Question

A cylindrical tank is filled with water to level of $3m$. A hole is opened at height of $52.5cm$ from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is $0.1$. The square of the speed with which water is coming out from the orifice is $(Takeg=10m{s}^{-2})$

• A. 50 m² s^-2
• B. 40 m²s^-2
• C. 51.5 m² s^-2
• D. 50.5 m² s^-2

Answer 1

Answer: (A)

Suppose A be the area of cross section of tank, a be the area of hole, v${}_e$ be the velocity of efflux, h be the height of liquid above the hole,
Let v be the speed with which the level decreases in the container. Using equation ofcontinuity, we get
av${}_e$ Av or $v=\frac{a{v}_e}{A}$
Using Bernoulli’s theorem, we have
$P_0+h\rho g+\frac{1}{2}\rho v^2=P_0+\frac{1}{2}\rho v_e^2$
$h\rho g+\frac{1}{2}\rho {\left(\frac{a{v}_e}{A}\right)}^{{}^2}=\frac{1}{2}\rho v_e^2$
or $v_e^2=\frac{2hg}{1-\left(a^2/A^2\right)}=\frac{2\times \left(3-0.525\right)\times 10}{1-{\left(0.1\right)}^2}$
= 50 m${}^2$ s${}^{-2}$