Question

A fixed container is filled with a liquid of density $\rho$ up to a height $4m$. A horizontal slit of small width but of area $=0.5m^2$ is made a height of $2m$ from bottom. The speed of top surface of the water level is ( Area of top surface of container is $4m^2$ and $g=10m/s^2$).

• A. $\sqrt{\frac{20}{63}}m/sec$
• B. $\sqrt{\frac{40}{63}}m/sec$
• C. $\sqrt{\frac{80}{63}}m/sec$
• D. $Noneofthese$

Answer 1

The correct option is B $\sqrt{\frac{40}{63}}m/sec$

Let us take the velocity of top of surface of water is $v_1$ and the water coming out from slit is $v_2$

Applying equation of continuity

$A_1{v}_1=A_2{v}_2$

$\Rightarrow v_2=\frac{A_1{v}_1}{A_2}=\frac{4v_1}{0.5}$

$\Rightarrow v_2=3v_1$

Now,

Applying Bernoulli's equation at top water surface and slit.

$3g{h}_1+\frac{1}{2}3v_1^2+P_1=3g{h}_2+\frac{1}{2}5v_2^2+P_2$

$3g4+\frac{1}{2}3v_1^2+P_0=3g2+\frac{1}{2}3(8v_1{)}^2+P_0$

$\Rightarrow \frac{643v_1^2}{2}-\frac{5v_1^2}{2}=25g$

$\Rightarrow 63v_1^2=4g$

$\Rightarrow v_1=\sqrt{\frac{4g}{63}}\Rightarrow v_1=\sqrt{\frac{40}{63}}$