The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbondioxide will be close to (ln 5 = 1.601, ln 2 = 0.693).

For a damped pendulum oscillating in a medium, the amplitude decays exponentially as $$A = A_0 e^{-bt/2m}$$, where $$b$$ is the damping coefficient proportional to the viscosity $$\eta$$ of the medium (by Stokes law, $$b \propto \eta$$).

In air, the amplitude decreases from 10 cm to 8 cm in 40 seconds. So $$8 = 10\, e^{-b_{\text{air}} \cdot 40 / 2m}$$, which gives $$e^{-20\, b_{\text{air}}/m} = 4/5$$. Taking the natural logarithm, $$\frac{20\, b_{\text{air}}}{m} = \ln\!\left(\frac{5}{4}\right) = \ln 5 - 2\ln 2 = 1.601 - 1.386 = 0.215$$.

In carbon dioxide, we need the time $$t$$ for the amplitude to reduce from 10 cm to 5 cm. So $$5 = 10\, e^{-b_{\text{CO}_2}\, t / 2m}$$, which gives $$\frac{b_{\text{CO}_2}\, t}{2m} = \ln 2 = 0.693$$.

Since $$\eta_{\text{air}} / \eta_{\text{CO}_2} = 1.3$$, we have $$b_{\text{CO}_2} = b_{\text{air}} / 1.3$$. Substituting, $$t = \frac{2m \times 0.693}{b_{\text{CO}_2}} = \frac{2m \times 0.693 \times 1.3}{b_{\text{air}}}$$. From the air data, $$b_{\text{air}}/m = 0.215/20 = 0.01075$$, so $$m/b_{\text{air}} = 1/0.01075$$.

Therefore, $$t = \frac{2 \times 0.693 \times 1.3}{0.01075} = \frac{1.8018}{0.01075} \approx 161 \text{ s}$$. The closest answer among the given options is $$161$$ s.