91Ó°ÊÓ

Given in the question,

Set $d=10\mathrm{cm},l=160\mathrm{cm}$and $\mathrm{ÒÏ}=5.0\mathrm{Î\copyright }\mathrm{m}$, and $I=100\mathrm{mA}$.

The area of the cross-section of the cylinder that is used to model the route between your hands is

$A=\frac{1}{4}\mathrm{Ï€}{d}^2$

The resistance of this cylinder is

$R=\mathrm{ÒÏ}\frac{l}{A}=\mathrm{ÒÏ}\frac{4l}{\mathrm{Ï€}{d}^2}$

The current and the potential difference are related by Ohm's law

$I=\frac{\mathrm{Δ}V}{R}$

Calculate the resistivity,

$R=\frac{\mathrm{ÒÏ}L}{A}$

The expression for the potential difference,

$\mathrm{Δ}V=\frac{4\mathrm{ÒÏ}IL}{\mathrm{Ï€}{d}^2}$

The diameter of the arms is $10\mathrm{cm}$.

Convert units of diameter from $cm$to $m,$

$d=10\mathrm{cm}$

$=(10\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=0.1\mathrm{m}$

The length of arm is $160\mathrm{cm}$.

Convert units of length from $cm$to $m,$

$L=160\mathrm{cm}$

$=(160\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=1.6\mathrm{m}$

Current across the chest is $100\mathrm{mA}$.

Convert units of current from $\mathrm{mA}$to $\mathrm{A}$.

$I=100\mathrm{mA}$

$=(100\mathrm{mA})\left(\frac{1\mathrm{A}}{1000\mathrm{mA}}\right)$

$=0.1\mathrm{A}$

Substitute $\frac{4\mathrm{ÒÏ}L}{\mathrm{Ï€}{d}^2}$for $R$in the equation $\mathrm{Δ}V=IR$and solve for $∆V,$

$\mathrm{Δ}V=I\left(\frac{4\mathrm{ÒÏ}L}{\mathrm{Ï€}{d}^2}\right)$

$=\frac{4\mathrm{ÒÏ}IL}{\mathrm{Ï€}{d}^2}$

Substitute the values,

role="math" localid="1649150118248" $\mathrm{Δ}V=\left(\frac{4(5.0\mathrm{Î\copyright }â‹\dots \mathrm{m})\left(0.100\mathrm{A}\right)\left(1.60\mathrm{m}\right)}{\mathrm{Ï€}(0.10\mathrm{m}{)}^2}\right)$

$=101.86\mathrm{V}$

Therefore, the potential difference required for lethal shock is$101.86V$