Question

A barometer kept in an elevator at rest reads $76\text{cm}$ of $Hg$. If the elevator starts accelerating downwards, then the atmospheric pressure measured by the barometer (w.r.t column of mercury) in the elevator is:

• A. $76\text{cm}$
• B. $<76\text{cm}$
• C. $>76\text{cm}$
• D. zero

Answer 1

The correct option is C $>76\text{cm}$

Let the atmospheric pressure measured by the barometer be $P_0^{\prime }$. The barometer set-up is shown in figure.


In a static fluid, pressure at each point in the fluid at the same level is equal.
i.e Pressure at point $A=$Pressure at point $B$
$\Rightarrow P_A=P_B$
$\Rightarrow P_0^{\prime }=P_B...\left(i\right)$
Since there is always vaccum,$(\text{pressure}=0)$ over the mercury column.
$\Rightarrow P_0^{\prime }=0+\rho g_{\text{eff}}{h}^{\prime }=\rho g_{\text{eff}}{h}^{\prime }...\left(ii\right)$
When barometer is kept in elevator accelerating downwards, $g_{\text{eff}}=g-a$
$\Rightarrow P_0^{\prime }=\rho g_{\text{eff}}{h}^{\prime }$
But $P_0^{\prime }=\rho gh$
[$\because h=76\text{cm}$ for barometer reading when lift is at rest]

Equating both,
$\rho g_{\text{eff}}{h}^{\prime }=\rho gh$
$\Rightarrow h^{\prime }=\left(\frac{g}{g-a}\right)h$
Clearly $\left(\frac{g}{g-a}\right)>1$
$\Rightarrow h^{\prime }>h$
$\therefore h^{\prime }>76\text{cm}$