Question

The pressure at the bottom of a water tank is 4P, where P is atmospheric pressure. If water is drawn out till the water level decrease by $\frac{3}{5}$ the, then pressure at the bottom of the tank is

• A. $\frac{3P}{8}$
• B. $\frac{7P}{6}$
• C. $\frac{11P}{5}$
• D. $\frac{9P}{4}$

Answer 1

The correct option is C $\frac{11P}{5}$

The tank is open from top.

Let initial height of water $=h$

Therefore, final given height $\left(h^{\prime }\right)=h-\frac{3}{5}h$

$h^{\prime }=\frac{2}{5}h$

Initial pressure at bottom $=4p$

$4P=\rho gh+P-\left(i\right)$

$\text{where}\rho \text{is density of water}\text{and}P\text{is atmospheric pressure}$

Final pressure $\left(p^{\prime }\right)=\frac{2}{5}\rho gh+P$

i.e, $p^{\prime }-p=\frac{2}{5}\rho gh$

upon solving (i) and (ii)

we get; ; $5p^{\prime }-5p=6p$

$p^{\prime }=\frac{11}{5}P$