Answer:
The net acceleration of the elevator accelerating downwards \[=g-a\] \[\therefore \] Pressure inside the elevator \[=h\rho (g-a)=\frac{76\times 13.6\times (g-a)}{13.6\times g}\text{ cm of Hg}\] Clearly, this pressure will be less than 76 cm of Hg.
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