A barometer kept in an elevator reads 76 , cm | vedclass

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Question

(D) The pressure measured by a barometer is given by $P = h \rho g_{eff}$,where $g_{eff}$ is the effective acceleration.When the elevator accelerates

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(D) The pressure measured by a barometer is given by $P = h \rho g_{eff}$,where $g_{eff}$ is the effective acceleration.When the elevator accelerates upwards with an acceleration $a$,the effective acceleration becomes $g_{eff} = g + a$.Since the barometer reads $76 \, cm$ of $Hg$ under this condition,the actual atmospheric pressure $P_0$ (which corresponds to the reading when $a = 0$) would be related to the observed reading $h'$ by $P_0 = h' \rho (g + a)$.However,the question asks for the pressure reading in the accelerating frame. If the barometer reads $76 \, cm$ while accelerating upwards,it implies that the effective weight of the mercury column has increased.Since $P = h \rho (g + a)$,for a fixed atmospheric pressure $P_0$,the height $h$ would decrease as acceleration increases. If the reading is $76 \, cm$ during upward acceleration,the actual atmospheric pressure must be higher than $76 \, cm$ of $Hg$ equivalent.Given the options,if the barometer is calibrated to read $76 \, cm$ at rest,and it reads $76 \, cm$ while accelerating,it suggests the pressure is effectively higher than the standard $76 \, cm$ reading at rest. Thus,the most likely value greater than $76 \, cm$ is $77 \, cm$.

$A$ barometer kept in an elevator reads $76 \, cm$ when the elevator is accelerating upwards. The most likely pressure inside the elevator (in $cm$ of $Hg$) is ........

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