Radius of the heel, r = \( \dfrac{ d }{ 2 } = 0.005\; m\)

Mass of the lady, m = 50 kg

Area of the heel, A = πr^{2} = π (0.005)^{2}

= 7.85 × 10^{-5 }m^{2}

Force on the floor due to the heel: F = mg

= 50 × 9.8 = 490 N

Pressure exerted by the heel on the floor:

P = \( \dfrac{ F }{ A }\) = \( \dfrac{ 490 }{ 7.85 \times 10^{ -5 }} \)

P = \( 6.24 \times 10^{ 6 }Nm ^{-2}\)

Answered by Abhisek | 1 year ago**Answer the following questions:-**

**(a)** According to the law of atmospheres density of air decreases with increase in height y as \( ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}\). Where ρ_{0} =1.25 kg m^{-3} is the density of air at sea level and y_{0 }is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

**(b)** A zeppelin of volume 1500 m^{3} is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10^{–2} N m^{–1}. Take the angle of contact to be zero and density of water to be 1.0 × 10^{3} kg m^{–3} (g = 9.8 m s^{–2}).

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m^{–1}. Density of mercury = 13.6 × 10^{3} kg m^{–3}

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius \( 2.0 × 10^{–5} m\) and density \( 1.2 × 10^3 kg m^{–3}\). Take the viscosity of air at the temperature of the experiment to be \( 1.8 × 10^{–5} Pa \). How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

A plane is in level flight at constant speed and each of its wings has an area of \( 25 m^2\). If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be \( 1 kg/m^3\)), g = \( 9.8 m/s^2\)