The maximum reduction in cross-sectional area per pass ( R ) of a cold wire drawing process is

                                                                         R= 1-e^-(n+1)

where n represents the strain hardening coefficient. For the case of a perfectly plastic material, R is

The correct option is 0.632.

To find the maximum reduction in cross-sectional area per pass ($R$) for a perfectly plastic material, we start with the given relationship:
$R=1-e^{-(n+1)}$
where $n$ represents the strain hardening coefficient of the material.

For a perfectly plastic material, there is no strain hardening, which means the strain hardening coefficient is zero:
$n=0$

Substituting $n=0$ into the equation for $R$:
$R=1-e^{-(0+1)}$
$R=1-e^{-1}$
$R=1-\frac{1}{e}$

Using the value of the mathematical constant $e\approx 2.71828$:
$e^{-1}\approx 0.36788$
Substituting this value back into the equation:
$R\approx 1-0.36788=0.63212$

Rounding to three decimal places, we get:
$R\approx 0.632$