Question

The area of cross-section of the wider tube shown in figure is $800{\text{cm}}^2$. If a mass of $12\text{kg}$ is placed on the massless piston, the difference in heights in the level of water in the two tubes is:

• A. $10\text{cm}$
• B. $6\text{cm}$
• C. $15\text{cm}$
• D. $2\text{cm}$

Answer 1

The correct option is C $15\text{cm}$

Excess pressure exepted by mass $12\text{kg}$ on the wider end $=\frac{F}{A}=\frac{mg}{A}=\left(\frac{12\times g}{800\times {10}^{-4}}\right)$
Pressure difference due to height difference $h=\rho gh$
Here, $\rho =$ Density of water $=1000{\text{kg/m}}^3$

As we know, for the same fluid at the same level, pressure should be same. Hence,
$P_0+\rho gh=P_0+\frac{12\times g}{800\times {10}^{-4}}$
$\Rightarrow 1000\times g\times h=\frac{12\times g}{800\times {10}^{-4}}$
$\Rightarrow h=(\frac{12}{800\times {10}^{-1}}\times 100)\text{cm}$
$\therefore h=15\text{cm}$