Answer
(a) The buoyant force on the chamber is $7.39\times 10^5~N$
(b) The tension in the cable is 9880 N
Work Step by Step
(a) We can find the volume $V$ of the chamber as:
$V = \frac{4}{3}\pi~r^3$
$V = \frac{4}{3}\pi~(2.60~m)^3$
$V = 73.6~m^3$
We can find the buoyant force which is equal to the weight of seawater that is displaced by the chamber. Let $\rho$ be the density of seawater.
$F_B = \rho~V~g$
$F_B = (1.025\times 10^3~kg/m^3)(73.6~m^3)(9.80~m/s^2)$
$F_B = 7.39\times 10^5~N$
The buoyant force on the chamber is $7.39\times 10^5~N$.
(b) The sum of the tension in the cable and the chamber's actual weight is equal to the buoyant force.
$T + M~g = F_B$
$T = F_B - M~g$
$T = (7.39\times 10^5~N) - (74,400~kg)(9.80~m/s^2)$
$T = 9880~N$
The tension in the cable is thus 9880 N.