The area of cross-section of the wider tube shown in figure is $800$ $cm^2$. If a mass of $12$ $ kg $ is placed on the massless piston, the difference in heights $h$ in the level of water in the two tubes is ........ $cm$
Diffcult
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Pressure due to difference in heights will be balanced by pressure due to $12 \mathrm{kg}$ block
$\Rightarrow \rho g h=\frac{120}{800 \times 10^{-4}}$
$10^{4} \mathrm{h}=\frac{120}{800} \times 10^{4}$
$\mathrm{h}=\frac{12}{80}=\frac{3}{20} \mathrm{m}=15 \mathrm{cm}$
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