The two pendulums shown in the figure below, each consist of a uniform solid ball of mass M supported by a rigid massless rod, but the ball for pendulum A is very tiny while the ball for pendulum B is much larger having a non-negligible radius R = 1.15 m. The length L in the figure is L = 1.6 m. Find the ratio of the ratio TB/TA of the period of pendulum B to the period of pendulum A.

P = 2 pi (I0 / M g L)^1/2 for object with inertia of I0

Note that if I0 = M L^2 you have P = 2 pi (L / g)^1/2 the simple pendulum

Radius of sphere = 1.15 = .719 L

I0 = M L^2 + 2/5 M (.719 L)^2 = 1.207 M L^2

P2 = 2 pi (1.207 M L^2 / M g L)^1/2 = 2 pi (1.207 L / g)^1/2

P2 / P1 = (1.207 / 1)^1/2 = 1.10     ratio of periods

i believe the answer would be:

d. violin, clarinet, french horn, and tuba