Question

A faulty barometer tube is 90 cm long and it contains some air above mercury. The
reading is 74.5 cm when the true atmospheric pressure is 76 cm. What will be the true atmopsheric pressure if the
reading on thisbarometer is 74 cm? (H = 10m of water column)​

Answer 1

AnSwer :

Let the area of cross- section of the tube be A cm² 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

According to Question,

P₁ = (76.0 - 74.5) = 1.5 cm of mercury

V₁ = A×(90 - 74.0) = A × 15.5cm³

P₂ = (H - 74.0) cm of mercury

V₂ = A × (90 - 74.0) = A × 16 cm³

Applying Boyle's law,

${ : \implies{ \sf{P_1 V_2 = P_2 V_2}}}$

${ : \implies{ \sf{(1.5 ) (A \times 15.5) = (H - 74)(A \times 16)}}}$

${ : \implies{ \sf{H - 74 = \dfrac{1.5 \times 15.5}{16} }}}$

${ : \implies{ \sf{H = 74 + 1.45}}}$

${ : \implies{ \boxed{ \sf{H = 75.45cm}}}}$

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