A manometer contains a liquid of density 5.44 g/cm is attached to a flask containing gas 'A as follows

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Detailed Solution

$P$ $gas A$ $$ $\approx$ $1.20$ $atm$

Calc

Density of manometer liquid $=5.44\ \text{g cm}^{-3}=5440\ \text{kg m}^{-3}$ $=$ $5.44$ $g cm$ $-3$ $=$ $5440$ $kg m$ $-3$
Height difference $h=38\ \text{cm}=0.38\ \text{m}$ $h$ $=$ $38$ $cm$ $=$ $0.38$ $m$
Excess pressure $=\rho g h=5440\times9.81\times0.38\approx2.03\times10^{4}\ \text{Pa}\approx0.20\ \text{atm}$ $=$ $ρgh$ $=$ $5440$ $\times$ $9.81$ $\times$ $0.38$ $\approx$ $2.03$ $\times$ $10$ $4$ $Pa$ $\approx$ $0.20$ $atm$
Open arm is at 1 atm and higher by 38 cm → gas side has higher pressure:
$P_{\text{gas}}=P_{\text{atm}}+\rho g h\approx1.00+0.20=1.20\ \text{atm}$ $P$ $gas$ $$ $=$ $P$ $atm$ $$ $+$ $ρgh$ $\approx$ $1.00$ $+$ $0.20$ $=$ $1.20$ $atm$