An inverted tube barometer is kept in a lift moving vertical upward with an acceleration -a- The density of mercury is - and acceleration due to gravity is g- If the atmospheric pressure be Po then

P_o = ρ(g+a)h ⇒ h = P_o/ρ (g+a) ...

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Standard XI

Physics

Atmospheric Pressure

An inverted t...

Question

An inverted tube barometer is kept in a lift moving vertical upward with an acceleration 'a'. The density of mercury is $ρ$ and acceleration due to gravity is g. If the atmospheric pressure be $P_{o}$ then

Height of the mercury column in the lift will be $\frac{P_{o}}{ρ (g + a)}$

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Height of the mercury column in the lift will be $\frac{P_{o}}{ρ (g - a)}$

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Height of the mercury column in the lift will be $\frac{P_{o}}{ρ g}$

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Height of the mercury column in the lift will be $\frac{P_{o}}{ρ a}$

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Solution

The correct option is A
Height of the mercury column in the lift will be $\frac{P_{o}}{ρ (g + a)}$

$P_{o}$ = $ρ$ (g+a)h

$\Rightarrow$ h = $\frac{P_{o}}{ρ (g + a)}$

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An inverted tube barometer is kept in a lift moving vertical upward with an acceleration 'a'. The density of mercury is ρ and acceleration due to gravity is g. If the atmospheric pressure be Po then

The correct option is A Height of the mercury column in the lift will be Poρ(g+a) Po = ρ(g+a)h ⇒ h = Poρ(g+a)

An inverted tube barometer is kept in a lift moving vertical upward with an acceleration 'a'. The density of mercury is $ρ$ and acceleration due to gravity is g. If the atmospheric pressure be $P_{o}$ then

The correct option is A
Height of the mercury column in the lift will be $\frac{P_{o}}{ρ (g + a)}$

$P_{o}$ = $ρ$ (g+a)h

$\Rightarrow$ h = $\frac{P_{o}}{ρ (g + a)}$