As depicted in the figure, a string is stretched by a block going...

The frequency of the $n$-th harmonic of a stretched string is given by:

$f = \dfrac{n}{2L} \sqrt{\dfrac{T}{\mu}}$

where $T$ is the tension in the string, $L$ is its length, and $\mu$ is its linear mass density.

Initially, the block is in the air, and the tension $T_1$ is equal to the weight of the block:

$T_1 = V \rho g$

where $V$ is the volume of the block and $\rho$ is its density. The string vibrates in its 10th harmonic, so the frequency is:

$f = \dfrac{10}{2L} \sqrt{\dfrac{V \rho g}{\mu}}$

When the block is completely submerged in water, the tension $T_2$ is reduced due to the buoyant force:

$T_2 = V \rho g - V \rho_w g = V (\rho - \rho_w) g$

where $\rho_w = 10^3 \text{ kg/m}^3$ is the density of water. The string now vibrates in its 11th harmonic with the same frequency:

$f = \dfrac{11}{2L} \sqrt{\dfrac{V (\rho - \rho_w) g}{\mu}}$

Equating the two expressions for frequency:

$\dfrac{10}{2L} \sqrt{\dfrac{V \rho g}{\mu}} = \dfrac{11}{2L} \sqrt{\dfrac{V (\rho - \rho_w) g}{\mu}}$

Squaring both sides, we get:

$100 \rho = 121 (\rho - \rho_w)$

Solving for $\rho$:

$121 \rho_w = 121 \rho - 100 \rho$

$21 \rho = 121 \rho_w$

$\rho = \dfrac{121}{21} \rho_w$

Substituting $\rho_w = 10^3 \text{ kg/m}^3$:

$\rho = \dfrac{121}{21} \times 10^3 \text{ kg/m}^3 \approx 5.76 \times 10^3 \text{ kg/m}^3$

Rounding to one decimal place, the density of the material of the block is $5.8 \times 10^3 \text{ kg/m}^3$.