$\text{f}_{10}=\frac{10}{2\text{l}}\sqrt{\frac{\text{T}-\rho_\text{w}\text{Vg}}{\text{m}}}=\frac{11}{2\text{l}}\sqrt{\frac{(\rho-\rho_\text{w})\text{Vg}}{\text{m}}}$
As the f of tuning fork is same,$\text{f}_{10}=\text{f}_{11}\Rightarrow\frac{10}{2\text{l}}\sqrt{\frac{\rho\text{Vg}}{\text{m}}}=\frac{11}{2\text{l}}\sqrt{\frac{(\rho-\rho_\text{w})\text{Vg}}{\text{m}}}$
$\Rightarrow\frac{10}{11}=\sqrt{\frac{\rho-\rho_\text{w}}{\text{m}}}$
$\Rightarrow\frac{\rho-1}{\rho}=\frac{100}{121}$ $\big($because, $\rho_\text{w}=1\text{gm/cc}\big)$
$\Rightarrow100\rho=121\rho-121$
$\Rightarrow5.8\times10^{3}\text{kg/m}^3$
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