Problem 2 - Master Resource Book in JEE Main Physics

Question

A \(50 \mathrm{~kg}\) girl wearing high heel shoes balances on a single heel. If the heel is circular with a diameter \(1.0 \mathrm{~cm}\). What is the pressure exerted on the horizontal floor? (a) \(6.9 \times 10^{6} \mathrm{~Pa}\) (b) \(6.2 \times 10^{6} \mathrm{~Pa}\) (c) \(9.6 \times 10^{6} \mathrm{~Pa}\) (d) \(9.0 \times 10^{6} \mathrm{~Pa}\)

Step-by-Step Solution

Answer

The pressure exerted is approximately \(6.2 \times 10^6 \mathrm{~Pa}\), option (b).

Step 1: Understanding Pressure

Pressure is defined as force per unit area. The formula to calculate pressure is \( P = \frac{F}{A} \), where \( P \) is the pressure, \( F \) is the force, and \( A \) is the area over which the force is distributed.

Step 2: Determine the Force Exerted

The force exerted by the girl is equal to her weight. Since weight \( W = mg \), where \( m \) is mass and \( g \) is the gravitational acceleration \( 9.8 \ m/s^2 \), we have:\[ F = 50 \ kg \times 9.8 \ m/s^2 = 490 \ N \]

Step 3: Calculate the Area of the Heel

The heel is circular with a diameter of \( 1.0 \ cm \), so the radius \( r = \frac{1.0}{2} = 0.5 \ cm = 0.005 \ m \). The area \( A \) of a circle is given by \( A = \pi r^2 \). Thus:\[ A = \pi (0.005 \ m)^2 = \pi \times 0.000025 \ m^2 \approx 0.0000785 \ m^2 \]

Step 4: Calculate the Pressure on the Floor

Substitute the values of force and area into the pressure formula:\[ P = \frac{490 \ N}{0.0000785 \ m^2} \approx 6.24 \times 10^6 \ Pa \]

Step 5: Choose the Closest Answer

Compare the calculated pressure value \(6.24 \times 10^6 \ Pa\) to the given options. The closest match is \(6.2 \times 10^6 \ Pa\).

Key Concepts

Force and Weight Relationships

When we talk about force and weight, it's important to understand that weight is essentially a force that results from gravity acting on a mass. This means that weight is a specialized force that is determined by the equation \( W = mg \), where \( m \) stands for mass and \( g \) represents gravitational acceleration. In simpler terms, weight is what you measure when you stand on a scale and it tells you how much the earth is pulling you down. For a given object, the heavier it is, the stronger the force it applies.

• Mass: The amount of matter in an object. For example, the mass of the girl is given as 50 kg.
• Gravitational Acceleration: On Earth, this is approximately \(9.8 \ m/s^2\).

Combining these, the weight or force exerted by our girl is calculated using her mass and Earth's gravity to be \( 490 \ N \). So every time you think about weight, remember that it's a force, just like when you push or pull something.

Area Calculations in Physics

In physics, calculating the area over which a force is distributed is crucial for understanding pressure. For a circular shape like the heel of the shoe, the area is calculated using the circle area formula \( A = \pi r^2 \), where \( r \) is the radius of the circle. In our example, the heel's diameter is given as 1.0 cm, so dividing by two gives us the radius:

• Radius: \( r = 0.5 \ cm = 0.005 \ m \)
• Convert radius to meters since other given values usually use SI units.
• Calculate area using the formula: \( A = \pi \times (0.005 \ m)^2 \approx 0.0000785 \ m^2 \)

Knowing the area of the heel is essential for determining the pressure it exerts, as pressure depends on how force is spread over this area. The smaller the area, the greater the pressure, much like how stepping on a tack with your heel would hurt versus doing so with your foot flat.

Gravitational Acceleration Concepts

Gravitational acceleration is a core concept in physics that explains how gravity acts to accelerate a mass towards the center of the Earth. This is approximately \( 9.8 \ m/s^2 \) on the surface of the Earth. It's a constant that represents how fast an object will speed up as it's pulled downwards by gravity.

• Universal Role: Gravitational acceleration is the same for all objects near Earth's surface, meaning a heavy rock and a light feather fall at the same rate if air resistance is negligible.
• Importance in Calculations: Anytime you're calculating weight (i.e., gravitational force), \( g \) is used as a multiplication factor with mass to find the force of gravity.

In this exercise, recognizing \( g \) helps you compute the weight or force exerted by the girl, which is a prerequisite for any pressure calculation. This fundamental concept helps bridge the gap between mass and force, making it easier to solve various physical problems related to gravity.