Question

It's possible to estimate the percentage of fat in the body by measuring the resistance of the upper leg rather than the upper arm; the calculation is similar. A person's leg meas-ures 40cm between the knee and the hip, with an average leg diameter (ignoring bone and other poorly conducting tissue) of 12cm . A potential difference of 0.79V causes a current of 1.6mA .

What are the fractions of muscle and fat in the leg?

Please show full work, and explantion, and units for full credit.

Answer 1

The resistance between the conductors will provide a measure of body fat between a pair of electrodes, since the resistance to electricity varies between adipose, muscular and skeletal tissue. Fat-free mass (muscle) is a good conductor as it contains a large amount of water (approximately 73%) and electrolytes, while fat is anhydrous and a poor conductor of electric current.

R = V/I =  0.79V / 1.6mA = 493.75 ohm

Now solving our prblem

R = resistivity * L / A

and i am taking  the resistivity is 13 ohms*cm and 25 ohms*cm for muscle and fat

let x be the fraction of muscle and (1-x) be the fraction of fat.

total resistivity of leg = (x*13) +((1-x)*25) = 25 -12x ohm*cm

L = 40
A = pi*r^2 = 113.09 cm^2

R = resistivity * L / A

493.75 =(25 -12x) *40*10^-2/113.09 *10^-4

4.9375 = 8.84251 - 4.2244*x

x=0.9112 ===>fraction of muscle that is 91.12 %

1-x=0.0888 ===>fraction of fat that is 8.88%

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