Height of the mercury column, $h_{1}=0.76 m$
Density of French wine, $\rho_{2}=984 kg / m ^{3}$
Height of the French wine column $=h_{2}$
Acceleration due to gravity, $g=9.8 m / s ^{2}$
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column
$=$ Pressure in the French wine column
$\rho_{1} h_{1} g =\rho_{2} h_{2} g$
$h_{2}=\frac{\rho_{1} h_{1}}{\rho_{2}}$
$=\frac{13.6 \times 10^{3} \times 0.76}{984}$
$=10.5 m$
Hence, the height of the French wine column for normal atmospheric pressure is $10.5\; m$

$1.$ If the piston is pushed at a speed of $5 \ mms ^{-1}$, the air comes out of the nozzle with a speed of
$(A)$ $0.1 \ ms ^{-1}$ $(B)$ $1 \ ms ^{-1}$ $(C)$ $2 \ ms ^{-1}$ $(D)$ $8 \ ms ^{-1}$
$2.$ If the density of air is $\rho_{ a }$ and that of the liquid $\rho_{\ell}$, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to
$(A)$ $\sqrt{\frac{\rho_{ a }}{\rho_{\ell}}}$ $(B)$ $\sqrt{\rho_a \rho_{\ell}}$ $(C)$ $\sqrt{\frac{\rho_{\ell}}{\rho_{ a }}}$ $(D)$ $\rho_{\ell}$
Give the answer question $1$ and $2.$

