The atmospheric pressure at a place is $10^5 \,Pa$. If tribromomethane (specific gravity $=2.9$ ) be employed as the barometric liquid, the barometric height is .......... $m$
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(a)
$1 \,atm \simeq 10^5 \,Pa =76 \,cm$ of $Hg$
Density of $Hg =13.6$
$\rho_{H g} \times g \times h_{H g}=\rho_{T B M} \times g \times h_{T B M}$ $\left\{\begin{aligned} \rho_{H g}= \text { density of } Hg \\ h_{H g}= \text { height of } Hg \\ \rho_{T B M}= \text { density of } \\ \text { tribromomethane } \\ h_{T B M}= \text { height of tribromomethane } \end{aligned}\right.$
Substituting values
$13.6 \times 76=2.9 \times h$
$3.52 \,m=h$
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