Question

For a polymer sample with a viscosity of $6 \times 10^{11}$ poise, if the apparent plateau modulus of $3 \times 10^6$ dyne cm$^{-2}$ drops to zero above a certain temperature, the relaxation time of the polymer is _________ days (rounded off to one decimal place).

Hint:

For polymer relaxation, the time is inversely proportional to the plateau modulus: higher modulus gives shorter relaxation times.

Answer 1

Correct Answer: 2.1

The relationship between the relaxation time (\( \tau \)) and the plateau modulus (\( G_0 \)) for a polymer sample is given by: \[ \tau = \frac{\eta}{G_0} \] Given: \[ \eta = 6 \times 10^{11}\ \text{poise} = 6 \times 10^{11}\ \text{dyne·s/cm}^2 \] \[ G_0 = 3 \times 10^6\ \text{dyne/cm}^2 \] Thus: \[ \tau = \frac{6 \times 10^{11}}{3 \times 10^6} = 2 \times 10^5\ \text{seconds} \] Convert to days: \[ \text{Days} = \frac{2 \times 10^5}{60 \times 60 \times 24} = \frac{2 \times 10^5}{86400} \approx 2.31\ \text{days} \] \[ \boxed{2.3\ \text{days}} \]