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The mass of a boy is $20 kg$ and the area of the sole of his foot is $0.01 m^{2}$ . If $g = 10 m / s^{2}$ . The pressure exerted on the floor when he is standing on his one feet is [[1]] $N / m^{2}$ .

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Detailed Solution

The mass of a boy is

20 kg

and the area of the sole of his foot is

0.01 m^{2}

. If

g = 10 m / s^{2}

. The pressure exerted on the floor when he is standing on his both feet is

20000

N / m^{2}

Pressure = \frac{Force}{Area}

Mass of a boy,

m = 20 kg

Force,

F = mg

= 20 \times 10

= 200 N

When the boy is standing on both feet,
Total area of feet

A = 0.01 m^{2}

Pressure,

P = \frac{200 N}{0.01 m^{2}}

= 20000 N / m^{2}

The pressure exerted on the floor when the boy is standing on his one feet is

20000 N / m^{2}

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