Question

A tube $1{\text{cm}}^2$ in cross section is attached to the top of a vessel $1\text{cm}$ high and of cross section $100{\text{cm}}^2$. Water is poured into the system filling it upto a height of $100\text{cm}$ above the bottom of the vessel as shown in the figure. Take $g=10{\text{m/s}}^2$. Find the correct statement. Take atmospheric pressure $P_0={10}^5\text{bar}$.

• A. The force exerted by the water against the bottom of the vessel is $1100\text{N}$.
• B. The weight of water in the system is $1.99\text{N}$.
• C. Both (a) and (b) are correct.
• D. Neither (a) nor (b) is correct.

Answer 1

The correct option is C Both (a) and (b) are correct.

Force exeted by the water against the bottom of the vesel $F_{\text{base}}=P_{\text{base}}\times$ Area of base
$P_{\text{base}}=P_0+{\rho }_{w}gh$
$={10}^5+{10}^3\times 10\times \left[\frac{(99+1)}{100}\right]$
[$\because h=(99+1)\text{cm}=1\text{m}$]
$=1.1\times {10}^5{\text{N/m}}^2$

$\text{Area of base}=100\times {10}^{-4}{\text{m}}^2$ [given]
$\therefore F_{\text{base}}=1.1\times {10}^5\times {10}^{-2}=1100\text{N}$


Weight of the water in the system
$=\text{Weight of water in tube of cross-sectional area}1{\text{cm}}^2\text{+ Weight of water in vessel of cross-sectional area}100{\text{cm}}^2$
i.e $W_{\text{system}}=W_{\text{tube}}+W_{\text{vessel}}$
$=\rho g{V}_1+\rho g{V}_2$
$={10}^3\times 10\times [V_1+V_2]$
$={10}^4\times [99\times 1+1\times 100]\times {10}^{-6}$
$=1.99\text{N}$