The area of the circular heel is calculated using the formula: \[ A = \pi r^2 \] Given the diameter of the heel is \( 2 \, \text{cm} \), the radius \( r = 1 \, \text{cm} = 0.01 \, \text{m} \). Therefore: \[ A = \pi (0.01)^2 = 3.14 \times 10^{-4} \, \text{m}^2 \] Step 3: The force exerted by the girl is: \[ F = mg = 31.4 \times 10 = 314 \, \text{N} \] Step 4: Now, the pressure is: \[ P = \frac{314}{3.14 \times 10^{-4}} = 10^6 \, \text{Pa} \] Thus, the correct answer is \( 10^6 \, \text{Pa} \).