|Take atmospheric pressure $=1.0 \times 10^5 \mathrm{~N} / \mathrm{m}^2$, density of water $=1000 \mathrm{~kg} / \mathrm{m}^3$ and $g=10 \mathrm{~m} / \mathrm{s}^2$. Neglect any effect of surface tension.]
$P+\rho g h=P_0$
$\text { or } P=10^5-(1000)(10)(0.2)=98 \times 10^3 N / m ^2$
$\text { now, } P_0 V_0=P V$
$\text { or } 10^5[A(0.5-H)]=98 \times 10^3[A(0.5-0.2)]$
where $A =$ cross-sectional area of a vessel.
$0.5-H=0.294$
$\Rightarrow H=0.206 \ m =206 \ mm$
The fall in height (in $mm$ ) of water level $=206-200=6$




$(A)$ $\beta=0$ when $a= g / \sqrt{2}$
$(B)$ $\beta>0$ when $a= g / \sqrt{2}$
$(C)$ $\beta=\frac{\sqrt{2}-1}{\sqrt{2}}$ when $a= g / 2$
$(D)$ $\beta=\frac{1}{\sqrt{2}}$ when $a= g / 2$