#### Answer

The tension in the string is 43.9 N

#### Work Step by Step

We can find the volume of the rock as:
$V = \frac{M}{\rho}$
$V = \frac{5.0~kg}{4800~kg/m^3}$
$V = 1.04\times 10^{-3}~m^3$
We can find the buoyant force on the rock. The buoyant force is equal to the weight of the water that is displaced by half the volume of the rock. Let $\rho_w$ be the density of water. Therefore;
$F_B = (\rho_w)~(\frac{V}{2})~(g)$
$F_B = (1000~kg/m^3)~(\frac{1.04\times 10^{-3}~m^3}{2})~(9.80~m/s^2)$
$F_B = 5.1~N$
The sum of the tension and the buoyant force is equal to the rock's weight. We can find the tension in the string as:
$T+F_B = Mg$
$T = Mg-F_B$
$T = (5.0~kg)(9.80~m/s^2)- 5.1~N$
$T = 43.9~N$
The tension in the string is thus 43.9 N.