Two simple pendulums are suspended from the same rigid support and...

Let $T_S$ and $T_L$ be the time periods of the shorter and longer pendulums respectively, and $L_S$ and $L_L$ be their lengths.

For the two pendulums to be in phase again at the mean position for the first time, they must have completed an integer number of oscillations, and the shorter pendulum must have completed exactly one more oscillation than the longer pendulum.

Given that the shorter pendulum completes $7$ oscillations, the longer pendulum must have completed $7 - 1 = 6$ oscillations in the same time interval.

Equating the total time taken:

$7 T_S = 6 T_L$

$\Rightarrow \dfrac{T_L}{T_S} = \dfrac{7}{6}$

The time period of a simple pendulum is given by $T = 2\pi \sqrt{\dfrac{L}{g}}$, which implies $L \propto T^2$.

Therefore, the ratio of their lengths is:

$\dfrac{L_L}{L_S} = \left(\dfrac{T_L}{T_S}\right)^2 = \left(\dfrac{7}{6}\right)^2 = \dfrac{49}{36}$

Answer: $49 : 36$