Question

A cylindrical tank is filled with water to level of 3$m$. A hole is opened at height of 52.5$cm$ from the 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 (Take g= 10$m{s}^{-2})$

• A. $50m^2{s}^{-2}$
• B. $40m^2{s}^{-2}$
• C. $51.5m^2{s}^{-2}$
• D. $50.5m^2{s}^{-2}$

Answer 1

Answer: (A)

$50m^2{s}^{-2}$

Given

$h=3-0.525$

$h=2.475$

$\frac{A_1}{A_2}=0.1$

$V_1=$ velocity of water from orifice.

$V_2=$ velocity of swrface of water

Acc. to continuity theorem

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

$v_1=10v_2$

Acc. to Bernouliis theorem

$P_0+\rho gh+\frac{1}{2}\rho v_L^2=P_0+0+\frac{1}{2}\rho v_1^2$

$gh+\frac{1}{2}\left(\frac{v_1^2}{100}\right)=\frac{1}{2}{v}_1^2$

$v_1^2\left(\frac{99}{100}\right)=2\times 10\times 2.475$

$v_1^2=\frac{4950}{99}=50m^2{s}^{-2}$

$\text{Answer:}\left(A\right)$