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Exercises · 9.5

Q.A 50 kg50\ \text{kg} girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm1.0\ \text{cm}. What is the pressure exerted by the heel on the horizontal floor?

Yanam CbseNCERTSubjective· 3mImportance★★★★★est
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The pressure exerted by an object is its weight divided by the area of contact. For the girl balancing on a single high heel, the pressure is calculated by dividing her weight by the circular area of the heel, resulting in approximately 6.24×106 Pa\boxed{6.24 \times 10^6\ \text{Pa}}.

When an object rests on a surface, it exerts a force perpendicular to that surface due to its weight. The effect of this force is described by pressure, which is defined as the force distributed over a given area. Understanding pressure is crucial because it explains why a sharp knife cuts easily (small area, high pressure) or why snowshoes prevent sinking into snow (large area, low pressure).

In this problem, a girl's entire weight is concentrated on the tiny area of a single high heel. This small contact area, combined with a significant force (her weight), will result in a very high pressure on the floor.

Here's how we calculate it:

  1. Identify the Force:

    The force exerted by the heel on the floor is the weight of the girl. Weight (WW) is calculated as mass (mm) times the acceleration due to gravity (gg). We'll use g=9.8 m/s2g = 9.8\ \text{m/s}^2.

    W=mgW = mg

    W=(50 kg)×(9.8 m/s2)W = (50\ \text{kg}) \times (9.8\ \text{m/s}^2)

    W=490 NW = 490\ \text{N}

  2. Calculate the Area of Contact:

    The heel is circular with a diameter of 1.0 cm1.0\ \text{cm}. We need to convert this diameter to meters and then calculate the area (AA) of the circle.

    Diameter d=1.0 cm=0.01 md = 1.0\ \text{cm} = 0.01\ \text{m}

    Radius r=d2=0.01 m2=0.005 mr = \frac{d}{2} = \frac{0.01\ \text{m}}{2} = 0.005\ \text{m}

    The area of a circle is given by A=πr2A = \pi r^2.

    A=π(0.005 m)2A = \pi (0.005\ \text{m})^2

    A=π(2.5×10−5) m2A = \pi (2.5 \times 10^{-5})\ \text{m}^2

    A≈7.85398×10−5 m2A \approx 7.85398 \times 10^{-5}\ \text{m}^2

    Watch out

    A common mistake is forgetting to convert units to SI units (meters for length) before calculation. Using centimeters directly would lead to an incorrect area and thus incorrect pressure. …

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