When at rest, a barometer kept inside an elevator shows a reading...

When the elevator is at rest, the atmospheric pressure $P_0$ is balanced by the mercury column of height $h_0 = 76 \text{ cm}$.

$P_0 = \rho g h_0$

When the elevator ascends with increasing speed, it moves with an upward acceleration $a$. The effective acceleration due to gravity inside the elevator becomes $g_{\text{eff}} = g + a$.

Assuming the atmospheric pressure $P_0$ remains constant, the new height $h$ of the mercury column is given by:

$P_0 = \rho g_{\text{eff}} h = \rho (g + a) h$

Equating the two expressions for $P_0$:

$\rho g h_0 = \rho (g + a) h$

$h = \dfrac{g}{g + a} h_0$

Since $g + a > g$, it follows that $h < h_0$.

Therefore, the reading of the barometer will be $< 76 \text{ cm}$.