(c) Pressure at the bottom $\rho = (h_{1}d_{1} + h_{2}d_{2})g$ $= \lbrack 250 \times 1 + 250 \times 0.85\rbrack g$ $= 250\lbrack 1.85\rbrack g$ $= 462.5\ g\ dyne/cm$
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Ph.D. in Physics with 19+ years of teaching experience.