To assemble a barometer, a tube of length 1 m is completely filled...

Length of the tube is $100 \text{ cm}$.

Initially, the tube contains $76 \text{ cm}$ of mercury and $24 \text{ cm}$ of air at atmospheric pressure, $P_0 = 76 \text{ cm}$ of Hg.

When the tube is inverted in the mercury cup and the cork is removed, the air bubbles up and gets trapped at the closed top end. Let the new height of the mercury column be $h$.

The length of the trapped air column becomes $(100 - h) \text{ cm}$.

The pressure of the trapped air is $P = P_0 - h = 76 - h$.

Applying Boyle's law for the trapped air at constant temperature:

$P_0 V_1 = P V_2$

$76 \times 24A = (76 - h)(100 - h)A$

Since the product on the left side is strictly positive, we must have $76 - h > 0$.

This implies $h < 76 \text{ cm}$.