A simple pendulum has a time period T when oscillating in air. When...

The time period of a simple pendulum in air is given by:
$T = 2\pi \sqrt{\dfrac{l}{g}}$

When immersed in a liquid, the effective acceleration due to gravity $g_{eff}$ is reduced due to buoyancy:
$g_{eff} = g \left(1 - \dfrac{\rho_w}{\rho_b}\right)$
where $\rho_w$ is the density of water and $\rho_b$ is the density of the bob.

The new time period is:
$T' = 2\pi \sqrt{\dfrac{l}{g_{eff}}}$

Given that $T' = \sqrt{2}T$, we have:
$2\pi \sqrt{\dfrac{l}{g_{eff}}} = \sqrt{2} \left(2\pi \sqrt{\dfrac{l}{g}}\right)$

Squaring both sides:
$\dfrac{l}{g_{eff}} = 2 \left(\dfrac{l}{g}\right)$
$g_{eff} = \dfrac{g}{2}$

Equating this to the expression for $g_{eff}$:
$g \left(1 - \dfrac{\rho_w}{\rho_b}\right) = \dfrac{g}{2}$
$1 - \dfrac{1000}{\rho_b} = \dfrac{1}{2}$
$\dfrac{1000}{\rho_b} = \dfrac{1}{2}$
$\rho_b = 2000 \text{ kg/m}^3$

Using the inverted density ratio would incorrectly yield $500 \text{ kg/m}^3$, and forgetting to square the time period ratio leads to an incorrect irrational density.

Answer: $2000 \text{ kg/m}^3$