Let the area of cross - section of the tube be A `cm^(2)` and true pressure be H cm of mercury. Since the temperature is constant, Boyle.s law can be applied to the air enclosed in the upper part of the barometer tube, thus `P_(1)` = (76.0 - 74.5) = 1.5 cm of mercury
`V_(1)=Axx(90-74.5)=Axx15.5cm^(3)`
`P_(2)=(H-74.0)cm` of mercury
`V_(2)=Axx(90-74.0)=Axx16cm^(3)`
Applying Boyle.s law `P_(1)V_(1)=P_(2)V_(2)`
`1.5xx(Axx15.5)=(H-74)xx(Axx16)`
`H-74=(1.5xx15.5)/16rArrH=74+1.45=75.45cm`