Chapter 15 - Fluids - Problems and Conceptual Exercises - Page 533: 49

$2.1cm$

We know that $\rho_{Hg}V_{submerged}=\rho_{lead}V$ $\rho_{Hg}(\frac{\pi d^2x}{4})=\rho_{lead}(\frac{\pi d^2h}{4})$ $\implies x=(\frac{\rho_{lead}}{\rho_{Hg}})h$ We plug in the known values to obtain: $x=\frac{11.3\times 10^3}{13.6\times 10^3}(2.5)=2.1cm$