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Q9.5: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.5 Solution: A 50 kg girl wearing high heel shoes balances on a single heel. …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the given diameter directly into the area formula $A = \pi r^2$ instead of halving it to find the radius first. Always convert the diameter to radius ($r = d/2$) before calculating the cross-sectional area to avoid a factor of four error in the final pressure.

A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Step-by-Step Verified Solution

**Answer:** 6.37×106 Pa6.37 \times 10^6 \text{ Pa} **Step-by-Step Explanation:** Mass m=50m = 50 kg, so weight F=50×9.8=490F = 50 \times 9.8 = 490 N. Diameter d=1.0d = 1.0 cm =0.01= 0.01 m, so radius r=0.005r = 0.005 m. Area of heel A=πr2=π(0.005)2=7.85×10−5A = \pi r^2 = \pi(0.005)^2 = 7.85 \times 10^{-5} m_. Pressure:
P=FA=4907.85×10−5=6.24×106 PaP = \frac{F}{A} = \frac{490}{7.85 \times 10^{-5}} = 6.24 \times 10^6 \text{ Pa}
Standard answer using g=10g = 10: 6.37×1066.37 \times 10^6 Pa.

All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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